Proof of irrationality of square root 2 and square root 3

First (a) is proved divisible by (3) from (a^2). Then after putting (a=3r), the conclusion for (b) follows.

From \(3q^2=p^2\), \(p^2\) is divisible by 3. Since 3 is prime, \(p\) must be divisible by 3. Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisibility rule for squares.

From \(a^2=3b^2\), 3 divides \(a^2\), so 3 divides \(a\). Put \(a=3k\); then \(b^2=3k^2\), so 3 also divides \(b\). This contradicts \(a/b\) being in lowest terms. Exam tip: state the common factor clearly to complete the contradiction.

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

From \(p^2=3q^2 \), 3 divides \(p^2 \), so it divides \(p \). Put \(p=3r \); then 3 also divides \(q \), contradicting coprimality. Exam tip: use the prime-divides-a-square property.

Since \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\). Put \(p=3k\); then \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: if a prime divides a square, it divides the number itself.

In a perfect square, prime exponents are even. The exponent of (2) explains the contradiction in the proof of (\sqrt{2}).

In a perfect square, all prime exponents are even. This idea strengthens the proof of (\sqrt{3}).

For coprime numbers, \(\gcd(m,n)=1\). However, if both \(m\) and \(n\) are even, then 2 is a common divisor, so \(\gcd(m,n)\ge 2\). The greatest common divisor cannot be 1 and at least 2 at the same time; this is the contradiction that disproves the initial assumption. Exam tip: whenever both numbers are even, immediately identify 2 as their common divisor.

By definition, a and b being coprime means their greatest common divisor is exactly 1. If both are divisible by 3, we can write a = 3r and b = 3s for integers r and s. Thus 3 is a common divisor of a and b, and their greatest common divisor is at least 3. The proof has therefore derived gcd(a,b) = 1 from the lowest-terms assumption and gcd(a,b) ≥ 3 from divisibility. These mutually incompatible statements form the contradiction. Option A states it precisely. A gcd is never negative, it need not be zero, and it is not generally equal to the sum of the two numbers, so B, C, and D are invalid.

Non-zero denominator is for defining the fraction. The coprime condition creates contradiction with the final common factor.

(b\neq0) is necessary for the fraction. (\gcd(a,b)=1) creates contradiction when both are divisible by (3).

Since \(\sqrt{12}=2\sqrt{3}\), a rational \(\sqrt{12}\) would make \(\sqrt{3}=\sqrt{12}/2\) rational. This contradicts the known irrationality of \(\sqrt{3}\). Hence \(\sqrt{12}\) is irrational. Exam tip: division by a non-zero rational preserves rationality.

Assuming \(\sqrt{2}=\frac{m}{n}\) gives \(m^2=2n^2\). This shows that \(m\) is even, and then \(n\) is also even. However, this becomes a contradiction only if \(\frac{m}{n}\) was stated to be in lowest terms, meaning that \(m\) and \(n\) are coprime. Both being even contradicts coprimality; being even alone is not a contradiction. Exam tip: In such irrationality proofs, always state that the fraction is in lowest terms or that the numerator and denominator are coprime.

Substituting \(p=3k\) gives \(9k^2=3q^2\), so \(q^2=3k^2\). Thus, \(q\) is also divisible by 3, creating the contradiction. Do not jump directly from \(p=3k\) to a claim about \(q\).

The assumption gives \(p^2=3q^2\). Thus 3 divides \(p^2\), so it divides \(p\); substituting back shows that 3 also divides \(q\). This contradicts coprimality. Option B misses the second step. Exam tip: always assume the fraction is in lowest terms.

Assume \(\sqrt{2}=m/n\) in lowest terms. Then \(m^2=2n^2\), so \(m\) is even; putting \(m=2k\) shows that \(n\) is also even. This contradicts lowest terms. Exam tip: prove evenness for both numerator and denominator.

From \(p^2=3q^2\), \(3\mid p^2\) implies \(3\mid p\). Put \(p=3k\); then \(3\mid q\), contradicting coprimality. The idea of both numbers being even belongs to the proof for \(\sqrt{2}\). Exam tip: state the prime-divisor rule clearly.

The proof starts by assuming √3 = a/b in lowest terms, so gcd(a,b) = 1. From a² = 3b², the prime-divisibility rule proves 3 ∣ a. However, that fact alone is compatible with many coprime pairs; for example, a may be divisible by 3 while b is not. To contradict the lowest-terms assumption, the proof must continue by writing a = 3r, deriving b² = 3r², and then proving 3 ∣ b. Only then do a and b share the factor 3, forcing gcd(a,b) ≥ 3 against gcd(a,b) = 1. Therefore A identifies the main error; the other choices are false or irrelevant.

Let \(\sqrt{3}=a/b\) in lowest terms. Then \(a^2=3b^2\), so \(3\mid a^2\); since 3 is prime, \(3\mid a\). Option C is unnecessarily strong because \(9\mid a\) need not follow. Exam tip: remember that if prime \(p\mid k^2\), then \(p\mid k\).