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 .

**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

P.S. Here’s a PDF version of the proof, if you’re interested.

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