Author: davidkeithbutler

The denominator of the rational roots

Introduction

In the previous post, I factorised a quadratic equation with a fraction coefficient $x^2+\tfrac16 x-2$ by thinking of numbers that add to 1/6 and multiply to -2. This is how I did it:

2/6+(-1/6)=1/6, 2/6×(-1/6)=1/3×(-1/6)=-1/18, which is not low enough.
3/6+(-2/6)=1/6, 3/6×(-2/6)=1/2×(-1/3)=-1/6, which is lower but not low enough, and I’ve got quite a long way to go.
7/6+(-6/6)=1/6, 7/6×(-6/6)=7/6×(-1)=-7/6, so much closer.
8/6+(-7/6)=1/6, 8/6×(-7/7)=… yeah that won’t work out right.
9/6+(-8/6)=1/6, 9/6×(-8/6)=3/2×(-4/3)=-2 yay!

But how could I be sure that counting in sixths would eventually get me to the right place? What if it was some other denominaror I needed?

Well, it turns out that if you want two numbers with specific rational sum and product, they are guaranteed to be able to be written with a specific denominator, so you will be able to try only numbers with that denominator.

The theorem

Theorem:
Let $a$, $b$, $c$ and $d$ be integers with $b$ and $d$ not zero. Suppose there are rational numbers $x$ and $y$ such that $x+y=\frac{a}{b}$ and $xy=\frac{c}{d}$. Then it is possible to write both $x$ and $y$ with denominator $bd$.

I would like to show my proof for this theorem. There might be an easier way than I did it, but I definitely enjoyed my way, so I want to share it. It’s actually two proofs, but the first one makes me feel a little uncomfortable. I’ll do that one first.

Proof 1

Since $x$ and $y$ are rational, let $x=\frac{p}{q}$ and $y=\frac{r}{s}$ for integers $p$, $q$, $r$, and $s$ with $q$ and $s$ not zero.

Then

\[\begin{aligned} x+y&=\frac{p}{q}+\frac{r}{s}\\ &=\frac{ps}{qs}+\frac{qr}{qs}\\ &=\frac{ps+qr}{qs}\end{aligned}\]

and

\[\begin{aligned}xy&=\frac{p}{q}\cdot\frac{r}{s} \\ &= \frac{pr}{qs}\end{aligned}\]

Thus both the sum and the product can be written with denominator $qs$. The numbers $x$ and $y$ themselves can also be written with that denominator, as

\[\begin{aligned} x&= \frac{p}{q} = \frac{ps}{qs}\\ y&= \frac{r}{s} = \frac{qr}{qs}\end{aligned}\]

In other words, the numbers $x$ and $y$ can be written with the same denominator as the common denominator of the sum and the product. The sum $\frac{a}{b}$ and the product $\frac{c}{d}$ have $bd$ as a common denominator, so that means $x$ and $y$ can be written with this denominator.

Interlude

I am certain this proof is watertight, except for maybe tidying up the idea that if a two fractions can be written with a common denominator, then they can be written with any of the common denominators.

But still it feels a bit uncomfortable somehow. It gives a whiff of circular reasoning, maybe, or at least I don’t directly talk about the original denominators $b$ and $d$ anywhere until I reveal them at the last moment, which feels sneaky. I’d much prefer a proof that begins with the equations and ends with solutions with the correct denominators. And so I have this proof too.

Proof 2

Consider the two equations:

\[\begin{aligned}x+y&=\frac{a}{b}\\xy&=\frac{c}{d}\end{aligned}\]

Multiply the first by $bdx$ and the second by $d$ to give

\[\begin{aligned}bdx^2+bdxy&=adx\\dxy&=c\end{aligned}\]

Subsitute the second equation into the first to give

\[\begin{aligned}bdx^2+bc=adx\\bdx^2-adx+bc=0\end{aligned}\]

Using completing the square or the quadratic formula,

\[x = \frac{ad\pm\sqrt{(ad)^2-4b^2cd}}{2bd}\]

Now $(ad)^2-4b^2cd$ is an integer, so $\sqrt{(ad)^2-4b^2cd}$ is either unreal, irrational or an integer. Since $x$ is rational, that means it must be an integer. Let it be $m$, so that

\[x = \frac{ad\pm m}{2bd}\]

Suppose $ad$ is odd. Then $(ad)^2$ is odd, and so is $(ad)^2-4b^2cd$, and therefore so is $\sqrt{(ad)^2-4b^2cd}=m$. But now $ad\pm m$ is even.

Alternatively suppose $ad$ is even. Then $(ad)^2$ is a multple of 4, and so is $(ad)^2-4b^2cd$, which means $\sqrt{(ad)^2-4b^2cd}=m$ is even. But now again $ad\pm m$ is even.

Hence $x$ can be written with an integer numerator and denominator as

\[x = \frac{ad\pm m}{2bd} = \frac{\left(\frac{ad\pm m}{2}\right)}{bd}\]

Because of the symmetry in the original equations, this is also the two solutions for $y$. Thus both $x$ and $y$ can be written with denominator $bd$.

Conclusion

I particularly enjoyed that second proof. I was convinced when I did it the first time that the denominator had to be $2bd$ but then I realised that the numerator had to be even and so it came down to $bd$ after all, which was very satisfying.

And I am very happy that this means I can now factorise monic quadratics with rational coefficients by directly working through numbers with one specific denominator that add to the x-coefficient. It just feels like such a bold move to me, and now I know it just seems bold, because it provably will definitely work.

26 April, 2026
Factorising quadratics by focusing on the sum first

This blog post is about my way of helping students factorise quadratic expressions by inspection, which is the opposite of how most people do it.

Factorising

When you multply out two monic linear factors to make a quadratic, the same thing always happens:

$\begin{aligned} & (x+a)(x+b) \\ &= x^2+ax+bx+ab \\ &=x^2+(a+b)x+ab \end{aligned}$

You end up with the sum of the two constant terms as the coefficient of $x$ , and the product of the two constant terms as the final constant term.

Therefore, if you want to do this process backwards – that is, to factorise a quadratic expression – then you need to think of two numbers that add to give the coefficient of $x$ and multiply to give the constant term.

For example, to factorise $x^2+5x+6$ , you need to think of two numbers that add to give 5 and multiply to give 6.

There are at least two ways you could go about doing this systematically.

One way is to think of pairs of numbers that multiply to give 6, and then test them to see if they also add to give 5. So, you’d think of 1×6 and test 1+6=7, which isn’t right. And you’d think of 2×3 and test 2+3=5, which is right. So your factorisation is $(x+2)(x+3)$ .

Another way is to think of pairs of numbers that add to give 5, and test them to see if they also multiply to give 6. So, you’d think of 1+4 and test 1×4=4, which isn’t right. And you’r think of 2+3, and test 2×3=6, which is right. So your factorisation is $(x+2)(x+3)$ .

Every maths teacher I’ve ever met tells students to list the product first and check the sum. I think that it’s much better to tell students to do the sum first and check the product.

Examples

Let me do several examples to compare the sum first approach with the product first approach.

