Question

Prove that
$\sqrt{5} - \sqrt{3}$
is irrational​

Answer 1

Answer

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

$\tt{\therefore \: \sqrt{5} - \sqrt{3} = \frac{p}{q}}$, where q and p are co-primes with no common factor other than 1

Squaring on both sides:

$\tt{(\sqrt{5} - \sqrt{3})^{2} = \frac{p^{2}}{q^{2}}}$

We get:

$\tt{(\sqrt{5} - \sqrt{3})^{2}q^{2} = p^{2}}$ ---(1)

Thus;

$\tt{(\sqrt{5} - \sqrt{3})^{2}}$ || $\tt{p^{2}}$

And so,

$\tt{(\sqrt{5} - \sqrt{3})^{2}}$ || p

Now, we know that:

p = $\tt{(\sqrt{5} - \sqrt{3})^{2} m}$

Squaring on both sides:

$\tt{p^{2} = (\sqrt{5} - \sqrt{3})^{4} m^{2}}$

Put (1):

$\tt{(\sqrt{5} - \sqrt{3})^{2} q^{2} = (\sqrt{5} - \sqrt{3})^{4}m^{2} }$

Crossing the same terms, we get:

$\tt{(\sqrt{5} - \sqrt{3})^{2}m^{2} = q^{2}}$

Thus;

$\tt{(\sqrt{5} - \sqrt{3})^{2}}$ || $\tt{q^{2}}$

And so,

$\tt{(\sqrt{5} - \sqrt{3})^{2}}$ || q

But p and q were co-primes.

This contradicts our assumption. As, p and q have common factor as $\tt{(\sqrt{5} - \sqrt{3})^{2}}$.

Thus, $\tt{\sqrt{5} - \sqrt{3} }$ is an irrational number.