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

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

The direct answer is option A. To prove that \(\sqrt{2}\) is irrational, begin by assuming the opposite: suppose \(\sqrt{2}\) is rational. Write it in lowest form as \(\sqrt{2}=p/q\), where \(p,q\) are integers, \(q\neq0\), and they have no common factor. Squaring gives \(2=p^2/q^2\), so \(p^2=2q^2\). Thus \(p^2\) is even, so \(p\) is even; write \(p=2k\). Substitution gives \(q^2=2k^2\), so \(q\) is also even. This contradicts the lowest-form condition. Therefore \(\sqrt{2}\) is irrational. Option A gives this correct short order. Option B is wrong because drawing a picture does not prove irrationality. Option C is wrong because there is no reason to assume the square root is zero. Option D is wrong because a decimal approximation or guess cannot be a proof. Memory cue: assume rational, square, show both even, contradict lowest form.