Example 1: $x^2+13x+40$

Product first

I need two numbers that multiply to give 40, which could be 1×40, 2×20, 4×10, 5×8 and I think that’s it. The matching sums are 41, 22, 14, 13. So the numbers I need are 5 and 8 and the factorisation is $(x+5)(x+8)$ .

Sum first

I need two numbers that add to give 13. I’ll start at 10+3, and the product is 10×3=30, which is too low. Now I’ll try 9+4, and the product is 9×4=36, which is higher but still too low. Now I’ll try 8+5, and the product is 8×5=40, which is just right. So the numbers I need are 8 and 5, and the factorisation is $(x+8)(x+5).$

Example 2: $x^2+20x+91$ .

Product first

I need two numbers that multiply to give 91. 1×91 obviously, and the matching sum is 1+91=92. So I need something else. What else? 2? Doesn’t go. 3? Doesn’t go. 5? Doesn’t go. 7? Oh yes that does work because 91=70+21, which is 10 and 3 sevens, so 91=7×13. The matching sum is 7+13=20, so that works. The factorisation is $(x+7)(x+13)$ .

Sum first

I need two numbers that add to give 20. My first thought is 10+10, and the matching product is 10×10=100, which is too high. Now 11+9=20, and 11×9=99, which is still too high, but lower. Next, 12+8=20 and 12×8=80+16=96, which is still too high, but lower. Next 13+7=20, and 13×7=70+21=91, which is just right. The factorisation is $(x+13)(x+7)$ .

Example 3: $x^2+30x+144$ .

Product first

I need two numbers that multiply to give 144. What goes into 144? It’s 12×12, so 1, 2, 3, 4, 6, 12 will all work. Have I missed anything? Oh 9, taking a 3 from each 12. Anything over 12 will go with one of the small numbers. Right, so what have we got?
1×144, but 1+144 is way too big.
2×72, but 2+72 is too big.
3×48, but 3+48 is too big.
4×36, but 4+36 is too big.
6×24, and 6+24 is just right.
So the factorisation is $(x+6)(x+24)$ .

Sum first

I need two numbers that add to 30. How about starting with 10 and 20?
10+20=30, 10×20=200, too big.
11+19=30, 11×19=110+99=209, that’s worse. I should be going the other way.
9+21=30, 9×21=189, still too big, but the right direction.
8+22=30, 8×22=160+16=176, closer.
7+23=30, 7×23=140+21=161, closer.
6+24=30, 6×24=120+24=144, just right.
So the factorisation is $(x+6)(x+24)$ .

Reasons for a sum first approach

The above examples point to the many reasons why I think focusing on the sum first is better than focusing on the product first. I’m going to list them, but they overlap quite a bit, so be prepared for me to repeat myself a lot in the explanations below each reason.

Reason 1

Doing the product first requires you to know or figure out what numbers divide into another. Doing the sum first doesn’t require any special knowledge about factors.

Look at Example 2. We had to figure out that 91 had 7 as a factor at all before we could get to the answer. With the sum, it just fell out along the way.

You may argue that guessing factors is a really important skill, and I don’t disagree, but honestly students don’t have much practice at that when they start factorising quadratics, and it’s a huge barrier to success. Focusing on the sum first allows them early success without the need for this skill. And you know what, they do a lot of multiplications along the way and might even notice what numbers tend to be multiples of what other numbers.

Also look at Example 3. The number 144 has a lot of factors, and you kind of need to find all of them to be able to have things to try to see if they come out to the right sum. Most worked examples for students dont even list all the options for factors, but just zero in on the magically right one, picking from an unspoken list in the teacher’s head. With the sums first approach, it doesn’t matter if you missed a factor.

And look, all the work you’ve done in the past to get good at seeing factors isn’t wasted! If one of the sums is 7+23 so you test 7×23 going for 144, you can actually say to yourself that 144 isn’t a multiple of 7 and just skip that one. I actually think developing this instinct for ways to shortcut the process can be quite an exciting idea to students.

Reason 2

With sums first, you can get started right away.

When you do product first, you have to think of some factors to begin with, and it’s very rare that 1×something is going to work, so there’s this job to do before you can even get started. When you do sums first, it’s not hard to think of a sum that works and you can just get on with it.

And there’s no wrong place to start either. You can just do a couple and you’ll know then if you’re going in the right direction. (See what happened with the one with 144.) So there’s no need to worry about your first inspiration – you can just get going.

Reason 3

With sums first, you feel like you’re getting somewhere.

When you investigate the sum first, you systematically change them by 1 each time and the product changes along with it, getting closer and closer to the right answer. There is a real feeling of progress, like the work is paying off. And to reinforce the previous Reason, this feeling happens right at the start, rather than having to wait for finding factors first.

I will concede that you can be systematic with the product first approach too, as you saw in my example with the 144. But to many students, the examples they see seem random, or worse, go straight to the right answer with no trial and improvement. If you do want to do product first, then I recommend being more systematic about it so that students can feel like they’re getting somewhere, rather than waiting for the lightning strike of the right one.

Note that the feeling of getting somewhere has another advantage: if you’re a long way away from the right result, it makes you feel safer to skip some steps to get there quicker. This way lies developing instincts for when some combinations of numbers are unlikely to work.

Reason 4

With sums first, there’s cool things you can help students to notice.

I personally think the experience of running through several possible sums and testing the products is some excellent fuel for helping students notice cool things, which are totally lost on a products-first approach.

For example, in the example with $20x$ , the highest possible product happened when the sum was 10+10=20. That is, when it was two of the same number. This is very cool and that way lies completing the square. Also the further apart the numbers were, the further away the next product is from this one. Indeed, the differences were two apart.

And I’ve already mentioned students noticing that a certain sum would require 144 to be a multiple of 7 and skipping it, that sort of thing makes the skill of noticing factors feel like a cheat code they’ve discovered, rather than a burden upon them. That sort of noticing is empowering for lots of students.

More examples

You’ve probably noticed that all the examples I’ve shown so far have had all positive coefficients, and they’re all monic (the coefficient of $x^2$ is 1). Well it’s time for some examples to deal with that. First I’ll deal with the negative coefficients, then later I’ll deal with non-monic quadratics. Mostly I’ll just do them straight using the sum first approach as if I didn’t know the answers yet, rather than compare them to a product first approach.

Example 4: $x^2-13x+40$

We have to think of two numbers that add to -13 and multiply to 40.

Positive numbers won’t add to a negative number, so I need two negative numbers, which will indeed multiply to a postive number.

(-10)+(-3)=-13, (-10)×(-3)=30, too low.
(-11)+(-2)=-13, (-11)×(-2)=22, even lower. Need to go the other way.
(-9)+(-4)=-13, (-9)×(-4)=36, higher.
(-8)+(-5)=-13, (-8)×(-5)=40, correct!

So the factorisation is $(x-8)(x-5)$ .

Example 5: $x^2+3x-40$

We have to think of two numbers that add to 3 and multiply to -40.

