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Prove that (√5) is an irrational number |
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Answer» Let √5be a rational number say p/q where q is not equal to zero and p and q are coprime.√5=p/qSquaring both side5=p²/q²5q²=p²----------(1)5q²is divisible by 5p² is divisible by 5p is divisible by 5let p=5aputting the value in equation 15q²=p²5q²=(5a)²5q²=25a²q²=5a²5a² is divisible by 5q² is divisible by 5q is divisible by 5 Therefore both p&q have common factor 5We supposed p&q are coprime.Our supposition is wrong √5 is irrational. let root 5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime]root 5=p/q=> root 5 ×q = psquaring on both sides=> 5×q×q = p×p ------> 1p×p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p×p = 25c×c --------- > 2sub p×p in 15×q×q = 25×c×cq×q = 5×c×c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso\xa0√5 is an irrational |
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