Irrationality of sqrt(2)

Irrationality of

\sqrt{2}

There's no $q \in Q$ such that $q^{2} = 2$ .

Proof
We'll do proof by contradiction. Assume for contradiction $\exists q \in Q$ where $q^{2} = 2$ . Then $\exists a, b \in Z$ where $q = \frac{a}{b}$ where $b \neq 0$ . We can assume further for $Q$ that we can reduce $\frac{a}{b}$ to reduced terms, so then $a, b$ have no common factors. Plugging it in:

\begin{aligned} q^{2} & = 2 \\ {(\frac{a}{b})}^{2} & = 2 \\ a^{2} & = 2 b^{2} \end{aligned}

Clearly $2 | a$ so then we can write $a = 2 k$ where $\exists k \in Z$ . Plugging it in:

\begin{aligned} a^{2} & = 2 b^{2} \\ (2 k)^{2} & = 2 b^{2} \\ 4 k^{2} & = 2 b^{2} \\ 2 k^{2} & = b^{2} \end{aligned}

So clearly $2 | b$ . But $a, b$ cannot share the factor of 2! Thus this is a contradiction, so then $∄ q \in Q$ where $q^{2} = 2$ . $\Rightarrow\Leftarrow$
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References

[[Abbott Real Analysis.pdf#page=15]]