Prove that $5\sqrt{3}$ and $3\sqrt{5}$ are irrational numbers.

Let $5\sqrt{3}$ be a rational number.

∴ $5\sqrt{3} = \dfrac{p}{q}$

where, p,q are integers and q ≠ 0.

⇒ $(5\sqrt{3})^2 = \Big(\dfrac{p}{q}\Big)^2$ (squaring both the sides)

⇒ $75 = \dfrac{p^2}{q^2}$

⇒ $p^2 = 75q^2$

As 3 divides 75q² , so 3 divides p² ans so p is also divisible by 3 ……………….(1)

So, let p = 3m for some integer m.

⇒ $p^2 = 9m^2$ (squaring both the sides)

⇒ $75q^2 = 9m^2$ ∵ $p^2 = 75q^2$

⇒ $25q^2 = 3m^2$

Since 3m² is divisible by 3, the right-hand side 25q² must also be divisible by 3.

But 25q² = 5²q² is not divisible by 3 unless q itself is divisible by 3.

Thus, 3 divides q. ……………….(2)

From 1 and 2, we get p and q both are divisible by 3 i.e., p and q have 3 as their common factor.

This contradicts our assumption that $\dfrac{p}{q}$ is rational i.e. p and q do not have any common factor other than unity (1).

⇒ $\dfrac{p}{q}$ is not rational.

⇒ $5\sqrt{3}$ is not rational i.e., $5\sqrt{3}$ is irrational.

Let $3\sqrt{5}$ be a rational number.

∴ $3\sqrt{5} = \dfrac{p}{q}$

where, p,q are integers and q ≠ 0.

⇒ $(3\sqrt{5})^2 = \Big(\dfrac{p}{q}\Big)^2$ (squaring both the sides)

⇒ $45 = \dfrac{p^2}{q^2}$

⇒ $p^2 = 45q^2$

As 5 divides 45q² , so 5 divides p² ans so p is also divisible by 5 ……………….(1)

So, let p = 5m for some integer m.

⇒ $p^2 = 25m^2$ (squaring both the sides)

⇒ $45q^2 = 25m^2$ ∵ $p^2 = 45q^2$

⇒ $9q^2 = 5m^2$

Since 5m² is divisible by 5, the right-hand side 9q² must also be divisible by 5.

But 9q² = 3²q² is not divisible by 5 unless q itself is divisible by 5.

Thus, 5 divides q. ……………….(2)

From 1 and 2, we get p and q both are divisible by 5 i.e., p and q have 5 as their common factor.

This contradicts our assumption that $\dfrac{p}{q}$ is rational i.e. p and q do not have any common factor other than unity (1).

⇒ $\dfrac{p}{q}$ is not rational.

⇒ $3\sqrt{5}$ is not rational i.e., $3\sqrt{5}$ is irrational.

Hence, $5\sqrt{3}$ and $3\sqrt{5}$ are irrational numbers.