Question

Prove that $5-2\sqrt{3}$ is an irrational number.

Answer 1

Hint: First assume that $5-2\sqrt{3}$ is a rational number and represent it in the form of $\dfrac{p}{q}$. Then arrange the values in the LHS equal to $\sqrt{3}$ and you will find a rational number in RHS equal to $\sqrt{3}$. That will contradict your assumption that $5-2\sqrt{3}$ is a rational number because $\sqrt{3}$ is an irrational number.

Complete step-by-step answer:

Let us assume that the given expression, $5-2\sqrt{3}$ is a rational number.

As, $5-2\sqrt{3}$ is a rational number.

So, $5-2\sqrt{3}$ can be written in the form of $\dfrac{p}{q}$ where p and q are integers, having no common factor except 1 and q≠0.

i.e. $5-2\sqrt{3}$= $\dfrac{p}{q}$ (p and q are co-prime and q≠0)

$\begin{align}

  & \Rightarrow -2\sqrt{3}=\dfrac{p}{q}-5 \\

 & \Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q} \\

 & \Rightarrow -\left( \dfrac{p-5q}{2q} \right)=\sqrt{3} \\

\end{align}$ $\begin{align}

  & \Rightarrow -2\sqrt{3}=\dfrac{p}{q}-5 \\

 & \Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q} \\

 & \Rightarrow \sqrt{3}=-\left( \dfrac{p-5q}{2q} \right) \\

\end{align}$

Since, $-\left( \dfrac{p-5q}{2q} \right)$ is a rational number as p and q are integers.

That means, $-\left( \dfrac{p-5q}{2q} \right)=\sqrt{3}$, should also be a rational number.

That gives, $\sqrt{3}$ is a rational number.

But we know that $\sqrt{3}$ is an irrational number.

We know that,

Irrational ≠ rational

So, our assumption is wrong.

We contradict the statement that $5-2\sqrt{3}$ is rational.

Therefore, $5-2\sqrt{3}$ is an irrational number.

Note: In this question, students do not need to prove $\sqrt{3}$ is an irrational number, students can use this information without proving it to prove the given expression $5-2\sqrt{3}$ is an irrational number.