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

2 thoughts on “if p is a prime number, then prove that rootp is irrational.​”

1. 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

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2. 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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