Question

show that √2 is a irratinal no

Answer 1

HEY FRIEND !

Given..

√2 is irrational number.

Let √2 = a / b wher a,b are integers b ≠ 0

we also suppose that a / b is written in the simplest form....

Now √2 = a / b ⇒ 2 = a2 / b2 ⇒   2b2 = a2
∴ 2b2 is divisible by 2
⇒  a2 is divisible by 2    
⇒  a is divisible by 2  
∴ let a = 2c
a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2
∴ 2c2  is divisible by 2
∴ b2  is divisible by 2
∴ b  is divisible by 2
∴a are b   are  divisible by 2
this contradicts our supposition that a/b is written in the simplest form..

Hence our supposition is wrong.....

∴ √2 is irrational number.

HOPE THIS HELPS U..

Answer 2

Hey preeti

Let us assume √2 as a rational number

As all rational numbers can be expressed in p/q form √2=p/q where p,q are integers, q isn't equal to 0 and p,q are co-primes

√2=p/q
Squaring on both sides
2=p²/q²
2q²=p²
q²=p²/2
As p²is divisible by 2
p is also divisible by 2
That is p iss a factor of 2

So let p=2c for some integer c
Squaring on both sides
p²=4c²
p²=2q²(from above)
So 2q²=4c²
2q²/4=c
q²/2=c
q² is divisible by 2
q is divisible by 2
That is q is a factor of 2

We can now conclude that both p and q have a common factor 2.
But this contradicts the fact that p,q are co primes
This contradiction arose as our assumption is wrong

So, √2 is irrational.