# Core 4 Proof Key Points

HideShow resource information

## Proof

Methods of Proof :

• Proof by Direct Argument - using algebra, find a formula that prooves your argument
• 1. Prove that the product of an even number and an odd number is always even.
• Proof by Exhaustion - for some conjectures it is possible to test all the possible cases.
• 2. Prove that when a two digit number is divisible by 3, reversing its digits will also give a number that is divisible by 3.
• Proof by Contradiction - in some cases it is possible to deduce a result by showing that the converse is impossible
• 3. prove that root 2 is irrational.
1 of 6

## Proof 2

1. Prove that the product of an odd number and an even number is always even

definition of an even number : 'divisible by two'

let m and n represent any two numbers, therefore 2m is even and 2n + 1 is odd.

2m x (2n + 1) = 2 [ m (2n + 1) ]

this is a multiple of two so must be even.

2 of 6

## Proof 3

2. Prove that when a two digit number is divisible by 3, reversing its digits will also give a number divisible by 3.

there are only 30 two-digit numbers divisible by 3.

12, 15, 18, 21, 24, ... 93, 96, 99

reversing these digits give:

21, 51, 81, 12, 42, ... 39, 69, 99

these numbers are also divisible by 3, so the conjecture has been proved.

3 of 6

## Proof 4

3. Prove that root 2 is irrational

first, assume that root 2 is rational, then disprove.

if it is rational then root 2 = m/n where m and n have no common factor.

squaring gives 2 = m^2 / n^2 which gives 2n^2 = m^2

since 2 n^2 it is a multiple of 2 so is even, therefore so is m^2

since m^2 is even so is m. let m = 2p

2 n^2 = (2p)^2 = 4 p^2 ..... so n^2 = 2p^2

2p^2 is even so n^2 is also even.

you have now shown that m and n are even, which contradicts the assumption that m and n have no common factor.

consequently root 2 is not rational, so is therefore irrational.

4 of 6

## Disproof

The Methods of Disproof are :

• Disproof by Direct Argument - similar to proof by direct argument
• Disproof by use of a counter example - finding one case where the conjecture is false is enough to disprove it.
• e.g. Is it true that any number whose square is the sum of two squares is itself the sum of two squares?
5 of 6

## Disproof 2

1. Is it true that if a number whose square is the sum of two square is itself the sum of two squares?

Pythagorean triples form the basis of the conjecture

5 = 1^2 + 2^2

13 = 2^2 + 3^2

17 = 1^2 + 4^2

however ....

15^2 = 9^2 + 12^2

15 itself cannot be written as the sum of two squares

here you have found a counter example, so disproving the conjecture.

6 of 6