Irrational Numbers

1. Sep 3, 2006

2. Sep 3, 2006

shmoe

"You do not have the permission to view this blog"

I guess you have to register to see it?

Can you type the problem here?

3. Sep 3, 2006

you should be able to see it now. changed the settings

4. Sep 3, 2006

shmoe

I'm getting a page that says it's your blog, but it also says "total entries: 0" and there's not much tehre besides a calender.

Is it that long of a problem that you couldn't just type it in here?

5. Sep 3, 2006

yeah, now it should work.

Last edited: Sep 3, 2006
6. Sep 3, 2006

7. Sep 3, 2006

shmoe

Alright. You get to p^2=n*q^2 and then assume that n=2*t? That's completely unjustified.

If n is not a perfect square then you can write it as n=r*m^2 where r has the property that if s is a prime and s divides r then s^2 does not divide r, r>1 obviously as well (you should prove this). Proceed from p^2=n*q^2 and see what happens.

You could also just look at the prime factorization of n if you have unique factorization at this point.

8. Sep 3, 2006

Ok, so $$p^{2} = rm^{2}q^{2}$$ and $$p^{2}$$ is divisible by $$rm^{2}$$. So then I have to show that $$q^{2}$$ is also divisible by $$rm^{2}$$. Is $$m$$ just any positive number?

Thanks

9. Sep 3, 2006

shmoe

It's probably easiest at this point to focus on some prime that divides r, say s. Then it will look very much like the usual proof sqrt(2) is irrational.

However we have a little snag. You are looking at p^2=r*(mq)^2 and you don't necessarily have mq and p relatively prime, but you should be able to fix this without much trouble.