In the proof of (\sqrt{2}), which condition is necessary along with assuming (\sqrt{2}=\frac{m}{n})?

Correct answer: (m,n) coprime and (n\neq0)

The direct answer is option A. In a proof by contradiction, write \(\sqrt{2}=m/n\) in lowest terms. This requires \(m,n\) to be integers, \(n\neq0\), and \(m,n\) to be coprime. Squaring gives \(m^2=2n^2\). Thus \(m^2\) is even, so \(m\) is even; writing \(m=2k\) then shows \(n\) is even too. That contradicts the coprime condition. Option A is correct because both the nonzero denominator and lowest-form condition are necessary. Option B is wrong because assuming both even at the start destroys the lowest-form setup and already assumes the contradiction. Option C is impossible because division by zero is undefined. Option D is not required: numerator and denominator need not be equal, and equality would not prove irrationality. The lowest-form condition is what makes “both even” a contradiction. Memory cue: rational fraction means coprime numerator and denominator, with denominator nonzero.