Given that $\sqrt{3}$ is an irrational number, prove that $\left( 2+\sqrt{3} \right)$ is an irrational number.

Hint: We will solve this question by assuming that the number $\left( 2+\sqrt{3} \right)$ is rational number r and then we will square both the sides and then we will arrange some terms so that $\sqrt{3}$ on one side and on the other side we will have some rational number. Now with the help of contradiction we can say that $\left( 2+\sqrt{3} \right)$ will be irrational.

Complete step-by-step solution -
Let’s start solving,
Let us assume that:
$\left( 2+\sqrt{3} \right)$ is a rational number.
Let, $\left( 2+\sqrt{3} \right)=r$ , where "r" is a rational number.
Now taking square on both the sides we get,
${{\left( 2+\sqrt{3} \right)}^{2}}={{r}^{2}}$
Now using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ we get,
$\begin{align}
  \Rightarrow & {{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+2\times 2\times \sqrt{3}={{r}^{2}} \\
 \Rightarrow & 4+3+4\sqrt{3}={{r}^{2}} \\
 \Rightarrow & 7+4\sqrt{3}={{r}^{2}} \\
 \Rightarrow & \sqrt{3}=\dfrac{{{r}^{2}}-7}{4}...........(1) \\
\end{align}$
So, we see that LHS which is $\sqrt{3}$ and it is purely irrational.
But, on the other side, RHS is rational.
Hence, this contradicts the fact that $\left( 2+\sqrt{3} \right)$ is rational.
Hence, our assumption was wrong.
Therefore, from this we can say that $\left( 2+\sqrt{3} \right)$ is an irrational number.
Hence proved.

Note: One can also solve this question by taking $\left( 2+\sqrt{3} \right)$ as $\dfrac{p}{q}$. Then we have to solve the expression so that $\sqrt{3}$ on one side and on the other side we will have some rational number and then by contradiction we can say that $\left( 2+\sqrt{3} \right)$ is irrational number. Both the methods are good and one can choose any one of them to solve this question.