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?
Tag: proofs
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Why mathematical induction is hard
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.
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Finding an inverse function
There is a procedure that people use and teach students to use for finding the inverse of a function. My problem with it is that it doesn’t make any sense, in two ways.
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David Butler and the Prisoner of Alhazen
Once upon a time, I did a PhD in projective geometry. It was all about objects called quadrals (a word I made up) – ovals, ovoids, conics, quadrics and their cones – and the lines associated with them – tangents, secants, external lines, generator lines. During the first two years, I did talks about my PhD research, which I could not resist calling “David Butler and the Philosopher’s Cone” and “David Butler and the Chamber of Secants”.
At that time, my use of JK Rowling’s titles had to stop because there was no suitable mathematical thing to insert into the third title. It’s been ten long years since “David Butler and the Chamber of Secants”, and finally I have found something to use. Hence, welcome to…
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Pretending not to know
Yesterday the Maths 1M students handed in an assignment question that asked them to prove a property of triangles using a vector-based argument. It’s not my job to help students do their assignment questions per se, but it is my job to help them learn skills to solve any future problem. This kind of problem, I find, is really hard to learn generalisable skills from.
For most students you really need to be there every step of the way as you try to solve the problem together, so that at the end you can look over what happened and figure out the sorts of things that made it possible to come up with a proof today. Students need to be hear the sorts of general self-questions you ask to help progress your thinking, even if they don’t lead anywhere straight away, and they need to see the dead-end paths you went down only to come back and go a different way.
The big problem is that if you’ve already seen the proof of this 15 times this week, it’s very very easy to guide students down a particular path that they could never possibly think of by themselves first go. It’s very easy to ask specific leading questions rather than general questions that might not lead anywhere. It’s very easy to push them away from the dead-end paths towards something that will give a result more quickly. You want to avoid doing that as much as possible, and the only way I know to do that is to pretend you haven’t seen the solution.
You’re going to have to pretend that you really don’t know how to do it and you really are just figuring it out with them today, and pretend to be surprised that something turned out nicely, and pretend to be frustrated when things don’t. It’s a real art and it takes a lot of practice and a lot of energy to pull it off.
I was very pleased the other day when I did pull it off. I was helping some students with this proof, and I said and did all the right things, including the dead-ends and everything.
After these students were happy with what we’d achieved and had a nice moral about problem-solving to take away, I turned to my other side to help the student who had been sitting there patiently. He had a whole different kind of proof to work on (mathematical induction), and I started as I often do by looking up the definition and writing that down, then saying “Now I’m not sure if this is going to help yet”. He responded to this by saying, “I don’t think I’ll ever believe you again when you say that.”
You see, I had helped him with the geometry proof only a couple of days before, and he had patiently sat there listening to the deja vu of me go through all the same things I went through with him. I looked him in the eye at that point and he said, “That was very impressive.” And he meant it. It’s nice when someone appreciates your craft.
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The crossed trapezium
Recently I started thinking about the properties of the following shape, which I like to call the “Crossed Trapezium”. It has two parallel edges, which are joined by two crossing lines.
You can read the rest of this blog post in PDF form here.
These comments were left on the original blog post:
Five Triangles 6 April 2016:
Related factoid: in the figure you dub a “crossed trapezium”, the two triangles that were removed from the original trapezium have equal area.
Also, this useful figure pops up in a lot of ratio problems we tweet about, such as “method 4” solution in this octagon problem:
https://casmusings.wordpress.com/2014/11/13/squares-and-octagons-a-compilation/David Butler 6 April 2016:
Interestingly, not only are they equal area, but their areas together are hab/(a+b). Most interesting.
Tom A 13 September 2017:
The relationship between the formulas for the ordinary and crossed trapezia is rather elegant. Unfortunately, the formula for the ordinary trapezium isn’t numerically well-behaved: if the two bases are the same length, then the a-b term is zero, and the result is undefined; if the two bases are nearly the same length, a-b is very small, and calculations on a computer may be inaccurate.
An alternative formula without that problem is:
a = 1/2 * ((b^2 + a^2 + 2ab) / (b + a)) * h
Which is still similar to the formula for the crossed trapezium, but doesn’t have an obvious (to me) geometric interpretation! Something about the triangles cut out of the sides?
Peter 19 September 2017:
Thanks for sharing this interesting work. I first started thinking of areas for crossed trapezia whilst preparing lessons on motion in a straight line for my students. I have been looking at the vel-time graph for a period of constant acceleration where the sign of the velocity changes. The area trapped between the graph and the t-axis is a crossed trapezium. Interestingly, if you consider the area in the first triangle to be postive and the second to be negative (consistent with calculating displacements), the formula for the area of the trapezium is unchanged from the canonical one.
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There is only one kind of function that distributes over plus
There is a very common thing that students do that causes pain, distress, confusion and depression in any maths educator who witnesses it. Both the error itself and the educator’s response to it are very clearly described by this excellent picture from the blog “Math with Bad Drawings”:
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The Sausage-Stacking Theorem
It’s no secret that the powers of two are some of my favourite numbers. There are so many interesting things to say about them that often I don’t know where to begin! (In case you’re not au fait with the terminology, the powers of two are the numbers you can make by starting with 1 and multiplying by 2 over and over. The list starts out 1, 2, 4, 8, 16, 64, 128 etc.)
Well, the other day, one of my staff (thanks Lyron) reminded me of one of my favourite facts about the powers of two. I like to call this fact “The Sausage-Stacking Theorem”.

