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Prove that √5 is irrational no.

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Prove that √5 is irrational no.
  • 4 answers
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Khushi Nagar 1 week, 1 day ago

Sorry it's mistake in this answer by me , Assume that √5 is a rational no.

Khushi Nagar 1 week, 1 day ago

Let we assume that √5 is an irrational no. So,√5 =p/q where p and q both are co-prime integers and q≠0 Hence,by squaring both the sides (√5)^2=p^2/q^2 5=p^2/q^2 q^2=p^2/5(this show that 5 is a divisor of p^2 and p also) Let p =5a q^2=(5a)^2/5 q^2=25a^2/5 q^2=5a^2 q^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.

Shreyansh Pandey 1 week, 1 day ago

We can use contradiction method You can get you answer in ncert book chapter no 1 easily We 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 no So , √ 5=p/q Squaring on both sides (√5)²= (p/q)² 5=(p)²/(q)² 5/q²=p²
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