Question

Prove that $\sqrt{3}$ is an irrational number. Hence, show that $7+2\sqrt{3}$ is also an irrational number.

Answer 1

Solution:

If possible , let $\sqrt{3}$ be a rational number and its simplest form be

$\frac{a}{b}$ then, $a$ and $b$ are integers having no common factor

other than $1$ and $b\ne 0.$

Now, $\sqrt{3}=\frac{a}{b}⟹3=\frac{a^2}{b^2}$ (On squaring both sides )

or, $3b^2=a^2$ .......(i)

$⟹3$ divides $a^2$ $(\because 3$ divides $3b^2)$

$⟹3$ divides $a$

Let $a=3c$ for some integer $c$

Putting $a=3c$ in (i), we get

or, $3b^2=9c^2⟹b^2=3c^2$

$⟹3$ divides $b^2$ $(\because 3$ divides $3c^2)$

$⟹3$ divides $a$

Thus $3$ is a common factor of $a$ and $b$

This contradicts the fact that $a$ and $b$ have no common factor other than $1.$

The contradiction arises by assuming $\sqrt{3}$ is a rational.

Hence, $\sqrt{3}$ is irrational.

$2^{nd}$ part

If possible, Let $(7+2\sqrt{3})$ be a rational number.

$⟹7-(7+2\sqrt{3})$ is a rational

$\therefore$ $-2\sqrt{3}$ is a rational.

This contradicts the fact that $-2\sqrt{3}$ is an irrational number.

Since, the contradiction arises by assuming $7+2\sqrt{3}$ is a rational.

Hence, $7+2\sqrt{3}$ is irrational.

Proved.