Prove that$\sqrt{2}$ is an irrational number.

Proof of $\sqrt{2}$ is an irrational numbers.

Assume, $\sqrt{2}$is a rational number, it can be written as $\frac{p}{q}$, in which $p$ and $q$are co-prime integers and $q\ne 0$,

i.e. $\sqrt{2}=\frac{p}{q}$. where, $p$ and $q$are coprime numbers, and $q\ne 0$.

On squaring both sides of the above equation;

$$\begin{gathered} {\left(\sqrt{2}\right)}^2={\left(\frac{p}{q}\right)}^2 \\ \Rightarrow 2=\frac{p^2}{q^2} \\ \Rightarrow 2q^2=p^2...\left(i\right) \\ \Rightarrow p^2\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}2 \\ \Rightarrow p\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}2...\left(ii\right) \end{gathered}$$

Since, $p$ is a multiple of two.

$$\begin{gathered} \therefore p=2m \\ \Rightarrow p²=4m²\dots \left(iii\right) \end{gathered}$$

Using equation$\left(i\right)$ into the equation $\left(iii\right)$, we get;

$$\begin{gathered} 2q²=4m² \\ \Rightarrow q²=2m² \\ \Rightarrow q^2\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}2 \\ \Rightarrow q\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}2...\left(iv\right) \end{gathered}$$

Equation $\left(ii\right)\mathrm{and}\left(\mathrm{iv}\right)$, implies that $p$ and $q$have a common factor $2$.

It contradicts the fact that they are co-primes which lead from our wrong assumption that $\sqrt{2}$is a rational number.

Hence, $\sqrt{2}$ is an irrational number(proved)