Question

if p is a prime number then prove that √p is irrational

Answer 1

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

So, We can find co-prime integers a and b (b ≠ 0)  such that √p = a/b 

=> √p b = a 

=> pb^2 = a^2 ….(i) [Squaring both the sides] 

=> a^2 is divisible by p 

=> a is divisible by p 

So, we can write a = pc for some integer c. 

Therefore,

a^2 = p^2c^2 ….[Squaring both the sides] 

=> pb^2 = p^2c^2 ….[From (i)] 

=> b^2 = pc^2 

=> b^2 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 has arisen because of our incorrect assumption that √p is rational. 

Therefore, √p is irrational.

Hope it helps !!!

Answer 2

Let √p be a rational number.

Let,√p=a/b,where a and b are integers and b is not = 0.

Then,√p=a/b

Squaring both sides,

(√p)2/1 = (a/b)2

p/1=a2/b2

pb2=a2. (Equation 1)

Therefore,p divides a2.

p divides a also.

Let,a=bq for some integer q.

Put a=bq in equation 1,

pb2=p2q2

b2=pq2

Therefore,p divides b2.

p divides b also.

Thus,p is a common factor of a and b.

So,our assumption is not correct.

Hence,√p is an irrational number.

Thank you.........