The square root of 2 is irrational
The first thing you might ask if you saw this definition is – are there any irrational numbers? How do we know that Ö2, say, is irrational?
To prove that Ö2 is irrational, suppose the contrary, namely that it is rational, so that there are integers p and q such that
We can also assume that p and q are mutually prime, that is, that they have no common divisors. Then p2 = 2q2 so that p2 is divisible by 2. This means there is another integer s such that
Now suppose p is odd. Then p=2r+1 for some integer r. Substituting into the formula for p2 and multiplying out we see that
But this isn’t possible, because the right hand side is odd, but the left hand side is even.
This contradiction means that p is even, so it can be written as p=2r for some integer r. Substituting into the formula
we get that
Dividing across by 2, this means that
so q2 is even, and q is even (just as we could prove p was). But this is a contradiction, because we assumed p and q had no common factors. This contradiction means that Ö2 is not rational.
Have fun, my friends!