After a long hiatus, I am dusting off my blog and I’ve moved it here to a new home. While I was going through the process of transferring everything here, I re-read the very first post I ever wrote, called Wisdom from the Dodecahedron. And I also found my drawing of the Dodecahedron that the original banner on the very old blog site was based on.
The Dodecahedron is a character from The Phantom Tollbooth by Norton Juster, and he lives in Digitopolis, the city of numbers. Many people prize mathematics for its cool logic and want to hold it up as emotionless as if that somehow makes it better. But here is the first person we meet in Digitopolis and he has on display twelve emotions!
And that appeals to me a lot, because I find that maths is full of emotion. Frustration, curiosity, surprise, satisfaction, pride, sadness, companionship, wonder, silliness, joy — they’re all there, sometimes in quick succession. Talk to any mathematician about their work and those emotional words are guaranteed to spill out. We’re human, and humans feel emotions, and humans do maths in a human way, which is an emotional way.
The Dodecahedron reminds me to feel my emotions as I do maths, and to make space for others to do the same.
NOTE: I wrote this article in September 2025 for the University’s Learning & Teaching News, and I thought I would reproduce it here so other interested people could read it.
Many people feel that their experience with maths up until now prevents them from approaching careers or courses that interest them. For example, a student in an Arts degree might want to become a teacher and must face the government’s numeracy test, or a student in a Music degree might be fascinated by the physics of sound, but intimidated by the calculations involved. The Maths Learning Centre provides an online course called Maths Foundations for just such people. I wanted to share the story of how Maths Foundations was created.
When I got funding in 2023 from PACE (Professional and Continuing Education) to make a course to support students’ maths skills, I knew exactly what I wanted to do. Lack of confidence with number and algebra is the very thing that blocks a lot of people from pursuing various goals, and I wanted a course that helped students build on their earliest maths experiences to foster that confidence.
I began to plan what would be in the course. Based on my extensive experience supporting students at the Maths Learning Centre, I brainstormed the number and algebra skills that would be most useful for people studying various sciences, and for people needing to pass numeracy tests for professional registration such as nurses and teachers. I also considered what would be necessary to move on to the MLC’s bridging course MathTrackX, which can be used in place of Year 12 Maths Methods as a prerequisite for various courses such as maths and engineering.
I took those topics and organised them into four main modules that would allow students to build their number skills and algebra skills at the same time, starting from addition of whole numbers, passing through calculations of areas and volumes and time, and ending with describing lines in the coordinate plane algebraically. The actual process here involved pieces of paper with concepts and skills written on them being physically placed and moved around on the floor and multiple desks. My colleagues and my family were most gracious in accommodating my physical planning process!
The final stage of planning was to use what I had learned from years of interacting with schoolteacher colleagues from early childhood to high school, and choose activities that would help students to learn concepts and practice skills. I was excited to include activities that I know are engaging for people with from a wide range of ages and backgrounds.
Once this planning was done, the hard yards of making the course began. The team from Learning Enhancement & Innovation (LEI) were champions to help me realise my vision. One particularly long process was making all of the videos I imagined. There were many explanatory videos with physical resources from huge fractal art pieces to play-dough to toy dinosaurs to a bottle of honey. I was able to use the Peer Assisted Study Sessions classroom – which was empty for the summer – to lay everything out ready to go and to practise, before recording in the Barr Smith Library recording studio. There were also over forty worked example videos, which were each recorded in one take including all the mistakes, to help students understand the messy process of problem-solving.
On top of the videos, I wrote many pages of text, drew many pictures, created interactive online graphs, designed engaging discussion activities, and coded about seventy randomised practice problem sets. It was a herculean effort, even with the indefatigable support of the LEI team. In April 2024, the course was launched on the AdelaideX platform. Anyone in the world can access the content for free for four weeks, and people can pay for extended access and the opportunity to test themselves with an exam. Now, in September 2025, thousands of students have enrolled, with about 40 so far choosing to complete an exam. There are multiple discussion board posts a day, and MLC lecturer Nicholas Crouch answers them with compassion and verve.
I am so proud of what I have built, so grateful for the support I had to build it, and so glad students are finding it useful.
Once upon a time (in 2016), I created a way to visualise where the complex points are in relation to the real plane, and then more recently (in 2022), I modified it to become the concept of i-arrows. I reread those blog posts recently while updating the blog to the new website, and I got all interested in them again. Here is what I’ve been working on over the last few weeks.
About i-arrows
Here’s a quick recap of how i-arrows work (copied from the 2022 series of blog posts)…
Every point with two complex coordinates is represented as an arrow from one place to another on top of the Cartesian plane.
Real points are dots on the Cartesian plane, the same as they have always been.
The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
In the picture below, there are three examples of i-arrows.
The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)
I have found some people are uncomfortable with drawing a point as an object that is shaped like a line, which is what an arrow feels like. And I do get it. If you are one of these people, it may be helpful to think of the arrow as not the point itself, but just as a thing that points to the point. In this way, it’s kind of like the only way to get to the complex point \(1+4i,2+i)\) is to start a journey from the real point \(1,2\) across some other reality that lies above ours in the distance and direaction described by the arrow \((4,1)\). (This is quite similar to my original idea of iplanes back in 2016.) I thought I’d mention it here in case it helps you.
I also want to add another bit of notation. I already mentioned that the complex point \((p+si,q+ti)\) is represented as an arrow starting at \((p,q)\) and extending along the vector \((s,t)\). To emphasise this even more, I will often write the complex point in “point-vector form” as \((p,q)+(s,t)i\), or even as \(P+\mathbf{p}i\), where \(P\) is the coordinates of the real base point and \(\mathbf{p}\) is the coordinates of the arrow. The arrowhead is then the point with coordinates \(P+\mathbf{p}\). This notation is very helpful for the algebra I want to do, and can make it a bit easier to see within the algebra some of the geometry happening.
The first thing that this notation helps with is complex conjugation. Given a real number \(p+si\), its complex conjugate is \(p-si\), found by switching the sign of the imaginary part. If you do complex conjugation to both coordinates of a complex point at once, you turn \((p+si,q+ti)\) into \((p-si,q-ti)\). That new complex point is the conjugate of the first one. If I write those in point-vector form, then the conjugate point to \((p,q)+(s,t)i\) is \((p,q)-(s,t)i\). This is the same as turning \(P+\mathbf{p}i\) into \(P-\mathbf{p}i\), which now looks just like conjugation looked in the plain complex numbers. I think this is nice.
Conjugate points have a very pleasant interpretation using i-arrows: two conjugate points have the same base, and their i-arrows are the same length, but they point in opposite directions. The original point \(P+\mathbf{p}i\) has its base at \(P\) and its arrowhead at \(P+\mathbf{p}\), and its conjugate point \(P-\mathbf{p}i\) has its base at \(P\) and its arrowhead at \(P-\mathbf{p}\). It’s one of the things I love about the i-arrow representation, the fact that complex conjugation is so easy to visualise.
I want to add one final bit of terminology: a real point is a point whose coordinates are both real numbers. An unreal point is a point whose coordinates are not both real numbers: if one or the other or both of the coordinates is not a real number, then it’s called an unreal point. In pictures, an unreal point is a point that is represented with an i-arrow as opposed to a dot. All the complex points are either real or unreal; every point in the complex coordinate plane that’s not real is unreal. (It’s a bit like how the real numbers that aren’t rational are called irrational.)
Okay. That’s enough setting up for now.
Lines in the complex coordinate plane
Last time I worked with i-arrows, I found there were three kinds of lines in the complex coordinate plane. A complex like can be:
A real line, with infinitely many real points, and whose unreal points are all the i-arrows with both base and arrowhead on the real line.
An unreal line parallel to a real line, with no real points, and whose unreal points are all the i-arrows with base on one real line and arrowhead on another real line parallel to it.
An unreal line not parallel to any real line, which has exactly one real point, and every other point in the real plane has one unreal point on the line with its i-arrow based there.
At the time, I really wanted a geometrical way to find the points on the line joining two complex points. I didn’t want to use algebra to find the equation of the line and then use the equation to find the points on it, I wanted to be able to construct them. If you take two real points, there is a real line joining them, and you can literally draw the other real points on the line by just putting your ruler on the page and drawing the line. The other real points on the line are revealed as you move your pencil along the ruler. I want something like that: a geometrical way to reveal the complex points on a line.
I did manage to do this before, but I wasn’t particularly satisfied at the time with the constructions I made. There were two:
Starting with an unreal point and a real point, construct new unreal points on the line that joins them.
Starting with two unreal points, construct the one real point on the line that joins them.
The constructions I had were complicated and hard for me to remember. Also I couldn’t prove easily that they did what they did every time. I’ve now found some better ones, and proved that they really work. But it will take me a while to lay it all out clearly.
Algebra for the points on the line joining two complex points
Consider the complex points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\). The i-arrow for the point \(P+\mathbf{p}i\) has base \(P\), arrow \(\mathbf{p}\) and arrowhead \(P+\mathbf{p}\). The i-arrow for the point \(Q+\mathbf{q}i\) has base \(Q\), arrow \(\mathbf{q}\) and arrowhead \(Q+\mathbf{q}\).
(In the picture above, both of the points are unreal, but the calculations below work fine even if they’re real.)
For real points, say \(A\) and \(B\), you can find all the other real points on the line joining them by calculating \(A+m(B-A)\) for every real number \(m\). It works the same way for complex points. Every complex point on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) is given by \(P+\mathbf{p}i + (m+\mu i)[(Q+\mathbf{q}i)- (P+\mathbf{p}i)]\) for some complex number \(m+\mu i\). Doing some algebra on this point…
The i-arrow for this point has base at \(P +m(Q-P) – \mu (\mathbf{q} -\mathbf{p})\). Its arrow is \(\mathbf{p} + \mu (Q- P) + m(\mathbf{q} – \mathbf{p})\). Its arrowhead is at \(P + m(Q-P) – \mu(\mathbf{q} -\mathbf{p}) + \mathbf{p }+ \mu(Q- P) + m(\mathbf{q} – \mathbf{p})\).
You may notice these all involve linear combinations of the vectors \(Q-P\) and \(\mathbf{q}-\mathbf{p}\). For the arrowhead, it will also be useful to rewrite it as a linear combination of a different pair of vectors… \[\begin{aligned}&P + m(Q-P) – \mu(\mathbf{q} -\mathbf{p}) + \mathbf{p} + \mu(Q- P) + m(\mathbf{q} – \mathbf{p})\\ &=P+\mathbf{p}+m(Q-P+\mathbf{q}-\mathbf{p})+\mu(Q-P-(\mathbf{q}-\mathbf{p}))\\ &=P+\mathbf{p}+m((Q+\mathbf{q})-(P+\mathbf{p}))+\mu((Q-\mathbf{q})-(P-\mathbf{p}))\end{aligned}\]
The vector \((Q+\mathbf{q})-(P+\mathbf{p})\) is the vector joining the arrowheads of \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\). And the other vector \((Q-\mathbf{q})-(P-\mathbf{p})\) joins the arrowheads of the conjugate points \(P-\mathbf{p}i\) and \(Q-\mathbf{q}i\).
In the picture, the bases \(P\) and \(Q\) are separate points, the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{q}\) are separate points, and the conjugate points’ arrowheads \(P-\mathbf{p}\) and \(Q-\mathbf{q}\) are separate points. Also, the three lines joining the bases, arrowheads and conjugate arrowheads are not parallel to each other. These things won’t always happen, but they do need to happen for the construction I have designed to work. For the sake of completeness, I am going to investigate what happens when some of those points are the same point or some of those lines are parallel. Before I do that, I want to do a little more algebra for a special case, that will display some of the reasoning that is used later, and also makes some nice things too.
Stepping along the base line and the arrowhead line
I’ve said this earlier, but it’s good to repeat it: Given two complex points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\), the complex points on the line joining them are given by the following for some complex number \(m+\mu i\):
The i-arrow for this point has base at \(P +m(Q-P) – \mu(\mathbf{q} -\mathbf{p})\). Its arrow is \(\mathbf{p} + \mu(Q- P) + m(\mathbf{q} – \mathbf{p})\). Its arrowhead is at \(P+\mathbf{p}+m((Q+\mathbf{q})-(P+\mathbf{p}))+\mu((Q-\mathbf{q})-(P-\mathbf{p}))\).
There are a lot of points in the description above: one for every combination of \(m\) and \(\mu\). I want to pick out just some of them, particularly those ones made when the complex number \(m+\mu i\) is actually a real number. That is, the points you make when \(\mu=0\).
These points are given by \(P + m(Q-P) + [\mathbf{p} + m(\mathbf{q} – \mathbf{p})]i\). The i-arrow for this point has base at \(P +m(Q-P)\). Its arrow is \(\mathbf{p} + m(\mathbf{q} – \mathbf{p})\). Its arrowhead is at \(P+\mathbf{p}+m((Q+\mathbf{q})-(P+\mathbf{p}))\).
Well would you look at that. If \(P\) and \(Q\) are different points, then \(P +m(Q-P)\) gives all the points on the real line joining them, stepping away from P some multiple of the vector from \(P\) to \(Q\). If \(Q+\mathbf{q}\) and \(P+\mathbf{p}\) are different points, then \(P+\mathbf{p}+m((Q+\mathbf{q})-(P+\mathbf{p}))\) gives all the points on the real line joining them. So this process produces complex points whose bases are on the line joining the bases of the original two points and whose arrowheads are on the line joining the arrowheads of the original two points. It’s not every arrow from one line to the other, because the multiple of \(Q-P\) that you step away from \(P\) has to be the same multiple of \((Q+\mathbf{q})-(P+\mathbf{p})\) that you step away from \(P+\mathbf{p}\). Here’s a picture:
I want to stress that these are not all of the points on the complex line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\). They are only the ones with bases at points along the line joining the bases \(P\) and \(Q\). There are many many other complex points on the complex line: for most pairs of unreal points, the line joining them has one complex point based at each point everywhere. However, I should say that if you choose your base points all on a specific real line, then the arrowheads will also fall along a line too, because of the exact construction here. This is one of the things I wondered about back in 2022 and never realised I already had the proof right in front of me!
There are a couple of specific situations where this construction is particularly interesting.
First, what if the bases \(P\) and \(Q\) are not different points? That is, what if \(Q=P\)? Then the complex points on the line made by choosing \(m+\mu i\) to be real become \(P + [\mathbf{p} + m(\mathbf{q} – \mathbf{p})]i\). The i-arrow for this point has base \(P\). It has arrow \(\mathbf{p} + m(\mathbf{q} – \mathbf{p})\). And it has arrowhead \(P+\mathbf{p}+m((P+\mathbf{q})-(P+\mathbf{p}))\).
All the bases are the same point! And the arrowheads are points along the line joining the two original arrowheads.
As I said before, this is only some of the complex points on the line joining \(P+\mathbf{p}i\) and \(P+\mathbf{q}i\). We’ll get to the other ones later.
