Question

prove that root 2 is irrational.​

Answer 1

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To prove :

√2 is irrational

prof:

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

$2 = (2k)^{2} /b^{2} \\ </p><p>2 = 4k^{2} /b ^{2} \\ </p><p>2*b ^{2} = 4k^{2} \\ </p><p>b ^{2} = 2k ^{2}$

This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.

hope it help ❣