Proof of irrationality of square root 2 and square root 3

Correct answer: A. If a prime divides a perfect square, it also divides its base.

Explanation: Since \(3\) is prime, \(3\mid p^2\) implies \(3\mid p\). Put \(p=3k\) in \(p^2=3q^2\) to get \(3\mid q\), contradicting coprimality. Exam tip: explicitly cite the prime-divisibility rule.

Correct answer: A. \(n\) is not a perfect square

Explanation: Every prime exponent in a perfect square is even. In \(12=2^2\times3\), 3 has an odd exponent, so \(\sqrt{12}\) is irrational. Being even or prime alone is insufficient; first check whether the number is a perfect square.

Correct answer: A. Both \(p\) and \(q\) are divisible by 3

Explanation: From \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 also divides \(q\). Thus they cannot be coprime. Exam tip: if a prime divides a square, it divides the number itself.

Correct answer: B. \(m\) and \(n\) are coprime

Explanation: A contradiction arises only when \(\frac{m}{n}\) is assumed to be in lowest terms, so \(m,n\) are coprime. Otherwise, both may be even, as in \(\frac{6}{8}\). Exam tip: always state the lowest-terms condition first.

Correct answer: A. Since \(a^2\) is even, \(a\) is even; hence write \(a=2k\), where \(k\) is an integer.

Explanation: From \(a^2=2b^2\), \(a^2\) is even because it is a multiple of 2. If the square of an integer is even, then the integer itself is even; therefore, write \(a=2k\), where \(k\) is an integer. Writing \(a=2b\) directly is incorrect because \(\sqrt{2b^2}=b\sqrt2\), not \(2b\). Exam tip: In irrationality proofs, first state that an even square implies an even integer.

Correct answer: B. 3 p को विभाजित करता है, इसलिए p=3k (जहाँ k एक पूर्णांक है) लिखना चाहिए; फिर प्रतिस्थापन से q भी 3 से विभाज्य होगा

Explanation: Since \(p^2=3q^2\), \(p^2\) is divisible by 3. As 3 is prime, if \(p^2\) is divisible by 3, then \(p\) must also be divisible by 3. Hence, write \(p=3r\), where \(r\) is an integer. Writing \(p=3q\) is unjustified because it incorrectly makes \(p\) directly three times \(q\). Exam tip: when a square is divisible by a prime, first conclude that its base is divisible by that prime.

Correct answer: B. From a/b, a smaller fraction divisible by 2 is obtained

Explanation: The proof begins by assuming that √2 can be written as a fraction a/b in lowest terms, where a and b have no common factor. Squaring gives a² = 2b², so a² is even and therefore a is even; write a = 2k. Substitution gives b² = 2k², so b is also even. Thus both numerator and denominator have a common factor 2, contradicting the lowest-terms assumption. This is the essence of infinite descent: the assumed fraction produces another equivalent representation with a smaller reducible pair, and the process cannot continue indefinitely. Option B captures this contradiction; the other options do not describe the proof.

Correct answer: A. From (\frac{p}{q}), a smaller fraction reducible by (3) can be formed

Explanation: The proof that \(\sqrt{3}\) is irrational can be described using infinite descent. Assume, for contradiction, that \(\sqrt{3}=p/q\) is a fraction in lowest terms, with integers p and q and \(q\ne0\). Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), so 3 divides p. Substituting \(p=3k\) then shows that 3 also divides q. Thus both numerator and denominator have a common factor 3.

Dividing both by 3 produces a smaller positive fraction representing the same number, which contradicts the assumption that the original fraction was already in lowest terms. If one repeatedly applies the same reasoning, it would create an endless chain of smaller positive integer pairs, which is impossible. Therefore option A captures the descent idea. Options B, C, and D do not follow from the proof and do not establish irrationality.

Correct answer: A. In a perfect square, the exponent of 2 must be even

Explanation: The relevant prime-factor principle is that every prime occurs to an even exponent in the prime factorisation of a perfect square. For example, if x = 2ᵏ times other prime factors, then x² contains 2²ᵏ, whose exponent is even. In the irrationality proof, assuming √2 = a/b in lowest terms leads to a² = 2b². Since the right side has one extra factor 2 beyond the square b², the parity of the exponent of 2 becomes impossible: the left side is a square and must have an even exponent, whereas the right side has an odd one. Hence option A states the governing fact. The other statements are false or irrelevant.

Correct answer: C. In a perfect square, the exponent of 3 must be even

Explanation: A perfect square has an even exponent for every prime in its prime factorisation. This applies to the prime 3 as well. In the usual contradiction proof, suppose √3 = p/q in lowest terms. Squaring gives p² = 3q². Because the right side is divisible by 3, p is divisible by 3; write p = 3k. Substitution gives q² = 3k², so q is also divisible by 3. The resulting common factor contradicts the assumption that p/q was in lowest terms. Equivalently, the square p² must have an even exponent of 3, while the factor 3q² introduces an odd exponent. Therefore option C is the precise useful statement; the remaining options are false.

Correct answer: A. \(3\mid a\)

Explanation: Since 3 is prime, \(3\mid a^2\) implies \(3\mid a\). Put \(a=3k\); then \(b^2=3k^2\), so \(3\mid b\), contradicting coprimality. Exam tip: use prime divisibility carefully.

