Here is some news of the possible breakthrough of the ABC conjecture. It might already be commonly known, but it is something I only recently discovered was going on. There are rumors that Shinichi Mochizuki from Kyoto university has solved the abc conjecture. You can find his proof here. Not that I understood anything of it, but I though a link would be appropriate. But don’t despair yet, I will have a few more links in this regard. Continue reading →

# Proofs

## Proving the fundamental lemma of Arithmetic

Last time I blogged I wrote about why proofs are so difficult. This time we should try to prove something a little more complicated and see if we can get some of the thoughts behind the proof while doing it to see if it really requires that amount of insights or part of a proof can be done by using more or less common sense. My guess is that we can come far with common sense, but we at some point need to rely on some insights or methods we have seen before. Continue reading →

## Why are proofs so frustrating?

I have often been sitting there swearing when I couldn’t figure out to how to prove something and I know for certain I am not the only one. But why are proofs so difficult when they seem so easy when you read them? This post is about the softer side of proofs, to encourage you to not loose faith when you meet something you can’t prove. Continue reading →

## A Short Treatise on a Vectors Concept

I have been in contact with Frederick Koh from Whitegroupmaths.com who kindly agreed that he would write a guest post for the blog to promote what he has to offer – Tutoring in A level maths in Singapore. So without further ado let me present you with the real content.

I have been asked on numerous occasions by students to provide a short effective mathematical proof verifying the fact that obtaining the vector product of the normals characterising two separate *non parallel planes* in 3 dimensional Cartesian space produces **the direction vector of the line arising from the intersection** of the above mentioned planes. I shall share this here:

(Note that the reader is assumed to possess knowledge of basic scalar and vector product operations) Continue reading →

## The pigeon hole principle

The pigeon hole principle is a counting argument stating something as simple as “if you have n items which are to be put into m < n boxes then there is at least 2 items in one of the boxes”. My first comment on that was “well duh!”, that is obvious, but it can actually be used to prove some less intuitive things.

So an example if you have 10 balls which are to be put into 9 boxes then at least 1 of the boxes contains at least 2 balls (unless you drop one of them…). But what can we use it for. I will show one of my favourite examples of it which I do not find intuitive at first glance. Continue reading →

## Proof method: Multidimensional induction

For the last couple of posts we have been dealing with mathematical induction. This post shall be no different. However, until now we have been limited to one variable, so this post will deal with induction on multiple variables – or in other words multidimensional induction as the headline suggests. I will treat the two dimensional case in this post, but the expansion to n dimensions should be relatively easy. So for now we will assume that we have a statement P(m,n) which we need to prove.

If we have the statement P(m,n) then we need to prove three things:

**Basecase: We need to prove a base case P(a,b) where a is the smallest value for which m is valid and b is the smallest value where n is valid.****Induction over m: We assume that P(k,b) is valid for some positive integer k. Then we need to prove that P(k+1,b) is valid.****Induction over n: We assume that P(h,k) is valid for some positive integers h,k. Then we need to prove that P(h,k+1) is valid.**

If we can prove these three things then we have proved that P(m,n) is true for all and . point 1 and 2 ensures that the statement is true for all P(m,a) just as in the in the single variable case. Once we have proven that then adding 3 ensures that it holds for all n as well. But let us take an example. Continue reading →

## Proof method: Strong Induction

As I promised in the Proof by induction post, I would follow up to elaborate on the proof by induction topic. Here is part of the follow up, known as the proof by strong induction. What I covered last time, is sometimes also known as weak induction.

In weak induction the induction step goes:

**Induction step: If P(k) is true then P(k+1) is true as well.**

Strong induction expands the concept to:

**Induction step: If P(m), P(m+1), P(m+2)… P(k) is true then P(k+1) is true as well for some k > m.**

The difference between the two methods is what assumptions we need to make in the induction step. To switch to the metaphor of dominoes again – in the weak induction we need to know that the previous domino is toppled, then the next one will topple as well. In strong induction we need to know that all the previous ones has toppled before we can ensure that the next one will fall as well.

Both methods have the same logical strength when we apply them, since in order to get to the k’th domino we need to topple 1,2,3,…k-1 anyway. However, sometimes strong induction makes the proof of the induction step easier.

