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Proof underroot N is not a rational number if \'n\' the s not a perfect square |
| Answer» Let , us assume that √n is rational .so, √n = a/b where a and b are integers and b is not equation to zero .Let, a/b are co- prime\xa0taking square both side ,we get\xa0=> n = a^2/b^2=> nb^2 = a^2 ......(1)so, n divide a^2\xa0it means n also divide a\xa0for some integer ca = nc\xa0now squaring both side\xa0a^2 = n^2c^2=> nb^2 = n^2c^2 [ from (1) ]=> b^2 = nc^2so , n divide b^2\xa0it means b also divide bso, a and b have n as a prime factor\xa0but this contradict the fact that a and b are co- prime .therefore , our assumption is wrong .hence, √n is irrational\xa0\xa0 | |