\(\sqrt{2}\) is irrational ◻

February 11th, 2020

\(\mathbf{Theorem}\)

\(\sqrt{2}\) is irrational.

\(\mathbf{Proof}\)

Let \(a,b \in \mathbb{Z} \quad, \frac{a}{b}=\sqrt{2},\quad b \neq 0, \quad (a,b)=1\)

Then \(a=\sqrt{2}b\)

\begin{align*} \implies & a^2=2 \times b^2 \quad (\Xi) \\ \implies & 2|a^2 \\ \implies & 2|a \\ \end{align*}

Then if \(a\) is even, \(\exists k \in \mathbb{Z} \ni a = 2k\)

Then substitute into \((\Xi)\), we get \((2 \times k)^2=2\times b^2\)

\begin{align*} \implies & 4 \times k^2 = 2 \times b^2 \\ \implies & 2 \times k^2 = b^2 \\ \implies & 2|b^2 \\ \implies & 2|b \end{align*}

If \(a\) and \(b\) are both even, it contradicts the initial condition \((a,b)=1\).

\(\therefore\) By Law of Contradiction, \(\sqrt{2}\) is irrational ◼︎