Proof of irrationality of square root 2 and square root 3

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}).

The direct answer is A. The key idea is that in the prime factorisation of a perfect square, every prime has an even exponent. For example, if a number is squared, each prime factor is used twice as many times: \\(3b^2\\) contains the factor 3 once, in addition to the even exponent already present in \\(b^2\\). Thus the exponent of 3 can become odd, so \\(3b^2\\) cannot be a perfect square. In the usual proof, assuming \\(\\sqrt{3}=a/b\\) leads to \\(a^2=3b^2\\). The left side is a square and must have even prime exponents, but the right side has an odd contribution from 3, producing the contradiction. Option A states this exact reason. Option B is wrong because many numbers, such as 1, 2, 4 and 5, are not divisible by 3. Option C is wrong because \\(\\sqrt{3}\\) is about 1.732, not 3. Option D is wrong because fractions can have many denominators, not always 3. Memory cue: a square has only even prime exponents.

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 assumes √3=a/b in lowest terms, so gcd(a,b)=1. From a²=3b², the prime-divisibility theorem correctly proves that 3 divides a. But this alone is not a contradiction: a can be divisible by 3 while b is not, and such a pair can still be coprime. The proof must continue by writing a=3r, substituting into the equation, and obtaining b²=3r². This second equation proves that 3 divides b as well. Then 3 is a common divisor of a and b, so gcd(a,b)≥3, contradicting gcd(a,b)=1. Therefore A identifies the error. The other choices either reject a valid step or introduce irrelevant and false claims.

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 \(p\mid k\).

An infinite decimal alone does not prove irrationality: \(0.\overline{3}=1/3\) is infinite but rational. For \(\sqrt{2}\), establish non-repetition or use contradiction. Exam tip: every recurring decimal is rational.

In a contradiction proof, first assume \(\sqrt{3}=\frac{m}{n}\) in lowest terms. Later, both \(m\) and \(n\) become divisible by 3, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

A rational representation in lowest terms is written as \(a/b\), with \(b\neq0\) and \(\gcd(a,b)=1\). In the proof of \(\sqrt{3}\), squaring the assumed equality gives \(a^2=3b^2\). Since 3 divides the square \(a^2\), it follows that 3 divides \(a\). Substituting this fact back into the equation then shows that 3 also divides \(b\).

Thus both numerator and denominator have the common factor 3. This directly contradicts the original choice of the fraction in lowest terms, namely \(\gcd(a,b)=1\). It does not mean that \(b=0\), that \(a=b\), or that \(\sqrt{3}=3\). Therefore option B identifies the exact assumption that is broken.

If \(7-\sqrt{3}\) were rational, then \(7-(7-\sqrt{3})=\sqrt{3}\) would also be rational, a contradiction. Also, \((\sqrt{3})^2=3\) is rational. In exams, use closure properties with rational numbers carefully.

An infinite decimal alone is insufficient because \(1/3=0.333\ldots\) is rational. For \(\sqrt{2}\), establish non-repetition or derive a contradiction from a lowest-term fraction. Exam tip: distinguish infinite decimals from non-repeating decimals.

A proof by contradiction must reach a statement that directly conflicts with an assumption. In the usual proof, write \(\sqrt{2}=m/n\) in lowest form. Squaring gives \(m^2=2n^2\). This first shows that \(m^2\), and therefore \(m\), is even. However, the fact that only \(m\) is even is not yet a contradiction, because there is no problem with one numerator being even.

Substituting \(m=2k\) back into the equation shows that \(n\) is also even. Now both numbers have a common factor 2, contradicting \(\gcd(m,n)=1\). Thus option A is the incomplete statement. Option B or C expresses the actual contradiction, while D is the conclusion drawn after it.

Only (a) being divisible by (3) is not the final contradiction. (b) must also be proved divisible by (3).

From \(a^2=3b^2\), \(3\mid a^2\), so prime-factor reasoning gives \(3\mid a\). Put \(a=3k\) to obtain \(b^2=3k^2\), hence \(3\mid b\), contradicting coprimality. Exam tip: always show the substitution step.

Under the assumption, both \(s\) and \(1/s=\sqrt{3}-\sqrt{2}\) are rational. Option A adds them to obtain \(2\sqrt{3}\), which would make \(\sqrt{3}\) rational—a contradiction. In B, subtraction gives \(2\sqrt{2}\), not \(2\sqrt{3}\). Exam tip: first check the product of conjugates.