Proof of irrationality of square root 2 and square root 3

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

Explanation: From \(3q^2=p^2\), \(p^2\), hence \(p\), is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts lowest terms. Exam tip: use prime divisibility of a square carefully.

Correct answer: A. The integer is divisible by 3.

Explanation: An integer leaves remainder 0, 1, or 2 on division by 3. The squares of remainders 1 and 2 both leave remainder 1, so a square divisible by 3 requires the integer itself to be divisible by 3. Exam tip: list possible square remainders modulo 3.

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

Explanation: From ext{\(p^2=3q^2\)} , ext{\(p^2\)} and hence ext{\(p\)} are divisible by 3. Put ext{\(p=3k\)} to get ext{\(q^2=3k^2\)} , so ext{\(q\)} is also divisible by 3. This contradicts coprimality. Exam tip: use the prime-divides-a-square property.

Correct answer: A. Both \(m\) and \(n\) are even

Explanation: Since \(m^2=2n^2\), \(m^2\) is even, so \(m=2k\). Substitution gives \(n^2=2k^2\), making \(n\) even too. This contradicts coprimality. Exam tip: establish that both numerator and denominator share 2.

Correct answer: A. \(2\sqrt{3}\)

Explanation: Since \(12=4\times3\), \(\sqrt{12}=\sqrt4\sqrt3=2\sqrt3\). As \(\sqrt3\) is irrational, multiplying it by non-zero rational 2 still gives an irrational number. \(3\sqrt2\) squares to 18, not 12. Exam tip: extract perfect-square factors first.

Correct answer: A. Prime divisor property

Explanation: Since 3 is prime,

b^2=3a^2 shows that 3 divides

b^2. By the prime divisor property, 3 must divide

b. This leads to a contradiction with the lowest-form assumption. In exams, state the prime-property step explicitly.

Correct answer: A. If \(3\mid p^2\), then \(3\mid p\); substituting \(p=3k\) then gives \(3\mid q\).

Explanation: From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, \(3\mid p\); put \(p=3k\) to obtain \(q^2=3k^2\), so \(3\mid q\) too. This contradicts coprimality. Exam tip: state the prime-divides-square rule first.

Correct answer: B. (\gcd(a,b)=1) and (\gcd(a,b)\ge3) cannot both hold

Explanation: If both have common factor (3), they cannot remain coprime. This makes the rational assumption false.

Correct answer: D. From m² = 2n², directly m = 2n

Explanation: The equation m² = 2n² shows that m² is even, and the parity theorem then shows that m is even. We may write m = 2k, but the equation does not imply the much stronger statement m = 2n. That direct equality is an unjustified shortcut, so option D is wrong. The other steps are valid parts of the contradiction proof.

Correct answer: D. Directly writing a = 3b from a² = 3b²

Explanation: The governing concepts are valid algebraic substitution and the prime-divisibility property of squares. From a²=3b², the right side is a multiple of 3, so a² is divisible by 3; A is valid. Since 3 is prime, 3 dividing a² implies 3 divides a, so B is also valid. Writing a=3r and substituting gives 9r²=3b², and division by 3 gives b²=3r²; therefore C is valid as well. However, a²=3b² does not imply a=3b. If a=3b, then squaring would produce a²=9b², which is different from 3b². Thus D is the wrong shortcut. It replaces a justified divisibility conclusion with an unsupported equality.

Correct answer: A. Both \(a\) and \(b\) are divisible by 3

Explanation: 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 coprimality. Exam tip: use the prime-divides-a-square rule carefully.

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

Explanation: If (a) has no factor (3), then (a^2) also has none. But (a^2=3b^2) makes (a^2) divisible by (3).

Correct answer: A. Assume rational \(\rightarrow\) \(m^2=2n^2\) \(\rightarrow\) (m) even \(\rightarrow\) (n) even \(\rightarrow\) contradiction

Explanation: The proof begins by assuming the opposite of what must be shown: suppose \(\sqrt{2}\) is rational and write it as \(m/n\) in lowest form, with \(n\neq0\). Squaring gives \(m^2=2n^2\). This equation shows that \(m^2\), and therefore \(m\), is even. Writing \(m=2k\) and substituting back shows that \(n^2\), and therefore \(n\), is also even.

