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root 3 and root 11 are irrational number by contradiction method

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Kishore answered 4 year(s) ago

Prove that root 3 is a irrational number

prove that root3 is a irrational number

Class-X Maths

person

Asked by Yash

Jul 2

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person

Kishore , Student

Member since Dec 10 2008

Let us assume that √3 is a rational number
That is, we can find integers a and b (≠ 0) such that √3 = (a/b)
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
√3b = a
⇒ 3b2=a2 (Squaring on both sides) → (1)
Therefore, a2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b2 =(3c)2
⇒ 3b2 = 9c2
∴ b2 = 3c2
This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 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 √3 is rational.
So, we conclude that √3 is irrational.

hence, similarly you can prove rt11 irrational.

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Answer 2

Let root 3be a rational no.. then it can be written in the form of p/q ,where p and q are integer and q not equal to zero Al's they are co primes.
So according to our supposition
P/Q=ROOT 3
now squaring in both sides
now replace q square in right side
NOW p square is divisible by 3
p is also divisible by 3


Now put p equal to 3x
we get . 9x square
now cut 3which is in right side and 9in left
we get 3x square=q square
then q is divisible by 3
q is also divisible by 3

So p and q have common factor as 3
but they are co primes
so our supposition is wrong that root .. 3is rational no ..
therefore it is an irrational no..