Question

Hey guys

Prove that $\bold{\sqrt {3}}$
is irrational using contradiction with proper explanation

No spam or copying from internet

Answer 1

Hey
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proof - √3 = rational number (let)

√3 =p/q (p,q)HCF =1
√3×q = p

squaring both sides

(√3q)^2 = p^2..........(1)
3 divides p^2 that is 3will also divide p

now, p =3c(let)
putting the value of p in (1)

3q^2 =3c^2
3q^2 =9c^2
q^2 =3c^2

3 divides q^2 that is 3 also divides q

now, HCf of (p,q) =3

hence, √3 is an irrational number prooved

hope helped
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Answer 2

Let us assume that it is rational.

Hence, √3 can be written as ratio of two numbers a and b as shown below:

⇒ √3 = a/b

⇒ √3b = a

On squaring both sides, we get

⇒ (√3b)^2 = a^2

⇒ 3b^2 = a^2   ---- (1)

∴ a^2 is a multiple of 3.Since 3 is prime, this implies a is a multiple of 3.

Let a = 3c.

On squaring both sides, we get

⇒ a^2 = 9c^2 ----- (2)

Substitute (2) in (1), we get

⇒ 3b^2 = 9c^2

⇒ b^2 = 3c^2.

∴ b^2 is also a multiple of 3, that means b is also a multiple of 3.

Hence, a and b have a factor 3.

Therefore, a & b are not co-prime.

By contradiction, √3 is irrational.

Hope it helps!