## What you need to know

Proof is at the heart of all the things you see in mathematics. That is, all the things that you use and take for granted, such as Pythagoras’ Theorem or the formula for the area of a circle, have, at some point over the last few thousand years, been proven to be true. You won’t have to delve into the proofs of everything you use (which is fortunate, because it would probably take a little while), instead this topic is intended as an introduction into how proofs work.

The idea of a proof is to make a universal statement – for example, you don’t just want to say that the angles in some triangles add up to 180, you want to say that the angles in all triangles add up to 180. This is a proof you actually do have to know, and you can see it here (https://mathsmadeeasy.co.uk/gcse-maths-revision/interior-and-exterior-angles-gcse-revision-and-worksheets/). The general structure of a proof is to begin with one statement, take a series of logical and mathematical steps, and end up at a desired conclusion. Of course, not everything we want can be proved true. In fact, all it takes to prove a statement false is to find a counterexample – a particular example for which the universal statement you’re trying to make doesn’t hold true.

Example: Hernan claims “if you square a number and add 1, the result is a prime number”. Find a counterexample to prove her statement wrong.

$1^2+1=1+1=2,\text{ which is prime.}$

$2^2+1=4+1=5,\text{ which is prime.}$

Now, at this point it might seem like her statement is true, but if we try the next square number:

$3^2+1=9+1=10,\text{ which is not prime.}$

This is a counterexample for her statement, so we have proved it to be false.

The proofs you see in this topic will involve a lot of algebraic expansion and collecting like terms together, so make sure you’re comfortable with that by going here (https://mathsmadeeasy.co.uk/gcse-maths-revision/multiplying-single-double-brackets/) before continuing.

Example: Prove that $(n+2)^2-(n-2)^2$ is divisible by 8 for any positive whole number $n$.

To do this, we need to show that $(n+2)^2-(n-2)^2$ can be written in some way that is clearly divisible by 8. To find a way to write an expression like this differently, we can try expanding it. So, the first bracket expands to

$(n+2)^2=n^2+2n+2n+4=n^2+4n+4$.

Then, the second bracket expands to

$(n-2)^2=n^2-2n-2n+4=n^2-4n+4$.

The expression in the question has the second bracket being subtracted from the first one. So, we will do this subtraction with the expansion of the two brackets:

$(n+2)^2-(n-2)^2=(n^2+4n+4)-(n^2-4n+4)$

We can see that the $n^2$ terms will cancel, as will the 4s, so all we’re left with is

$(n^2+4n+4)-(n^2-4n+4)=4n-(-4n)=8n$

So, the whole expression simplifies to $8n$. Now, if $n$ is a whole number, then $8n$ must be divisible by 8 (if we divide it by 8, we get the answer $n$). Since $8n$ is equivalent to the expression we started with, it must be the case that $(n+2)^2-(n-2)^2$ is divisible by 8 for any positive whole number $n$ – so the statement is now universal. Thus, we have completed the proof.

So, this proof gave us an expression to simplify and write in a way that satisfied the conclusion we wanted. Sometimes, the proof is phrased in words and we have to figure out how to turn those words into something we can do maths with.

Example: Prove that the square of an odd number is also odd.

Okay, so if we’re going to make a statement about odd numbers and we want to use algebra to prove the statement, we need some way to express an odd number with algebra. To do this, let’s consider even numbers: an even number is a multiple of 2, by definition, so if $n$ is any whole number, then $2n$ must be an even number. Indeed, any even number can be expressed by $2n$ if you pick the right $n$. Furthermore, any odd number is always the next one along from an even number. So, we can express an odd number like

$2n+1$.

So, now that we have our algebraic expression of an odd number, we can really get going. The question asks about the square of an odd number, so let’s square our expression for a general odd number:

$(2n+1)^2=4n^2+2n+2n+1=4n^2+4n+1$.

The question is now: how do we know this is odd? Well, we can’t take a factor of 2 out of the whole thing, but we can take a factor of 2 out of the first two terms (if you’re not sure where this is going that’s okay, just bear with me). Doing so, we get

$4n^2+4n+1=2(2n^2+2n)+1$

Now, $2n^2+2n$ might seem like a reasonably complicated expression but importantly, it must be a whole number. This is because $n$ is a whole number, and if you square a whole number/multiply it by other whole numbers, the result must still be a whole number. Already, I’ve said the phrase “whole number” far too many times here, but the whole point is that

$2(2n^2+2n)+1\,\,\,\text{ is just }\,\,\, 2\times(\text{some whole number})+1$.

