Prove that √3 is irrational.
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Say √3 is rational.
Then √3 can be represented as a/b, where a and b have no common factors.
So 3= a^2/ b^2 and 3b^2= a^2
a^2 must be divisible by 3, but then so must a.
3b^2=3k^2 and
3b^2= 9k^2
The right hand side is odd and we have found a contradiction, therefore our hypothesis is false.
√3 is irrational.
Hence Proved.
Then √3 can be represented as a/b, where a and b have no common factors.
So 3= a^2/ b^2 and 3b^2= a^2
a^2 must be divisible by 3, but then so must a.
3b^2=3k^2 and
3b^2= 9k^2
The right hand side is odd and we have found a contradiction, therefore our hypothesis is false.
√3 is irrational.
Hence Proved.
sujathasrnvsnpczdb5:
how is the right hand side odd
9k^2
if k=2 ,then rhs is even
this is a proving equation
you cannot put values
thank you
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Sol: 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 3Hence ‘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 = 3c2This 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.
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