Proof Questions | Worksheets and Revision | MME

# Proof Questions, Worksheets and Revision

Level 8 Level 9

## What you need to know

### Proof

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.

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 (interior and exterior angles revision)

Having a good understanding of the following topics will help with proof

### Learning Proofs

Students often think you can’t teach or learn how to do proof questions as they are all different, but they do often use the same logic. The following logic will help with different algebraic proofs questions:

• $n$ is a number so $2n$ must always be even
• $2n+1$ must therefore always be odd
• $(2n+1)(2n+1)$ will always produce an odd number because $odd\times odd = odd$

Note: Factorising and expanding brackets can often help to identify rules so keep this in mind.

### Example 1: Algebraic Proof Counter 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.

To make life easy, we’ll start with the smaller numbers.

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

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

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.

### Example 2: Algebraic Proof

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

We can start by expanding brackets to see if this helps. 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.

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

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. Since $8n$.Thus, we have completed the proof.

### Example 3: Algebraic Proof

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

So if $n$ is any whole number, then $2n$ must be an even number and an odd number must be

$2n+1$

The question asks about the square of an odd number, so let’s square our expression:

$(2n+1)^2$

Now expand the brackets

$4n^2+2n+2n+1$

Simplify to get

$4n^2+4n+1$

How do we know this is odd? Well, we can’t take a factor of 2 out of the whole thing, but we can factorise part of the expression to get

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

Making sense of this

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

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, we have completed the algebraic proof.

### Example Questions

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

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

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

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.

For any 3 consecutive odd numbers: $2n+1, 2n+3,$ and $2n+5$, adding them together gives us,

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

Now, to show that this number is odd 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 an odd number and so the result must be odd. Thus, we have completed the proof.

To show that the left and right hand sides of the equation are identical we expand the brackets on the left hand side,

\begin{aligned}5(3x-5)-2(2x+9)&=15x-25-4x-18 \\ &=11x-43\end{aligned}

Hence we have show the identity is true.

For any 3 consecutive even numbers: $2n, (2n+2)$ and $(2n+4)$, adding them together gives us,

$2n+(2n+2)+(2n+4)=6n+6$

Now, to show that this number is divisible by 6 we simply take out a factor of 6, then we get

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

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