Question

Verify by the method of contradiction $P:\sqrt{7}$ is irrational.

Answer 1

Let $\sqrt{7}$ be rational

It must in the form of (a/b) where a, b have no common factor

Let $\sqrt{7}=\frac{a}{b}$(a, b have no common factor)

$\Rightarrow 7=\frac{a^2}{b^2}$

$\Rightarrow a^2=7b^2$

$\Rightarrow$ $7$ divide $a$

$\Rightarrow$Therefore there exists an integer ${}^{\prime }{c}^{\prime }$ such that $a=7c$

$\Rightarrow a^2=49c^2$

$\Rightarrow 49c^2=7b^2$

$\Rightarrow b^2=7c^2$

$\Rightarrow$$7$ also divides $b$

$\Rightarrow$we got that $7$ divides both ${}^{\prime }{a}^{\prime }and{}^{\prime }{b}^{\prime }$

But we assumed that $a,b$ have no common factor

This is a contradiction and what we assumed is incorrect

$\therefore \sqrt{7}$ is irrational