Second, what if the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{q}\) aren’t different points? That is, what if \(P+\mathbf{p}=Q+\mathbf{q}\)? Then \(\mathbf{p}-\mathbf{q}=Q-P\) and the complex points on the line made by choosing \(m+\mu i\) to be real become \(P + m(Q-P) + [\mathbf{p} – m(Q-P)]i\). The i-arrow for this point has base at \(P +m(Q-P)\). Its arrow is \(\mathbf{p} – m(Q-P)\). Its arrowhead is at \(P+\mathbf{p}\).
All the arrowheads are the same point! And all the bases are points on the line joining the original bases \(P\) and \(Q\).
Finally, what if the first point is real? That is, \(\mathbf{p}=\mathbf{0}\), so that \(P+\mathbf{p}i\) is actually just \(P\)? Then the complex points made by choosing the multiplier \(m+\mu i\) to be real become \(P + m(Q-P) + m\mathbf{q}i\). The i-arrow for this point has base at \(P +m(Q-P)\). Its arrow is \(m\mathbf{q}\). Its arrowhead is at \(P+\mathbf{p}+m((Q+\mathbf{q})-(P+\mathbf{p}))\).
Look at the formula for the arrow. All of these arrows are multiples of the unreal point’s arrow, so they’re all parallel to the unreal point’s arrow! In fact, any arrow parallel to the unreal point’s arrow that goes from the base line to the arrowhead line is the i-arrow for a complex point on this line. (And the base line and the arrowhead line both pass through the real point \(P\), because its base is the same as its arrowhead, since there is no arrow.) Look at the picture of this! It’s so satisfying.
I’ll return later to what happens when you create the complex line joining a general unreal point and a real point. Right now, I’ll make good on my promise to investigate special arrangements of complex points that produce special complex lines.
Arrangements of two i-arrows that result in a real line when you join them
As I said earlier, I am going to describe a specific construction from two unreal points, but they will have to be arranged so that various points are not the same and various lines are not the same and not parallel. I’ll investigate now what happens when I don’t have them arranged the way I want.
The first arrangements of complex points are those where the line joining the bases is the same as the line joining the arrowheads. In all of these situations, the line joining the two complex points is a real line.
Two real points
If you imagine real points as i-arrows, then their bases are equal to their arrowheads. So if you have two real points, the line joining their bases is definitely the same as the line joining their arrowheads. And of course when you join two real points you get a real line. But I want to know about all the complex points on this line too.
Consider two real points \(P\) and \(Q\). If I think of them as complex points, they are \(P+\mathbf{0}i\) and \(Q+\mathbf{0}i\), and so the rest of the complex points on the line are:
The base is \(P+m(Q-P)\), which is any of the points on the real line joining \(P\) and \(Q\). The arrow is \(\mu(Q-P)\), which is any vector parallel to the vector from \(P\) to \(Q\). Note that if \(\mu=0\), then there is no arrow so the point is just its base and is a real point. The arrowhead is \(P+m(Q-P)+\mu(Q-P)=P+(m+\mu)(Q-P)\), which could be any point on the line
So the complex points on this line include all the real points on the real line, and also all the unreal points whose i-arrows go from one point on the line to another. This makes sense: the unreal points on the plane are arrows joining two points of the plane, it should be that the unreal points on a line are arrows joining two points on the line. It’s very satisfying.
Since all the unreal points on a real line are arrows along the line itself, this means that there is only one real line that an unreal point can be on – it has to be the real line it is already along! I knew this already, but it somehow still has such a deep appeal for me.
Finally, I want to think about conjugate points. All the conjugate points on this line will be arrows pointing the opposite way, and so they will still be aligned with the real line. This makes sense: the conjugate of a real number is the same real number again, so the conjugate of a real line ought to still be the same real line!
One unreal point whose i-arrow is aligned with a real point
The above construction shows that a real line contains all the complex points whose i-arrows lie along the real line, and so an unreal point can only be on one real line, which is the one its i-arrow lies along. That means that if an unreal point has its i-arrow pointing at or pointing away from or passing through any specific real point, then the line joining the unreal point and the real point is the real line the i-arrow lies along.
But I do want to be sure, so consider real point \(Q\) and unreal point \(P+\mathbf{p}i\) aligned in such a way that \(Q\) is on the extended line through the i-arrow for \(P+\mathbf{p}i\).
That extended line is given by all the points \(P+k\mathbf{p}\) for real number \(k\). So let \(Q=P+k\mathbf{p}\) for a specific \(k\). The points on the line joining \(P+\mathbf{p}i\) and \(Q=P+k\mathbf{p}+\mathbf{0}i \) are given by:
The base of such a point is \(P+(mk+\mu)\mathbf{p}\), which is a point on the extended line through the i-arrow for \(P+\mathbf{p}i\). The arrowhead is
This is also a point on that same extended line. So all the complex points with i-arrows from one point on the line to another are here.
Importantly, you can choose \(m=1+\mu k\) so that the arrow is \(0\mathbf{p}=\mathbf{0}\), and the point is real. Substituting this value of \(m\) into the base gives
That final division is valid because \(k^2+1\) can’t be zero, so there really is a value of \(\mu\) that will produce any point \(P+r\mathbf{p}\) on the line. That is, all the real points on the entire real line aligned with the i-arrow for \(P+\mathbf{p}i\) are indeed here. Which I knew already, but it’s nice for the algebra to bear it out.
Two unreal points whose i-arrows are aligned
If two unreal points have i-arrows that both lie along the same real line, then that real line must be the unique complex line that joins them, since it is a line they are both on, and two points can only be on one line together.
But again, I want to be sure the algebra checks out, so consider two unreal points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) such that the i-arrow for \(Q+\mathbf{q}i\) lies along the extended line through the i-arrow for \(P+\mathbf{p}i\). This means that \(\mathbf{q}\) must be a scalar multiple of \(\mathbf{p}\), and \(Q\) must be a multiple of \(p\) away from \(\mathbf{P}\). That is, \(\mathbf{q}=a\mathbf{p}\) for some real number \(a\) and \(Q=P+k\mathbf{p}\) for some real number \(k\). It’s worth noting that you can’t have both \(a=1\) and \(k=0\) or \(Q+\mathbf{q}i\) would be the same as \(P+\mathbf{p}i\).
The points on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i=P+k\mathbf{p}+a\mathbf{p}i\) are given by:
This is another point on that same real line. So all the complex points with i-arrows from one point on the line to another are here.
The arrow is \((1+\mu k+m(a-1))\mathbf{p}\), which means the point will be real when \(\mu k+m(a-1)=-1\). Since not both \(k\) and \(a-1\) is zero, there are infinitely many solutions for \(m\) and \(\mu\). Thus there are infinitely many real points on this complex line. Consider the base points given by \(P+(mk-\mu(a-1))\mathbf{p}\), and let \(r\) be any real number. For the point \(P+r\mathbf{p}\) to be on the line, it needs to be possible for \(\mu k+m(a-1)=-1\) and \(mk-\mu(a-1)=r\) simultaneously.
Writing both equations to refer to \(m\) and \(\mu\) as the unknowns:
That division is valid since \((a-1)\) and \(k\) are not both zero, so \((a-1)^2+k^2\) can’t be zero. Similar operations produce this value for μ:
\[\mu=\frac{-k-r(a-1)}{(a-1)^2+k^2 }\]
So there is a unique choice of \(m\) and \(\mu\) to produce every real point on the real line that the i-arrows for \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) lie along. That was very satisfying. It wasn’t really necessary, since I already knew all of it was true, but as I said before, it is nice for the algebra to bear it out.
I want to point out that this was the first time two unreal points have been used to make a line. In this case, the line joining the bases is the same as the line joining the arrowheads. If those two are different lines, you won’t get a real line, and I’ll investigate a special case of that next.
Arrangements of two i-arrows that result in an unreal line parallel to a real line when you join them
The next arrangements of two complex points I want to investigate are those where the resulting complex line has no real points at all, but instead all the complex points on it have i-arrows joining a base on one real line to an arrowhead on another real line parallel to it. You could say the complex line is parallel to those real lines.
Two unreal points whose i-arrows share a base
Consider two unreal points whose i-arrows share a base and point in different directions. (If they point in the same or opposite directions, then we have already seen that situation above.) Write them \(P+\mathbf{p}i\) and \(P+\mathbf{q}i\).
The complex points on the line joining \(P+\mathbf{p}i\) and \(P+\mathbf{q}i\) are given by
This point is on a line parallel to the line joining the arrowheads but passing through the point \(P\).
Since neither \(\mathbf{p}\) nor \(\mathbf{q}\) is zero, the two lines are not equal, and every point has a proper i-arrow and is an unreal point. That is, this line has no real points at all.
Two unreal points whose i-arrows share an arrowhead
Consider two unreal points whose i-arrows share an arrowhead and point in different directions. (If they point in the same or opposite directions, then we have already seen that situation above.) Write them \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\).
The complex points on the complex line joining P+pi and Q+qi are given by
Since the two points share i-arrow arrowheads, that means \(Q+\mathbf{q}=P+\mathbf{p}\), and so \(Q-P=\mathbf{p}-\mathbf{q}=-(\mathbf{q}-\mathbf{p})\), which means the points on the complex line are given by
This can be any point on the line joining the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{p}\). The second expression of the working also shows that this line is parallel to the line joining \(P\) and \(Q\), since you arrive at new points by travelling multiples of the vector \(Q-P\).
Once more, all the points on this complex line have i-arrows drawn from one real line to another real line parallel to it.
Two unreal points whose i-arrows have base line parallel to their arrowhead line
The above situation, where the two i-arrows have the same arrow vector, forced the line joining the bases of the i-arrows to be parallel to the line joining the arrowheads. Whenever this happens, even if the arrows aren’t the same, you also get one of these unreal lines parallel to a real line.
Consider two unreal points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\), whose i-arrows are arranged so that the line joining the bases \(P\) and \(Q\) is parallel to the line joining the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{q}\).
Since the two lines are parallel, that means the vector \(Q-P\) is parallel to the vector \((Q+\mathbf{q})-(P+\mathbf{p})=(Q-P)+(\mathbf{q}-\mathbf{p})\). That is, \((Q-P)+(\mathbf{q}-\mathbf{p})\) is a scalar multiple of \(Q-P\). Let \((Q-P)+(\mathbf{q}-\mathbf{p})=k(Q-P)\) for some real number \(k\), which implies that \((\mathbf{q}-\mathbf{p})=(k-1)(Q-P)\).
It is worth stopping at this point to explicitly notice that \(\mathbf{q}-\mathbf{p}\) being parallel to \(Q-P\) goes hand-in-hand with the line joining the bases of the i-arrows being parallel to the line joining the arrowheads.
The points on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) are given by
The base of such a point is \(P+(m-μ(k-1))(Q-P)\), which is a point on the line joining the bases \(P\) and \(Q\).
The arrowhead of such a point is \(P+\mathbf{p}+(m-\mu(k-1)+\mu+m(k-1))(Q-P)\), which is a point on the line through \(P+\mathbf{p}\) parallel to the line joining \(P\) and \(Q\). But the line joining the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{q}\) is the line through \(P+\mathbf{p}\) parallel to the line joining \(P\) and \(Q\)! So once again the complex points on this line have i-arrows drawn from the line joining the bases to the line joining the arrowheads.
Two unreal points whose conjugate points have i-arrows with the same arrowhead
There is one final special case to consider, which will shortly turn out to be one I’ve already covered, but I want to deal with it separately, just to make absolutely certain I can do the construction I want to do.
Consider two unreal points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) whose conjugate points \(P-\mathbf{p}i\) and \(Q-\mathbf{q}i\) have i-arrows that share an arrowhead.
The arrowheads of the two conjugate points are \(Q-\mathbf{q}\) and \(P-\mathbf{p}\), and so:
That vector is parallel to \(Q-P\), and so the line joining the arrowheads is parallel to the line joining the bases, and so the complex line joining the two unreal points is parallel to these real lines, as shown above.
Note I could have used similar triangles rather than vector calculations to show that the two lines were parallel. The triangle made of the red-and-blue arrows and the dashed line is similar to the one made by just the blue arrows and the dot-and-dashed line since the sides are half as long and they share an angle, and therefore the two lines are parallel. It’s lovely.
That’s all the scenarios where joining two complex points results in a real line or an unreal line parallel to a real line.
Arrangements of two i-arrows that result in an unreal line not parallel to any real line when you join them
Finally, I come to the place I really wanted to be, which is having either an unreal point and a real point, or a pair of unreal points, whose i-arrows are arranged in such a way that the complex line joining them is neither a real line nor parallel to a real line.
For an unreal point and a real point, that means the i-arrow for the unreal point is not pointing directly towards or away from or through the real point.
For two unreal points, that means the bases of the i-arrows are different, and the arrowheads of the i-arrows are different, and the arrowheads of the i-arrows for the conjugate points are different, and the lines joining the bases and the arrowheads are not parallel.
Algebraically, the unreal points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) are arranged just right when the vectors \(Q-P\) and \(\mathbf{q}-\mathbf{p}\) are not zero, and are not scalar multiples of each other. That is, when they are linearly independent vectors.
I can use algebra to show that the line joining these points is definitely not a real line nor parallel to a real line, because it has the wrong number of real points. The complex points on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) are of the form:
That is, when \(\mathbf{p}\) is a linear combination of \(Q-P\) and \(\mathbf{q}-\mathbf{p}\). If those vectors are linearly independent, they must form a basis since there’s two of them, so there is exactly one way to write every vector, including p, as a linear combination of them, and so this line has exactly one real point. Importantly, it doesn’t have infinitely many real points, like a real line does, and it doesn’t have zero real points, like an unreal line parallel to a real line.
Also, since every vector is a unique linear combination of \(Q-P\) and \(\mathbf{q}-\mathbf{p}\), that means every point in the real plane can be written as \(P + m(Q-P) – \mu(\mathbf{q} -\mathbf{p})\) for some unique combination of \(m\) and \(\mu\). And so each point in the real plane is the base for the i-arrow of exactly one complex point on this line.
The question is, how do I find those i-arrows that have their base at each point? This is much easier if I have the real point on the line, so I’ll start there.
Finding complex points on the line joining an unreal point and a real point
Consider a real point \(P\) and an unreal point \(Q+\mathbf{q}i\), arranged so that the arrow for the unreal point is not pointing at, away from, or through the real point
I want to create geometrical processes that will find the other unreal points on the line joining these two.
Drawing the i-arrow with a specific base
Using the algebra from earlier, all the complex points on the line are given by the following, for real numbers \(m\) and \(\mu\):
The i-arrow for this point has base at \(P +m(Q-P)-\mu q=P+m(Q-P)+\mu(-\mathbf{q})\). Its arrow is \(\mu(Q- P) + m\mathbf{q}\). Its arrowhead is at \(P+m((Q+\mathbf{q})-P)+\mu((Q-\mathbf{q})-P)\).
