HaCkEd By KaSpEr511

Irrationality of Square Root of Two

June 17th, 2009

We want to show that $\sqrt{2}$ is irrational by contradiction. Suppose that it is rational. Then we can write $\sqrt{2}$ as
$\sqrt{2}=\frac{p}{q}$ and without of lost of generality we can assume that the fraction $p/q$ is in the lowest term, i.e., $p$ and $q$ has no common factor.

If we square both sides we have
$\begin{aligned} 2 &= \frac{p^2}{q^2}\\ 2q^2&=p^2 \end{aligned}$ .

It follows that $p^2$ is even and hence $p$ is also even (if $p$ is odd then $p^2$ is odd).
Write $p=2s$ and substitute to the last equation to get
$\begin{aligned}2q^2 &= 4s^2\\ q^2&= 2s^2 \end{aligned}$
Thus by the same reasoning as before, we conclude that $q$ is even. So here assuming that $\sqrt{2}$ is rational we arrive at the conclusion that $p$ and $q$ are even. But this is a contradiction since from the beginning we assume that $p,q$ has no common factor.
Therefore we must conclude that $\sqrt{2}$ is irrational.

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Irrationality of Square Root of Two

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