If you think about the product first, there’s twice as many options as there were before, because while 4×10=40, both (-4)×10 and 4×(-10) are -40 and you have to decide which one. If you only think about the product long enough to realise you need one positive and one negative, then you can start your search with sums that add to 3 like this:
4+(-1)=3, 4×(-1)=-4, not low enough.
5+(-2)=3, 5×(-2)=-10, lower, so I’m going the right way.
6+(-3)=3, 6×(-3)=-18, lower.
7+(-4)=3, 7×(-4)=-28, getting there.
8+(-5)=3, 8×(-5)=-40, and we’re there.

I probably could have skipped a couple since there was a long way to go, but it was so pleasant watching it get closer.
Anyway, the factorisation is $(x+8)(x-5)$ .

Example 6: $x^2-11x-26$

We have to think of numbers that add to -11 and multiply to give -26.

I’ll need a positive and a negative number to get a negative product, so let me start with -12+1.

-12+1=-11, (-12)×1=-12, not low enough
-13+2=-11, (-13)×2=-26, correct!

So the factorisation is $(x-13)(x+2)$ .

What if I had decided to start with something less obvious, like -20+9?

-20+9=-11, (-20)×9=-180, way too low.
-21+10=-11, (-21)×10=-210, even lower, so I need to go the other way.
I’ll skip some since I was so far away.
-15+4=-11, (-15)×4=-60, getting closer.
-14+3=-11, (-14)×3=-42, getting closer.
-13+2=-13, (-13)×2=-26, just right!

Example 7: $x^2+11x+26$

We have to think of numbers that add to 11 and multiply to give 26.

Two positive numbers will work, so I’ll start with 10+1.

10+1=11, 10×1=10, too low.
9+2=11, 9×2=18, higher so I’m going the right way.
8+3=11, 8×3=24, closer.
7+4=11, 7×4=28, too big.

So there’s definitely a factorisation that will work with roots somewhere between 7 and 8 and between 3 and 4, but there’s not one with integers.

Example 8: $x^2+8x+20$

We have to think of numbers that add to 8 and multiply to give 20.

I’ll start with 1+7=8, 1×7=7, too low.
2+6=8, 2×6=12, too low, but closer, so I’m going the right way.
3+5=8, 3×5=15, too low.
4+4=8, 4×4=16, too low.
But there’s nowhere else to go from here. I’ll never get to 20.
So this one doesn’t factorise at all.

Interlude

It’s time to stop for a short break. I’m hoping that this set of examples has convinced you that this approach has some merit for helping students understand how quadratic equations work, and indeed making the process a bit more playful.

I also sneakily wanted to cover some objections people have brought up, such as how you could be sure it doesn’t factorise if there’s infinitely many choices for numbers that add to the x-coefficient.

I just have one more thing to deal with, which is what do do with a non-monic quadratic. I’m just going to do one example in two ways.

Two more examples

Example 9a: $6x^2+x-12$

There is this method called by many “the ac method” which allows you to factorise a non-monic polynomial. I didn’t learn it at school, so I don’t think of it first, but it’s always something people bring up when I talk about factorising quadratics.

The way it works is you multiply the constant term and the leading coefficient, and then think of two numbers that add to the x-coefficient and multiply to give this new answer. (If your quadratic was $ax^2+bx+c$ , that means making the sum $b$ and the product $ac$ , hence the name of “ac method”.) Then you split the x-term into two parts with these numbers as the coefficents and continue from there.

(For a proof, consider the product of two linear factors:

$\begin{aligned} & (ax+b)(cx+d) \\ &= acx^2+adx+bcx+bd \\ &= (ax)x^2+(ad+bc)x+(bd) \end{aligned}$

Notice how the numbers $ad$ and $bc$ add to give the x-coefficient and multiply to the same answer as the x²-coefficient times the constant term. There can only be one pair of numbers with a specific sum and product, so if you find these numbers, they will be $ad$ and $bc$ and you will be able to do that algebra in reverse. )

Anyway, this still requires you to find two numbers with a specific sum and a specific product, so you can still do sum first.

The quadratic is $6x^2+x-12$ . So I need numbers that multiply to give 6×(-12)=-72 and add to give 1. I’ll need a positive and a negative.

2+(-1)=1, 2×(-1)=-2, which is way too high.
3+(-2)=1, 3×(-2)=-6, which is lower, so I’m going the right way, but I have a long way to go. I’ll skip some.
6+(-5)=1, 6×(-5)=-30, which still has a long way. I’ll skip some more.
10+(-9)=1, 10×(-9)=-90, which is too far, but quite close.
9+(-8)=1, 9×(-8)=-72, which is just right

So I need to split the $x$ into $9x$ and $-8x$ .

(You could argue that if I went product first, I might have realised immediately that 8 and 9 would be right, but I can guarantee you that a heap of students would not realise that. This way, they’ll get there in the end.)

So,

$\begin{aligned} & 6x^2+x-12 \\ &= 6x^2+9x -8x-12\\ &= 3x(2x+3)-4(2x+3)\\ &= (3x-4)(2x+3) \end{aligned}$

If you wanted this in fully factorised form so that it shows the roots, you’d have to pull out a 3 from one factor and a 2 from the other to get

$\begin{aligned} & (3x-4)(2x+3)\\ &= 3\left(x-\tfrac43\right)\times 2\left(x+\tfrac32\right)\\ &=6\left(x-\tfrac43\right)\left(x+\tfrac32\right) \end{aligned}$

(It’s worth noting that for many people, this “splitting the middle term and then factorising twice” thing is the way that you’re supposed to do all quadratic factorisations, including the monic ones, which I can see the appeal of if I’m honest. But I’m not rewriting my entire set of examples now.)

Example 9b: $6x^2+x-12$

There is a far more prosaic approach than the ac method, which is just to do what I’ve been doing all along but with fractions. Let me show you:

$\begin{aligned} & 6x^2+x-12 \\ &= 6\left(x^2+\tfrac16 x-2\right) \end{aligned}$

Now I’ll factorise the monic quadratric in the brackets there. I need two numbers that add to give 1/6 and multiply to give -2. They’ll have to be a positive and a negative.

2/6+(-1/6)=1/6, 2/6×(-1/6)=1/3×(-1/6)=-1/18, which is not low enough.
3/6+(-2/6)=1/6, 3/6×(-2/6)=1/2×(-1/3)=-1/6, which is lower but not low enough, and I’ve got quite a long way to go, so I’ll skip some.
7/6+(-6/6)=1/6, 7/6×(-6/6)=7/6×(-1)=-7/6, so much closer.
8/6+(-7/6)=1/6, 8/6×(-7/7)=… yeah that won’t work out right.
9/6+(-8/6)=1/6, 9/6×(-8/6)=3/2×(-4/3)=-2 yay!

So the factorisation is $6\left(x+\frac32\right)\left(x-\frac43\right)$ .

I have to say I prefer this one to the other one in a lot of ways. But yes a big fly in the ointment is the fraction arithmetic. But honestly this seems to me to be quite low stakes, and it certainly gives a lot of practice! You have to decide how you want to play it.

Oh, and why did I choose to count in sixths? Well it turns out that in a monic quadratic with rational coefficients, if there are any rational solutions, they’ll be able to be written with the common denominator of the coefficients. (But that’s another story and shall be told at another time.)

