prove root3 is irrational
Answers
Answer:
Prove that sqaure root of 3 is an irrational number.
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 = p^2/q^2 (Squaring on both the sides)
⇒ 3q^2 = p^2………………………………..(1)
This means that 3 divides p2. This means that 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
Where q^2 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.
Answer:
To Prove root (3) is irrational.
Step-by-step explanation:
Let root(3) be rational
root(3) = p/q where p and q are co-prime and q is not equal t zero
squaring both sides - 3 = p*p/q*q
p*p=3q*q------1
as 3 divides p*p
3 divides p as well
p = 3c (where c is an no zero integer)------2
substitute 2 in 1
9c^2 = 3q^2
3c^2 = q^2
as 3 divides q^2
3 divides q as well
therefore p and q have a common factor 3
but this is not possible as p and q are co-prime
this problem occured due to our wrong assumption that root(3) is rationla
therefore root(3) is irrational
HOPE IT HELPS:););)