Pure Year 2 Overview

Proof by contradiction

Proof by contraction is where a statement is assumed to be false, then a contradiction is found, showing that the original statement must be true.

Assume that $2$ is rational, so $2 = \frac{a}{b}$ for some integers $a$ , $b$ , where $a$ and $b$ are coprime.
Squaring gives $2 = \frac{a ^{2}}{b ^{2}}$ , so $2 b^{2} = a^{2}$ .
Because $a^{2} = 2 b^{2}$ , $a^{2}$ is even, so $a$ itself is even.
If $a$ is even, then $a = 2 k$ for some integer $k$ .
Then, $2 b^{2} = a^{2} ⟹ 2 b^{2} = (2 k)^{2} = 4 k^{2}$ , so $b^{2} = 2 k^{2}$ .
Because $b^{2} = 2 k^{2}$ , $b^{2}$ is even, so $b$ itself is even.
$a$ and $b$ are both even, but this contradicts the assumption that $a$ and $b$ are coprime, as they both share a factor of 2. Thus, the assumption must be false, so $2$ is irrational.

Assume that there is a finite list of primes, so there is a largest prime number $p_{n}$ .
Consider multiplying all the prime numbers together, $Q = p_{1} \times p_{2} \times \dots \times p_{n}$ .
Consider $Q + 1$ . As it is 1 more than a multiple of every prime, it is not divisible by any prime (dividing by any prime leaves a remainder of 1).
Thus, $Q + 1$ is either prime, or divisible by a prime larger than $p_{n}$ (all numbers are either prime or have prime factors).
This contradicts the assumption that there are a finite number of primes, as there is some prime number larger than $p_{n}$ . Thus, the assumption must be false, so there are an infinite number of primes.