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Gerry-mean-dering

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.
{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.
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.

Posted in Isn't maths cool?, One Hundred Factorial | Tagged , ,

Making the lie true

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.

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Introducing Digit Disguises with a small game

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.

1. 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.
2. 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.
3. 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.
4. 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.)
5. 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.
6. 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.)
7. 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.
8. 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.
9. 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.
10. 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.
11. 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.
12. Celebrate the win, and thank the One Who Knows for their help.
13. 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?
14. Split them into teams and put the teams into pairs by whatever method you like best.
15. Give them the Digit Disguises game board handout.
16. Remind them of the rules.
• 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.
17. Let them play and circulate to hear their awesome thinking and point it out.
18. 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!
19. 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.

Posted in Isn't maths cool?, One Hundred Factorial | Tagged , ,

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.
Some reasons why students find mathematical induction difficult.
These come from a […]

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Space to enter

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 […]

Posted in Being a good teacher, Other MLC stuff | Tagged

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 […]

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Four levels of listening

Introduction
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 […]

Posted in Being a good teacher, Education reading |

Other(ing) Explanations

Most people who teach mathematics are aware that it’s useful to have alternative explanations for concepts, and useful to have different ways to approach problems. Given enough time, you are guaranteed to come across students for whom the standard explanation isn’t working today (as long as you give students a chance to tell you about […]

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Arbitrary mnemonics

A mnemonic is a mental trick to help you remember things. People use them all the time for all sorts of things, like the traditional colours of the rainbow (ROY G BIV), the order of the letters in the English alphabet (a song to the tune of Twinkle Twinkle Little Star), the order of operations […]

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