Show that the square root of 3 is not rational

This proof is similar to the one for , and it's done by contradiction.

Suppose that is rational. So, there must exist a fraction such that

Suppose that the fraction is already reduced, i.e. a and b are coprime.

Squaring both members give

So, is divisible by 3, and thus so is , which means that we can write for some

So, the expression becomes

So, is also divisible by 3, and thus so is .

Here is the contradiction, becuase we assumed that a and b were coprime, but it turned out that they are both divisible by 3.

the function makes a strait line horizontally with no x intercept but a y intercept at 9.