Correct answer: C. (p^2) should not be divisible by (3), but the equation shows it is divisible

Explanation: If (p) has no factor (3), then (p^2) also has no factor (3). But the equation shows (p^2) divisible by (3).

Correct answer: A. An infinite decimal expansion alone does not prove irrationality; recurring infinite decimals can be rational.

Explanation: An infinite decimal is not sufficient evidence: \(0.333\ldots=\frac13\) is rational. To prove \(\sqrt{3}\) irrational, assume \(\frac pq\) is in lowest terms and derive a divisibility contradiction. Exam tip: distinguish recurring from non-recurring decimals.

Correct answer: A. To ensure that \(p\) and \(q\) cannot both be even

Explanation: In lowest terms, \(p\) and \(q\) are coprime. If the proof shows both are even, 2 becomes a common factor, giving a contradiction; they need not both be odd. Exam tip: lowest terms means coprime.

Correct answer: B. (b\neq0) keeps the fraction defined and (\gcd(a,b)=1) is the basis of contradiction

Explanation: The two conditions serve different purposes. The statement \(b\neq0\) is required because \(a/b\) must be a defined fraction; division by zero has no meaning. It does not say that the fraction is reduced or that its numerator and denominator have no common factor. The condition \(\gcd(a,b)=1\), on the other hand, says that the fraction is in lowest form.

In the proof, assume \(\sqrt{2}=a/b\) with these conditions. The algebra eventually shows that both \(a\) and \(b\) are even, so they share the factor 2. That contradicts \(\gcd(a,b)=1\), producing the desired contradiction. Thus option B is correct: one condition keeps the fraction defined, while the other supports the contradiction.

Correct answer: C. (q\neq0) keeps the fraction defined and (\gcd(p,q)=1) gives final contradiction

Explanation: (q\neq0) is necessary for a rational fraction. (\gcd(p,q)=1) breaks when both are divisible by (3).

Correct answer: B. Assume rational \(\rightarrow\) \(a^2=2b^2\) \(\rightarrow\) (a) even \(\rightarrow\) (b) even \(\rightarrow\) contradiction

Explanation: The standard proof starts with rational assumption and derives contradiction that both are even. This is the correct order.

Correct answer: A. Both \(p\) and \(q\) are divisible by 3.

Explanation: Since \(p^2=3q^2\), 3 divides \(p\). On writing \(p=3k\), we get \(q^2=3k^2\), so 3 divides \(q\) too. This contradicts coprimality. Tip: prove divisibility of \(p\) before substituting.

Correct answer: A. It is always irrational.

Explanation: If \(r\sqrt{3}=q\) were rational, then since \(r\ne0\), \(\sqrt{3}=q/r\) would also be rational, a contradiction. Thus A is correct. Tip: a non-zero rational multiplier preserves irrationality.

Correct answer: B. Contradiction when both become divisible by (3)

Explanation: Lowest form makes (p) and (q) coprime. This condition makes both divisible by (3) a contradiction.

Correct answer: A. \(a\) और \(b\) दोनों 3 से विभाज्य हैं

Explanation: From \(a^2=3b^2\), 3 divides \(a^2\), so the prime-divisibility property gives \(3\mid a\). Put \(a=3k\) to obtain \(3\mid b\) too, contradicting coprimality. In exams, state the prime-divisibility step clearly.

Correct answer: C. Because initially (p,q) are coprime in lowest form

Explanation: Both divisible by (3) is the contradiction found at the end. Initially (\gcd(p,q)=1) is assumed.

Correct answer: A. If \(a\sqrt{3}\) were rational, dividing by \(a\) would make \(\sqrt{3}\) rational, which is a contradiction.

Explanation: Assume \(a\sqrt{3}\) is rational, where \(a\ne0\) is rational. Then \(\frac{a\sqrt{3}}{a}=\sqrt{3}\) would be rational, contradicting its irrationality. Exam tip: always check that division is by a non-zero number.

Correct answer: A. (\sqrt{3}) is rational

Explanation: A proof by contradiction temporarily assumes the opposite of the statement that is to be proved. Here the target is to show that \(\sqrt{3}\) is irrational, so the proof begins by assuming that \(\sqrt{3}\) is rational. It is then written as a fraction \(p/q\) in lowest terms, with \(q\neq0\). The algebraic argument eventually forces both \(p\) and \(q\) to be divisible by 3.

That conclusion conflicts with the choice of a lowest-terms fraction, so the temporary assumption cannot be true. The contradiction does not reject the fact that \(\sqrt{3}>0\), that it is real, or that the denominator is non-zero; those facts remain valid. It rejects only the assumption of rationality. Therefore option A is the correct choice.

Correct answer: A. If \(5+\sqrt{3}\) were rational, subtracting 5 would make \(\sqrt{3}\) rational, which is impossible.

Explanation: Let \(5+\sqrt{3}=r\) be rational. Then \(\sqrt{3}=r-5\) is rational, a contradiction. Adding rational 5 cannot make it rational. Exam tip: state the subtraction step.