How do you prove that square root 15 is irrational?

Sep 20, 2015

See explanation...

Explanation:

This proof uses the unique prime factorisation theorem that every positive integer has a unique factorisation as a product of positive prime numbers.

Suppose $\sqrt{15} = \frac{p}{q}$ for some $p , q \in \mathbb{N}$. and that $p$ and $q$ are the smallest such positive integers.

Then ${p}^{2} = 15 {q}^{2}$

The right hand side has factors of $3$ and $5$, so ${p}^{2}$ must be divisible by $3$ and by $5$. By the unique prime factorisation theorem, $p$ must also be divisible by $3$ and $5$.

So $p = 3 \cdot 5 \cdot k = 15 k$ for some $k \in \mathbb{N}$.

Then we have:

$15 {q}^{2} = {p}^{2} = {\left(15 k\right)}^{2} = 15 \cdot \left(15 {k}^{2}\right)$

Divide both ends by $15$ to find:

${q}^{2} = 15 {k}^{2}$

So $15 = {q}^{2} / {k}^{2}$ and $\sqrt{15} = \frac{q}{k}$

Now $k < q < p$ contradicting our assertion that $p , q$ is the smallest pair of values such that $\sqrt{15} = \frac{p}{q}$.

So our initial assertion was false and there is no such pair of integers.