Notice how the base is a specific linear combination of \(Q-P\) and -\(\mathbf{q}\) away from \(P\), and the arrowhead is a specific linear combination of \(Q+\mathbf{q}-P\) and \(Q-\mathbf{q}-P\) away from \(P\), and the coefficients are the same! That means if I can locate the base’s coordinates according to the coordinate axes defined by \(Q-P\) and \(-\mathbf{q}\) (I’ll call them the base axes), then what I need to do is locate the point with the same coordinates according to the axes defined by \(Q+\mathbf{q}-P\) and \(Q-\mathbf{q}-P\) (I’ll call them the arrowhead axes), and that will be the arrowhead.
Here’s a picture of the construction of the coordinate axes. The axis defined by \(Q-P\), I’ve called \(x_\text{base}\) and the axis defined by \(-\mathbf{q}\), I’ve called \(y_\text{base}\) . The axis defined by\( Q+\mathbf{q}-P\), I’ve called \(x_\text{arrowhead}\) and the axis defined by \(Q-\mathbf{q}-P\), I’ve called \(y_\text{arrowhead}\) .
Given any point \(M\), I can find its coordinates relative to the base axes by following lines parallel to those axes and comparing where they meet to the lengths of the basis vectors.
Somehow, we have to figure out where the matching coordinates are on the arrowhead axes, and then we can do a similar parallel line process to find the right arrowhead. This is actually much easier than I expected! It turns out I just need to follow the line that told me the base x-coordinate of \(M\) to where it meets the arrowhead x-axis, and the similar for the y-axes, and it will work.
Let me do the algebra.
Suppose \(M=P+m(Q-P)+\mu(-\mathbf{q} )\), for some specific real numbers \(m\) and \(μ\), and consider the line through \(M\) parallel to the base y-axis. The direction of that axis is \(-\mathbf{q}\), and so the points on this line are of the form \(P+m(Q-P)+\mu(-\mathbf{q })+a(-\mathbf{q})\) for real numbers \(a\). This line meets the arrowhead x-axis at a point \(P+b((Q+\mathbf{q})-P)\) for some real number \(b\). This is the same as \(P+b(Q-P+\mathbf{q})=P+b(Q-P)-b(-\mathbf{q})\), which is supposed to be equal to \(P+m(Q-P)+(\mu+a)(-\mathbf{q} )\). Since \(Q-P\) and \(-\mathbf{q}\) are independent, their coefficients much match, and so \(b=m\). That is, this point has the same coordinate on the arrowhead x-axis as \(M\) had on the base x-axis.
Similarly, consider the line through \(M\) parallel to the base x-axis. The direction of that x-axis is \(Q-P\), so the points on the line are of the form \(P+m(Q-P)+\mu(-\mathbf{q} )+c(Q-P)\) for real numbers \(c\). This line meets the arrowhead y-axis at a point \(P+d((Q-\mathbf{q})-P)\) for some real number \(d\). This is the same as \(P+d(Q-P-\mathbf{q})=P+d(Q-P) +d(-\mathbf{q})\), which is supposed to be equal to \(P+(m+c)(Q-P)+\mu(-\mathbf{q} )\). Since \(Q-P\) and \(-\mathbf{q}\) are independent their coefficients much match, and so \(d=\mu\). That is, this point has the same coordinate on the arrowhead y-axis as \(M\) had on the base y-axis.
Now that the correct points on the arrowhead axes have been found, their coordinates can be combined by following lines from each point parallel to the other axis.
So now we have a geometrical process for finding the i-arrow for any complex point on the line joining \(P\) and \(Q+\mathbf{q}i\), starting with its base. You can investigate it further using this GeoGebra activity.
This process is much simpler than the process I had back in 2022, which required finding a lot of other i-arrows before I got to the one I really wanted. Now the first i-arrow I draw is the arrow I want. Plus, once I set up the axes, I can repeat the process as many times as I like to get as many i-arrows as I like.
Complex points whose i-arrows have their bases on one of the axes
If you play around with that GeoGebra activity, you might notice something interesting. The four axes divide the plane into eight slices centred on the point \(P\). An arrow with base in any slice has its arrowhead in the next slice around, in the direction (anticlockwise or clockwise) indicated by the original i-arrow used to make the axes. You can see this in the diagram above.
In particular, if the base of the i-arrow is on one of the axes, then its arrowhead is exactly on the next axis around. Even more, its arrow is parallel to the opposite matching axis. For example, if the base is on the arrowhead y-axis, then the arrowhead is on the base x-axis and is parallel to the arrowhead x-axisaxis; if the base is on the base x-axis, then the arrowhead is on the arrowhead x-axis, and is parallel to the base y-axis.
I can prove this. The points on the complex line joining \(P\) and \(Q+\mathbf{q}i\) are given by
The i-arrow for this point has base at \(P+m(Q-P)+\mu(-\mathbf{q})\). Its arrow is \(\mu(Q- P) + m\mathbf{q}\). Its arrowhead is at \(P+m((Q+\mathbf{q})-P)+\mu((Q-\mathbf{q})-P)\).
If the point has its base on the base x-axis, then its base is \(P +m(Q-P)\) for some \(m\). So its arrow is \(m\mathbf{q}\), which is parallel to the base y-axis. And its arrowhead is at \(P+m((Q+\mathbf{q})-P)\), which is a point on the arrowhead x-axis.
If the point has its base on the arrowhead x-axis, then its base is \(P+m((Q+\mathbf{q})-P)\) for some \(m\), which is equal to \(P+m(Q-P+\mathbf{q})=P+m(Q-P)-m(-\mathbf{q})\). So its arrow is \(-m(Q- P)+ m\mathbf{q}=-m(Q-P-\mathbf{q})=-m((Q-\mathbf{q})-P)\), which is parallel to the arrowhead y-axis. And its arrowhead is at \(P+m((Q+\mathbf{q})-P)-m((Q-\mathbf{q})-P)=P+m(2\mathbf{q})\), which is on the base y-axis.
If the point has its base on the base y-axis, then its base is \(P+\mu(-\mathbf{q})\) for some \(\mu\). So its arrow is \(\mu(Q- P)\), which is parallel to the base x-axis. And its arrowhead is at \(P+\mu((Q-\mathbf{q})-P)\), which is a point on the arrowhead y-axis.
If the point has its base on the arrowhead y-axis, then its base is \(P+\mu((Q-\mathbf{q})-P)\) for some \(\mu\), which is equal to \(P+\mu(Q-P-\mathbf{q})=P+\mu(Q-P)+\mu(-\mathbf{q})\). So its arrow is \(\mu(Q- P) +\mu\mathbf{q}=\mu(Q-P+\mathbf{q})=\mu((Q+\mathbf{q})-P)\), which is parallel to the arrowhead x-axis. And its arrowhead is at \(P+\mu((Q+\mathbf{q})-P)+\mu((Q-\mathbf{q})-P)=P+m(2(Q-P))\), which is on the base x-axis.
So yes, for all four axes, if the i-arrow base is on one axis, then it must be parallel to the other matching axis and its arrowhead must be on the next axis around.
Making spirals
One of the nice constructions that happened in 2022 was making strings of i-arrows on the same complex line all arranged head-to-tail. They would form very nice spirals, moving outwards away from the real point on the line. The four axes can simplify the process of making those spirals: just keep drawing arrows to the next axis around, parallel to the opposite axis to the one you started on. And you can do it backwards too, drawing arrows from the previous axis around, parallel to its opposite axis and arriving at the start of the arrow you’ve got already.
It’s actually quite hard to draw these spirals because the i-arrows get very long and very far from the centre very quickly. In fact, I can figure out exactly how long and how far and how quickly.
The points on the complex line joining \(P\) and \(Q+\mathbf{q}i\) are given by
The i-arrow for this point has base at \(P+m(Q-P)+\mu(-\mathbf{q})\). Its arrow is \(\mu(Q- P) + m\mathbf{q}\). Its arrowhead is at \(P+m((Q+\mathbf{q})-P)+\mu((Q-\mathbf{q})-P)\).
To make a spiral, the base of each new i-arrow is the arrowhead of the previous i-arrow. The first i-arrow is the one for \(Q+\mathbf{q}i\) itself, which has base \(Q\), arrow \(\mathbf{q}\), and arrowhead \(Q+\mathbf{q}\).
The second i-arrow has base \(Q+\mathbf{q}=P+(Q-P)-(-\mathbf{q})\), so its arrow is \(-(Q-P)+\mathbf{q}\), and its arrowhead is at \(P+((Q+\mathbf{q})-P)-((Q-\mathbf{q})-P)=P+2\mathbf{q}\).
The third i-arrow has base \(P+2\mathbf{q}=P+0(Q-P)-2(-\mathbf{q})\), so its arrow is \(-2(Q-P)\), and its arrowhead is at \(P-2((Q-\mathbf{q})-P)=P-2(Q-P)+2\mathbf{q}\).
The fourth i-arrow has base \(P-2(Q-P)+2\mathbf{q}=P-2(Q-P)-2(-\mathbf{q})\), so its arrow is \(2(Q-P)+2\mathbf{q}\), and its arrowhead is at \(P-2((Q+\mathbf{q})-P)-2((Q-\mathbf{q})-P)=P-4(Q-P)\).
The fifth i-arrow has base \(P-4(Q-P)=P-4(Q-P)+0(-\mathbf{q})\), so its arrow is \(0(Q-P)-4\mathbf{q}=-4\mathbf{q}\).
So the fifth i-arrow has an arrow that is four times as long as the first i-arrow, and its base is four times as far from P as the base of the first i-arrow! Repeating this process until the spiral comes back to the first slice again (that is, the ninth i-arrow), the i-arrow is now sixteen times as long as the first arrow, and its base sixteen times as far from P. They do indeed get very long and very far very quickly!
I should stress at this point that this one spiral is not all the complex points on this complex line. Remember there is a complex point with i-arrow at every point in the real plane. Each one of those i-arrows will form its own spiral, so the whole complex line is made of infinitely many spirals all going around the same direction outwards and away from the real point.
Parallelograms and simultaneous spirals
I want to finish this section with a way to draw four of these spirals simultaneously starting with a real point and an unreal point’s i-arrow.
The first stage is to notice that the four axis directions are the edges and diagonals of a parallelogram. Go all the way back to where the four axes were defined from the real point \(P\) and the unreal point \(Q+\mathbf{q}i\). The vector \(Q-P\) and the vector \(\mathbf{q}\) define a parallelogram, and if you put it on either side of the line segment from \(P\) to \(Q\), the arrowhead axes are the two different diagonals of this parallelogram!
If you put more copies on the other side of the base y-axis, you’ll make an even bigger parallelogram. The lines through its centre are the base axes, and its diagonals are the arrowhead axes. These four lines and their directions are all that is needed to draw any i-arrow on the line, so you can use this parallelogram as a tool to draw i-arrows on a complex line.
Here’s the process. It’s based on the arguments earlier about arrows starting on an axis being parallel to the opposite axis and finishing on the next axis around.
Begin with a parallelogram. Draw its two diagonals, and also the two lines through its centre parallel to its edges. These four lines are the base and arrowhead axes, and the centre point is the real point on this complex line.
Draw an arrow along half of each edge of the parallelogram, with the base at the midpoint of the edge and the arrowhead at the corner, and so that the arrows all face the same direction clockwise or anticlockwise. These arrows are the i-arrows for complex points that are all on the same complex line through the central real point.
Connect the bases of these four arrows to make a smaller parallelogram. Notice how this parallelogram also has its diagonals drawn in, and the lines through its centre parallel to the edges.
Draw an arrow along half of each edge of this new parallelogram, with base at the midpoint of the edge and arrowhead at the base of the existing arrow on the edge of the bigger parallelogram. These arrows are the i-arrows for more complex points on the same complex line.
Connect the bases of the arrows just drawn to make another smaller parallelogram. Notice how this parallelogram also has its diagonals drawn in, and the lines through its centre parallel to the edges.
Draw an arrow along half of each edge of this new parallelogram with base at the midpoint of the edge and arrowhead at the base of the existing arrow on the edge of the bigger parallelogram. These arrows are the i-arrows for more complex points on the same complex line.
Continue this process as far inwards as you like.
(Note the process can also be continued outwards, but you have to make a bigger parallelogram with its edge midpoints at the arrowheads of the biggest arrows, and that’s a bit harder.)
I really do love how this is a purely geometrical process you could give someone as an art project and they would never know (unless you told them) that they were drawing part of a complex line. In fact, as an art project, you don’t actually have to draw in each smaller parallelogram. Just lining your ruler up from one base point to the next and drawing the arrow from the axis line to the base of the next arrow will do the trick. One final thing to comment on: All the arrows only use half of a parallelogram edge. What’s going on with the other half of each edge? Well, the arrows along those edges are on the conjugate line! Those spirals still go outwards, but in the opposite direction. This is deeply pleasant to me.
Finding the real point on an unreal line
Now I have various ways to find the complex points on a complex line joining a real point and an unreal point. As long as I have a real point on a complex line, I can find all the other complex points. But what if I don’t have a real point yet?
It’s now finally time to describe the construction from two complex points (whose complex line isn’t parallel to a real line) that finds the real point on the complex line that joins them. Here it is…
Start with the i-arrows of two complex points \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\). (If the bases are equal or the arrowheads are equal, then the complex line joining them is parallel to the real line that joins the other end of the arrows. This will be a real line if it contains both arrows. It will be an unreal line with no real points if it does not contain the arrows.)
Draw the real lines connecting the bases and the real line connecting the arrowheads. (If the base line and the arrowhead lines are the same line, then the line joining the two complex points is that real line. If the base line and the arrowhead lines are parallel, then the line joining the two complex points is an unreal line parallel to both of them and containing no real points.)
Draw i-arrows for the conjugate points. These i-arrows are the same as the ones already there, but pointing in the opposite direction. Draw the real line connecting their arrowheads.
Draw an arrow along the line joining the conjugate arrowheads, from the base line to the line joining the original arrowheads. This arrow is one of the complex points on the original line. Also draw an arrow along the line joining the original arrowheads, from the base line to the line joining the conjugate arrowheads. This arrow is one of the complex points on the conjugate line.
From the base of each of the arrows you just drew, draw a copy of the other arrow going in the opposite direction. The point these two new arrows are pointing at is the real point on the complex line joining the original two complex points. (Note that the two copies of arrows you drew last aren’t actually points on either the conjugate or original line, they just help to find the real point.)
This is quite different to my old construction that find the real point, and I like it much better. It’s simpler than the old one, and also highlights that given a complex line with one real point, its conjugate line has that same real point too. You can play around with moving the two unreal points to see how it changes the location of the real point on the complex line using this GeoGebra activity.
