if p is a prime number, then prove that rootp is irrational.​

Question

if p is a prime number, then prove that rootp is irrational.​

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Iris 1 month 2021-09-18T16:42:31+00:00 2 Answers 0 views 0

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    0
    2021-09-18T16:43:37+00:00

    Step-by-step explanation:

    given than p is prime number

    let us assume that √p is a rational number

    √p=a/b

    where a and b are coprimes and b not equal to 0

    now,

    squaring both sides we get

    p=a^2/b^2

    now bring the b^2 to other side

    => pb^2=a^2——————–(1)

    this shows that p divides a^2

    therefore, p divides a

    => a=p×c —————-(2)

    substitute (2) in (1) ,we get

    pb^2=(pc)^2

    => pb^2=p^2×c^2

    cancelling p we get

    b^2= p×c^2

    from this we can conclude that

    p divides b^2

    therefore p divides b

    => p divides both a and b. => a and b have at least p as a common factor. But this contradicts the fact that a and b are coprimes. This contradiction arises because we have assumed that √p is rational. Therefore, √p is irrational.

    hence proved

    0
    2021-09-18T16:43:40+00:00

    Let us assume, to the contrary, that √p is rational.

    So, we can find coprime integers a and b(b ≠ 0) such that √p = a/b

    => √p b = a

    => pb2 = a2 ….(i) [Squaring both the sides]

    => a2 is divisible by p

    => a is divisible by p So, we can write a = pc for some integer c.

    Therefore, a2 = p2c2 ….[Squaring both the sides]

    => pb2 = p2c2 ….[From (i)]

    => b2 = pc2

    => b2 is divisible by p

    => b is divisible by p

    => p divides both a and b.

    => a and b have at least p as a common factor.

    But this contradicts the fact that a and b are coprime.

    This contradiction arises because we have assumed that √p is rational.

    Therefore, √p is irrational.

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