Question

If $7$ is prime, then prove that $\sqrt{7}$ is irrational.

Answer 1

Assume $\sqrt{7}$ is a rational number.
Let $\sqrt{7}=\frac{a}{b}$, where g.c.d. $(a,b)$ $=1;a\epsilon N,b\epsilon N$
$\Rightarrow a^2=7b^2$ ........ (1)
$\Rightarrow 7|a^2$
$\Rightarrow 7|a$
Let $a=7$, $k$ $\epsilon$ $N$
Therefore, $49k^2=7b^2$ ....... [Subs. $a=7k$ in (1)]
$\Rightarrow b^2=7k^2$
$\Rightarrow 7|b^2$
$\Rightarrow 7|b$
Thus $7|a$ and $7|b$
This is a contradiction, because g.c.d. $(a,b)=1$.
Thus, our assumption is wrong .
Hence, $\sqrt{7}$ is irrational.