Prove that√3 is a irrational number?
Answers
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)
⇒ 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.
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)
⇒ 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.