Question

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

Answer 1

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.

Answer 2

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