Question

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

Answer 1

If possible, let $5\sqrt{3}$ is rational.

Let its simplest form be $5\sqrt{3}=\frac{a}{b}$, where a and b are positive integers having no common factor other than 1.Then,

$5\sqrt{3}=\frac{a}{b}$

$\sqrt{3}=\frac{a}{5b}$

Since, $a$ and $5b$ are non-zero integers, so $\frac{a}{5b}$ is rational.

Thus, $\sqrt{3}$ is rational.

Let simplest form of $\sqrt{3}$ be $\frac{a}{b}$.

Then, a and b are integers having no common factor other than 1, and b$\ne$0.

Now,

$\sqrt{3}=\frac{a}{b}$

$3b^2=a^2$

3 divides a${}^2$. [3 divides 3b${}^2$]

$3$ divides $a$.

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

Therefore,

$3b^2=9c^2$

$b^2=3c^2$

$3$ divides b${}^2$. [3 divides 3c${}^2$]

$3$ divides $b$.

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

But this contradicts the fact that $a$ nad $b$ have no common factor other than 1.

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

Hence, $5\sqrt{3}$ is irrational.