Proof of irrationality of square root 2 and square root 3

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.

In a perfect square every prime exponent is even. In (2b^2), the exponent of (2) becoming odd explains the contradiction.

Assume \(\sqrt{2}=p/q\) with coprime \(p,q\). From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

Coprime \(m,n\) have no common prime factor. If both are divisible by 3, then \(\gcd(m,n)\ge3\), so \(\frac{m}{n}\) was not in lowest terms. Exam tip: check the claimed common factor.

The diagonal is \(\sqrt{2}\). Assuming \(\sqrt{2}=m/n\) in lowest terms gives \(m^2=2n^2\). Put \(m=2k\); then \(n\) is also even, contradicting coprimality. Option C is wrong because an infinite decimal may repeat. Exam tip: state the lowest-terms assumption first.

Since \(d^2=2\), \(d=\sqrt{2}\), which is irrational. B is false: a rational square need not have a rational square root. Exam tip: find \(d^2\) first using Pythagoras.

Only (p) being divisible by (3) is not a contradiction to coprime condition. (q) must also be divisible by (3).

From \(3\mid p^2\), the prime-square rule gives \(3\mid p\), so \(p=3k\). Substitution yields \(q^2=3k^2\), hence \(3\mid q\); assuming it directly is invalid. Exam tip: write both divisibility steps.

Since \(a^2=2b^2\), \(a\) is even. Writing \(a=2k\) gives \(b^2=2k^2\), hence \(b\) is even. This contradicts coprimality. Tip: begin with the fraction in lowest terms.

When a number is assumed rational, it is represented by a fraction \(a/b\) with \(b\neq0\). This first condition has a basic meaning: the denominator must not be zero, because a fraction with denominator zero is undefined. It says nothing about whether the numerator is even. The separate condition \(\gcd(a,b)=1\) chooses the fraction in lowest terms.

In the proof, \(\sqrt{2}=a/b\) leads to \(a^2=2b^2\). Therefore \(a\) is even; writing \(a=2k\) and substituting back shows that \(b\) is even too. The two numbers then share the factor 2, contradicting \(\gcd(a,b)=1\). Thus option A correctly identifies the denominator condition and the source of the contradiction.

(q\neq0) keeps the fraction defined. (\gcd(p,q)=1) gives contradiction when both are divisible by (3).

From \(p^2=3q^2\), \(3\mid p^2\); since 3 is prime, \(3\mid p\). Put \(p=3k\) to obtain \(3\mid q\), a contradiction. C fails at \(n=3\). Tip: use the prime-divisor rule.

The fraction must be in lowest terms. From \(2q^2=p^2\), \(p\) is even; substituting back shows that \(q\) is even, contradicting coprimality. Assuming both even already presumes the result. Exam tip: state \(q\ne0\).

\(3\mid p\) alone does not contradict coprimality. Put \(p=3k\): \(9k^2=3q^2\), so \(q^2=3k^2\) and \(3\mid q\). Thus both have a common factor 3. Exam tip: prove divisibility for both terms.

If \(a\) is odd, then its square \(a^2\) is also odd. On the other hand, \(2b^2\) is even for every integer value of \(b\), since it has 2 as a factor. Thus, the equation would make the same number both odd and even, which is the contradiction. Option B is not valid because the equation does not prove that \(b^2\) is odd. Exam tip: the square of an odd integer is odd, and the square of an even integer is even.

The correct answer is A. If p is not divisible by 3, then its prime factorisation contains no factor 3. Squaring p does not introduce a new prime factor, so p² is also not divisible by 3. However, the equation p² = 3q² writes the left side as three times an integer square, so the right-hand expression is divisible by 3; consequently p² must be divisible by 3. This is a direct contradiction. In the standard irrationality proof, one first assumes √3 = p/q in lowest terms, with q nonzero and p, q coprime. The contradiction forces both numbers to be divisible by 3, violating the lowest-form assumption. The other options do not follow from the equation.

The correct answer is A because it follows the contradiction proof in the required logical order. Assume √2 is rational and write it as a/b in lowest terms, where a and b are coprime integers and b is nonzero. Squaring gives a²/b² = 2, hence a² = 2b². Therefore a² is even, so a is even; write a = 2k. Substitution gives 4k² = 2b², or b² = 2k², so b is also even. Thus both a and b have a common factor 2, contradicting the assumption that the fraction was in lowest terms. The other sequences omit the essential algebra or replace proof with an unsupported decimal guess.

The correct answer is A. Begin by assuming that √3 is rational and express it as p/q in lowest terms, with p and q coprime and q nonzero. Squaring produces p² = 3q², so p² is divisible by 3. Since 3 is prime, divisibility of p² by 3 implies that p itself is divisible by 3; put p = 3k. Substituting and cancelling gives q² = 3k², so q is also divisible by 3. This contradicts the assertion that p and q have no common factor. The proof therefore establishes irrationality. The other options do not derive the required divisibility chain and cannot produce a valid contradiction.

For a prime \(p\), \(p\mid n^2\Rightarrow p\mid n\). From \(a^2=2b^2\), 2 divides \(a\), and then \(b\). But \(p^2\mid n\) need not hold. Exam tip: state this lemma first.

\(\sqrt{12}=2\sqrt3\). A non-zero rational times an irrational is irrational, so A works. B wrongly uses \(\sqrt{a+b}=\sqrt a+\sqrt b\). Exam tip: extract perfect-square factors first.