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

The only square roots that are rational numbers are those who are perfect squares. `sqrt16` for example is a rational number because it equals 4 and 4 is an integer. `sqrt14`=3.74, which is not an integer and therefore is an irrational number.

Suppose `sqrt(14)` is rational.

Then `sqrt(14) = p/q` for some positive integers `p, q` with `q != 0`.

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

`(p/q)^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 = 2k` for some positive integer `k`.

So:

`14 q^2 = (2k)^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 = 7m` for some positive integer `m`.

`7 q^2 = 2 (7m)^2 = 7*14m^2`

Divide both sides by `7` to find:

`q^2 = 14 m^2`

So `14 = q^2/m^2 = (q/m)^2`

So `sqrt(14) = q/m`

Now `m < q < p`, contradicting our supposition that `p` and `q` are the smallest pair of positive integers such that `sqrt(14) = p/q`.

So our supposition is false and therefore our hypothesis that `sqrt(14)` is rational is also false.