Prove that

We have to prove that $\sqrt{7}$ is an irrational number

We will prove this by contradiction:

Let $\sqrt{7}$ be an irrational number such that $\sqrt{7}=\frac{x}{y}$ where $x$ and $y$ are co prime

So' $\sqrt{7}=\frac{x}{y}$

$\Rightarrow 7=\frac{x^{2}}{y^{2}}$

$\Rightarrow \quad 7 y^{2}=x^{2}$

$\Rightarrow \quad 7$ divides $x^{2}$

$\Rightarrow \quad 7$ divieds $x$.......(1)

This means that:

$x=7 k$ where $\mathrm{k}$ is any positive integer

$\Rightarrow \quad x^{2}=49 k^{2}$

$\Rightarrow \quad 7 y^{2}=49 k^{2}$

$\Rightarrow \quad y^{2}=7 k^{2}$

$\Rightarrow \quad 7$ divides $y^{2}$

$\Rightarrow \quad 7$ divides $y$.............(2)

From equation (1) and (2) we see that 7 is a common factor of x and y. This contradicts the fact that x and y have no common factor

Hence is an irrational number.