I was able to explicitly prove that the construction works using my algebra process for making points on complex lines. Here is the proof…
Let \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) be two complex points so that the points \(P\) and \(Q\) are different, the points \(P+\mathbf{p}\) and \(Q+\mathbf{q}\) are different, and the points \(P-\mathbf{p}\) and \(Q-\mathbf{q}\) are different, and the three lines joining those pairs of points are not the same lines and not parallel. This happens when the vectors \(Q-P\) and \(\mathbf{q}-\mathbf{p}\) are linearly independent. The complex points on the complex line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) are of the form
for some real numbers \(m\) and \(\mu\). This form will be used multiple times in the proof.
Let \(S\) be the point where the line through the bases \(P\) and \(Q\) meets the line through the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{q}\). Let \(T\) be the point where the line through the bases \(P\) and \(Q\) meets the line through the conjugate arrowheads \(P-\mathbf{p}\) and \(Q-\mathbf{q}\). Finally let \(U\) be the point where the line through the arrowheads \(P+\mathbf{p}\) and \(Q+\mathbf{q}\) meets the line through the conjugate arrowheads \(P-\mathbf{p}\) and \(Q-\mathbf{q}\). (It is possible that some of these are the same point.)
Consider the complex point on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) whose i-arrow has base \(T\). Let this point be \(T+\mathbf{t}i\). The base of any complex point on this line is \(P + m(Q-P) – \mu(\mathbf{q }-\mathbf{p})\) for some real numbers \(m\) and \(\mu\). However, \(T\) is a real point on the line joining \(P\) and \(Q\), so \(T=P+a(Q-P)\) for some specific real number \(a\). Hence
But \(Q-P\) and \(\mathbf{q}-\mathbf{p}\) are independent, so we can equate coefficients to get \(m=a\) and \(\mu=0\).
The arrow for this point is \(\mathbf{t}=\mathbf{p} + \mu(Q- P) + m(\mathbf{q} – \mathbf{p})=\mathbf{p} +a(\mathbf{q} – \mathbf{p})\). The arrowhead is \(P+a(Q-P)+\mathbf{p}+a(\mathbf{q}-\mathbf{p})=P+\mathbf{p}+a(Q+\mathbf{q}-(P+\mathbf{p}))\). This is a point on the line joining the original arrowheads \(Q+\mathbf{q}\) and \(P+\mathbf{p}\).
Now \(T\) is also a point on the line joining the conjugate arrowheads \(P-\mathbf{p}\) and \(Q-\mathbf{q}\), which means for some specific real number \(b\),
This arrow is parallel to the line joining \(P-\mathbf{p}\) and \(Q-\mathbf{q}\), and since it starts at \(T\), which is on this line, it lies along the line itself.
Therefore, the complex point on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) whose i-arrow has base \(T\) in fact has its arrow lying along the line joining the conjugate arrowheads \(P-\mathbf{p}\) and \(Q-\mathbf{q}\), and its arrowhead lies on the line joining the original arrowheads \(Q+\mathbf{q}\) and \(P+\mathbf{p}\). That is, the arrowhead must be the intersection of those two lines, which is the point \(U\) in the diagram above.
Now consider the point \(S\). This point is on the line joining \(P\) and \(Q\), so for some specific real number \(c\),
\[S=P+c(Q-P)\]
But also, \(S\) is on the line joining the original arrowheads \(Q+\mathbf{q}\) and \(P+\mathbf{p}\), so for some specific real number \(d\),
Since \(Q-P\) and \(\mathbf{q}-\mathbf{p}\) are independent, we can equate coefficients to get \(d=b\) and \(c-d=b-a\). Substituting the first into the second gives: \(c-b=b-a\), and so \(c=2b-a\).
Now consider the complex point on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) with base at \(R\). Let this point be \(R+\mathbf{r}i\). Again, the base will be \(P + m(Q-P) – \mu(q -p)\) for some real numbers \(m\) and \(\mu\). This base is \(R\), so
So actually the point on the line joining \(P+\mathbf{p}i\) and \(Q+\mathbf{q}i\) with base at \(R\) is a real point and is just \(R\) itself!
Note that the conjugate line joining \(P-\mathbf{p}i\) and \(Q-\mathbf{q}i\) must have the same real point. Thus, following the same argument starting with \(S\) first instead of \(T\) will show that if \(\mathbf{s}=U-S\), then the point \(S+\mathbf{s}i\) is on the conjugate line and \(T-\mathbf{s}\) is also equal to \(R\).
I have to say I was very surprised this all worked out and I had to check multiple times I wasn’t using circular reasoning. But I really believe it works and I’m so happy, because now I have the algebra to back up what I know to be true.
Conclusion
What a ride it has been over the last few weeks working through all of this! Time to take stock of what I’ve done.
I’ve brought all my calculations for finding new points on complex lines under one overarching algebra system with consistent notation.
I’ve redone the algebra for the move for finding more points on a complex line of stepping bases of i-arrows along a line and stepping the arrowheads along another.
I’ve explicitly used algebra to prove that various arrangements of complex points, both real and unreal, resulting in lines parallel or equal to real lines.
I’ve created a new simpler construction that finds the i-arrow for any complex point on the line joining a real point and an unreal point, and used algebra to prove that it works.
I’ve created a new simpler construction that draws points on a complex line in spirals of i-arrows joined head-to-tail, and pitched it as an art activity.
I’ve created a new simpler construction that finds the real point on a complex line joining two unreal points, and used algebra to prove that it works.
I’m proud of all of it, and I have trouble deciding which part I like best, so I won’t. I’m also very happy to have scratched an itch that has been in the back of my mind for two years. It’s now time to leave the i-arrows to their spiralling for now. If you’ve come this far, I hope you enjoyed it.
Acknowlegements
Thank you to everyone who has listened to me wax lyrical about all of this over the last several weeks, and everyone who has had to wait for other tasks while I thought about and wrote up this maths when I should have been doing those tasks.
A recent video from Howie Hua showed how if you split a collection of numbers into equal-sized groups, then find the mean of each group, then find the mean of those means, it turns out this final answer is the same as the mean of the original collection. He was careful to say it usually does not work if the groups were different sizes. Which got me to wondering: just how much of an effect on the final mean-of-means can you have by splitting a collection of numbers into different-sized groups?
This is how I asked the question:
You have the whole numbers from 1 to 12. You split them into however many groups you like of whatever sizes you like, so every number is used exactly once.
For example, you could split them like this: {2,5}, {8},{1,6,10,11},{3,4,7,9,12}.
Now find the mean of each of these groups.
In the example, those means are (2+5)/2=3.5, 8, (1+6+10+11)/4=7, (3+4+7+9+12)/5=7.
Then find the mean of those means.
In the example, this mean-of-means is (3.5+8+7+7)/4=6.375.
What is the smallest you can make the final mean-of-means, by choosing your groups just right?
Having played with this question online for a day, and in person with others at two sessions of One Hundred Factorial, it has now become one of my favourite problems. It has been particularly fascinating watching other people think about the problem. It seems to defy instincts to begin with, but as people try things out, their instincts develop and they quite quickly have a conjecture about the smallest possible answer. Proving this really is the smallest possible answer was actually quite difficult, though I have to say I really enjoyed the process. The ideas that produced the proof came from multiple people, including me, Lyron Winderbaum, Alex Mackay and James Fitton-Gum..
I’m going to include a proof here, because I love the ideas in it so much, and because I happen to find a certain pleasure in writing up a proof.
While writing it, I was struck anew by how different the process of coming up with a proof is from writing one. You can know a thing is true, and know all the reasons why, and be able to explain to someone why, but it’s not the same thing at all as a formal written proof. A formal written proof is a thing all its own, which requires its own special thought processes different from just coming up with or explaining the ideas in a proof to someone live. For a start, a formal proof usually does not run by just giving suggestive examples like an explanation can. You have to define everything carefully and choose variables and terminology, and choose an efficient and logical order for the statements you make, which usually takes reordering along the way. I happen to like this process, even though it can be hard work.
Anyway, the proof is done and I have written it below. It is slightly less formal than the traditional formal proof, but I like it. if you don’t want to see a proof right now, please look away!
In all that follows, assume that n is a fixed natural number.
Definition 0: Consider a collection of disjoint sets of numbers. When you calculate the mean of each set, and then calculate the mean of those means, the final result is called the mean-of-means of the collection of sets.
For example, the mean-of-means of the collection of sets {2,5},{8},{1,6,10,11},{3,4,7,9,12} is 6.375.
Of particular interest here are the collections of sets that are a partition of the set {1,…,n}. A partition of a set is a collection of disjoint subsets which together contain all the elements of the original set. We are interested in which partition of {1,…,n} has the lowest mean-of-means.
Since the set {1,…,n} is finite, it has only finitely many possible partitions, and so there are only finitely many means-of-means. Therefore there must actually be a lowest mean-of-means.
Lemma 1: Consider a collection of natural numbers that add to n. Of all partitions of the set {1,…,n} into subsets of those sizes, choose one with the lowest mean-of-means. In such a partition, if two numbers are in different-sized subsets, then the larger number is in the larger subset.
For example, if you partition {1,…,12} into subsets of sizes 2,3,3,4, then, out of all the ways to arrange the numbers into sets of those sizes, a way to do it to get the smallest mean-of-means is this: {1,2},{3,4,5},{6,7,8},{9,10,11,12}. A different arrangement of the numbers into the same sized subsets is: {1,2},{3,9,5},{6,7,8},{4,10,11,12}. And that has a bigger mean-of-means.
In the first case, the mean-of-means is 1/4( 1/2 (1+2) + 1/3 (3+4+5) + 1/3 (6+7+8) + 1/4 (9+10+11+12) ) = 1/4( 1/2+2/2+3/3+4/3+5/3+6/3+7/3+8/3+9/4+10/4+11/4+12/4) In the second case, the mean of means is 1/4( 1/2 (1+2) + 1/3 (3+9+5) + 1/3 (6+7+8) + 1/4 (4+10+11+12) ) = 1/4( 1/2+2/2+3/3+9/3+5/3+6/3+7/3+8/3+4/4+10/4+11/4+12/4)
Comparing these calculations, everything is the same except that these two terms in the first expression 4/3+9/4 become these two terms in the second 9/3+4/4. Changing that 4/3 to 9/3 makes the total go up by 5/3. Changing that 9/4 to 4/4 makes the total go down by 5/4. But 5/3 is more than 5/4 and so the total has gone up overall.
This is the argument I’ll use in the proof, though I’ll go the other way, from a partition with the numbers in the wrong order to one with the numbers in the right order.
Proof of Lemma 1: Suppose a partition of {1,…,n} into k subsets has the number p in a subset of size t and the number q>p in a different subset of size s<t. That is, the larger number is in the smaller subset.
The contribution of p and q to the mean-of-means is 1/k (p/t+q/s).
If everything in the partition is kept the same except that p and q are switched, then all numbers will make the same contribution to the mean-of-means except for p and q, whose contribution is now 1/k (p/s+q/t).
So, the change in the mean-of-means is 1/k (p/s+q/t-p/t-q/s) = 1/k (q/t-p/t -q/s+p/s) = 1/k ((q-p)/t-(q-p)/s) = 1/k (q-p)(1/t-1/s)
Now t>s, so 1/t<1/s. This means 1/t-1/s is negative, and since (q-p) is positive, the change in the mean-of-means is negative. That is, the mean-of-means has decreased.
Therefore, if the partition is to produce the lowest mean-of-means, then for every pair of numbers in subsets of different sizes, the greater number must be in the larger subset. Otherwise the mean-of-means could be reduced. End of proof!
Note that technically this proof does not show that there have to be consecutive numbers in each subset. For example, {1,2},{3,5,8},{4,6,7},{9,10,11,12} has the same mean-of-means as {1,2},{3,4,5},{6,7,8},{9,10,11,12}. But it doesn’t have a properly lower mean-of-means, so that’s ok.
Lemma 2: Consider a collection of natural numbers that add to n. Of all partitions of the set {1,…,n} into subsets of those sizes, there is one with lowest mean-of-means where every subset has consecutive numbers.
Proof of Lemma 2: Suppose a specific partition does have the lowest mean-of-means and consider the subsets of size s. Let p be the smallest number in all the subsets of size s and let q be the largest number in all the subsets of size s.
Consider a number c with p<= c <=q. The number c cannot be in a subset larger than s, or it would have to be larger than q, by Lemma 1. The number c cannot be in a subset smaller than s, or it would have to be smaller than p, also by Lemma 1. So c must be in a subset of size s. So, every number from p to q is in a subset of size s.
If there were only one subset of size s, this shows it contains consecutive numbers. If there is more than one subset of size s, consider again the number c in one of these sets.
If there are k subsets in total, then the contribution of c to the mean-of-means is c/(sk) regardless of which subset of size s it is in. So, the numbers in the subsets of size s can be rearranged into other collections of subsets of size s without changing the mean-of-means. Therefore, it is possible to arrange them with consecutive numbers in each subset of size s. End of proof!!
The next thing I want to prove uses a move like the following. Start with these two sets {3,4,5},{6,7,8}. Their means are 4 and 7. Move the 5 from one set to the other to get the new sets {3,4},{5,6,7,8}. Their means are 3.5 and 6.5.
By moving the largest number in one set to become the smallest number in the other, both means have been reduced.
It is always true that if you remove the largest number from a set (that has more than one element), the mean reduces, and it is always true that if you include an extra number in a set smaller than all the ones already there, the mean reduces, but I just feel a need to prove them to myself.
Lemma 3: Suppose a set of numbers S has a number p larger than all the others. Then the set S\{p} has a lower mean than S.
Proof of Lemma 3: Let s be the size of S and the mean of S be σ.
The s-1 numbers in S other than p are all less than p. Therefore their total is less than (s-1)p. So the total of S is less than (s-1)p+p=sp. Hence the mean of S is less than sp/s=p. That is, σ<p.
The total of S is sσ, so the total of S/{p} is sσ-p. Since p>σ, that implies sσ-p<sσ-σ=(s-1)σ,
The mean of S\{p} is (sσ−p)/(s-1) <(s-1)σ/(s-1) =σ. That is, the mean of S\{p} is less than the mean of S. End of proof!
Lemma 4: Suppose T is a set of numbers and p is a number less than every number in T. Then the mean of {p}UT is less than the mean of T.
Proof of Lemma 4: Let the size of T be t and the mean of T be τ. All of the t numbers in T are greater than p, so the total of T is greater than tp. Therefore the mean of T is greater than tp/t=p. That is, τ>p.
The total of T is tτ, so the total of {p}UT is tτ+p. Since p<τ, this implies tτ+p< tτ+τ=(t+1)τ.
The mean of {p}UT is (tτ+p)/(t+1) <(t+1)τ/(t+1) =τ. That is, the mean of {p}UT is less than the mean of T. End of proof!
Now I am ready to prove a different way to reduce the mean-of-means.
Lemma 5: Let k be a natural number with k<=n. Of all the partitions of {1,…,n} into k subsets, the one with the lowest mean-of-means has the numbers k to n all together in the same subset, and all numbers less than k each in their own subset of size 1.
