Question

# Prove that √2 is an irrational number.

Solution

## Suppose, to the contrary,let us assume that √2 is rational. Then √2 =m/n for some integers m, n in lowest terms, i.e., m and n have no common factors. Then 2=m2n2, which implies that m2=2n2. Hence m2 is even, which implies that m is even.  Then m=2k for some integer k. So 2=(2k)2n2, but then 2n2=4k2, or n2=2k2. So n2 is even.  But this means that n must be even, because the square of an odd number cannot be even. We have just showed that both m and n are even, which contradicts the fact that m, n are in lowest terms. Thus our original assumption (that √2 is rational) is false, so the √2 must be irrational.  Mathematics

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