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- Thread starter courtrigrad
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shmoe

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"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?

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

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shmoe

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

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yeah, now it should work.

Last edited:

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StatusX

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shmoe

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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.

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Thanks

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shmoe

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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.courtrigrad said:

Thanks

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.

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