Conclusion

So, I’ve given a lot of examples to show how the reasoning works when you factorise quadratic expressions by first focusing on the sum and checking the product, rather than the other way around as is more traditional. And I’ve tried to describe why I think it has a lot of advantages. I hope you give it some consideration when you next help students with their factorising.

(And one little addendum: I think it’s worth considering this for all of your students first, rather than just reserving it for students who struggle with doing products first. Don’t let it become an othering explanation.)

25 April, 2026
Home & Away: Geometric Arithmetic

Introduction

This blog post is about a way to define addition and multiplication on a number line using the geometry of the plane that surrounds the line.

Recently I talked about how numbers have multiple purposes and one of those purposes is locating. When you draw a number line, that’s pretty much what you’re doing – saying that the numbers are exactly locations on that line. There is an issue with this idea, which is this: how do you add or multiply locations?

The usual way of dealing with this is to think of numbers as not just locations but journeys too, so that when we see something like 10+3, the first number is a location and the second one is a journey and the answer 13 is the location we arrive at by beginning at 10 and travelling onwards 3. This is fine and I like it very much, actually, but there is still the huge problem of how to multiply locations or indeed how to multiply journeys. You can say that 3×10 is three lots of a journey of 10, which makes sense, but now the 3 is neither a journey nor a location but a literal amount and we have three different things so far that the numbers mean. Is there a way that preserves the location-ness of everything?

Yes, there is.

There is indeed a way to define addition and multiplication of locations on a number line that preserves their fundamental location-ness, by looking outwards to the other lines of the plane your line is part of. The system doesn’t directly use lengths or angles, but only uses the most fundamental geometrical actions of drawing lines through two points, finding where two lines meet, and drawing lines parallel to other lines. This is my favourite thing about it, that it’s fully based on the relationships between the points and the lines as objects. As a pure mathematician and a finite geometer specifically, geometry isn’t really about measurements at all but is all about relationships, so something that focuses on the relationships deeply appeals to me.

I created this method in October 2020, heavily based on the method invented by Marshall Hall Jr in the late 1950s. Hall’s method works by adding coordinates to all the points in the plane including the ones on your number line, making equations for the lines, and referring to the coordinates of specific points on specific lines to define the arithmetic. I studied his method in 2001 while doing the honours year in my undergraduate maths degree, but it wasn’t until 19 years later that I realised there was a way to do it without referring to coordinates, by focusing my attention on the number line itself rather than the coordinate axes. My method for multiplication also has striking similarities to René Descartes’ original definition for the multiplication of lengths published in the 1630s, even though I didn’t mean it to and only found this out later. An important difference between them is that his method has the two factors on different sides of a triangle instead of along the same line. I find it very interesting that Hall’s book is called “The Theory of Groups” and Descartes’ book is called “Geometry”, highlighting the deep connection between geometry and algebra which all three methods point to.

Anyway, enough historical notes. Let’s get to it.

How to do geometric arithmetic

You can watch me doing both processes live in a video here, or you can read a text description and see screenshots from the video below.

Setting up

To do parallel line arithmetic, you need to set up a few things.

First, choose a line to be your number line. The points on the number line are your numbers.

Next, you’ll need to choose two different points on the line to call 0 and 1. The point 0 will be important for defining addition and both 0 and 1 will be important for defining multiplication.

Finally, you’ll need to choose two lines other than the number line itself that are not parallel to the number line and not parallel to each other. It doesn’t really matter where they are because the specific lines themselves aren’t important, only their directions, since during the constructions for addition and multiplication, you won’t be making them intersect with any lines. Instead you will draw several lines parallel to each of these lines. I call one direction “home” and the other “away”. (The reason why I chose these words specifically rather than “in” and “out” for example is because I am Australian and “Home and Away” means something to me.) To make it easier to focus when I’m drawing diagrams for the arithmetic, I usually put my home and away lines a bit away from the part of my number line I drew.

Addition

To add two numbers geometrically, you follow this process.

First, you need to have your two points a and b, and the point 0 on your number line. You don’t need a and b to be different from each other or different from 0, but I’ve drawn them as different to make it easier to show how the process works.

Create a journey from 0 to a first following the away direction and then following the home direction. That is, draw a line through 0 parallel to the away line.

Then draw a line parallel to the home line that passes through a.

Find the point where these two lines meet. If you follow the journey from 0 to a first along the away direction and then along the home direction, this point is where the journey turns from going away to going home.

Now draw a line parallel to the number line, through this turning point. This is the turning line for all additions that go some number plus a.

Now you are ready to do b+a.

Draw a line through b parallel to the away line and find where it meets the turning line. This is where the journey from b will turn and return home to the number line.

Now draw a line through this turning point parallel to the home line, and find where it meets the number line.

This point is the point b+a.

Multiplication

To multiply two numbers geometrically, you follow this process, which is similar to the process for addition, but with two very important differences.

First, you need to have your two points a and b, and the two points 0 and 1 on your number line. Just like before, you don’t need a and b to be different from each other or different from 0 or 1, but I have chosen them that way to make it easier to show how the process works.

Create a journey from 1 to a first following the away direction and then following the home direction. (This is the first point of difference between multiplication and addition, that the journey begins at 1 and not 0.)

That is, draw a line through 1 parallel to the away line.

Then draw a line parallel to the home line that passes through a.

Find the point where these two lines meet. If you follow the journey from 1 to a first along the away direction and then along the home direction, this point is where the journey turns from going away to going home.

Now draw a line that passes through both this turning point and 0. This is the turning line for all multiplications that go some number times a. (This is the second point of difference between multiplication and addition, that the turning line passes through 0 rather than being parallel to the number line.)

Now you are ready to do b×a.

Draw a line through b parallel to the away line and find where it meets the turning line. This is where the journey from b will turn and return home to the number line.

Now draw a line through this turning point parallel to the home line, and find where it meets the number line.

This point is the point b×a.

Thoughts

So that’s David Butler’s methods of geometric addition and multiplication. I love it so much, especially the multiplication. Addition is a little more complicated than you’d expect if you are used to adding numbers by joining journeys head to tail, but multiplication somehow feels so much simpler than it has a right to be.

My favourite thing to do is to convince myself that various algebraic properties of numbers have to be true using these two definitions of addition and multiplication. For example, this diagram shows that a+b = b+a (at least for positive numbers).

And this diagram shows that b+b = b×2 (at least for numbers more than 1).

And this diagram shows that when you multiply two negative numbers, you get a positive number. (I know those numbers are negative because they’re on the opposite side of 0 from 1.)

Yeah none of them are formal proofs, but I still love them.

The truly remarkable thing is that it doesn’t matter how you choose your home and away directions, it will still work! For example, here are three diagrams showing 1+1=2 and 2×2=4.

The geometry of the real plane is so neatly structured that it works every time. If our number line was in a plane with a different structure, this might not work the same every time. Indeed, you may end up with entirely different rules for how addition or multiplication work, such as multiplication not being associative.(That is, (a×b)×c not being the same as a×(b×c), which is very annoying!)

