Question

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

Answer 1

Let us prove that $\sqrt{3}$ is irrational. The question will automatically follow.

We prove it by contradiction.

Assume $\sqrt{3}$ is a rational number.

Thus, $\sqrt{3}=\frac{p}{q}$ where p,q are co-prime integers.

Thus $⟹3q^2=p^2$

This means $p^2$ is a multiple of 3.

As p is an integer, p also must have a factor 3.

We can say $p=3\lambda$, where $\lambda$ is a constant

Thus $(3\lambda {)}^2=3q^2$,

$3{\lambda }^2=q^2$

Again, we see $q^2|3$ which means $q|3$.

What do we see?

Both p and q have a common factor 3!

This is not in agreement to our initial assumption that p and q must be co-prime.

So, $\sqrt{3}$ must be irrational.

Multiplying a rational number(in this case 5) with an irrational, makes the whole number irrational.

So $5\sqrt{3}$ is irrational.