# Question #627f1

Nov 26, 2014

Assume: $1 + \sqrt{2}$ is a rational number.

$\implies$ There exist integers $m$ and $n$ such that $1 + \sqrt{2} = \frac{m}{n}$.

$\implies \sqrt{2} = \frac{m}{n} - 1 = \frac{m - n}{n}$,

which means that $\sqrt{2}$ is a rational number.

$\implies$ There exist integers $a$ and $b$ such that $\sqrt{2} = \frac{a}{b}$,

which is reduced to lowest terms.

$\implies {\left(\sqrt{2}\right)}^{2} = {\left(\frac{a}{b}\right)}^{2} \implies 2 = {a}^{2} / {b}^{2} \implies 2 {b}^{2} = {a}^{2}$

$\implies {a}^{2}$ is even $\implies a$ is even

$\implies$ There exists an integer $k$ such that $a = 2 k$.

$\implies 2 {b}^{2} = {\left(2 k\right)}^{2} \implies 2 {b}^{2} = 4 {k}^{2} \implies {b}^{2} = 2 {k}^{2}$

$\implies {b}^{2}$ is even $\implies b$ is even

Since both $a$ and $b$ are even, $\frac{a}{b}$ is NOT in lowest terms, which is a contradiction.

Hence, $1 + \sqrt{2}$ is an irrational number.

I hope that this was helpful.