I may do future blog posts about some of that other stuff, but for now, I’m enjoying just revelling in the coolness that is the existence of a method for multiplying locations, and the even higher coolness of watching geometry cause the algebraic properties of numbers. I hope you enjoyed it too.

10 April, 2026
Too many presents

Once upon a time when my daughter was very young, she was given a lot of presents – it was a birthday or Christmas but I can’t remember which. What I do remember is that she played with just one present all day long, leaving all the others untouched.

I think we sometimes do the same thing to our students that friends and family did to Charlotte: we give them too many presents and then get upset when they don’t play with them all.

Let me explain.

Playing with it is one of the main ways to get a deep understanding of concepts and to get fluency with procedures. You ask yourself, “What would happen if…?” and say to yourself, “I wonder…”, and you try things out in different combinations. It’s awesome when it happens and you feel all sorts of positive feelings like curiosity and joy and satisfaction. Even teachers who don’t consciously subscribe to a play-based approach are usually happy when they see this sort of thing happenning. Many of the people who become university lecturers had similar experiences when they were students and assume their students also play with the ideas in their courses.

And the students actually do. It’s amazing how often even the struggling students are trying to explore. And the students who were engaged with the content long before they joined your course are sometimes aching for chances to explore that aren’t being given to them. But there’s a big problem: there’s just not enough time.

A university course has multiple new concepts and procedures every week, and there’s just too many of them to play with all of them. Yet the assignment questions tend to assume a level of familiarity with every single thing in the course that only comes with a decent amount of playing with every one. There’s just too many things in the course to be able to give all of them the time they need. And if a student gets nerdsniped and goes on a deep dive on one thing, they are forced to sacrifice play time with the other things.

I see it most clearly in two places.

First, in a course like Nursing where students are expected to be fluent in all the various ways to do calculations with multiplication and division quickly without a calculator. This fluency comes to most people through years of play: trying new problems, seeing how others do them, noticing strategies worth trying, and noticing when they’re not worth trying, storing away relationships between numbers. But in a first-year Nursing course with students who have past traumatic experiences with maths, there is literally not enough time for this kind of play, even if a student realised that the play was the thing that helped them be better at getting the answer, because they also have to play with how to listen to patients and what all the drugs do and any number of other things that go into becoming a nursing professional.

Second, in a pure maths course for students who chose a pure maths degree because they were interested in pure maths. These students deeply want to play, but there are so many concepts coming at them, there is just not enough time to play with them all, and they feel overwhelmed. Especially when their lecturer assumes unconsciously they have already done the play just because they’ve got previous experience with maths.

My great hope is actually to give students less to play with, so they have time to play with each of them as they go. But failing that, we need to at least not get upset when they don’t play with everything, and definitely not assume that they are lazy or uninterested or ungrateful. They’re just toddlers with too many presents.

7 April, 2026
Three types of infinity
Numbers have multiple purposes, some of which are
- Locating things in a list, such as the first person to walk on the moon, the second star to the right, the seventh film in the Fast and Furious franchise. These are all answers to the question, “Which one?”
- Counting things, such as 1 green sheep, 7 dwarfs, 101 dalmations. These are all answers to the question, “How many?”
- Locating things in time or space, such as 6.5 km away, 500 miles away, 5 years ago. These are all answers to the questions, “Where?” or “When?”
There are other purposes, such as measuring or comparing or having something to think about, but they aren’t what I feel like highlighting right now.

The purposes of number give you various models to help think about what numbers are doing when you operate on them, and even if a number began life as a location, you can think about it as a size if that helps. So, even though you know that starting at the 10th person in the list then moving forward three people and starting at the 3rd person in the list and moving forward 10 people are very different, if you think of them as combining collections of people passed so far, you can tell they’re the same.

But infinity is different. The three purposes of listing, counting, and locating in space don’t line up so neatly with each other when you’ve stopped using ordinary numbers you can reach in time. That causes the infinities that go with different purposes to be different to each other.

[Disclaimer: I am not going to go into all the mathematical or philosophical detail possible here, and I’m not going to use all the formal terminology. That’s not the point of this post. So if you’re a mathematician or philosopher who knows about this stuff, think carefully about what I’m trying to achieve here before getting upset at me. And if you want to find out more, I’m sure a quick internet search will provide all the details you need.]

Infinity of location

First, the infinity represented by the symbol ∞. That infinity is a location. When you draw a line and mark it out with numbers, you’re using numbers as locations. You can go forwards using positive numbers and backwards using negative numbers. And what if you go forwards forever, following all the way to the end of the line? That location is called “infinity” and we label it with the symbol ∞. If you go backwards forever you get to the location “minus infinity” or -∞. So with a number line going both ways, there are actually two infinite locations of ∞ and -∞. These infinities aren’t numbers in quite the same way as other numbers, because really they are only locations and can’t also represent counts of how many things there are

So, thinking of numbers as locations in space produces ∞ and -∞. What if you think of numbers as serving the purposes of counting or listing? What infinites do you get then?

Infinity of counting

When you count literal things like sheep, dwarfs, and dogs, there is always a perfectly good number to count them with. Those numbers are called the natural numbers. But some abstract things don’t have a nice natural number for how many there are, most notably, the very numbers you use to count with. How many natural numbers are there? This is another type of infinity. And its name is… however many natural numbers there are. There isn’t a name, really, though when someone says “infinitely many”, the size of the collection natural numbers usually what they mean. Many mathematicians use the Hebrew letter aleph with a zero as subscript – ℵ₀ – to represent this quantity.

This isn’t the only quantity that is infinite. Think about all the sets of natural numbers such as the set with just 1 in it, the set of just the digits 1 to 9, the set of even numbers, or the set of prime numbers. How many possible sets of natural numbers are there? That’s another infinity that’s even bigger than ℵ₀.

These infinities are called the transfinite cardinals. This is in reference to the word cardinality, which is the size of a set.

Infinity of listing

What about the first purpose of numbers I mentioned earlier – listing things and knowing where in the list you are? Well there’s infinities for that too.

Imagine listing things in order: first, second, third, fourth, … and your list was somehow longer than all the natural numbers you had available, what number would you say after you’d gotten through all the natural numbers and still needed another one? Well some mathematicians call that ω, a lower case omega. What about the next thing in your list? Well that would be ω+1. It doesn’t have it’s own name, it’s just the number after ω. The number after that is ω+2.

That’s the fundamental thing about numbers used for listing: every number has a number right after it, and that’s exactly what “+1” means. In that sense, the two numbers in an addition mean different things. When I say “10+3”, that means, start at the 10th object in the list, and then go 3 positions after that, which arrives at the 13th object in the list. The first number is where to start, and the second number is how many positions forward in the list to go, and the final answer is where you end up. With this interpretation, it’s actually surprising that 10+3 is the same as 3+10, though if your numbers are sizes of collections of things, it’s not surprising.

