Mother nature has been working to get southern California’s rainfall back up to average, and I’ve been working overtime to deal with it.  Consequently, I’m tired.  Jan made a squirrel, and I’ve dusted off an old favorite proof.

I was first introduced to this line of reasoning as a proof that is irrational.  However, the concept is easily generalized to all non-square radicals.  The proof follows the same reductio ad absurdum model, with the added assumption that .

Assumptions
This proof begins where a Pythagorean would have preferred it to end, by saying some square root is equal to a ratio of integers.

To make sure this isn’t trivialized, we need to throw in a couple additional caveats. Namely, , , and must all be integers,

the square root of can’t be an integer,

and must be fully reduced; that is, if is a factor of , it can’t be a factor of .

Proof
We start with a little resorting of the first assumption; saying is the quotient of and is the same as saying is the product of  and .

If we then square both sides of this equation, we see that is a multiple of .

Thing is, squaring any integer will square all of its factors. In that way, should be a factor of , except isn’t an integer, so it can’t be a factor of . Naturally, the only way could be a factor of is for it to be a factor of .

By that same reasoning, if is a factor of , then is a factor of , must be a factor of , and must, therefore, be a factor of .

Unfortunately, we assumed that and couldn’t have common factors, so we have a contradiction. The only conclusion we can draw is that can’t be rational if it isn’t an integer.

~Nick