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

From lowest (\frac{p}{q}) a smaller fraction reducible by (3) is obtained

Infinite descent expresses the contradiction as an impossible chain of ever-smaller positive integer examples. Assume that \(\sqrt{3}=p/q\) has been written in lowest terms, with \(q\neq0\) and positive denominator if needed. Squaring gives \(p^2=3q^2\). The divisibility argument shows that 3 divides \(p\), and substituting \(p=3k\) shows that 3 also divides \(q\).

Consequently, the fraction \(p/q\) can be reduced by cancelling a factor 3. The new numerator and denominator are smaller positive integers, yet their ratio is still \(\sqrt{3}\). Repeating the same reasoning would produce an endless sequence of smaller positive denominators, which cannot exist. Hence the assumed lowest fraction is impossible, exactly as stated in option A.

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