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October 27th, 2017, 10:54 AM   #1
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Square root of multiplication of 2 primes is irrational

**Before you start reading I must say I don't know how to use math symbols so sorry if it's really messy, and if you have a guide somewhere on how to write maths here I'd appreciate
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I'm trying to prove that sqrt(ab) is irrational, if a and b are different primes
Using contradiction:
Assume sqrt is rational:
sqrt(ab) = x/y, where x/y is irreducible fraction
then I do power of 2 to both sides to get:
ab = (x^2)/(y^2), then:
(y^2)*ab=x^2, so ab is divisor of x^2, and thus divisor of x.
so there is a k that is k*ab = x, put that back in:
(y^2)*ab=(k*ab)^2, divide both sides by (ab):
(y^2)=(k^2)*ab,
so ab is also divisor of y^2, and also of y!
So we get that ab is divisor of x and y, contradiction to the assumption that x/y is irreducible fraction. Q.E.D

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Is this OK?
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October 27th, 2017, 11:23 AM   #2
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Why does "ab is divisor of x²" imply that ab is a divisor of x?
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October 27th, 2017, 11:29 AM   #3
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I was thinking about Euclid's lemma but not sure if it applies here too because this lemma speaks of a single prime number and here we have multiplication of 2 primes. So I guess I am wrong
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October 27th, 2017, 11:41 AM   #4
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Sorry, I was using Euclid''s lemma wrong!
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October 28th, 2017, 06:29 AM   #5
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Anyone? I can't find out
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October 28th, 2017, 09:13 AM   #6
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I've seen proofs that the square root of any non-negative integer that isn't itself a perfect square is irrational. Unfortunately, I can't recall it but I don't think they were particularly difficult. Can you prove that the square root of 2 is irrational and then extend that to all non-square integers?
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October 28th, 2017, 11:31 AM   #7
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I think I did it finally
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