Math, asked by karthik4894, 1 year ago

prove that root2 is irrational

Answers

Answered by iHelper
13
Hello!

Let √2 = \dfrac{\sf a}{\sf b} --( a and b are co-prime & b ≠ 0 )

Then,

√2 = \dfrac{\sf a}{\sf b}

⇒ 2 = \dfrac{\sf a^{2}}{\sf b^{2}}

⇒ 2b² = a²

Since, a² is divisible by 2
Therefore, \textbf{\sf a \:is\: also\: divisible\: by\: 2}

Now,
Let a = 2c 

Squaring both sides :

a² = 4c²

⇒ 2b² = 4c²

⇒ b² = 2c² 

Since, b²  is divisible by 2
Therefore, \textbf{\sf b\: is\: also\: divisible \:by\: 2} 

That means,

‘a’ are ‘b’ are  divisible by 2 . 
But, ‘a’ and ‘b’ were co-prime!

This contradiction arised due to our wrong assumption that √2 is Rational 

\boxed{\sf \sqrt{2} \:is \:Irrational}...

Cheers!
Answered by Pratik021205
2

Answer:

here's Ur answer buddy

√2 = p/q

2 = (p/q)^2

2 = p^2/q^2

p^2 = 2q^2

thus p^2 is even. the only way this can be true is that p itself is even . but then p^2 is actually divisible by 4 . hence q^2 and therefore q must be even . so p and q both are even which is a contradiction to our assumption that they have no common factors. the square root of 2 cannot be rational.

hope it helps u

plzz mark as brainliest

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