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

Assume rational then square then contradiction

To prove that \(\sqrt{2}\) is irrational, we use proof by contradiction. We begin by assuming the opposite of what we want to prove: suppose \(\sqrt{2}\) is rational and can be written in lowest terms as \(p/q\), with nonzero integers p and q having no common factor. Squaring then gives a relation that forces both p and q to be even, contradicting their being in lowest terms.

Thus the logical order is to assume rationality first, square the expression, and then derive a contradiction from the parity of the integers. More specifically, \(2=p^2/q^2\) gives \(p^2=2q^2\), so p is even; writing \(p=2k\) then shows q is also even. This contradicts the assumption that the fraction was reduced. Therefore option A states the correct order. The other choices do not describe this standard proof.

This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Proof of irrationality of square root 2 and square root 3.