For example, when you split {1,…,12} into 4 subsets, the lowest mean-of-means belongs to {1},{2},{3},{4,…,12}.
Proof of Lemma 5: Suppose {1,…,n} is partitioned into k subsets in such a way that there is a subset S with more than one number in it, and there is another subset T all of whose numbers are greater than those in S.
Let p be the largest number in S. Create a new partition by replacing S with S\{p}, replacing T with {p}UT, and leaving all other subsets unchanged. That is, move the number p from S to T.
Since the largest number p has been removed, the mean of S\{p} must be lower than that for S, by Lemma 3. Since a smaller number has been included, the mean of {p}UT must be lower than that for T, by Lemma 4.
Consider the list of k means from the sets in the new partition and compare to the list of k means from the sets in the old partition. All the means are the same except two of them, which are lower in the new partition. Therefore the total of the k means is lower in the new partition compared to the old partition. Since both lists have k means and the total has reduced, the mean-of-means must have reduced also.
Therefore, if there are any subsets with more than one number, whose numbers are below all the numbers in another subset, the mean-of-means is not the lowest possible for partitions with k subsets.
Suppose we do have the partition with k subsets that has the lowest mean-of-means out of all partitions with k subsets. It is necessarily the lowest mean-of-means for all partitions with its particular list of subset sizes.
By Lemma 1, in such a partition, all the biggest numbers are in the biggest subsets. If there are any other subsets, they must have exactly one number each, otherwise the mean-of-means could be reduced by the procedure above.
If there is more than one largest subset, the numbers in them can be rearranged without changing the mean-of-means, as shown in the proof of Lemma 2, so that the set containing n has consecutive numbers. In that case, this partition could not have the lowest mean-of-means for partitions with k subsets as it could again be reduced by the procedure above.
Since there are k subsets in total and all but one of them has size 1, there are k-1 subsets of size 1. The remaining subset contains the remaining largest numbers. End of proof!
It’s worth noting at this point that the arguments above can actually work for any collection of distinct numbers, not just {1,…,n}. You can even modify them to deal with repeated numbers in the list.
I also realise that you could probably argue this in fewer lemmas, or different lemmas, but I just loved some of the arguments there so much I couldn’t bring myself to whittle it down and risk losing them. Sometimes the most efficient proof is not the most instructive or inspiring.
The final step, which does depend on the actual set being {1,…,n} is to find the optimum number of sets in the partition. To get my first answer for that, I used some calculus. But before I used calculus, I had to figure out a function for the lowest mean-of-means based on the number of subsets.
Lemma 6: Let k be a natural number and consider the partition of the set {1,…,n} which has the numbers k to n all together in the same subset, and all numbers less than k each in their own subset of size 1. That is, the partition {1},…,{k-1},{k,…,n}. The mean-of-means of this partition is 1/2 (k +n/k).
Proof of Lemma 6: The sum of the numbers 1,…,k-1 is 1/2 (k-1)k. The mean of the set {k,…,n} is (k+n)/2. So the total of the means of the subsets in the partition {1},…,{k-1},{k,…,n} is 1/2 (k-1)k+(k+n)/2 = 1/2 ( (k-1)k+k+n) = 1/2 ( k²-k+k+n) = 1/2 (k²+n)
Therefore the mean-of-means of the partition is 1/k ∙1/2 (k²+n) = 1/2 (k+n/k) End of Proof!
Now I am ready to begin finding the number of subsets that has the lowest mean-of-means.
Lemma 7: For each natural number k, consider the partition of the set {1,…,n} which has the numbers k to n all together in the same subset, and all numbers less than k each in their own subset of size 1. That is, the partition {1},…,{k-1},{k,…,n}. The value of k with the lowest mean-of-means is either √n if it is an integer or it is one of the integers on either side of √n.
For example, when you partition {1,…,16}, the lowest mean-of-means happens with √16=4 subsets arranged as {1},{2},{3},{4,…,16}. When you partition {1,…,12}, since √12 is between 3 and 4, the lowest mean-of-means happens with one of either 3 groups or 4 groups.
Proof of Lemma 7: By Lemma 6, the mean-of-means of such a partition is 1/2 (k+n/k)
Consider the function m(k) = 1/2 (k+n/k) for real number k>0. The minimum of this function will happen when its derivative is 0.
dm/dk= 1/2 (1-n/k²) =0 1=n/k² k²=n k=√n (since k is positive)
When k<√n, n/k²>1 so dm/dk=1-n/k² is negative. When k>√n, n/k²<1 so dm/dk=1-n/k² is positive. Thus as k increases, the function decreases until k=√n and then increases again.
If √n is itself an integer, then the minimum possible value of m occurs when k=√n.
If √n is not an integer, then the minimum value of m for integer k will occur at one of the nearest integers to √n. End of proof!
The important question is, if √n is not an integer, which of the two integers on either side is the correct choice for the number of groups? Looking at the list of answers for several values of n, it appears that the correct k is actually the nearest integer to √n. That is, to find the correct k, round off √n in the usual way.
I wasn’t able to use derivatives to prove this, so I went back and re-proved the previous lemma in a different way that let me figure it out. This meant I should probably have removed Lemma 7 from the argument, but I really liked the argument I made, so I have left it in.
Lemma 8: For each natural number k, consider the partition of the set {1,…,n} which has the numbers k to n all together in the same subset, and all numbers less than k each in their own subset of size 1. That is, the partition {1},…,{k-1},{k,…,n}. The value of k with the lowest mean-of-means is ⌊√n⌉, the nearest integer to √n.
For example, when you partition {1,…,12}, since √12≈3.46 which rounds to 3, the lowest mean-of-means happens with 3 groups.
Proof of Lemma 8: By Lemma 6, the mean-of-means for this partition is 1/2 (k+n/k). Let m(k) = 1/2 (k+n/k) for natural numbers k.
Given a natural number k, consider how much the value of m changes between k and the next number: m(k+1)-m(k) = 1/2 ( (k+1)+n/(k+1) ) – 1/2 (k+n/k) = 1/2 ( k+1 + n/(k+1) – k -n/k) = 1/2 (1 + n/(k+1)-n/k) = 1/2 (1+n(1/(k+1)-1/k) ) = 1/2 (1-n/(k(k+1))
This is negative if k(k+1)<n and positive if k(k+1)>n. So the next value of m(k) is lower until k(k+1) ≥ n, and then after that, the next value is the same or higher.
That means the value of k with the lowest value of m(k) is the first natural value of k with k(k+1)≥n.
k(k+1) ≥ n (k+1/2-1/2)(k+1/2+1/2) ≥ n (k+1/2)²-(1/2)² ≥ n (k+1/2)²-1/4 ≥ n (k+1/2)² ≥ n +1/4 k+1/2 ≥ √(n+1/4) k ≥ √(n+1/4) -1/2
So we are looking for the first integer equal to or greater than √(n+1/4) -1/2. Let r= √(n+1/4) -1/2. Then we are looking for ⌈r⌉, the ceiling of r.
Note that ceiling of r must be between r and r+1, so we will find bounds for r and r+1.
First notice r=√(n+1/4)-1/2 > √n-1/2 Now ⌈r⌉ >= r >√n-1/2 So ⌈r⌉>√n-1/2.
Also, from the working above that found the value of r, (r+1/2)²=n+1/4 <n+2√n∙1/2+1 =(√n+1/2)² So r+1/2<√n+1/2 So r<√n
Finally, begin again with (r+1/2)²=n+1/4 (r+1-1/2)²=n+1/4 (r+1)²-2(r+1)∙1/2+1/4=n+1/4 (r+1)²-(r+1)=n (r+1)²=n+r-1 <n+√n-1 <n+√n+1/4 =(√n+1/2)² So r+1<√n+1/2 Now ⌈r⌉<r+1<√n+1/2 So ⌈r⌉<√n+1/2
Therefore √n-1/2<⌈r⌉<√n+1/2.
The integer ⌈r⌉ is within 1/2 of √n, which means it must be the nearest integer to √n. That is, ⌈r⌉= ⌊√n⌉. That is, the value of k with the lowest mean-of-means is ⌊√n⌉. End of proof!
Later, I came up with a different proof of Lemma 8, but I enjoyed the inequality reasoning and algebra from the proof I first made so much I didn’t want to remove it. So here is the alternative as a bonus.
Alternative Proof of Lemma 8: By Lemma 6, the mean-of-means for this partition is 1/2 (k+n/k). Let m(k) = 1/2 (k+n/k) for natural numbers k.
Given a natural number k, consider how much the value of m changes between k and the next number: m(k+1)-m(k) = 1/2 ( (k+1)+n/(k+1) ) – 1/2 (k+n/k) = 1/2 ( k+1 + n/(k+1) – k -n/k) = 1/2 (1 + n/(k+1)-n/k) = 1/2 (1+n(1/(k+1)-1/k) ) = 1/2 (1-n/(k(k+1))
This is negative if k(k+1)<n and positive if k(k+1)>n. So the next value of m(k) is lower until k(k+1) ≥ n, and then after that, the next value is the same or higher.
That means the value of k with the lowest value of m(k) is the first natural value of k with k(k+1)≥n. (This is the same as the previous proof up until here.)
Suppose k is the first natural number with k(k+1)≥n. Then k²+k≥n k≥n-k² n-k²≤k
So, if n is above k² then it is within k of k².
Now (k+1)² = k²+2k+1 = k²+k+k+1 ≥ n+k+1 So (k+1)² – n ≥ k+1
Therefore, n is below (k+1)² and the nearest it can be to (k+1)² is k+1. Hence n is closer to k² than to (k+1)².
If k is the first natural number with k(k+1)≥n, then k(k-1)<n. k²-k < n k²-k ≤ n-1 (since both k²-k and n are natural numbers) k²-k+1 ≤ n k²-n ≤ k-1
So if n is below k², then it is within k-1 of k².
Now (k-1)² = k²-2k+1 = k²-k-k+1 ≤ n-1-k+1 = n-k So (k-1)²-n ≤ -k So n-(k-1)² ≥ k.
So n is above (k-1)² and the nearest it can be to (k-1)² is k away. Hence n is closer to k² than to (k-1)².
Therefore, k² is the nearest perfect square to n, and so k is the nearest whole number to √n. End of proof!
The common part of the two proofs above has something interesting worth noting. When n=k(k+1), then m(k+1)-m(k)=0, so that means the two nearest values of k to √n actually both give the same value for the mean-of-means. For example, the number 12 is 3∙(3+1) so actually the lowest mean-of-means for n=12 happens both for 3 groups and 4 groups.
Now that we know what the partition with the lowest mean-of-means is, all that remains is to find the actual lowest value for the mean-of-means.
Lemma 9: Let k= ⌊√n⌉ and let d = n-k². That is, let d be the signed distance from the nearest perfect square to n. Consider the partition of the set {1,…,n} which has the numbers k to n all together in the same subset, with all numbers less than k each in their own subset of size 1. That is, the partition {1},…,{k-1},{k,…,n}. The mean-of-means of this partition is k+d/(2k).
For example, the nearest perfect square to 12 is 9=3², and the distance from 9 to 12 is 3. So the mean-of-means of the partition {1},{2},{3,…,12} is 3+3/6 = 3.5. For another example, the nearest perfect square to 21 is 25=5², and the distance from 25 to 21 is -4. So the mean of means of the partition {1},{2},{3},{4},{5,…,21} is 5-4/10=4.6.
Proof of Lemma 9: By Lemma 6, the mean of means of the partition {1},…{k-1},{k,…,n} is m=1/2 (k+n/k) = 1/2 (k²+n)/k = (k²+n)/(2k) = (k²+k²+n-k²)/(2k) = (2k²+d)/(2k) = k + d/(2k) End of proof!
It is worth noting that if n is itself a perfect square, then the k in Lemma 9 is exactly √n, and the distance d is 0, so the sum-of-sums of the partition {1},…,{k-1},{k,…,n} is exactly k=√n.
Now that the final answers have been found, I will put it all together into a theorem.
Theorem 10: Out of all partitions of {1,…,n} into any number of subsets, the lowest mean-of-means possible occurs when the number of subsets is k=⌊√n ⌉, the nearest integer to the square root of n. The lowest mean-of-means occurs for the partition which has the numbers k to n all together in the same subset, and all numbers less than k each in their own subset of size 1. That is, the partition is {1},…,{k-1},{k,…,n}. If d = n-k², the signed distance from the nearest perfect square to n, then the lowest mean-of-means is k+d/(2k).
Proof of Theorem 10: The proof is given by all the previous lemmas. I will recap here.
Lemma 1: Out of all partitions with a specific list of subset sizes, the ones with the lowest mean-of-means have larger numbers in larger subsets
Lemma 2: Out of all partitions with a specific list of subset sizes, there is one with the lowest mean-of-means that has consecutive numbers in every subset (proven using Lemma 1).
At this point, we know how to choose a partition to get the lowest mean-of-means for any specific list of subset sizes. The next step is to find the lowest mean-of-means out of all partitions with a specific number of subsets.
Lemma 3: Removing the largest number from a set reduces its mean.
Lemma 4: Including a new smallest number in a set reduces its mean.
Lemma 5: Out of all partitions with a specific number k of subsets, the one with the lowest mean-of-means has the numbers k to n all together in the same subset, and all numbers less than k each in their own subset of size 1 (proven using Lemmas 2, 3 and 4). That is, the partition is {1},…,{k-1},{k,…,n}.
At this point, we have found a specific kind of partition which has the lowest mean-of-means for all the partitions using each specific number of subsets. The overall lowest mean-of-means must belong to one of these few partitions.
Lemma 6: The partition {1},…,{k-1},{k,…,n} has mean-of-means is 1/2 (k+n/k).
Lemma 7: Of all partitions of the form {1},…,{k-1},{k,…,n}, the lowest mean-of-means happens when k is either √n if it is an integer or it is one of the integers on either side of √n (proved using Lemma 6).
Lemma 8: Of all partitions of the form {1},…,{k-1},{k,…,n}, the lowest mean-of-means happens when k=⌊√n⌉ , the nearest integer to the square root of n (proved using Lemma 6).
This now means that the lowest mean-of-means possible for all partitions happens for this specific kind of partition with this specific number of subsets.
Lemma 9: For the partition {1},…,{k-1},{k,…,n},with k=⌊√n⌉, the mean of means is k+d/(2k), where d=n-k², the signed distance from the nearest perfect square to n.
Since this specific partition with this specific number of subsets is the one with the lowest mean-of-means, this is indeed the way to calculate the lowest mean-of-means. End of proof!!
A list of examples to round it out.
The lowest mean-of-means for partitions of {1,…,4} happens with 2 sets {1},{2,3,4} and the lowest mean-of-means is 2.
The lowest mean-of-means for partitions of {1,…,5} happens with 2 sets {1},{2,3,4,5} and the lowest mean-of-means is 2+1/4=2.25.
