Question

Prove that √3 is an irrational number.​

Answer 1

Answer:

Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.

Answer 2

Step-by-step explanation:

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)

⇒ 3q^2 = p^2………………………………..(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.

⇒ p^2 = 9r^2………………………………..(2)

from equation (1) and (2)

⇒ 3q^2 = 9r^^2

⇒ q^2 = 3r^2

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