# Why is the square root of 5 an irrational number?

Mar 30, 2016

See explanation...

#### Explanation:

Here's a sketch of a proof by contradiction:

Suppose $\sqrt{5} = \frac{p}{q}$ for some positive integers $p$ and $q$.

Without loss of generality, we may suppose that $p , q$ are the smallest such numbers.

Then by definition:

$5 = {\left(\frac{p}{q}\right)}^{2} = {p}^{2} / {q}^{2}$

Multiply both ends by ${q}^{2}$ to get:

$5 {q}^{2} = {p}^{2}$

So ${p}^{2}$ is divisible by $5$.

Then since $5$ is prime, $p$ must be divisible by $5$ too.

So $p = 5 m$ for some positive integer $m$.

So we have:

$5 {q}^{2} = {p}^{2} = {\left(5 m\right)}^{2} = 5 \cdot 5 \cdot {m}^{2}$

Divide both ends by $5$ to get:

${q}^{2} = 5 {m}^{2}$

Divide both ends by ${m}^{2}$ to get:

$5 = {q}^{2} / {m}^{2} = {\left(\frac{q}{m}\right)}^{2}$

So $\sqrt{5} = \frac{q}{m}$

Now $p > q > m$, so $q , m$ is a smaller pair of integers whose quotient is $\sqrt{5}$, contradicting our hypothesis.

So our hypothesis that $\sqrt{5}$ can be represented by $\frac{p}{q}$ for some integers $p$ and $q$ is false. That is, $\sqrt{5}$ is not rational. That is, $\sqrt{5}$ is irrational.