An Irrationally Radical Idea

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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