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

Use proof by contradiction...

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.