Which option gives the correct order in the proof of irrationality of (\sqrt{2})?

Correct answer: Assume rational then square then contradiction of both even

The direct answer is option A. In a proof by contradiction, we begin by temporarily assuming the opposite of what we want to prove: suppose \(\sqrt{2}\) is rational. Then write it in lowest terms as \(a/b\), where \(a\) and \(b\) have no common factor. Squaring gives \(2=a^2/b^2\), so \(a^2=2b^2\). Thus \(a^2\) is even, so \(a\) is even; write \(a=2k\). Substitution shows \(b^2=2k^2\), so \(b\) is also even. Both then have a common factor 2, contradicting lowest terms. Therefore \(\sqrt{2}\) is irrational. Option A lists the correct order: assume rational, square, and obtain the both-even contradiction. Option B is insufficient because a decimal expansion alone does not prove irrationality. Option C is a construction activity, not this proof. Option D assumes an unrelated equality and supplies no contradiction. Exam cue: always include the lowest-terms assumption and the final contradiction.