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Prove that √5 is irrational no. |
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Answer» Google kar Sorry it\'s mistake in this answer by me ,Assume that √5 is a rational no. Let we assume that √5 is an irrational no.So,√5 =p/q where p and q both are co-prime integers and q≠0Hence,by squaring both the sides (√5)^2=p^2/q^25=p^2/q^2q^2=p^2/5(this show that 5 is a divisor of p^2 and p also)Let p =5a q^2=(5a)^2/5q^2=25a^2/5q^2=5a^2q^2/5=a^2(this show that 5 is a divisor of q^2 and q also)Hence ,5 is a divisor of p and q both This is because of our wrong assumption of taking √5 as rational no.Hence,√5 is an iraational no. We can use contradiction method You can get you answer in ncert book chapter no 1 easilyWe can assume that √5 is a rational no Then we can equate √ 5 as or in the form of P/q( where p and q are co prime ) means both no is not divisible by a single noSo ,√ 5=p/qSquaring on both sides(√5)²= (p/q)²5=(p)²/(q)²5/q²=p² |
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