In the proof of (\sqrt{2}), if (a,b) are assumed coprime and later (a=2r) and (b=2s) are obtained, which conclusion is most precise?

Correct answer: (\frac{a}{b}) can be reduced to (\frac{r}{s}), so the initial lowest form was impossible

The direct answer is A. In a proof by contradiction, we first suppose that \\(\\sqrt{2}=a/b\\), where a and b have no common factor and b is not zero. Squaring gives \\(a^2=2b^2\\). This shows that a is even, so write \\(a=2r\\). Substitution gives \\(4r^2=2b^2\\), hence \\(b^2=2r^2\\), so b is also even and can be written \\(b=2s\\). Therefore both a and b have the common factor 2. But that contradicts the original statement that the fraction was already in lowest terms. Equivalently, \\(a/b=(2r)/(2s)=r/s\\), so it could be reduced. Option A expresses this precise conclusion. B is wrong because the argument proves that \\(\\sqrt{2}\\) is not an integer. C is wrong because b is a denominator and must not be zero; the proof does not show b=0. D is wrong because nothing proves a and b are equal. Exam cue: a common factor obtained after assuming lowest form is the contradiction.