## Sqrt(2) is irrational

Definitions:

1.  An integer is called even if and only if it is divisible by 2, that is some integer $\displaystyle r$ is even if for some other integer $\displaystyle s$ the statement $\displaystyle r = 2s$ is true.
2. An integer is called odd if and only if it is not divisible by 2, that is some integer $\displaystyle n$ is odd if for some other integer $\displaystyle m$ the statement $\displaystyle n = 2m + 1$ is true.
3. A number is called rational if it is in the form of $\displaystyle \frac{p}{q}$ where $\displaystyle p$ and $\displaystyle q$ are integers. Further, it is said to be in lowest terms if $\displaystyle p$ and $\displaystyle q$ are not commonly divisible by any integer save the identity 1. A number that is not rational is called irrational.
4. A number $\displaystyle x$ is called a square root if it satisfies $\displaystyle x^2 = y$ for some number $\displaystyle y$ where for our purposes $\displaystyle y$ is a positive integer. “The square root of $\displaystyle y$” is denoted by $\displaystyle \sqrt{y}$.

Theorem 1 The product of two even integers is even.
Proof For any two even integers $\displaystyle r = 2k$ and $\displaystyle s = 2j$, we have

$\displaystyle rs = (2k)(2j) = 2(2kj)$

which is even. QED.

Theorem 2 The product of two odd integers is odd.
Proof For any two even integers $\displaystyle p = 2m + 1$ and $\displaystyle q = 2n + 1$, we have

$\displaystyle pq = (2m + 1)(2n + 1) = 4mn + 2m + 2n +1 = 2(2mn + m + n) + 1$

which is odd. QED.

Corollary 1 If the square of a number is even, then the number itself is even. If the square of the number is odd, the number is likewise odd.

Theorem 3 $\displaystyle \sqrt{2}$ is irrational.
Proof Assume that $\displaystyle \sqrt{2}$ is rational and in lowest terms, such that

$\displaystyle 2 = \frac{p^2}{q^2}$

or

$\displaystyle p^2 = 2q^2$

that is $\displaystyle p^2$ is even. Then, by Corollary 1, $\displaystyle p$ is even as well. That is,

$\displaystyle p = 2n$

for some integer $\displaystyle n$. Therefore,

$\displaystyle (2n)^2 = 4n^2 = 2q^2$

which implies

$\displaystyle q^2 = 2n^2$

and thus $\displaystyle q^2$ is even and therefore $\displaystyle q$ is even, again by Corollary 1. Therefore, both $\displaystyle p$ and $\displaystyle q$ are even which contradicts our initial assumption that $\displaystyle \sqrt{2}$ was rational. Therefore, the assumption was false and $\displaystyle \sqrt{2}$ is irrational via proof by contradiction. QED.