The lowest mean-of-means for partitions of {1,…,6} happens with 2 sets {1},{2,…,6} and 3 sets {1},{2},{3,…,6} and the lowest mean-of-means is 2+2/4=3-3/6=2.5.
The lowest mean-of-means for partitions of {1,…,7} happens with 3 sets {1},{2},{3,…,7} and the lowest mean-of-means is 3-2/6≈2.67.
The lowest mean-of-means for partitions of {1,…,8} happens with 3 sets {1},{2},{3,…,8} and the lowest mean-of-means is 3-1/6≈2.83.
The lowest mean-of-means for partitions of {1,…,9} happens with 3 sets {1},{2},{3,…9} and the lowest mean-of-means is 3.
The lowest mean-of-means for partitions of {1,…,10} happens with 3 sets {1},{2},{3,…,10} and the lowest mean-of-means is 3+1/6≈3.17.
The lowest mean-of-means for partitions of {1,…,11} happens with 3 sets {1},{2},{3,…,11} and the lowest mean-of-means is 3+2/6≈3.33.
The lowest mean-of-means for partitions of {1,…,12} happens with 3 sets {1},{2},{3,…,12} and 4 sets {1},{2},{3},{4,…,12} and the lowest mean-of-means is 3+3/6=4-4/8=3.5.
The lowest mean-of-means for partitions of {1,…,13} happens with 4 sets {1},{2},{3},{4,…,13} and the lowest mean-of-means is 4-3/8=3.625.
The lowest mean-of-means for partitions of {1,…,14} happens with 4 sets {1},{2},{3},{4,…,14} and the lowest mean-of-means is 4-2/8=3.75.
The lowest mean-of-means for partitions of {1,…,15} happens with 4 sets {1},{2},{3},{4,…,15} and the lowest mean-of-means is 4-1/8=3.875.
The lowest mean-of-means for partitions of {1,…,16} happens with 4 sets {1},{2},{3},{4,…,16} and the lowest mean-of-means is 4.
The lowest mean-of-means for partitions of {1,…,17} happens with 4 sets {1},{2},{3},{4,…,17} and the lowest mean-of-means is 4+1/8=4.125.
That’s enough. Thanks for reading, and I hope you enjoyed it.
These comments were left on the original blog post:
Jim Propp 14 August 2023
Very fun problem, nice solution! (Noticed a typo between Lemma 9 and its proof: 3+3/6 is 3.5, not 3/5.)
David Butler 15 August 2023
Thanks James! And thanks for catching the typo — getting other people to read something is the easiest way to catch typos. 🙂
Calvin Lin 19 August 2023
Note that because of Lemma 2, you do not yet have a proof of the complete description of the optimal sets, even though the list of examples at the end seems to suggest this is what you’re thinking of.
To fill in that gap, we just have to make the obvious observation that for each of those partitions P, there is no other partition P* that can result in P after going through the manipulations described in Lemma 2. (In fact, this is true for the partitions described in Lemma 6.)
David Butler 21 August 2023
I’m not 100% sure what you’re getting at. The lowest mean-of-means for a specific collection of subset sizes happens with bigger numbers in bigger subsets, and the proof of Lemma 2 shows that if there’s more than one set with the same size in that lowest mean-of-means partition then all the rearrangements of those numbers in sets of the same size have the same mean-of-means.
If you have the lowest mean-of-means for a specific number of subsets, it has to be the lowest mean-of-means for its own collection of subset sizes, so must be arranged like I described in Lemma 5.
I have edited the proof of Lemma 5 to make it clearer how the argument works, and this may be what you were talking about after all.
Anonymous 4 September 2023
>> (…) when you split {1,…,12} into 4 subsets, the lowest mean-of-means belongs to {1},{2},{3},{5,…,12}.
This statement is false. Your split does not contain number 4.
David Butler 15 August 2023 Oops. Just a typo! I’ll go fix it now
We at my university regularly sell quite a big lie.
At Open Day and the Ingenuity STEM Showcase and any number of outreach activities, students do puzzles and play with construction toys and walk around with ropes and draw curves on balloons. Whether we say it explicitly or not, there is a message there that says: here at this University, maths is fun. This is a lie.
Maths at university is not fun. There are hours of video content to watch where the presentation is basically slides or handwritten examples. The classes are presentations, possibly with little quizzes breaking them up, or they consist of doing maths problems similar to the relentless weekly quizzes and assignments. Pictures are rare, making sense by manipulating something with your hands is much much rarer, making sense by moving your body is non-existent. The chances to chase your curiosity are few. The chances to have your own thinking validated and celebrated are fewer. It is very far removed from the experience of university maths the prospective students get when they visit us.
We are lying to our prospective students. The experience they have of university maths at our events is a lie.
I do understand that learning does not have to be “fun”, and expecting it to be so all the time is unreasonable and unhealthy. I also understand that ordinary everyday problem-solving and figuring out can feel fun. I understand as well that play, which is essential to learning deeply, is not the same thing as fun. But there is no denying that the activities we do with prospective students are indeed fun, and that experience is not what it will be like at university.
Do I want to change the activities we do with prospective students to look as boring as life will be at uni? Of course not. But there is another way to not lie, and it’s to make your lie true.
One way I make the lie true is to provide One Hundred Factorial, a weekly games, art and puzzle session where students can experience mathematical play without having to be assessed on it. The sorts of things that happen as a one-off at outreach events happen every week at One Hundred Factorial, and I think it would be a good thing to tell prospective students that this exists. (Writing this blog post is partly to help myself pluck up the courage to suggest to the academics in Maths here that they can do so.)
Another way is to actually include some of the features in your outreach activities actually in your teaching. I’ve seen the maths academics do an awesome job of running engaging activities and helping students feel like their efforts are meaningful and valued. They’re good at it. What I want to say to them is this: Perhaps you can actually include some whole-body movement or physical models in your university classes, or at very least in your videos. Perhaps you can actually have some free exploration of new ideas without having to immediately write an assignment about it. Perhaps you can keep the idea of celebrating students’ mathematical thought in the very front of your mind more often when they are doing everyday maths problems or answering questions in the lecture. Even just a little more of any of these things might make university maths a little more like the outreach activities you do so well.
The experience prospective students have in your outreach activities doesn’t have to be a lie. You can make the lie true.
Because [reasons], my game Digit Disguises has been on my mind recently, and reading the original blog post from 2019, I suddenly realised I had never shared my ideas on how to introduce the game to a whole class at once. This blog post fixes that. To keep in the spirit of it, I have not put a link to the original blog post about the game yet, and will introduce it here the way I would in a class. It reads like a recipe, but of course you as the teacher can make your own professional decisions about what to do. I just find that approximately this plan works and want to share it.
Tell the class that you’re going to be playing a game called Digit Disguises in small groups, but first we’ll play a smaller version as a class.
Choose one volunteer to be the One Who Knows by whatever method you like best. You might like to choose someone who would get a boost from succeeding publicly. In my experience, even someone not confident with algebra will totally understand what to do as the One Who Knows, and will get a kick out of being the one with the secret knowledge.
Give the One Who Knows a sheet with A, B, C, D and spaces next to them. Show everyone what the sheet looks like before you hand it over to the One Who Knows. There is a printable version of the sheet at this link.
Ask the One Who Knows to secretly write the numbers 1, 2, 3, 4 next to the letters A, B, C, D in any order they like, but maybe mix them up a bit. Tell them to show nobody, not even you. (I have chosen the numbers 1, 2, 3, 4 for a reason. I deliberately did not include 0 because I think it’s good for people to have the success of come up with the strategies around 0 for themselves later during a real game. Also I need four different numbers so that there is just enough to have to tease apart the logic in this game.)
Tell everyone that One Who Knows has disguised the numbers 1, 2, 3, 4 as letters and their goal is to collectively figure out which number is disguised as which letter. The way they’re going to do this is to ask the One Who Knows to do a calculation using two different letters and +, -, × or ÷. Then the One Who Knows will tell us all what letter is the answer. It’s important that the One Who Knows never says a number, only a letter.
To start them off, ask the One Who Knows for A+B, and remind them that they are not allowed to say a number out loud. There’s a 2/3 chance that the answer won’t be one of the letters, and they’ll ask you what to do. You can say they’ve already done it: if it’s not one of the letters, then just say it’s not one of the letters. (If it doesn’t come up here, you can deal with the “not a letter” thing whenever it first comes up.)
Regardless of what happens with that first question, write down the question and the response on the board/screen/document camera, and then ask the class if they can say what that means for which number is disguised as which letter.
Now ask the class to suggest things to ask, and get the One Who Knows to respond. Write the questions and the answers up as you go. Each time, ask everyone what the response means for what you know about which number is disguised as which letter. You can also ask the person who suggested the calculation why they suggested what they did.
Note: There’s no need to push them too hard on the “what does it mean”. Just one idea is enough each time. However I do think it’s important to make sure they realise at some point that a response of “not a letter” is not actually failure but can give them information.
Note: It is very likely for someone to suggest early on something like A÷A. Tell the One Who Knows not to answer and ask the person why they suggested that, to which they are likely to say that the answer is 1, and then they’ll know what letter 1 is. Celebrate their thinking, because it’s a really important thing about numbers that they’re using there. But then say that actually the rules of the game say you have to use two different letters. In fact, if nobody does suggest a move like A÷A, then it might be worth asking at some point why the rules say you have to use two different letters, and then someone will possibly suggest this as a reason.
At some point, you will have figured out which number is disguised as which letter, and at that point, you should ask ask the class if they are sure. It’s a good idea to ask someone to go through the reasoning so far that got you to the point you are at, and then ask again if they’re sure. When they are, ask the One Who Knows to reveal the sheet and tell everyone if the class is right.
Celebrate the win, and thank the One Who Knows for their help.
Now it’s time to play the game yourselves, but there will be some differences…
This time, you won’t just be disguising 1 to 4, but the digits from 0 to 9. How might that be different?
This time, instead of everyone and the One Who Knows, it will be two teams. Both teams will disguise the digits as letters, and both will take turns ask questions to figure out the other team’s disguises. How might that be different?
This time, when you think you know the other team’s letters and numbers, you can ask them all by saying which letter is which number, but you’ll only get one chance. If you’re right for all of them, you’ll win. If you’re wrong for any of them, you’ll lose. How might that be different?
Split them into teams and put the teams into pairs by whatever method you like best.
Your questions are calculations with two different letters.
Respond with the letter that the answer is disguised as, or say “not a letter”.
Nobody ever says a number …
…until the very last turn when you guess them all.
Feel free to write down whatever you want as you go to help you figure it out.
The handout for the game has all the instructions if you forget.
Let them play and circulate to hear their awesome thinking and point it out.
Stop them after everyone has had a few turns but before anyone has finished the game to have a discussion about how the game is going so far, and to highlight some great thinking you’ve seen. It’s important to remember you don’t have to wait until the end of the game to get some good learning out!
Whenever you choose to end, have a debrief. You may want to ask the class if they have any questions about the game they want to investigate or discuss. Some possible questions are listed in the original Digit Disguises blog post. But basically it’s up to you and your class what you do with the game now.
I hope this is helpful to you. It will certainly help future me when I next introduce Digit Disguises to a group.
PS: Special thanks go to the participants in my 2021 MASA conference session who lived through the first version of this, and to Michaela Epstein, who I discussed it with while originally planning that conference session.
Students find mathematical induction hard, and there is a complex interplay of reasons why. Some years ago I wrote an answer on the Maths Education Stack Exchange describing these and it’s still something I come back to regularly. I’ve decided to post it here too.
Some reasons why students find mathematical induction difficult
These come from a combination of reading various research articles and my own experience helping students in the Maths Learning Centre.
Many students don’t know what proof is.
Many students don’t realise it’s actually about statements.
Many students don’t have experience in manipulating inequalities and divisibilities.
Many students are philosophically uncomfortable with it.
Many students find analogies unhelpful.
Many students don’t know what to focus on to come up with something to prove.
Below I give further explanation of each point.
1. Many students don’t know what proof is
For many students, the problem with induction proofs is wrapped up in their general problem with proofs: they just don’t know what a proof is or why you need one.
Most students starting out in formal maths understand that a proof convinces someone that something is true, but they use the same reasoning that convinces them that everyday things are true: empirical reasoning. That is, in everyday life they can use several pertinent examples to convince themselves something is true, so they do the same with maths. If you ask them to prove that, say, all prime numbers beyond 3 are one more or one less than a multiple of 6, they’ll say “7 = 6+1, 11 = 12-1, 13 = 12+1, 17 = 18-1, so yes it’s true”. They transfer this approach to mathematical induction too [2].
After some experience with maths teachers, students become aware that this is unacceptable. For many it is literally unacceptable in the sense that “my maths teacher doesn’t accept this sort of thing”. So, they learn to apply other methodologies, such as manipulating symbols [2], and think that now this constitutes proof, because it is what is convincing to their teachers. They will attempt to apply symbol manipulations to an induction problem, which often won’t work, because the logical structure of induction is different to their previous experience.
It’s probably a good idea to get the students to discuss what constitutes a proof, and bring out ideas of why it’s necessary to use certain ways of arguing in maths.
2. Many students don’t realise it’s actually about statements.
You can define mathematical induction as being sure the statement “true for n=1” is the truth, and being able to transform the statement of “true for n=k” into the statement “true for n=k+1”. As such, it’s actually something you do to statements, rather than objects or numbers per se. That means it’s much closer to propositional calculus than, say, a geometry or algebra proof. If your students have learned any propositional calculus, this has a chance of help them [1], (though in my experience, sometimes it gets in the way).
In any case, knowing it’s about statements helps them to structure their proof appropriately because they know it hangs upon certain statements in certain places. This is something many students don’t realise about proofs in general, and I find it really helps them to focus their attention when they prove almost anything.
3. Many students don’t have experience in manipulating inequalities and divisibilities
Whether students are attacking induction proofs operationally or with an understanding of how the idea actually works, they often don’t have the operational skill with the types of algebraic manipulations required [3]. Standard fare for induction proofs are inequalities and divisibilities, and most students I meet simply don’t know how to reason with these, even if they’ve done the highest level of maths at school.
For example, consider a sequence of working like this: (n+1)² =n²+2n+1 >2n+2n+1 ≥2n+6+1, since n≥3 >2n+2 =2(n+1) Most students don’t understand that each equality/inequality symbol only refers to the step immediately before it, so that the final statement is actually (n+1)²>2(n+1). They also don’t realise you can replace the “6+1” with “2” because you want to make a statement that one number is bigger than another. Moreover, they don’t have well-developed instincts for what they should choose to do next to advance towards the goal.
Divisibility proofs generally involve rewriting things so that they have the divisor out the front, both for the k and the k+1 case. Many students don’t realise this is what divisibility means, and also have trouble seeing how to split up the expression to sub in the induction hypothesis.
