prove death root 3 is irrational number
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Say √3 is rational. Then √3 can be represented as ab, where a and b have no common factors. ... Now a2 must be divisible by 3, but then so must a (fundamental theorem of arithmetic). So we have 3b2=(3k)2 and 3b2=9k2 or even b2=3k2 and now we have a contradiction.
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