Question

Prove that root 5 - root 3 is an irrational number

Answer 1

We can prove this by "Proof by contradiction", and we can find a contradiction through arithmetic basis.

$\text{Assume } \sqrt{5} - \sqrt{3} \text{ is rational.}$
$\text{That is, } \sqrt{5} - \sqrt{3} = \frac{a}{b}\text{, where a and b do not have any common factors.}$

$\text{Squaring both sides: } (\sqrt{5} - \sqrt{3})^{2} = \frac{a^{2}}{b^{2}}$
$5 - 2\sqrt{15} + 3 = \frac{a^{2}}{b^{2}}$
$8 - 2\sqrt{15} = \frac{a^{2}}{b^{2}}$
$2(4 - \sqrt{15}) = \frac{a^{2}}{b^{2}}$

We can prove by contradiction based on the fact that the square root of 15 is irrational. We've made our assumption that we can write √5 - √3 in fraction form. By making √15 the subject, we would have contradicted our original assumption.

$4 - \sqrt{15} = \frac{a^{2}}{2b^{2}}$
$\sqrt{15} = 4 - \frac{a^{2}}{2b^{2}}$

Now we've hit our jackpot. Since we can write √15 in a rational form, we've contradicted ourselves. This implies that our original assumption was wrong, which was that we can write √5 - √3 in fraction form. This further implicates that √5 - √3 cannot be rewritten in simplified fraction form, which means that √5 - √3 is irrational.

Thus, our proof is complete.