And since $2\times(\text{some whole number})$ is how we defined an even number, it follows that $2\times(\text{some whole number})+1$ is how we define an odd number. Therefore, this expression – and thus, the square of any odd number – must be odd. This is our universal statement, and we have completed the proof.

Proof is quite different to a lot of the things you’ve seen in maths before so don’t worry if it’s tricky to get your head around. The more you see and practice following logical arguments like these ones, the less foreign they will seem. Don’t worry if you keep getting stuck at first – even the greatest mathematicians get stuck on proofs on a very regular basis.

## Example Questions

#### 1) Luke claims that if you square an even number and add 3, the answer is prime. Find a counterexample to show that Luke’s statement is false.

We will try the first few even numbers (squaring them and adding 3) until we find an example that isn’t prime. So, we get

$2^2+3=4+3=7,\text{ which is prime}.$

$4^2+3=16+3=19,\text{ which is prime}.$

$6^2+3=36+3=39,\text{ which is not prime}.$

Since 39 is divisible by 3, it must not be prime, so we have proved Luke’s statement to be false.

Note: there are many, many counter examples to Luke’s statement, and any one of them is an acceptable answer to this question.

#### 2) Prove that $(3n+1)^2+(n-1)^2$ is always even for any positive whole number $n$.

To answer this question, we will need to expand and simplify the expression given to us, so we can hopefully write it in a way that shows it is clearly divisible by 2 (since that’s the definition of even). So, expanding the first bracket, we get

$(3n+1)^2=9n^2+3n+3n+1=9n^2+6n+1$.

Then, expanding the second bracket, we get

$(n-1)^2=n^2-n-n+1=n^2-2n+1$.

Now, the expression in the question has the two brackets added together. So, adding the expansions together, we get

$(9n^2+6n+1)+(n^2-2n+1)=10n^2+4n+2$

Is this an even number? Well, if we take a factor of 2 out of the expression:

$2(5n^2+2n+1)$,

we see that since $5n^2+2n+1$ is a whole number because $n$ is a whole number, the expression in question is equal to $2\times(\text{some whole number})$ and so must be even. Thus, we have completed the proof.

#### 3) Prove that the sum of 3 consecutive odd numbers is an odd number. (HINT: if $2n+1$ is how we express an odd number, how can we express the next odd number along?)

Answering the question in the hint is the key to getting going with this question. So, we know that for any whole number $n$, $2n$ must be an even number and $2n+1$ must be an odd number. So, if we have one odd number and want to get to the next one, all we must do is add 2. Therefore, the next odd number after $2n+1$ is

$(2n+1)+2=2n+3$

Furthermore, for the next odd number after this one, we must add 2 again:

$(2n+3)+2=2n+5$.

So, now we have expressions for 3 consecutive odd numbers: $2n+1, 2n+3,$ and $2n+5$, all that remains is to add these up and show that the result is always odd. Adding them together gives us

$(2n+1)+(2n+3)+(2n+5)=6n+9$

Now, to show that this number is odd we have to be a little clever. If we write this expression instead like $6n+8+1$, then we get

$6n+8+1=2(3n+4)+1$

Since $n$ is a whole number, clearly $3n+4$ must also be a whole number, so we get that our expression is of the form $2\times\text{(some whole number)}+1$, which is precisely how we expressed an odd number in the first place, and so the result must be odd. Thus, we have completed the proof.

This is a tough question, so if you managed to get through any of the steps – good job. If you really had no idea, that’s okay too. Give the proof another read through to make sure you fully understood each step, and then next time a question similar comes up you’ll have a better idea.

## Proof Revision and Worksheets

Proofs
Level 8-9
Proofs 2
Level 8-9
Proofs Test
Level 8-9

Proof and algebraic proof are one of the most difficult topics to teach and yet there are very few great resources out there to help tutors and teachers to deliver this topic. Thankfully at Maths Made Easy we have taken the time to research the best resources out there as well as add in much of our own to help create a super resource for proof.