Question

Prove that $2\sqrt{3}-4$ is an irrational number.

Answer 1

Let us assume that $\sqrt{3}$ be a rational number which can be expressed in the form of $\frac{p}{q}$ where $p$ and $q$ are integers, $q\ne 0$ and $p$ and $q$ are co prime that is $HCF(p,q)=1$.

We have,

$\sqrt{3}=\frac{p}{q}\Rightarrow \sqrt{3}q=p.....\left(1\right)\Rightarrow 3q^2=p^2\left(squaringbothsides\right)$

$\Rightarrow p^2$ is divisible by $3$

$\Rightarrow p$ is divisible by $3$$......\left(2\right)$

Now, for any integer $r$,

$Let,p=3r\Rightarrow \sqrt{3}q=3r\left(from\right(1\left)\right)\Rightarrow 3q^2=9r^2\left(squaringbothsides\right)\Rightarrow q^2=\frac{9}{3}{r}^2\Rightarrow q^2=3r^2$

$\Rightarrow q^2$ is divisible by $3$

$\Rightarrow q$ is divisible by $3$$......\left(3\right)$

From equation $\left(2\right)$ and $\left(3\right)$, we get that $3$ is the common factor of $p$ and $q$ which contradicts that $p$ and $q$ are co prime. This means that our assumption was wrong.

Thus $\sqrt{3}$ is an irrational number which implies that $2\sqrt{3}$ is also an irrational number.

We know that the subtraction of an irrational number and a rational number is an irrational number.

Hence $2\sqrt{3}-4$ is an irrational number.