Question

Prove √3 is irrational

Answer 1

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 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.

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Answer 2

$\huge{\underline{\bf{\red{Solution:-}}}}$

Let √3 be rational number

So,

√3 = a/b

where a and b are intigers and b ≠ 0.

Then,

√3= a/b

squaring on both sides

$: \implies \sf3=\frac{a^2}{b^2}$

$: \implies \sf3b^2=a^2..........(i)$

$: \implies \sf3\:\:divides \:\:a^2[\therefore\:3\:divides\:3b^2]$

$: \implies \sf3\:divides \:a$

So,

3 is a prime and divides b²

Then, 3 also divides b.

Let a = 3c for some intiger c.

putting a = 3c in (i)

$: \implies \sf3b^2=9c^2$

$: \implies \sf\:b^2=3c^2$

$: \implies \sf3\: divides\:b^2$so,3 also divides 3c².

$: \implies \sf3\: divides\:b$

So,

3 is a prime and 3 divides b² and b also.

Then, 3 is a common factor of a and b.

but, this contradicts the fact that a and b have no common factor other then 1.

So, this contradiction is arissen because we assume that √3 is rational .

Hence,

√3 is irrational.

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