Thus both \(m\) and \(n\) have 2 as a common factor. That contradicts the assumption that \(m/n\) was in lowest form, meaning the two integers were coprime. This contradiction proves that the original assumption was false. Hence option A gives the correct logical chain; the other options omit the essential algebra and contradiction.

Correct answer: A. परिमेय मानना \(\rightarrow\) \(a^2=3b^2\) \(\rightarrow\) \(a\) \(3\) से विभाज्य \(\rightarrow\) \(b\) \(3\) से विभाज्य \(\rightarrow\) विरोधाभास

Explanation: The proof of \(\sqrt{3}\) follows a chain of divisibility by \(3\). Finally the coprime assumption breaks.

Correct answer: A. The contradiction will not be clear when both become even

Explanation: The lowest-form condition states coprimality. Without it, both even will not form a decisive contradiction.

Correct answer: A. The sum of a rational number and an irrational number is irrational.

Explanation: If \(7+\sqrt{2}\) were rational, subtracting the rational number 7 would make \(\sqrt{2}\) rational. This contradicts the proven irrationality of \(\sqrt{2}\). Exam tip: rational ± irrational is always irrational.

Correct answer: A. Both \(m\) and \(n\) are even

Explanation: From \(m=2k\), \(m\) is even. Also, \(n^2=2k^2\) shows that \(n^2\) is even; if the square of an integer is even, the integer itself must be even. Hence \(n\) is also even, so both \(m\) and \(n\) are even. This contradicts the assumption that the fraction \(m/n\) was in lowest terms. Exam tip: In irrationality proofs, remember that an even square implies an even integer.

Correct answer: A. \(p\) and \(q\) are coprime

Explanation: The fraction is taken in lowest terms, so \(p\) and \(q\) must be coprime. The proof then shows that both are divisible by 3, contradicting this condition. Exam tip: explicitly state that the fraction is in lowest terms before starting the proof.

Correct answer: A. दावा सत्य है, क्योंकि 3 से विभाज्य न होने वाले पूर्णांक का वर्ग 3 से विभाज्य नहीं हो सकता।

Explanation: The claim is true. On division by 3, an integer leaves remainder 0, 1, or 2. Squaring remainders 1 and 2 gives remainders 1 and 4, i.e. 1 modulo 3, never 0. Hence a square divisible by 3 has a base divisible by 3. Exam tip: use remainders to test divisibility claims quickly.

Correct answer: A. 3 is prime

Explanation: The basis is the prime-divisor property: if a prime p divides x², then p divides x. Here p = 3, so 3 ∣ a² implies 3 ∣ a. This can be seen from prime factorisation. If 3 did not divide a, then no factor 3 would occur in the factorisation of a, and consequently no factor 3 could occur in a², contradicting 3 ∣ a². In the √3 proof, this result permits writing a = 3r; substitution then leads to b² = 3r² and eventually proves 3 ∣ b. Thus the primality of 3 is essential. Options B, C, and D are not assumptions of the proof and do not justify the implication.

Correct answer: A. The conclusion is correct, but the reason is false.

Explanation: Assume \(r=\sqrt{2}+\sqrt{3}\) is rational. Then \(r^2=5+2\sqrt6\), which would make \(\sqrt6\) rational, an impossibility. Thus the conclusion is correct. However, two irrational numbers need not always have an irrational sum; avoid this overgeneralisation.

Correct answer: A. First (a) must be proved divisible by (3) and (a=3r) must be used

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

Correct answer: A. \(p\) is divisible by 3

Explanation: 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.

Correct answer: A. A fraction in lowest terms cannot have a common factor such as 3 in its numerator and denominator.

Explanation: 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.

Correct answer: A. \(p\) is divisible by 3, and consequently \(q\) is also divisible by 3

Explanation: 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.