Subject: Irrational numbers
Date: Wed, 20 Jan 1999 22:52:55 -0500
If you could help me out with this I would appreciate it. Prove that the square root of 17 is irrational.
Thank you for your help.
Here is a proof that the root of 2 is irrational, taken from the notes of a course I taught last semester. We use what is called a proof by contradiction. The essential idea is that we make an assumption and if we follow that up with a number of logical steps and end up with a false statement, it must be that our assumption was false.
where p and q are relatively prime integers. (p and q relatively prime means that they have no common factor except 1, hence p/q cannot be simplified.) Thus,
Here's where we use our number theory sense (on divisibility properties)-- we know 2 divides the left side of the equation above hence 2 must divide the right side. But this means that the right side must be even. The only way this can happen is that p itself is even. For simplicity's sake let's write p = 2m. Then we have
But now 2 divides the right side of the equation above and hence 2 divides the left side. The only way this can happen is that q is even also! But then we have both p and q being even when we've already assumed that they were relatively prime. This can't be. Thus, something has gone wrong -- it is our assumption that
where p and q are relatively prime integers, i.e. that cannot be rational and thus must be irrational.
Now repeat this argument replacing 2 by 17. The key here is that 2 and 17 are both primes so you need to watch for the place in this argument where it is crutial that 17 is prime.
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