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Whow thatr's anJong divigian mehedly |
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Answer» Assume that sqrt(6) is rational. Then sqrt(6) = p/q where p and q are coprime integers. sqrt(6)^2 = 6 = p²/q² p² = 6q² therefore p² is an even number since an even number multiplied by any other integer is also an even number. If p² is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd. So we can replace p with 2k where k is an integer. (2k)² = 6q²4k² = 6q²2k² = 3q² Now we see that 3q² is even. For 3q² to be even, q² must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if q² is even then q is even. So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational. |
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