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.
Irrationality of Non-Square Radicals
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 .
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 .
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.
P.S. Here’s a PDF version of the proof, if you’re interested.