Which statement creates the actual contradiction with the coprime condition in the proof of (\sqrt{2})?

Correct answer: Both (a) and (b) are even

The direct answer is C, both a and b are even. In the contradiction proof, assume
\(\sqrt{2}=a/b\), where a and b are integers, b is not zero, and the fraction is in lowest form, so a and b are coprime. Squaring gives \(2b^2=a^2\). Therefore \(a^2\) is even, and an integer whose square is even must itself be even, so write \(a=2k\). Substitution gives \(b^2=2k^2\), so b is also even. Thus both a and b have the common factor 2, contradicting their being coprime. Option A is only an intermediate result. Option B is another intermediate result, not the complete contradiction. Option C states the complete contradiction and is correct. Option D, \(\sqrt{2}>0\), is true but does not conflict with the lowest-form assumption. Memory cue: the contradiction is not merely “one number is even”; it is “both numerator and denominator are even.”