First I need to set up the situation. Imagine a nice neat stack of sausages. It has a row of sausages on the bottom layer, and another row on top of that with one less sausage because they sit in the furrows between the sausages on the bottom layer. There may be other layers on top of this, each with one less sausage than the layer below.
The Sausage-Stacking Theorem is this:
A power of two sausages cannot be arranged into a nice neat sausage stack.
No matter how you try, it simply can’t be done; there will always be sausages left over or not enough sausages to fill in the top row. (In mathspeak, you could say that a power of two cannot be written as the sum of two or more consecutive natural numbers, but the sausages are cuter.)
What Lyron encouraged me think about was a way to prove this theorem without resorting to the usual mathematical notation a trained mathematician like me would normally use. Well, using the proof he showed me as inspiration, I have come up with something that is strongly visual, and it goes like this…
Consider a number of sausages that can be made into a nice neat stack. We’ll show that it must be possible to divide this number evenly by an odd number. We’ll split it into two cases: if the stack has an even number of layers or if the stack has an odd number of layers.

Suppose the stack has an even number of layers (my diagram has four, looking end-on). Imagine removing the top half of the layers, flipping them over, and joining them to the rest of the stack. (I say “imagine” because you’ll need the sausages to hang in mid-air to do this physically.) Now you have a parallelogram made of several layers, each with the same number of sausages. The number of sausages in the bottom layer was made by joining the bottom layer and the top layer of the original stack. Since the original stack had an even number of layers, the top and bottom layer are an even number and an odd number. So when you add them together you’ll get an odd number in total. Thus the new parallelogram stack has several layers, each with the same odd number of sausages. So, it’s possible to divide the total number of sausages by this odd number.

Now suppose the stack has an odd number of layers (my diagram has three). Imagine slicing your sausage stack down the centre, flipping one side upside-down and sticking it back on. Because there are an odd number of layers, the half-sausages will join together neatly and you’ll get a parallelogram made of several layers, each with the same number of sausages. There are still an odd number of layers, so it’s possible to divide the total number of sausages by this odd number.
So in either case, it’s possible to divide your number of sausages by an odd number. A power of two can’t be divided by any odd numbers at all, so it can’t possibly be the number of sausages in a nice neat stack.
How cool is that?!
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The square root of two
In first year maths, they briefly study the five families of number: the natural numbers N, the integers Z, the rational numbers Q, the real numbers R, and the complex numbers C. In particular, they focus on the distinction between the rational numbers and the real numbers. A classic proof they are given at this time is one that the number √2 is irrational. This blog post is about some alternative proofs.
You can read the rest of this blog post in PDF form here.
The following comments were left on the original blog post:
Jim Propp 25 May 2017:
I really like the second proof (though some purists might object that its use of facts about reduction of fractions amounts to a tacit appeal to the fundamental theorem of arithmetic that makes the proof less elementary than the other two). For one thing, the second proof readily generalizes to show that for all positive integers n, not just n=2, the square root of n cannot be rational unless it is an integer. Nice!David Butler 25 May 2017:
Yes I rather liked that you could use it to show that square roots can only be rational if they’re integers.
