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प्रश्न
Prove that for any prime positive integer p, \[\sqrt{p}\]
is an irrational number.
उत्तर
Let us assume that `sqrtq` is rational .Then, there exist positive co primes a and b such that
`sqrtq=a/b`
`p= (a/b)^2`
`⇒ p = a^2/b^2`
`⇒ pb^2=a^2`
`⇒ pb^2=a^2`
`⇒ p|a^2`
`⇒ p|a`
`⇒ a= pc`for some positive intger c
`⇒ b^2p=a^2`
`⇒ b^2 p = p^2c^2(because a= pc)`
`⇒ p|b^2(since p|c^2p)`
`⇒ p|b`
`⇒ p|a and p|b`
This contradicts the fact that a and b are co priumes
Hence `sqrtp`is irrational
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