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If a/b=sqrt(6), then a2=6b2

On the other hand, given integers ''a'' and ''b'', because the valuation (i.e., highest power of 2 dividing a number) of 6b2 is odd, while the valuation of a2 is even, they must be distinct integers. Contradiction.

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โˆ™ 2012-08-10 10:15:00
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Q: Prove that square root of 6 is irrational number?
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