Question

is √3 a rational number prove it​

Answer 1

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Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

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Answer 2

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If possible , let 3 be a rational number and its simplest form be 

ba then, a and b are integers having no common factor 

other than 1 and b=0.

Now, 3=ba⟹3=b2a2    (On squaring both sides )

or, 3b2=a2         .......(i)

⟹3 divides a2   (∵3 divides 3b2)

⟹3 divides a

Let a=3c for some integer c

Putting a=3c in (i), we get

or, 3b2=9c2⟹b2=3c2

⟹3 divides b2   (∵3 divides 3c2)

⟹3 divides a

Thus 3 is a common factor of a and b

This contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming 3 is a rational.

Hence, 3 is irrational.