Solve it, √3 is irrational
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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
⇒ 3b^2=a^2 (Squaring on both sides) → (1)
Therefore, a^2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b^2 =(3c)^2
⇒ 3b^2= 9c^2
∴ b^2 = 3c^2
This means that b^2 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.
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
⇒ 3b^2=a^2 (Squaring on both sides) → (1)
Therefore, a^2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b^2 =(3c)^2
⇒ 3b^2= 9c^2
∴ b^2 = 3c^2
This means that b^2 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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