If you show any examples of doing a proof by induction in these situations, you’re going to have to be extremely explicit about your ordinary algebraic reasoning at every step, and how you made the decision of what to do on every line, so that you can help them develop these skills.
4. Many students are philosophically uncomfortable with it
Some students, no matter how many analogies of dominoes or ladders you describe, will still just feel that it’s somehow all a bit too unbelievable. You start by assuming it’s true, and prove something with an unknown value of n, and then somehow the thing is proved for all values of n? It seems a bit too much like Baron Munchausen pulling himself up by his own hair. A common phrase the students utter is “it feels like magic”. They can’t even consider giving it a go with an actual proof assignment question when they feel so philosophically uncomfortable with it.
Partly this is due to them not understanding that you actually aren’t assuming it’s true for any specific value of n at all. You’re doing a thought experiment of what would happen if it was true for n=k [3]. I use the phrase “suppose true when n=k” rather than “assume true when n=k” to emphasise this.
Even if they do realise this, some students still want to somehow prove to themselves that it’s ok to just say “therefore it’s true for all n”. A mathematically mature student will often understand things by proving to themselves that they work (such as deriving the quadratic formula). However, you can’t derive mathematical induction. It’s an axiom. Telling them this removes the burden of proof and they can feel slightly better about using it. Alternatively, you could prove it from the well-ordering of the natural numbers, and that might make them feel better if they think the well-ordering is a more obvious sort of axiom to use.
5. Many students find analogies unhelpful
Everyone always advocates using the domino analogy for teaching induction, but there are problems with this. One is that some students simply will have never seen dominoes set up in a line to knock down! If you’re going to use this analogy, bring actual dominoes!
A second problem is that if you are not very specific about exactly what in the domino analogy corresponds to what in the induction proof, students will develop unhelpful ideas about induction. This of course goes for any analogy you might use, such as climbing a ladder or a game of whispers [4].
Worse than this is the problem that no amount of very salient analogies will actually help them to attack an actual problem when faced with it [5]. Analogies are really only good for helping them understand how induction works in a general way. You’re going to have to give them examples and experiences of the act of coming up with a proof because that is a separate skill from describing how induction works in a general way.
6. Many students don’t know what to focus on to come up with something to prove
Students can actually become quite successful in solving your standard identity, inequality and divisibility induction proofs. But anything other than this leaves them completely stumped. Mostly this is because they have learned some standard classes of problems and how to deal with them. (Many students actually think this is what mathematics is — a list of problems and how to solve each one!)
When faced with something entirely different (eg the Tower of Hanoi, or a visual pattern), they don’t know what to focus on. First, you need to help them find where the natural number is that describes the possible situations [3]. Induction only works on natural numbers, and is the gold standard of proving things for all natural numbers, so you really need to look for a natural number! But it has to describe the situation well.
After this, some authors suggest getting them to focus on a process pattern generalisation [2,3]. That is, figure out how you would move from size 1 to size 2, from size 2 to size 3 etc, and see if you can figure out if the process of moving from one size to the next follows a pattern. This pattern is the shape of your proof of “true for n=k implies true for n=k+1”.
An unhelpful thing to focus on is verifying that the formula works for each number in turn. This is just reinforcing their empirical reasoning, which won’t help them figure out how to prove! [2]
References:
[1] Dubinsky, E. (1986) Teaching Mathematical Induction I, Journal of Mathematical Behavior, 5, 305-317
[2] Harel, G. (2002) The Development of Mathematical Induction as a Proof Scheme: A DNR-Based Instruction, in Campbell, S. & Zazkis, R. (Eds.) Learning and Teaching Number Theory: Research in Cognition and Instruction, Greenwood Publishing Group
[3] Palla, M., Potari, D. & Spyrou, P. (2012) Secondary school students’ understanding of mathematical induction: structural characteristics and the process of proof construction, International Journal of Science and Mathematics Education, 10, 1023-1045
[4] Ron, G. & Dreyfus, T. (2004) The use of models in teaching proof by mathematical induction, in Proceedings of the 28th Conference of the International Group for the Psychology of Mathematics Education, 113-120
[5] Segal, J. (1998) Learners’ difficulties with induction proofs, Journal of Mathematical Education in Science and technology, 29, 159-177
This comment was left on the original blog post:
Thoa 6 June 2023
I remember trying to learn what proofs were around this time last year. I want to expand on number 6. I did not know what I was looking at when I was watching the proof happen. There’s a bunch of statements and quite often I don’t know what I am trying to prove. I have seen another student this year also not quite understanding the formula given by watching them think what the first term was. It wasn’t as complicated as they thought it was. This. This happened to me. Soon after that, I was able to do some but not all. It just comes with practice after all.
This is a photo of the entrance to my Maths Learning Centre. What do you notice?
There are many many things to notice in that photo, and if you ever want to ask me about any of them, please do. Today, the thing I want to focus your attention on is the empty space right at the front as you walk in. Every so often someone asks me why I leave that space empty and I don’t put an extra table there, and there are a couple of very good reasons why.
First, the space isn’t empty: it has the floor graph in it. The floor graph used to be at the back between the % and the 3 on the wall, but one day I realised that I could have a bit of extra space for it if I put it in the entrance. I also hoped it might send the message to people arriving not only that maths is a thing that happens here, but also that we do things a little differently to your regular university maths classroom. One day I will write about the floor graph to tell you all about how we use it, but there is one purpose I want to tell you today: the floor graph helps us to break students out of staring forlornly at their page or screen. The open space on the floor graph gives a sense of physical freedom, which can translate to a sense of mental freedom.
The openness of the floor graph space was the main reason I moved it to the entrance, actually, because it makes the space easier to enter, for several different kinds of students:
We regularly get tours of new students or prospective students come past the MLC, and with an open space in the entrance, we can bring those tour groups right into the MLC, rather than standing outside and pointing. The experience is so much realer if you can come right in and stand surrounded by the art and whiteboards. They can remember that we asked them to come all the way in. Without the space in the entrance, we’re just pointing from outside and they miss so much.
Students who are familiar with the MLC stand on the floor graph for a moment when they arrive and look around to find a good table, or other students they recognise. The empty space allows them to take a moment to make a choice, and to prepare themselves for working in the space.
Students who are not familiar with the MLC have a place to stand looking lost. (We tutors even call it the “lost soul zone”.) When there was a table in the entrance, people wouldn’t want to be too close to the students already studying there, so newcomers would do their unsure dithering stance outside where we couldn’t see them, and more often than not they would just leave without us ever knowing. The emptiness of the space now means that they can come in without feeling like they’re encroaching on the work of the students and staff already in the room. Just like our regulars, they can prepare themselves for asking for help while actually being in the room. And since they’re in the room, we can see them and go up to them to ask what they are looking for.
Without an empty space in the entrance, we would not be able to welcome as many new students to the MLC as we do. The emptiness is important to provide space for the complex process of deciding whether and how to engage with us. I am so happy I managed to created the space to enter.
This blog post is about the book You’re Not Listening by Kate Murphy, and in particular my reactions to it from a teacher’s perspective.
First, I want to apologise to Chelsea Avard for borrowing the book from her little student leadership library and holding onto it for a whole year while I got round to reading it and then got round to writing about it. Thanks for your patience and thanks in advance for forgiving me for the slightly battered state the book will be when I return it.
Second, I want to say what an excellent book it is. It has a lot to say about what listening is and is not and how it makes a difference to us in everyday life. It’s full of vivid stories to illustrate and lovely turns of phrase, and it is very clear that Murphy clearly researched it extensively. I would recommend anyone to read it.
Ok, now on to the actual purpose of the blog post. The book is not a teaching book, and it doesn’t even mention teachers as far as I recall, but I am a teacher and I can’t help reading it from that perspective. As a teacher who mostly works one-on-one with students, what Murphy had to tell me felt particularly relevant. And yet I’ve had trouble organising it into a cohesive blog post. I have decided to give up and just list some things I learned. Most of them are about teaching. Some of them are just life lessons. I may or may not distinguish them.
Thoughts from the book
Your brain thinks faster than people speak. Use it to listen better.
In Chapter 6, Murphy clearly describes exactly how conversations feel to me, which is that the person is talking, but my mind is buzzing all around thinking about all sorts of things. Almost all of those things are related to what the other person is saying – memories their stories brought up, how I feel about the way they’re talking, what I am going to say next – but only a small amount is actually about the other person’s meaning.
Murphy asserts this is totally normal because your brain thinks at least twice as fast as the other person speaks. Reading that, I was so glad that I wasn’t some sort of freak. However, she recognised that the problem with this normal state is that you can get wrapped up in your thoughts and miss what the other person is saying. Also, she points out that a lot of what people say isn’t contained in the words they speak, but in their silences and their body language and the things they look at.
What she suggests is to use all that extra mental bandwidth to listen better. Instead of letting your mind wander or worrying about what to say next, direct your attention to all those extra things that aren’t audible that might tell you more about what they mean.
This spoke to me powerfully as a teacher, particularly the bit about missing important things when you are planning what to say. In the MLC with a student, my mind can be very much occupied with what explanation or example I’m going to give the student in response to what they’re saying. But in worrying about this while they’re talking, I am missing important things. I’m missing what they’re telling me about how they feel about their maths. I’m missing the pauses that tell me when they’re figuring things out or struggling to articulate their thoughts. I could be a much more responsive teacher if I used my mental powers to think about their meaning while they are talking rather than think about mine.
You don’t have to respond right away.
The big problem in my line of work with the advice above to use your brain’s bandwidth to focus even more on listening and watching is that you really do feel like it’s your responsibility to respond immediately as soon as the student stops talking with useful explanations or advice. The desire to plan your explanation while they are still talking isn’t just your mind wandering, it’s motivated by a fear that you need to be immediately helpful and clever.
However, Murphy argues that actually silences are really important in conversation. Space between speakers allows everyone time to process. And most importantly, it’s a sign of respect for the speaker that the listener takes time to process. She relates stories of many different cultures around the world where they have an expectation of silence in conversation as a sign of respect. I know students have said to me they appreciate me telling them “just let me think about that for a minute” because then they know I’m working on something good.
I’ve heard this advice about silence before, for example in Making Number Talks Matter, and also in a recent UX Research Methods training. But not until reading Murphy’s book did I connect it to the idea of giving yourself the space to really listen in the first place. Knowing that it’s ok and actually respectful to think quietly before responding to students means that you are free from the expectation to have something ready and you can actually focus on listening interpretively. It has taken a huge worry off my shoulders as a teacher that I wasn’t even aware I was worrying about before.
Assumptions stop you listening.
Murphy has a whole chapter titled “I know what you’re going to say: assumptions as earplugs”, and that title really does sum it up extremely well. It just shows her skill as a writer that she can pack so much meaning into just a title.
The idea is that when you assume things about people, it stops you listening to them clearly, and sometimes stops you listening at all. If you literally think you know what someone will say, then you think you have no need to hear it.
The most shocking thing Murphy does is explain how finishing someone else’s sentences is proof that you’re not actually listening. Yes two people being able to finish each other’s sentences is often used as proof that two people are in sync, but she related stories from marriage counsellors who noted that couples who are really familiar with each other just assume the other person doesn’t need to be told things or doesn’t need to tell them things and so people never find out important stuff.
This idea that assumptions stop you listening really gave me pause, because I have seen it first-hand in my own work. A student asks me a question about Question 3 on their assignment, and I assume they are struggling in the same place I would struggle, so I tell them how to deal with that part, even though it might not be the place they need help at all. A student is studying engineering, so I assume they don’t want to know the theory behind the maths and I only tell them procedures, even though they might actually want to make sense of it. A student asks a question about the topic at hand and I just respond to the key words and the question I assume they probably have, but their actual question is about quite a different thing. A student is slow to answer my question, so I assume they don’t know and I answer for them, but actually they are just taking time to figure out how to say what they want to say.
There are so many times when I assume what a student wants or needs rather than listening to their actual words and demeanour or seeking more information. I really need to turn off my assumptions and find out more about the actual situation before responding.
People won’t tell you things if they don’t think you want to hear them.
There were a couple of quotes close together that really spoke deeply to me.
Researchers at the University of Utah found that when talking to inattentive listeners, speakers remembered less information and were less articulate in the information they conveyed. Conversely, they found that attentive listeners elicited more information, relevant detail, and elaboration from speakers, even when the listeners didn’t ask any questions. So if you’re barely listening to someone because you think that person is boring or not worth your time, you will actually make it so.
Think of how you, yourself, might tell different people different things. It doesn’t necessarily have to do with the type of relationship you have with them or the degree of closeness. You might have once told a stranger something you hadn’t told anyone else. What you tell, and how much you tell, depends on how you perceive the listener at that moment. And if someone is listening superficially, listening to find fault, or only listening to jump in with an opinion, then you’re unlikely to make any kind of meaningful disclosure and vice versa.
The things that spoke most deeply to me were the things that can derail the students I work with saying anything useful:
Not worth my time Phew this one is hard. All those second- and third-year students who I have said are second and third priority after first-year students, what does knowing they’re not my priority do for them sharing things that are important? But they still have deeply important things about studying and learning maths to discuss, even if I don’t know their content.
Listening superficially I can do this so easily. I can wait until they say a key word and then launch into an explanation in response, and totally ignore the things they’re saying about how they feel about maths and their experiences studying it, or just ignore how what they say tells me about how much they already understand.
Listening to find fault How many students’ experience of maths in the past is that someone is waiting for them to be wrong? Knowing this, how might they shut themselves up if the first experience with me is telling them something they did wrong?
Listening to jump in with an opinion Phew this is so easy to do as a teacher. They’ve come for help, so I feel the need to provide it as soon as possible, so I wait until they say something I could comment on and make the comment. But they may not want a comment on that thing and they are unlikely to give me what they really want because they know I’m not really listening.
I believe that to be the best teacher and support worker I can be, I need to know the students’ context and know what they already know, but if I really do believe that, then I need to get out of the way so that they will actually tell me what I need to know in order to help them.
Other people’s thoughts really are more interesting than anything you have pre-prepared to say.
The previous section was about how the way you (don’t) listen affects how much people tell you, and I made it all about how I need information to help people the best I can. Except the thing Is, I think people would prefer someone who actually was interested in what they have to say for its own sake, rather than just as a means to an end. I have been working on fostering the belief that all students have existing thoughts worth listening to, and this book confirmed how true and important this belief is.
This quote says it clearly:
The most valuable lesson I’ve learned as a journalist is that everybody is interesting if you ask the right questions. If someone is dull or uninteresting, it’s on you.
Also it’s just a vibe running through the whole book: it’s not just that it’s good to listen to people, but that people are worth listening to. Listening can be hard work, but it is worth the work, because everyone has something interesting to say, something you can learn from.
It’s not about you: shifting versus supporting.
