Try to prove the irrationality of $\sqrt{3}$ using the approach of proof by contradiction. Will the same approach work for $\sqrt{5}$, $\sqrt{7}$, or $\sqrt{10}$?

Proof of irrationality of $\sqrt{3}$ by contradiction :

Assume $\sqrt{3}$ is rational. Then $\sqrt{3} = \dfrac{p}{q}$, where p and q are integers with q ≠ 0 and gcd(p, q) = 1.

Squaring both sides :

$\Rightarrow \sqrt{3} = \dfrac{p}{q} \\[1em] \Rightarrow 3 = \dfrac{p^2}{q^2} \\[1em] \Rightarrow 3q^2 = p^2.$

So, p² is divisible by 3, which means p is also divisible by 3.

Let p = 3k for some integer k. Then :

$\Rightarrow 3q^2 = (3k)^2 \\[1em] \Rightarrow 3q^2 = 9k^2 \\[1em] \Rightarrow q^2 = 3k^2.$

So, q² is divisible by 3, which means q is also divisible by 3.

Both p and q being divisible by 3 contradicts gcd(p, q) = 1.

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

Yes, the same approach works for $\sqrt{5}$, $\sqrt{7}$ and $\sqrt{10}$, since 5, 7 and 10 are not perfect squares.

The proof relies on the fact that if a prime number p divides n², then p divides n. This works for any non-perfect square integer.

Hence, $\sqrt{3}$ is irrational, and the same approach proves $\sqrt{5}$, $\sqrt{7}$ and $\sqrt{10}$ are also irrational.