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Prove that underroot 5 irrational no by contradiction method |
| Answer» Let √5 be a rational number say,p/q where q is not equal to zero and p&q are co prime.√5=p/qSquaring both side5=p²/q²5q²=p²--------------(1)5q² is divisible by 5q²is divisible by 5q is divisible by 5Let q=5aPutting the value in (1)5q²=(5a)²5q²=25a²q²=5a²5a² is divisible by 5q²is divisible by 5q is divisible by 5p&q has common factor 5But we assumed they are co prime.It\'s a contradiction. Our supposition is wrong. Therefore √5 is irrational. | |