# How do you prove that the square root of 14 is irrational?

Oct 10, 2015

A rational number is expressed by ratio of integers.

#### Explanation:

The only square roots that are rational numbers are those who are perfect squares. $\sqrt{16}$ for example is a rational number because it equals 4 and 4 is an integer. $\sqrt{14}$=3.74, which is not an integer and therefore is an irrational number.

Oct 10, 2015

#### Explanation:

Suppose $\sqrt{14}$ is rational.

Then $\sqrt{14} = \frac{p}{q}$ for some positive integers $p , q$ with $q \ne 0$.

Without loss of generality, we can suppose that $p$ and $q$ are the smallest such pair of integers.

${\left(\frac{p}{q}\right)}^{2} = 14$

So:

${p}^{2} = 14 {q}^{2}$

In particular, ${p}^{2}$ is even.

If ${p}^{2}$ is even, then $p$ must be even too, so $p = 2 k$ for some positive integer $k$.

So:

$14 {q}^{2} = {\left(2 k\right)}^{2} = 4 {k}^{2}$

Dividing both sides by $2$, we get:

$7 {q}^{2} = 2 {k}^{2}$

So ${k}^{2}$ must be divisible by $7$. So $k$ must be divisible by $7$ too. So $k = 7 m$ for some positive integer $m$.

$7 {q}^{2} = 2 {\left(7 m\right)}^{2} = 7 \cdot 14 {m}^{2}$

Divide both sides by $7$ to find:

${q}^{2} = 14 {m}^{2}$

So $14 = {q}^{2} / {m}^{2} = {\left(\frac{q}{m}\right)}^{2}$

So $\sqrt{14} = \frac{q}{m}$

Now $m < q < p$, contradicting our supposition that $p$ and $q$ are the smallest pair of positive integers such that $\sqrt{14} = \frac{p}{q}$.

So our supposition is false and therefore our hypothesis that $\sqrt{14}$ is rational is also false.