## Strong Induction Example 1

I have shamelessly stolen this example from Hammack since I think it brilliantly shows when strong induction is better to use. But lets first see what happens if we try to use weak induction on the following:

**Proposition**: if , then 12|(n^{4} – n^{2})

### Weak induction

**Base case: **we need to prove that 12|(1^{4} – 1^{2}) = 12|(1- 1) = 0, which is divisible by 12 by definition.

**Induction step: **We assume that the 12|(k^{4} – k^{2}) is true such that (n^{4} – n^{2}) = 12a for some . We then need to show that ((k+1)^{4} – (k+1)^{2}) = 12b for some .

My approach would be to a direct proof such that

(k+1)^{4} – (k+1)^{2} = k^{4} + 4k^{3} + 6k^{2} + 4k + 1 – (k^{2} + 2k + 1) = (k^{4} – k^{2}) + 4k^{3} + 6k^{2} + 2k = 12a + 4k^{3} + 6k^{2} + 2k

How do we proceed from there? I don’t have a clue. If you can show me how to continue of this road I would be glad to hear it.

### Strong induction

The weak induction method failed. However, we can show that n = k-5 implies that the statement is true for k+1, so we need to expand the base case to include everything up to n = 6.

**Base case: **

n = 1: 12|(1^{4} – 1^{2}) = 12|(1- 1) = 0 = 0*12
n = 2: 12|(2 ^{4} – 2^{2}) = 12|16-12 = 12 = 1*12

n = 3: 12|(3

^{4}– 3

^{2}) = 12|(81 – 9) = 72 = 6

*12*

n = 4: 12|(4

n = 5: 12|(512

n = 4: 12|(4

^{4}– 4^{2}) = 12|(256- 16) = 240 = 20 * 12n = 5: 12|(5

^{4}– 5^{2}) = 12|(625- 25) = 600 = 50n = 6: 12|(6

^{4}– 6

^{2}) = 12|(1296- 36) = 1260 = 105*12

So far it fits really well.

**Induction step: **let and assume that 12|(m^{4} – m^{2}) for , now we need to prove that 12|((k+1)^{4} – (k+1)^{2}) is true as well.

Let us define l = k-5 for which we assume the proposition to be true such that (l^{4} -l^{2}) = 12a for some value of a. We need to show that 12|((l+6)^{4} – (l+6)^{2}) is true. So let us try with a direct approach

(l+6)^{4} – (l+6)^{2}= (l^{4} +24l^{3} + 180l^{2} + 864l + 1296) – (l^{2} + 12l + 36) = (l^{4} – l^{2}) +24l^{3} + 180l^{2} + 852l + 1260 = 12a +12(2l^{3} + 15l^{2} + 71l + 105).

This statement is clearly divisible by 12, and thus proofs the proposition.

QED

So in order to prove it for n = k+1 = 7 we need n = k-5 = 1 to prove it, but that is handled in the base case, same goes for all the n = 8,9,10,11,12 and then we start to rely on the fact that we can prove n= 8 through the induction step.

Image Credits

The metaphor of dominoes also gave rise to the chosen post photo, which was kindly shared under the creative common license by Malkav. My crop of the photo is of course shared under the same conditions.

## Proof methods: Proof by mathematical induction

It has been a while since I last posted something about proof methods, but lets dig that up again and take a look at a fourth method. The first three were direct proof, proof by contradiction and contrapositive proofs. Proof by induction is a somewhat different nature.

I have been reading quite a few blog posts recently and they all seem to be witty and clever, so I really wanted to add a joke right here on induction. But I honestly couldn’t find one I think was funny enough. So if you could just laugh or smirk for a few seconds before reading on, my day is saved. Continue reading →

## Contrapositive proofs – part 2

Once I finished up the post on contrapositive proofs I spend the better part of an hour feeling I wasn’t quite finished with the topic. I still had a couple of things to explore. The first one is a contrapositive proof that puzzled me, the other thing is De Morgan’s Laws which tells us how to negate a statement. Continue reading →

## Proof Method: Contrapositive proof

On my journey to improve my mathematical rigour I have covered direct proofs and Proof by Contradiction. In this post I will cover the third method for proving theorems.

Reading up on different methods for proving things I realized that a contrapositive proof is a really clever thing to used and often a better way to prove things than a proof by contradiction. However, I love the proof by contradiction so much that I wanted to cover it first. Continue reading →