Is $\sqrt{2}$ a rational number?

Hint: We first assume that $\sqrt{2}$ is a rational number to make the contradiction. We take the square value of the equation $\dfrac{p}{q}=\sqrt{2}$ to find the combination of odd and even for the numbers p and q. We find the contradiction and prove that $\sqrt{2}$ an irrational number.

Complete step by step solution:
$\sqrt{2}$ is an irrational number.
We use the method of contradiction to prove that $\sqrt{2}$ is an irrational number.
We assume that $\sqrt{2}$ is a rational number.
We can express any rational number in the form of $\dfrac{p}{q},\left[ p,q\in \mathbb{Z},q\ne 0 \right]$.
The terms p and q are in their simplest form and that’s why they both can’t be even. One or both of them has to be odd.
Therefore, $\dfrac{p}{q}=\sqrt{2}$. We now take squares on both sides of the equation.
So, $\dfrac{{{p}^{2}}}{{{q}^{2}}}=2\Rightarrow {{p}^{2}}=2{{q}^{2}}$.
We know that square of any odd/even number will remain odd/even respectively and the inverse statement is also true.
Now in the equation of ${{p}^{2}}=2{{q}^{2}}$, the value of ${{p}^{2}}$ is even as it is the multiple of 2 with the value of ${{q}^{2}}$. This gives that the value of p is even also.
We assume that $p=2k,k\in \mathbb{Z}$. We substitute the value to get
\[\begin{align}
  & \dfrac{{{p}^{2}}}{{{q}^{2}}}=2 \\
 & \Rightarrow 2=\dfrac{{{\left( 2k \right)}^{2}}}{{{q}^{2}}}=\dfrac{4{{k}^{2}}}{{{q}^{2}}} \\
 & \Rightarrow {{q}^{2}}=2{{k}^{2}} \\
\end{align}\]
With similar argument for the equation of \[{{q}^{2}}=2{{k}^{2}}\], we have that the value of ${{q}^{2}}$ is even as it is the multiple of 2 with the value of ${{k}^{2}}$. This gives that the value of q is even also.
We get that the values of p and q both are even which is not possible. Hence the contradiction.
Then the assumption of $\sqrt{2}$ being a rational number is wrong. $\sqrt{2}$ is an irrational number.

Note: We need to simplify the rational number representation of $\dfrac{p}{q}$. If they are not in their simplest form then we take the G.C.D of the denominator and the numerator. If it’s 1 then it’s already in its simplified form and if the G.C.D of the denominator and the numerator is any other number d then we need to divide the denominator and the numerator with d and get the simplified fraction form as $\dfrac{{}^{p}/{}_{d}}{{}^{q}/{}_{d}}$.