# Proof

A LevelAQAEdexcelOCR

## Proof

You may be asked to prove something in mathematics. There are a number of different types of proof questions that you will encounter.

There are 4 skills for proof that you need to understand.

A LevelAQAEdexcelOCR

## Skill 1: Proof Notation

The following notation will be used throughout proofs, but will also be relevant to topics later in the course.

A set is a collection of objects or numbers (called elements). A set is denoted using a capital letter, and curly brackets are used to show what is in the set, e.g. $A = \{3,4,5 \}$.

You can write sets in various different ways:

• as a list of elements: e.g. $\{3,4,5\}$
• as a rule: e.g. $\{\text{prime numbers}$$\text{ between 0 and 10}\}$
• as mathematical notation: e.g. $\{x : x > 5 \}$ – this means “the set of values such that $x$ is greater than $5$

There are different variations of an $=$ sign that you need to be aware of:

• $\neq$ means “not equal to”
• $\approx$ means “approximately equal to”
• $\equiv$ means two things are “equivalent”. It is called the identity symbol.

There are logic symbols that you need to understand:

• $p \Rightarrow q$” means “if $p$ then $q$” or “$p$ implies $q$
• $p \iff q$” means “$p$ if and only if $q$” or “$p$ implies $q$ and $q$ implies $p$

Note: “if and only if” can be written as “iff”

A LevelAQAEdexcelOCR

## Skill 2: Direct Proof

A direct proof (or proof by deduction) is a proof where a statement is proven to be true using fundamental mathematical principles.

Example: Prove that $n^2 - 6n + 11$ is positive for any integer.

Completing the square gives

$\textcolor{red}{(n-3)^2} + \textcolor{blue}{2}$

$\textcolor{red}{(n-3)^2}$ is always positive, since it is a square number.

Therefore, if we add $\textcolor{blue}{2}$ to the square number, then the result is still positive.

Hence, $n^2 - 6n + 11$ is positive for any integer.

A LevelAQAEdexcelOCR

@mmerevise

## Skill 3: Proof by Exhaustion

In a proof by exhaustion, you will need to break situations down into two or more cases and try all possibilities to prove that the statement holds true for each case.

Example: Prove the following statement:

“For any integer $x$, $f(x)=3x^2+3x-2$ is an even integer”

We need to split the statement into two cases: if $x$ is an even number, or if $x$ is an odd number.

Case 1: If $x$ is even, then $\textcolor{red}{x=2n}$ for some integer $n$.

Then,

\begin{aligned} f(2n) &= 3(2n)^2+3(2n)-2 \\ &= 12n^2 + 6n - 2 \\ &= 2(6n^2+3n-1) \end{aligned}

$n$ is an integer, therefore $6n^2+3n-1$ is an integer.

Hence, $2(6n^2+3n-1)$ is an even integer.

So, $f(x)$ is even when $x$ is even.

Case 2: If $x$ is an odd number, then $\textcolor{blue}{x = 2m+1}$ for some integer $m$.

Then,

\begin{aligned} f(2m+1) &= 3(2m+1)^2 + 3(2m+1) - 2 \\ &= 12m^2 + 12m + 3 + 6m + 3 - 2 \\ &= 12m^2 + 18m + 4 \\ &= 2(6m^2 + 9m + 2) \end{aligned}

$m$ is an integer, therefore $6m^2 + 9m + 2$ is an integer.

Hence, $2(6m^2 + 9m + 2)$ is an even integer.

So, $f(x)$ is even when $x$ is odd.

Hence, $f(x)$ is even for any integer $x$, and the statement is true.

A LevelAQAEdexcelOCR

## Skill 4: Disproof by Counter-Example

To disprove a statement by a counter-example, all you need to do is show that the statement is false for one case.

Example: Disprove the following statement:

“For any pair of real numbers $x$ and $y$, if $x^2=y^2$ then $x=y$

We only need to find one set of values such that the statement is false.

For example, take $x=3$ and $y = -3$

Then,

$x^2 = 3^2 = 9$ and $y^2 = (-3)^2 = 9$

So, $\textcolor{limegreen}{x^2 = y^2}$

However, $\textcolor{red}{x \neq y}$, and so the statement is not true.

A LevelAQAEdexcelOCR

## Proof Example Questions

a) $\{ -4, 4\}$

b) $\{1,2,3,6,9,18\}$

We only need to give one case for a counter-example, if it is false.

The only case in which the statement isn’t true is when $x=0$ and $y=0$. This gives $0^2+0^2=0$, and clearly $0$ is not greater than $0$

Take any two rational numbers, and call them $a$ and $b$

By definition of rational numbers, $a$ can be written in the form $a = \dfrac{p}{q}$ and $b = \dfrac{r}{s}$, where $q$ and $s$ are not $0$.

Then either:

$\dfrac{p}{q} + \dfrac{r}{q} = \dfrac{p+r}{q}$ if the denominators are the same

$\dfrac{p}{q} + \dfrac{r}{s} = \dfrac{ps + qr}{qs}$ if the denominators are different

$ps$, $qr$ and $qs$ are all products of integers, so they must also be integers.

$p+r$ and $ps + qr$ are all sums of integers, so they must also be integers.

Since $q$ and $s$ are both non-zero, $qs$ must also be non-zero.

Hence, $\dfrac{p+r}{q}$ and $\dfrac{ps + qr}{qs}$ are rational numbers.

Therefore, $a+b$ must be rational.

So, the statement is true.

By definition, three consecutive even numbers are $2n, (2n+2)$ and $(2n+4)$.

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

$n+1$ is an integer, so $6(n+1)$ is divisible by $6$

We need to prove by exhaustion that the statement is true for all integers in the interval $3 \leq n \leq 6$

$n=3$:   $3^2 - 2(3) = 3 > 2$

$n=4$:   $4^2 - 2(4) = 8 > 2$

$n=5$:   $5^2 - 2(5) = 15 > 2$

$n=6$:   $6^2 - 2(6) = 24 > 2$

Hence, the statement is true.

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