If a proof assumes √3 rational and finally gets both p and q divisible by 3, with which initial condition is the contradiction?

Correct answer: gcd(p,q) = 1

The contradiction comes from the way the rational number is initially represented. In a standard irrationality proof, assume √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. The last condition is expressed as gcd(p,q) = 1. If the algebra later proves that both p and q are divisible by 3, then 3 is a common divisor and gcd(p,q) is at least 3, not 1. This directly contradicts the initial lowest-form condition. Positivity of √3 is true but irrelevant, p = q was never assumed, and q = 0 is forbidden rather than an initial condition. Hence option C is correct.