How can the proof of (\sqrt{3}) be understood in the language of infinite descent?

From (\frac{p}{q}), a smaller fraction reducible by (3) can be formed

The proof that \(\sqrt{3}\) is irrational can be described using infinite descent. Assume, for contradiction, that \(\sqrt{3}=p/q\) is a fraction in lowest terms, with integers p and q and \(q\ne0\). Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), so 3 divides p. Substituting \(p=3k\) then shows that 3 also divides q. Thus both numerator and denominator have a common factor 3.

Dividing both by 3 produces a smaller positive fraction representing the same number, which contradicts the assumption that the original fraction was already in lowest terms. If one repeatedly applies the same reasoning, it would create an endless chain of smaller positive integer pairs, which is impossible. Therefore option A captures the descent idea. Options B, C, and D do not follow from the proof and do not establish irrationality.

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