√3 is irrational.
To prove this, we assume that √3 is rational and can be written as a fraction in the form p/q, where p and q are integers with no common factors other than 1, and q is not equal to 0.
Squaring both sides of the equation, we have:
(√3)^2 = (p/q)^2
3 = (p^2)/(q^2)
Rearranging the equation, we have:
3q^2 = p^2
This implies that p^2 is divisible by 3, and therefore p must also be divisible by 3. Let's assume p = 3k, where k is an integer.
Substituting this value into the original equation:
3q^2 = (3k)^2
3q^2 = 9k^2
q^2 = 3k^2
This implies that q^2 must also be divisible by 3, and therefore q must be divisible by 3.
However, we assumed initially that p and q have no common factors other than 1, which is contradictory since both p and q are divisible by 3.
Therefore, our assumption that √3 is rational is false, and √3 must be irrational.
Determine if √3 is rational or irrational and give a reason for your answer
1 answer