Square Root of 2 is Irrational

Theorem:
$$\sqrt{2} \,\!$$ is irrational.

Theorem A
for any $$ k, p > 1, \in \mathbb{N}$$, there exists $$k' \in \mathbb{Z}$$ Such that $$2k = p^2 \Rightarrow 2k' = p$$ That is, if a number is even, it's square root is also even

Proof
For any given integer, $$m, \exists n $$ such that $$ m = 2n $$ or $$m = 2n + 1$$ (That is to say, a number is either odd, or even)

if $$m = 2n + 1 $$ then $$m^2 = 4n^2 + 4n + 1$$ which is not even

if $$m = 2n $$ then $$m^2 = 4n^2$$ which is even

Therefore, if it is not the case that a number is even, then it is not the case that its square is even. Conversely, if it is the case that a number is even, then it is the case that its square is even.

Or $$ k, p > 1, \in \mathbb{N}$$, there exists $$k' \in \mathbb{Z}$$ Such that $$2k = p^2 \Rightarrow 2k' = p$$

Theorem B
$$2|p^2 \Rightarrow 2|p \,\!$$

Proof
be definition $$2 | p^2 \iff \exists q \in \mathbb{Z}$$ such that $$2q = p^2 $$

This theorem is then proved by Theorem A

Proof by Contradiction:
Assume that $$\sqrt{2} \,\!$$ is rational; so $$\sqrt{2}={\frac{p}{q}}$$ for some $$p,q \in \mathbb{Z}$$ and $$gcd(p,q)=1\,\!$$.

Squaring both sides yields $$2=\frac{p^2}{q^2} \iff p^2=2q^2\,\!$$.

Therefore, $$2|p^2 \Rightarrow 2|p \,\!$$; (see Theorem B) that is, $$p \,\!$$ is an even number. So, $$p = 2k \,\!$$ for some $$k \in \mathbb{Z}$$.

Thus, $$2q^2 = p^2 = (2k)^2 = 4k^2 \Rightarrow q^2 = 2k^2 \,\!$$, so by the same reasoning, $$2|q^2 \Rightarrow 2|q \,\!$$, contradicting our assumption that $$gcd(p,q)=1\,\!$$, since $$2|p,q \,\!$$.

Therefore,$$\sqrt{2} \,\!$$ is irrational.

Q.E.D.

Note: this is a special case of the result that the square root of any prime is irrational.