Prove that $\sqrt{5}$ is an irrational number.

Hint: Use the contradiction method to prove $\sqrt{5}$ as an irrational number i.e. suppose $\sqrt{5}$ is rational. Any rational number can be represented in the form of $\dfrac{p}{q}$ , where $q\ne 0$ . So, represent $\sqrt{5}$ as $\dfrac{p}{q}$ and suppose p, q are co-primes. Now, use the property that if ${{n}^{2}}$ is a multiple of ‘a’. then ‘n’ is also a multiple of ‘a’. And hence contradict your assumption taken initially.
Complete step-by-step answer:
Let us prove that $\sqrt{5}$ is an irrational number, by using the contradiction method.
So, say that $\sqrt{5}$ is a rational number and as we know any rational number can be expressed in form of $\dfrac{p}{q}$ with $q\ne 0$ .So let $\sqrt{5}$ equals to $\dfrac{p}{q}$
So, we get $\sqrt{5}=\dfrac{p}{q}$
Where p, q are co-prime integers i.e. they do not have any common factor except ‘1’. It means $\dfrac{p}{q}$ is the simple form of fraction because p and q do not have any common factor.
So, we have $\sqrt{5}=\dfrac{p}{q}\left( q\ne 0 \right)$
On squaring both sides of the above equation, we get
$\begin{align}
  & {{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}} \\
 & \dfrac{5}{1}=\dfrac{{{p}^{2}}}{{{q}^{2}}} \\
\end{align}$
On cross-multiplying the above equation, we get
${{p}^{2}}=5{{q}^{2}}......\left( i \right)$
Now, we can observe that ${{p}^{2}}$ is a multiple of ‘5’ because ${{p}^{2}}$ is expressed as $5\times {{q}^{2}}$ . As we know that ‘n’ will be multiple of any number ‘a’ if ${{n}^{2}}$ is a multiple of ‘a’ which is the fundamental theorem of arithmetic.
So, if ${{p}^{2}}$ is a multiple of ‘5’, then p will also be a multiple of 5.
Hence, we can express ‘p’ as a multiple of 5 in a following way as
$p=5m......\left( ii \right)$
Where m is any positive integer.
On squaring both sides of equation (ii), we get
${{p}^{2}}={{\left( 5m \right)}^{2}}=25{{m}^{2}}........\left( iii \right)$
Put value of ${{p}^{2}}$ as $5{{q}^{2}}$ from the equation (i) to the equation equation(iii) we get
$\begin{align}
  & 5{{q}^{2}}=25{{m}^{2}} \\
 & {{q}^{2}}=5{{m}^{2}}........\left( iv \right) \\
\end{align}$
Now, we can observe that ${{q}^{2}}$ is a multiple of ‘5’, it means q will also be a multiple of ‘5’.
So, we get that ‘p’ and ‘q’ both have ‘5’ as a common factor from equation(i) and (iv), but we supposed earlier that p and q are co-primes which have ‘1’ as a common factor. So, it contradicts our assumption that $\dfrac{p}{q}$ supposed will not be a rational number. Hence $\sqrt{5}$ is an irrational number.

Note: ‘p’ and ‘q’ are co-prime numbers that there are no common factors of ‘p’ and ‘q’ except ‘1’. It was the key point of the solution.
One may get confused with the statement that p will also be a multiple of 5 if ${{p}^{2}}$ is also multiple of 5. It is the Arithmetic fundamental property in mathematics.
Example : 25 is multiple of 5 then $\sqrt{25}=5$ is also a multiple of 5; 49 is a multiple of 7, then $\sqrt{49}=7$ is also a multiple of 7. So, it can be observed from the several examples as well. So, don’t be confused with this point.