Question

Prove that root 7 is irrational by the method of contradiction

Answer 1

Let assume that $\sqrt7$ is a rational number. Therefore it can be expressed as a fraction $\dfrac{a}{b}$ where$a,b\in\mathbb{Z}$ and $\text{gcd}(a,b)=1$.

$\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2$

This means that $a^2$ is divisible by 7, and therefore also $a$ is divisible by 7.

So, $a=7k$ where $k\in\mathbb{Z}$

$(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2$

Analogically to $a^2=7b^2$ ------- $b^2$ is divisible by 7 and therefore so is $b$.

But if both numbers $a$ and $b$ are divisible by 7, then $\text{gcd}(a,b)=7$ which contradicts our earlier assumption that $\text{gcd}(a,b)=1$.

Therefore $\sqrt7$ is an irrational number.