With infinite listing numbers, the connecton to sizes is broken, and so it’s not actually true that you can do addition in any order you want. For example, ω+1 is the number after ω. But 1+ω is the number ω steps beyond 1, which is in fact just ω. It doesn’t matter where you start in the natural numbers, you still have infinitely many of them to go to get past them all, so going that far just gets you to the end, not any further. (Yes I know how confusing that sentence is. Sorry. Infinities are confusing.) So 1+ω=ω ≠ ω+1.

These numbers with their weird order-matters addition are called the transfinite ordinals. That goes with the fact that they started life as ordinal numbers: first, second, third etc.

It’s worth noting that this weirdness with addition having different meanings in different orders doesn’t happen with the transfinite cardinals, because combining sets definitely doesn’t have an order. (Though there is other weirdness: ℵ₀+1 and 1+ℵ₀ are the same, yes, but they’re both the same as ℵ₀. This is why you can still fit an extra person in Hotel Infinity even though it’s full.) And with ∞, well you can’t add anything to that at all. It’s a place that isn’t really a number.

Conclusion

So there you go. There’s at least three different infinities to describe the number just beyond the ones we’re familiar with: ω, ℵ₀, and ∞. They are the next number in the list, the size of the counting numbers, and the location at the end of the number line. We need them because each purpose of number has a different way to imagine what the numbers are and so a different way to imagine what is beyond beyond the numbers. And that different purpose dictates the sorts of operations that are possible on these infinities too.

This deep connection to purpose and metaphor in maths is something that really appeals to me.
24 March, 2026
When perimeter is equal to area

Some triangles have the same perimeter (in cm) as area (in cm²). For example, the 6-8-10 right-angled triangle:

What properties of the triangle might signal that this sort of alignment of perimeter and area is possible?

One answer I discovered to this question is completely surprising and delightful to me:

If any of the side lengths is 4 cm or less, then the perimiter (in cm) can’t be equal to the area (in cm²).

It seems remarkable that knowing about just one side of the triangle could rule out a property that ostensibly requires all three sides to confirm, but it’s definitely true, because I proved it. Indeed, the proof itself is one of the reasons I love this fact so much.

You may want to attempt to prove it yourself before I show you my proof…

Ok, here’s my proof.

Consider a triangle with side lengths $a$ , $b$ and $c$ in cm. It’s possible to arrange the sides in order of size, so choose the labels so that $a \leq b \leq c$ .

First we work with the perimeter.

$\begin{aligned} \text{Perimeter} &= a + b + c \\ &\gt b + c \\ &\geq b + b \\ &= 2b\end{aligned}$

That is, the perimieter is strictly greater than $2b$ .

Now we work with the area. The area of a triangle is half the base times the height. Choose the side with length $b$ as the base and let the height be $h$ .

The side with length $a$ goes from the end of the side with length $b$ to the apex of the triangle. If it goes straight upwards, then the height is equal to $a$ , but otherwise, the height will necessarily be less than $a$ . Either way $h \leq a$ .

So, when working with the area,

$\begin{aligned} \text{Area} &= \frac12 bh \\ &\leq \frac12 b a \\ \end{aligned}$

Suppose that at least one of the sides is 4 units or less. Then the shortest side must be 4 cm or less. That is $a \leq 4$ . And so

$\begin{aligned} \text{Area} &= \frac12 bh \\ &\leq \frac12 b a \\ &\leq \frac12 b \times 4\\ &\leq 2b \end{aligned}$

That is, the area is less than or equal to $2b$ .

Combining these two facts gives

$\text{Area} \leq 2b < \text{Perimeter}$

And so the perimeter is strictly greater than the area. In particular, they can’t be equal.

End of proof! Yay!

There are so many things about this proof that I love. I love that it uses so much inequality reasoning, which I have a particular fondness for. I love that you decide whether the area and perimeter are equal based on comparing them both to a third thing rather than to each other directly. This feels like such a ninja move. I love that the thing you compare them to is just the middle-length side. Indeed, the longest side doesn’t really feature much in the argument, which is surprising. I also love that the 4 appears more-or-less to cancel out in just the right way with the half in the area formula. A different number wouldn’t have worked, so 4 is the perfect one.

I hope you like my little fact and its proof as much as I do.

9 February, 2026
It’s just subbing into formulas

(This blog post is a slightly expanded copy of a thread I wrote on Twitter in 2021.)

I have met people before who say that statistics and applied maths are easy for students because it’s just subbing numbers into the right formulas. Sometimes it’s not quite so overt, and I just see people express frustration because the sudents just aren’t succeeding, even though all they need to do is sub things into formulas. Either way, it makes me sad and angry, because “just” subbing numbers into formulas is not easy, in multiple different ways.

I will now describe a number of those ways, numbered idiosyncratically in the order I thought of them.

1. For many students studying especially intro stats, they have very little experience with maths and what they have has been traumatic. The very act or substituting things into formulas is actually hard for them, if not triggering. Just looking at the formula will create a rush of feelings that make it hard to think.

2. For the students with little or traumatic experience with maths, one thing that really holds them back from succeeding is not knowing the order of operations. They don’t know that addition is after multiplication, or that their calculator will in general know the correct order. This means even if they put the numbers in the right places, they won’t get the right answer every time.

3. The formulas often use letters that the students have never seen before and don’t know what they’re called. I can’t count the number of students I’ve met who don’t know the name of the letter μ. There is nothing like not being able to read the formula to make you feel like an idiot.

4. Even if you can read the formula, finding which numbers in the problem go into it is often difficult. It’s like one of those reading comprehension tasks that ask about technical details you probably missed. You have to go searching again to find the numbers you need hidden in the text.

5. The search for the right numbers also requires you to know what the words mean. If your formula has a standard deviation, then it’s useful to know what standard deviation means, if only to know how people talk about it so you can find it in the problem text. This is further complicated by the fact that the context can change the meaning. If it’s implied in the text that the standard deviation came from the data not somehow from the population, that changes everything.

6. Some might argue that you don’t need to know what the words mean. Can’t they just find the numbers next to those words in the text and put those numbers into the formula? But imagine having six meaningless words to look for every time. That’s exhausting. And do you really want students using formulas without meaning? That doesn’t seem like a good idea to me even in the short term.

6b. Many applied maths courses introduce new applications in every assignment, with a whole raft of brand-new terminologies that have letters and/or numbers to go with them, and it’s a big load to deal with even if you decide not to try to understand the context. It’s one of those difficult technical reading comprehension tasks again.

7. If there are even a handful of formulas to choose from, and they’re spread out through the course materials, it’s actually a large task to search for the right one. Flicking through all the materials, you might easily miss one, and even if you have a handy cheat sheet with them all there, you still have to find the right one among the list.

8. In order to choose which formula is the right one, you need to be able to distinguish between different situations that formulas might apply to. What features do the situations have that make them different enough to need different formulas? What features are irrelevant? This is actually hard to notice and to a novice is not obvious at all. It’s also hard to notice because it’s rare for teachers to explicitly point out those features.

8b. Indeed, to notice how the situations are similar and different, you need to have enough examples so that you can compare them to each other, and many courses only have one or two examples of each situation/formula. That’s just not enough to glean the unspoken rules of deciding between them.

