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Mathematics
General Math
Proof that the square root of 2 is irrational
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[QUOTE="andrewkirk, post: 5802206, member: 265790"] \ We are using (without citing it) the fact that ##p,q## are positive integers and that the set of positive integers is [B]well-ordered[/B], meaning that every subset has a least element. That guarantees that there exists a unique representation of the fraction in lowest terms, because the set S of all numerators of representations of the fraction is a subset of the positive integers and so must have a least element m. We choose the subset of S that have numerators equal to m. Then we observe that if m/a and m/b are both representations of the fraction, we have m/a=m/b and neither of a nor b is zero, so that a=b, so there is only one representation with numerator m. BTW, we don't need to assume p and q are on lowest terms. All we need to assume to get the proof to work is that 2 is not a common factor of both, ie that ##2\not| \ \gcd(p,q)##. If we don't assume that, as would be the case if ##\gcd(p,q)=2##, we can't get our contradiction. We could probably recover it by then entering into a long induction, but there's no need to do that since we can just assume lowest terms, which implies ##2\not| \ \gcd(p,q)##. [/QUOTE]
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Mathematics
General Math
Proof that the square root of 2 is irrational
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