Question

Prove that √3 is irrational.
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Answer 1

We can easily prove irrationality of any square root number by using Contradiction Method.

TO PROVE :- Root 3 is irrational

PROOF :-

Let root 3 be a rational number.

That means root three can be written in fractional form .

$let \: \sqrt{3} = \frac{p}{q}$

Where p & q are co-primes. [Integers]

Squaring both the sides.

$3 = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 3 {q}^{2} = {p}^{2} ...(eq.01) \\ \\ \implies {q}^{2} = \frac{ {p}^{2} }{3} \\ \\ \\ \implies {p}^{2} \: is \: divisible \: by \: 3 \\ \\ \implies \: p \: is \: divisible \: by \: 3 \\ \\ \\ let \: p \: = \: 3s \\ \\ \\ \implies {p}^{2} = 9 {s}^{2} \\ \\ \\ \implies 3 {q}^{2} = 9 {s}^{2} \\ \\ \\ \implies {s}^{2} = \frac{ {q}^{2} }{3} \\ \\ \\ \implies \: {q}^{2} \: is \: divisible \: by \: 3 \\ \\ \implies \: q \: is \: divisible \: by \: 3$

Now both p and q are divisible by 3 while we assumed that p and q are co-primes.

Hence, our assumption is wrong root 3 is not rational i.e. irrational. (ANS)

Answer 2

Answer:

Let √3 be a rational number

√3 = a/b (a and b are integers and co-primes and b ≠ 0)

On squaring both the sides, 3 = a²/b²

⟹ 3b² = a²

⟹ a² is divisible by 3

⟹ a is divisible by 3

We can write a = 3c for some integer c.

⟹ a² = 9c²

⟹ 3b² = 9c²

⟹ b² = 3c²

⟹ b² is divisible by 3

⟹ b is divisible by 3

From (i) and (ii), we get 3 as a factor of ‘a’ and ‘b’ which is contradicting the fact that a and b are co-primes.

Hence our assumption that√3 is an rational number is false.

∴ √3 is an irrational number.