9. When they’ve chosen the right formula and put the right numbers in it and gotten an answer, they still have to know what to do with the answer. This is a separate skill, often requiring more understanding of the context and also skills with written language.

10. Language is a whole set of problems of its own. In the genre of maths assignment questions, information is presented in terse sentences that assume you can tell the difference between technical words and ordinary words and keywords that indicate expectations. And you can’t always do that, especially on the fly without specific instruction.

11. Many many students actually do want to understand, despite you telling them they don’t need to. They want to make connections between ideas and have reasons why the choices were made the way they were. If this is unrequited, they come to resent the whole thing.

11b. They don’t necessarily need the whole mathematical theory, but they do need reasons and connections. And actually you yourself decide between methods based on these reasons and connections anyway.

I’ll stop there. But I hope I’ve made it clear that choosing and using formulas is a long long way from easy, so we need to support our students and give them the credit they deserve when they struggle with it. Because it is actually hard and they’re trying.

28 January, 2026
Jenga Views

This blog post is about a sequence of visual perception and geometry puzzles I have created called Jenga Views. You can download a file here with 39 puzzles, roughly in order of difficulty.

In my previous blog post, I showed two variations on traditional Jenga that I think are more interesting and more fun. But long before those were thought of, I had already been doing something non-traditional with Jenga blocks.

Back in May 2018, I saw a tweet that shared a maths/art activity called Fun With Orthoprojections by JD Hamkins. It was a list of challenges where 1 by 2 by 4 wooden blocks had to be arranged to match views from the top and two sides. I thought it was really cool, but I didn’t have any blocks with those proportions. I did have a quadruple set of Jenga blocks, but Jenga blocks have very different proportions to 1:2:4, so in order to use them I’d have to make my own.

After a furious effort in Inkscape and Word over one day and night, I had made the first version of Jenga Views, where Jenga blocks had to be arranged to match views of the shape from the top and two sides. There were 20 puzzles in total: the first eleven were based on JD Hamkins’s Fun with Orthoprojections, but the later ones were all my own. (The SVG Inkscape file with the carefully measured block faces is here, if you’re ever interested in making any yourself.)

I put them out at One Hundred Factorial the next day, where they were an immediate success. Check out these two students working on one of the challenges on that very first day:

In general, over the years, I have found that the Jenga Views challenges are strangely compelling to anyone for whom they spark interest. Some people aren’t intrigued by them, but those who are intrigued tend to just stick around and do all of them, one after the other.

One thing I particularly like about them is that you don’t need an answer key, because you can just look at your construction and tell if it looks right or not. There’s something empowering about being able to check it yourself. (I’m sure there’s a lesson there for maths problems we give to students for practice, but I’m not going to explore that thought more right now.) I love watching students crouching down to see their construction from the correct angle and holding up the pictures to compare.

One of those people who found it compelling was my younger daughter Charlotte. In April 2020, she was 11 years old and only a couple of months into Year 6 (which here in South Australia is the final year of Primary School) and suddenly we all went into pandemic lockdown. My wife, who is a trained teacher, cobbled together her own school away from school curriculum for Charlotte before Charlotte’s school sorted anything out, and I provided some maths activities I had brought home with me from university. I had brought a selection of things from One Hundred Factorial home so I didn’t go insane and so I could try to run One Hundred Factorial from lockdown. One of those things was Jenga Views.

And Charlotte loved it. Because of the quadruple set of Jenga blocks, she actually had enough to do all 20 challenges separately and lay out all the answers at once.

(Note that the file has been updated since this image was taken, so don’t use it as an answer key, because almost all the numbers refer to different challenges now! Also you can tell if your solution really is one without an answer key, as I said earlier.)

Charlotte was so very proud of herself and I was proud of her. I shared the photo and the Jenga Views document on Twitter, and it got a lot of attention. There were a lot of people who were also in lockdown who were extremely grateful for something fun and mathematical to do using resources they had in their house already.

I was all inspired too, and I created more puzzles, adding five more the next day and then five more the day after that, bringing the total to 30. I also created a print-and-cut net that would allow people to make their own blocks in the Jenga proportions, since someone complained they didn’t have any in their house, as well as a video of how to fold it up into a Jenga block. (Two nets are on the last page of the Jenga Views document, if you want something better quality than this picture.)

Eventually, the interest settled down on Twitter. But over the years I’ve pulled out Jenga Views at One Hundred Factorial regularly, and every time, there’s always people who do what Charlotte did and just work through all of them.

Now we’re here five and a half years later (and seven and a half years after its original creation), and I’ve finally gotten around to writing all this up as a blog post.

To reward everyone who waited – including me – I have created nine more puzzles. They are mixed in among the 30 that were already there, so be careful if you’ve downloaded it before because almost all the numbers have changed! Also at the end of the document there are four empty challenges for you to draw your own, and the net of the make-your-own blocks too.

As an aside, the reason I did nine more and not ten is because then you can print them two-to-a-page or three-to-a-page including the title page and they will line up neatly without annoying blanks. It’s been something that’s bothered me every time I’ve printed them over the last seven years.

So there’s the whole story of Jenga Views. I hope you enjoyed reading about it, and I hope you enjoy doing the challenges yourself.

11 December, 2025
Jenga Variations
In this blog post I will tell you about two variations to traditional Jenga that I find fun, and also reflect on what these say about what “fun” means to me.

Original Jenga

It’s best to discuss original Jenga before talking about the variations, just in case there are important things about Jenga you don’t know, but also to explain why I might want variations at all.

Original Jenga is a family game of dexterity. Rectangular blocks are stacked in layers of three into a tower, with blocks in each layer at right angles to the layer below. Players take turns to pull a block out of the tower from anywhere except the last completed layer, then lay it on the top to complete new layers. If the tower falls during a player’s turn or in the few seconds afterwards, then that player loses and the game is over.

One thing that is kind of ambiguous is exactly what counts as “falling over”. Does just one block falling to the table count as falling over? If not, then how many blocks does it have to be to count as fallen? If you catch the falling blocks so they don’t hit the ground, does it count as falling? Can you catch the tower as a whole and straighten it to avoid losing? So many questions! But actually I love that there’s so many questions. It’s fun for me to have the discussion and decide together, and sometimes to have to change that decision because a situation comes up that we didn’t think of. In the end, when rules are ambiguous or missing, I follow Dan Finkel’s philosophy: “Choose what seems like the most fun way to resolve questions about the rules”. And having to resolve them is part of the fun.

Other than that little ambiguity, that’s pretty much it. You choose blocks to pull out, you pull them out and put them on top, and if you’re lucky and/or skilled, you don’t knock it over. The suspense is pretty thrilling near the end of the game, and the tower falling always sparks a cheer. But after a while, if I’m honest, it’s just not that interesting.

I think I know why: everything’s the same and there’s no creativity. All the layers of the tower are the same as each other and there are only two ways to pull out a block – pull out a side one or pull out a middle one – and putting it back on top is similarly restricted. Yes there’s suspense, but it’s always the same suspense every time, where you fear the tower will fall over. I like games to have a little more variety, and have a little more scope for creativity in the choices you make on your turn. The two alternatives below solve those two problems in different ways.