Murphy spends a lot of time talking about how much time we spend talking and not listening, even when we think we’re listening. She describes two main types of response a listener makes: the shift response and the support response. When you make a shift response, you shift the focus to yourself; when you make a support response, you keep the focus on the other person, supporting them to continue and to share what’s important to them. This is one of her examples:
Sue: I watched this really good documentary about turtles last night. Bob: I’m not big on documentaries. I’m more of an action-film kind of guy (shift response)
Sue: I watched this really good documentary about turtles last night. Bob: Turtles? How did you happen to see that? Are you into turtles? (support response)
I have to say I am repeatedly guilty of the shift response. Just now I responded to someone telling me about something that happened to them with a story about my own experience, when I could have so easily asked them more about theirs. Not to say you should never talk about your own story in a conversation, since of course a conversation moves both ways, but you miss out on so much if you do it too early.
Support responses tend to be questions, where you seek more information from the speaker, but there are lots of questions that are still shift responses. There are ones where you set up what you want them to say such as, “Wouldn’t you agree that …?”, and the ones which really just describe your own thoughts or shift the conversation to an entirely new topic that was on their mind already. I found it appropriate that the example of this cited in the book was from an academic at a conference.
In my work in the MLC, I am guilty of shift responses so often, especially when the conversation turns to study skills or experience in classes. I always end up just saying what my experience is, when I could learn so much more by finding out about the student’s. You could argue that the students need to know they’re not alone in their frustration, and you could argue that the might appreciate knowing how someone else dealt with similar situations. However, listening just a little longer will also let them know their feelings are valid, and will tell you whether they need advice at all or just want to vent. (Not to mention that my experience as a student is now 20 years old, and might not be as relevant as I feel it is.)
Also I am suddenly reminded of the distinction between focusing and funnelling questions: a focusing question while a student is doing problem-solving helps a them student focus on the relevant details at hand so they can use it in their problem-solving, whereas a funnelling question pushes them towards a path you the teacher have in your head. This is very similar to the concept of the shift and the support response. Both shift responses and funnelling responses are about you, whereas support responses and focusing responses are about them.
The listening itself is what helps people.
The final thing that stood out to me is that people don’t usually want or need your advice. The very act of listening supportively helps people to achieve clarity and sort out what they want to do. Here’s a useful couple of quotes:
Being aware of someone’s troubles does not mean you need to fix them. People usually aren’t looking for solutions from you anyway; they just want a sounding board. Moreover, you shut people down when you start telling them what they should do or how they should feel. … Your answer to someone else’s deepest difficulties merely reflects what you would do if you were that person, which you are not.
The solutions to problems are often already within people, and just by listening, you help them access how best to handle things, now and also in the future. … If you jump in to fix, advise, correct, or distract, you are communicating that the other person doesn’t have the ability to handle the situation.
This is a really hard thing to hear because we are so used to providing advice as the way to help people. It’s particularly hard to hear for someone in my line of work, where people are literally talking to me because they do actually want help. But if I really do believe that people have thoughts of their own and I really do believe that all people are capable of figuring stuff out, then the best way to show that belief is to listen to them. And if I’m honest, whenever I’ve let them, they really do surprise both me and themselves with what they figure out on their own.
Conclusion
This book was full of such interesting and compelling stuff. I’d recommend it to anyone to read. I think I have listening on my mind even more than before after reading it, both in my life and in my work in the MLC. To sum up, here are my titles from above again, to hold onto for the future:
Your brain thinks faster than people speak. Use it to listen better.
You don’t have to respond right away.
Assumptions stop you listening.
People won’t tell you things if they don’t think you want to hear them.
Other people’s thoughts really are more interesting than anything you have pre-prepared to say.
It’s not about you: shifting versus supporting.
The listening itself is what helps people.
Seeing them all together really highlights to me how much they interact with each other, and how much they all hang off the idea that other people have a lot to say that is worth hearing. Thanks for reading.
Listening is one of the most important aspects – no, scratch that – the most important aspect of my work in the Maths Learning Centre.
It is not obvious to people starting out tutoring in the MLC that this should be the case. To a beginning tutor, it seems that it’s their job to explain things to the students, and to show them how to do stuff. But even if the actual goal was to explain, you can be much surer which explanation to give the student if you first listen to their current understanding. More importantly, you can never improve as a teacher unless at some point you listen to the students to see how well your explanation has gone.
But how do you go about doing the business of listening? This blog post is about my interpretation of a framework that describes different levels of listening for the purposes of teaching, which I read about in two papers:
Davis, B (1997) Listening for Differences: An Evolving Conception of Mathematics Teaching, Journal of Research in Mathematics Education, 28, 255-376
Yackel, E, Stephan, M, Rasmussen C and Underwood, D (2003) Didactising: Continuing the work of Leen Streefland, Educational Studies in Mathematics, 54, 101-126
I spoke at a conference about this framework some years ago, and I have been meaning to write about it ever since. I am finally actually writing about it now (and you are reading it). My thinking has evolved a little since then, so you get the updated and extended version.
The papers
Davis 1997
In the first paper, Davis tells us about how he and schoolteacher Wendy reflected on the types of listening Wendy did in her classroom, and how they were related to her beliefs about what mathematics is and what the teacher’s role is in helping students learn it. It is a truly fascinating and powerful paper and I recommend everyone read it.
Davis notes from previous research that “the quality of student articulations seemed to be as closely related to teachers’ modes of attending as to their teaching styles”, which is a very deep observation. Before, I said that at the very least a teacher needs to listen in order to figure out what to do next, but this says even more that the way you listen may change the very things the students say. Davis goes on to give three vignettes from Wendy’s classroom to display three types of listening.
1. Evaluative listening
When a teacher is listening evaluatively, their reason for listening is to evaluate the correctness of what the student is saying. Ultimately, they are “listening for something in particular, rather than listening to the speaker”. The vignette describes a whole-class discussion where student responses are dismissed until the exact right one was finally accepted. Even right responses that were perceived to be in the wrong form were dismissed. This reminded me of so many times when I had been a frustrated student in such class discussions (and several times I had been a teacher leading one).
I had never quite put my finger on why this felt frustrating until reading this quote from the paper: “No one is attending to the answer in a way that will make a difference to the course of subsequent events”. The teacher in such discussions is waiting for the right response in order to continue on their pre-planned course. As Davis says of Wendy’s vignette, it was “a teaching sequence that seemed impervious to student input”. It sounds harsh, but Davis was more forgiving than that. He noted that Wendy was indeed seeking information from the students. She could see that the students were or were not able to give the responses she hoped for, and she could see that her lesson was more or less successful based on how quickly students could use her explanations to produce the right answers. The listening was doing exactly what she wanted it to do: evaluating.
2. Interpretive Listening
When a teacher is listening interpretively, their reason for listening is to interpret what ideas are actually happening inside the student’s minds. They are still usually seeking to bring students to the understanding they perceive as the correct one, but now they do it through figuring out how to talk about ideas in shared ways that move students forwards in their thinking.
Davis notes that teaching sequences with an interpretive listening stance need to have materials that “serve as a commonplace for learners to talk about ideas, enabling the process of re-presentation and revision”. For example, in the vignette, Wendy used two-coloured chips to help her students talk about adding and subtracting negative numbers. In my own teaching in the MLC, drawings or play dough often play this role.
3. Hermeneutic Listening
When a teacher is listening hermeneutically, they are listening not only to interpret what their students are thinking, but also to understand how their own thinking relates to that, and how the group as a whole understands. This is my description of it, anyway. Davis has several long paragraphs discussing philosophical and theoretical standpoints, which are a bit heavy (though his style makes it much lighter than I’m sure it could have been). The two main takeaways for me are that understanding isn’t only something that lives in one person, but lives in the shared communication of many, and that teachers listen not just to help students grow their understanding but also to change the teacher’s own understanding. A relevant quote: “Instead of seeking to prod learners toward particular predetermined understandings, Wendy seems to have engaged, along with her students, in the process of revising her own knowledge of mathematics.”
Davis notes that this type of listening seems to go hand in hand with a teacher’s conception of what mathematics itself is. You have to be prepared to believe that mathematics concepts have multiple valid ways to understand and describe them, and that mathematics is at least in part a construction of a community, all of whom (including novices) have a part to play in the construction. Otherwise you won’t be ready to listen in this way.
A final comment on the terminology… The word “hermeneutic”, no matter how often I look up definitions, still remains more-or-less meaningless to me. It seems to refer to a type of inquiry that in itself seems difficult to describe and has different meanings in different disciplines, so I can’t borrow meaning from whatever it means elsewhere to make it meaningful in this context, like Davis seems to have done for himself. This makes it hard for me to hold onto the framework.
Yackel et al 2003
In the second paper, the authors are thinking about how teachers structure and restructure their instruction, a process they call “didactising” after Leen Streefland. The reason the paper is here in a post about listening is that the way that you get information about what needs reworking in your instruction is to listen to the students.
I have used the word instruction, as opposed to teaching, because it’s the word the authors used. And they really do seem to be thinking about instruction, in the sense of a sequence of explanations and activities you do with students. The main theme of the paper is about how listening to students helps design these instructional sequences, which I do not question the importance of. It’s just that the overall feeling I get is that students aren’t quite real people but sources of data, and that making good instructional sequences is a good in an of itself, as opposed to being something for the students. I’ve been a bit too dramatic there, and it’s not really as bad as I’ve made it sound, but still my feeling is that it dances a little too far from viewing the students as people.
Anyway, the most useful thing in the paper for me was a new terminology for what Davis called hermeneutic listening; these authors call it “generative listening”.
3. Generative listening
These authors decide to use the word generative rather than hermeneutic because it’s easier to process for their purposes. They say, “Listening in this way can generate or transform one’s own mathematical understandings and it can generate a new space of instructional activities.” While Davis was more focused on the way that hermeneutic listening changes the listener and the community’s understandings, these authors are more focused on the way generative listening generates new instructional activities. I’m happy to have both in my life. I think it’s important to recognise that teachers still have to decide what to do each day and that listening can help them make those decisions!
I’ll finish off with three questions the authors list to help people focus on generative listening, which really do bring it back to the students as people at the last moment: “How does student thinking suggest alternative ways of thinking about particular mathematical ideas? How does student thinking suggest what mathematical ideas are experientially real for them? How can the instructional sequence be redesigned to capitalize on the fresh points of view that students offer?”
Some thoughts
From these two papers, we have a framework with three levels of listening: evaluative, interpretive, and generative. The authors of those papers focused a lot on the mindset of the teacher, and how this makes a difference to how you attend to what the students are doing and saying. Davis talks a bit about the kinds of questions people ask when they have those mindsets. But it occurred to me that even if you have a particular mindset, if you ask the wrong questions, you still won’t get the information you need. So yes the kind of question you ask is evidence for the kind of listening you hope to do, but also the kind of question you ask can also dictate the kind of listening you have to do, because you will only get certain kinds of responses.
For example, if you ask a yes-or-no question (eg Is this a subspace or not?) or a direct question about factual information (eg What is the definition of subspace?), you are unlikely to get much information about what a student is thinking, even if that’s what you hoped for. You will have no choice but to simply evaluate their response.
And there is one question that is famous for giving you no choice for what to listen to: “Does that make sense?” If you ask this of a whole class, students will usually give no response at all. If you ask it of a single student, they’ll say “yeah ok”. So basically it tells you nothing at all: to ask this question is to give no opportunity for you to listen. So actually there is another lower level of listening: not listening.
And if we’re talking about not listening, then there is something worse than asking “Does that make sense?”, which at least shows you think things ought to make sense, and theoretically has a chance of a student saying “no” and so giving you some information to work with. What’s worse is asking no questions of any kind. It is amazing how often a maths teacher, even one-on-one, will speak continuously for half an hour with no opportunity for the student to say anything. I always feel such a sense of shock and shame when I realise I’ve done this and that I have absolutely no idea how the student is going.
I think sometimes the impulse to talk continuously comes from a belief that it’s your main job to provide explanations, and sometimes it comes from believing in the power of an explanation you’ve worked hard to perfect. However, even if maths teaching were transmission, that process can’t possibly be perfect, and so you really do need to check in every so often! As Davis says, “Implicit in the act of questioning is a certain lack of faith in the transmission process.” I think everyone needs to have that certain lack of faith.
So it’s good not to have total faith in the power of a single explanation. But what should you have faith in? I think you need to have faith that students actually are thinking. Implicit in the interpretive listening stance is the assumption that there is something to listen to. You have to believe that students have ideas if you seek to interpret them. If you don’t believe they do have ideas already, then of course you don’t seek to listen to them. For me, this is a huge part of working in the MLC that changes the whole approach. The next level above this is to believe that students have ideas that can change your own, which is where generative listening lives.
My version of the framework
So, finally, this is my interpretation of the listening framework of Davis (with the third level renamed by Yackel et al). There are a lot more aspects to this, such as the nature of the teacher’s role, but this version helps me think about what I am doing with students on the fly. You can download a handout PDF version of the framework if you want.
Level 0: Not listening
Goal:
Tell what the teacher thinks is important
Give clear explanations
Types of questions:
Not asking questions
“Does that make sense?”
Beliefs:
Faith in the power of the teacher’s explanation
Students are waiting for your ideas
Level 1: Evaluative listening
Goal:
Judge student responses against a standard
Get a specific response so you can continue the plan
Types of questions:
Yes/no questions
Direct questions about raw information
Results of calculations
Beliefs:
The teacher’s explanation is not perfect
Students are waiting for your ideas
Level 2: Interpretive listening
Goal:
Decipher the sense that students are making
Understand student thinking
Create a shared language to describe thinking
Types of questions:
Open-ended questions about thinking or process
Beliefs:
Students are reasoning
Student ideas are worth listening to
Level 3: Generative listening
Goal:
Jointly explore ideas
Discover new ways to think about or to learn concepts
Types of questions:
Open-ended questions about thinking or process
What-if questions and I-wonder questions
Beliefs:
Students are reasoning
Student ideas are worth listening to
Student ideas can change yours
Final thoughts
I have deliberately numbered the types of listening and called them levels, because I wanted to explicitly say to myself that some are higher than others. However, I don’t want to say that you should never seek to provide clear explanations and never listen evaluatively. Of course you should explain things when you need to, and of course there are times when you need to know students can do things in a standard way. And I also don’t want to say you should spend all your time listening generatively. That would be exhausting for everyone. It’s just that the types of listening definitely do progress in how student ideas shape what happens, and it is definitely a good thing for students to feel that what they think and do makes a difference to the outcome.
What I want is to always be open to the opportunity of finding out how students think and possibly having it change the way I think. I also know that while beliefs definitely guide actions, it also works the other way too. If I spend all my time talking, I may come to believe implicitly that the students have nothing to say. If I spend all my time evaluating against a standard, I may come to believe implicitly that the students have nothing wonderful to say. I need to actively work in opportunities to listen at the higher levels, so that I never go too long without them.
In daily work, where I spend most of my time one-on-one with students, this is even more important. Because when you’re right there next to the student, what a waste it would be to never hear the wonderful things they have to share, or to never make something wonderful together.
