Proof of irrationality of square root 2 and square root 3

First (p) is proved divisible by (3) from (p^2). Then putting (p=3k) gives the conclusion for (q).

Every rational number can be written as a/b with integer a and non-zero integer b, but the irrationality proof needs a special, reduced representation. We choose a/b in lowest terms, expressed by gcd(a,b)=1. Assuming √2=a/b then gives a²=2b². The parity argument forces a and b both to be even, which contradicts the reduced-form condition because they would share the factor 2. Thus option A states the proof-ready form precisely. A zero denominator is not allowed in a fraction, and requiring both entries to be even would destroy the contradiction rather than create it. Equal numerator and denominator is also irrelevant and would describe only a restricted fraction, not every rational number.

If \(\sqrt{n}=p/q\), then \(nq^2=p^2\). Since every prime exponent in a square is even, \(n\) must be a perfect square. Thus \(\sqrt{2}\) is irrational. Exam tip: first check whether the integer is a perfect square.

Both proofs have the same structure, but the prime factors are (2) and (3) respectively. Mention this in comparison.

Assume \(5+\sqrt{3}\) is rational. Subtracting 5 makes \(\sqrt{3}\) rational, a contradiction. Thus 5 being rational is insufficient. Exam tip: rationals are closed under subtraction.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, if it divides a square, it must divide its base \(p\); hence \(3\mid p\). Only after writing \(p=3k\) can we show \(3\mid q\), contradicting coprimality. Exam tip: use the prime-divisor property carefully.

The final contradiction completes only when both are divisible by (2). This breaks the coprime assumption.

The governing concept is proof by contradiction together with the lowest-form condition for a rational number. Assume √3 = p/q, where p and q are integers, q ≠ 0, and gcd(p,q) = 1. Squaring gives p² = 3q². From this equation, 3 divides p, so p = 3k; substitution then shows that 3 also divides q. Thus p and q have a common factor 3, contradicting gcd(p,q) = 1. Therefore the rational assumption must be rejected. Option A is incomplete because divisibility of p alone does not contradict lowest form; options C and D do not produce the required contradiction.

In a square, exponents of prime factors are even. A situation like an odd power of (2) creates contradiction.

In a perfect square, every prime exponent is even. This principle strengthens the proof of (\sqrt{3}).

From \(p^2=3q^2\), \(p^2\) is divisible by 3. Since 3 is prime, divisibility of \(p^2\) by 3 implies divisibility of \(p\) by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-factor property for squares.

Since \(\sqrt{8}=2\sqrt{2}\), \(\frac{\sqrt{2}}{\sqrt{8}}=\frac12\), which is rational. But \(\sqrt{2}\sqrt{3}=\sqrt{6}\) is irrational and \(\sqrt{3}-\sqrt{3}=0\). Tip: simplify radicals first.

From \(p^2=3q^2\), \(p^2\), and hence \(p\), is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: look for a common factor contradiction.

Non-zero denominator is for definition. Coprime condition creates contradiction with the final common factor.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-divisibility rule.

Only (p) being divisible by (3) is not enough. The contradiction completes only when (q) is also divisible by (3).

A rational number can be represented as a/b, but for this contradiction proof the fraction must be chosen in lowest terms, meaning gcd(a,b) = 1 and b ≠ 0. Starting with √2 = a/b and squaring gives a² = 2b². The parity argument then shows that a is even and b is also even. That conclusion is a contradiction only because the original representation was assumed to have no common factor. Without explicitly stating gcd(a,b) = 1, the proof has not established the condition that the final conclusion violates. Therefore option A is the best mathematical comment. The omission does not prove b = 0 or rationality, and it does not make the proof complete.

If \(r\) were rational, \(r^2\) would also be rational. From \(r^2=5+2\sqrt6\), \(\sqrt6=(r^2-5)/2\) would be rational, impossible because 6 is not a perfect square. Exam tip: retain the cross term \(2\sqrt6\).

If both are divisible by (2), numerator and denominator can be reduced by (2). This opposes lowest form.

A non-terminating decimal is not sufficient: \(\frac{1}{3}=0.333...\) is non-terminating but rational. An irrational decimal is non-terminating and non-recurring. Exam tip: check recurrence too.