Before I talk about variations, I just want to mention two properties of the blocks that are both interesting and relevant.

The first is how many blocks there are. When it was first released and for a long time, a Jenga set was 51 blocks, which is how I know that 51 is not prime – it has to be a multiple of three since each layer has three blocks. Nowadays a Jenga set is 54 blocks, so one layer higher.

The second thing is the shape of the blocks. Because the orthogonal layers have to fit together, the blocks have to be three times as long as they are wide. There is no particular need for the thickness to have a special relationship with either of the other dimensions, but it just so happens that the thickness of brand-name Jenga blocks is one fifth the length.

(Off-brand stacking block games don’t have this thickness-to-length proportion. Most are three times as long as they are wide, but not all. The ones that aren’t manage it by having gaps between the blocks.)

It’s these proportions that inspired the first variation…

Cursed Jenga

As I said earlier, brand-name Jenga blocks have the property that the longest edge is a multiple of both the other dimensions. That means it’s actually possible to make a square out of any of the three rectangular faces, with the blocks in different orientations. I wondered what it would be like to play Jenga with the blocks in these different orientations, and Cursed Jenga was born. (A student gave it this name during the first ever game at a welcome event for new students in February 2024 where I provided the activities.)

In the traditional orientation, there are three blocks making up a square. If you put them on edge, you can line up five to make a square. If you stand them on their ends, you need to put them in a three-by-five grid of 15 blocks to make the square. Cursed Jenga alternates between these three arrangements for its layers.

An important question is whether you can actually alternate the orientations with the 51 blocks you have. If you have all three, that’s 3+5+15, which is 23 blocks, and another set of those is 46 blocks. You can achieve 51 blocks with one more layer of 5. That leaves two options: 5+3+15+5+3+15+5 or 5+15+3+5+15+3+5. You can decide which you prefer, but the first one does begin the game filling in a 3-layer on the top, which is a bit easier. If you want to go to 54 blocks, you’ll need another 3-layer which again gives you two options: 5+3+15+5+3+15+5+3 or 3+5+15+3+5+15+3+5. This is another reason to start with 5+3, because then it will work with either 51 or 54 blocks and you don’t have to change the whole thing if you discover later it’s a different number of blocks after all. (There’s another argument for why there have to be three 5-layers: the 15-layer has a multiple of 3 blocks. A Jenga set has a multiple of 3 blocks. So you must have 3 sets of 5 to make a multiple of 3 in total.)

Another important question is how to orient two successive layers relative to each other. It matches the original spirit of the game for the long lines between blocks in the 3-layers and 5-layers to be at right angles, but what do we do between a 15-layer and the other layers? Well for structural reasons, you want to avoid lines between blocks in two successive layers being parallel on top of each other. Or stated another way, you want there to be blocks in the 15-layer that touch more than one block in the layer above or below. That effectively makes your choice for you. I very much like that having a principle just decides it for you without having to remember a specific orientation.

So here’s 51 blocks all set up ready to play Cursed Jenga:

You can probably already see one of the reasons I like this version of Jenga – just setting it up involves a whole lot of fun problem-solving!

And now to play! You can keep the same rules as ordinary Jenga, but you quickly run into an issue caused mainly by that layer of 15 vertical blocks: it’s very easy to make more blocks fall out than you intended! You can’t end the game just because more blocks fell out, because you’d have to restart all the time. My solution to this involved two aspects.

First, I changed the rule for what counts as “falling over”. The definition is now that the tower has fallen when any block in the top completed layer or above touches the table/floor. You could choose a different rule, and feel free to do so, but most of the time it doesn’t really matter because usually it’s pretty clear to everyone involved that it’s all over. Indeed, another possible rule is just to let everyone vote and if the majority agree it’s fallen, then it’s fallen.

Second, I made the rule that all blocks that fall out are passed one at a time to the future players in the order of play. Each of those players can choose to just add a block they’re holding to the top rather than take a new block out. Effectively, the goal of a turn becomes to add a block to the top, and if you don’t happen to have one, you have to take it out of the tower. Alternatively, you could leave blocks that fall out where they are, seeing them all as part of the bottom layer for the purposes of the “pulling a block out” part of the turn. They’re not quite the same, because the second one doesn’t allow specific players to save up blocks for later, but still they both mean the punishment for pulling out too many blocks at once is to make it easier for the players after you. It is important to allow people to take a block out of the tower if they want, because that is indeed a fun part of the game and you don’t want to deny them that!

And that’s it. It’s more interesting during play than traditional Jenga for a number of reasons. There are more different options for the types of blocks you can pull out at any moment, and you can sometimes make the choice of whether to pull out a block at all. More variety for the choices you can make always makes a game more interesting to me. Also, the tower is much more precarious than before, making the end of the game a lot more suspenseful. Plus, it just looks weird so people come to ask what’s going on or at least do a double-take on the way past. I do like giving people something to wonder about.

Ceiling Jenga

This variant of Jenga was originally suggested by one of my MLC tutors Alexander Mackay in December 2024, and I’ve added some minor tweaks since then.

You set up the tower as usual, except the top three layers only put two blocks each at the sides, like this:

(It will become clear shortly why this is important.)

The rules are the same traditional Jenga except for two important changes:
1. When you put a block into the tower, you can put it anywhere and in any orientation as long as at least one part of the block is higher than it was before.
2. The tower as a whole is not allowed to get any higher than it started. (This is the “ceiling” referred to in the name.)
That second rule is why I set up the top three layers like I did, so that there are places to actually put the blocks that you move upwards! The first time we played, we just pushed other blocks a little out of the way to fit the new ones in, and if you feel comfortable with that, then go for it.

You’ll also have to decide about what counts as fallen down, because the concept of “completed top layer” ceases to mean quite as much as it did before. My usual rule is that the tower has fallen when any of the blocks at the highest level touch the ground, but as I said, most of the time nobody quibbles about whether the tower has fallen.

Here’s a picture of a game in action.

And here’s a different game in action. This one is with giant blocks that don’t have the thickness proportion of brand-name blocks, but that’s fine because this variant of Jenga doesn’t actually require the thickness to be anything in particular.

I really love this game. One reason is that it just looks so bizarre. Cursed Jenga looks like you’ve set it up wrong, but Ceiling Jenga looks like you’re playing it wrong. Which you are, according to the regular rules. And if people ask, you can tell them all about it.

The other reason is on every single turn you have scope to do something creative. Want to think ahead to set up gaps to put future blocks into? You can, and feel clever. Want to put a block hanging out the side, or up a different way? You can, and feel inspired. Want to see how far you can push the definitions and argue your case to the other players? You can, and feel empowered. These are feelings you can’t have in regular Jenga, and I love how such a simple-to-state rule can make them possible.

Conclusion

So there you go: Cursed Jenga and Ceiling Jenga are two variations on traditional Jenga that allow for more variation in what happens during the game, and so make Jenga more interesting and more fun. Give them a try, and do let me know how they go for you.
1 December, 2025
More wisdom from the Dodecahedron

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.

28 November, 2025