Proof of irrationality of square root 2 and square root 3

Correct answer: B. If \(n\) is not a perfect square, then \(\sqrt{n}\) is irrational.

Explanation: Option B is correct. If \(n\) is not a perfect square, its prime factorisation has at least one odd exponent, so \(\sqrt{n}\) cannot be rational. For example, \(12=2^2\times3\). Exam tip: first check whether the number is a perfect square.

Correct answer: B. Because (3) is prime and divides (u^2)

Explanation: If prime (3) divides a square, it divides the original number too. Therefore (u) is a multiple of (3).

Correct answer: A. \(q\) must also be divisible by 3

Explanation: Let \(p=3k\). Substituting in \(p^2=3q^2\) gives \(9k^2=3q^2\), so \(q^2=3k^2\). Hence \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime. Exam tip: if a prime divides a square, it divides the number itself.

Correct answer: A. दोनों 3 से विभाज्य होते हैं

Explanation: Assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), 3 divides \(p\); putting \(p=3k\) then shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: if a prime divides a square, it divides the number itself.

Correct answer: C. Both (x) and (y) are even while they are coprime

Explanation: The contradiction completes when numerator and denominator both have common factor (2). This opposes the coprime condition.

Correct answer: A. ताकि \(p\) और \(q\) दोनों के सम होने से प्राप्त विरोधाभास स्पष्ट हो सके।

Explanation: In lowest terms, \(p\) and \(q\) have no common factor. From \(p^2=2q^2\), \(p\) is even; substituting \(p=2k\) shows that \(q\) is also even, contradicting coprimality. Exam tip: always begin with a fraction in lowest form.

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

Explanation: From \(a^2=3b^2\), \(a^2\) is divisible by 3. If the square of an integer is divisible by 3, the integer itself is divisible by 3, so write \(a=3k\). Substitution then makes \(b\) divisible by 3 too, contradicting coprimality. Exam tip: check squares of remainders 0, 1, and 2 modulo 3.

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

Explanation: \(2+\sqrt{3}\) is irrational. If it were rational, subtracting the rational number 2 would make \(\sqrt{3}\) rational, which is impossible. Also, \(\sqrt{9}=3\), and terminating or recurring decimals are rational. Exam tip: adding a rational number to an irrational number gives an irrational result.

Correct answer: A. \(p\) और \(q\) दोनों 3 से विभाज्य हैं

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 coprimality. Exam tip: check common divisibility of both terms.

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

Explanation: From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: an even square always has an even base.

Correct answer: A. यदि किसी अभाज्य संख्या से \(p^2\) विभाज्य है, तो वह \(p\) को भी विभाजित करती है।

Explanation: From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, \(3\mid p\); putting \(p=3k\) then gives \(3\mid q\) too. Thus \(p,q\) are not coprime. Exam tip: use the prime-divisor property in such proofs.

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

Explanation: The direct answer is A. If u is not divisible by 3, then its square u squared is also not divisible by 3, because squaring cannot create a missing prime factor. But the equation u squared equals 3v squared has an explicit factor 3 on the right-hand side, so the right side is divisible by 3. Equality would then force u squared to be divisible by 3, producing a contradiction. Option A correctly states this conflict. Option B, v=0, does not follow; in a fraction denominator is nonzero anyway. Option C, u=v, is unrelated to the equation and is not required. Option D, square root of 3 equals zero, is false and is not the contradiction used. The proof works by comparing divisibility facts, not by claiming arbitrary values are zero or equal. Memory cue: no factor 3 in u means no factor 3 in u squared, but the equation supplies one.

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

Explanation: From \(p^2=3q^2\), \(p^2\) is divisible by 3. Since 3 is prime, \(p\) must also be divisible by 3. Let \(p=3k\). Then \(9k^2=3q^2\), so \(q^2=3k^2\); hence \(q\) is also divisible by 3. This contradicts the fact that \(p\) and \(q\) are coprime, and this contradiction is used to prove that \(\sqrt{3}\) is irrational. Option B is incomplete because once \(p\) is divisible by 3, \(q\) must be divisible by 3 as well. Exam tip: If a prime divides a square, it also divides the original number.

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

Explanation: Since \(3\) is prime, if it divides the square \(p^2\), it must divide \(p\); hence \(3\mid p\). On putting \(p=3k\), we also obtain \(3\mid q\), contradicting coprimality. Exam tip: state this prime-divides-a-square property explicitly.

Correct answer: D. \(3\mid p\)

Explanation: Since 3 is prime, the prime-divisor property gives \(3\mid p\) whenever \(3\mid p^2\). Substituting \(p=3k\) later shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: state the prime-divisor step explicitly.

Correct answer: A. Because the contradiction is not clear when both become divisible by (3)

Explanation: The lowest coprime form creates contradiction with the final common factor (3). Write it in exams.

Correct answer: C. \(p\) and \(q\) are both divisible by 3

Explanation: From \(3q^2=p^2\), \(p^2\), and hence \(p\), is divisible by 3. Put \(p=3k\) to show that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: begin such proofs with a fraction in lowest terms.

Correct answer: B. Both are divisible by 3

Explanation: Since u=3t, u is divisible by 3. Substituting this in u²=3v² gives 9t²=3v², so v²=3t². Hence 3 divides v². As 3 is prime, if it divides v², it must also divide v. Therefore, both u and v are divisible by 3. Exam tip: If a prime p divides x², then p divides x.

Correct answer: C. 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 3 also divides
\(q\). This contradicts coprimality. Exam tip: identify the common prime factor causing the contradiction.

Correct answer: A. If 3 divides u², then 3 divides u

Explanation: The governing concept is the prime-divisor property of a square. If a prime p divides the square of an integer u, then p must divide u itself. This follows from prime factorisation: every prime factor in u² appears with twice its exponent, so the presence of 3 in u² means that 3 already occurs in the factorisation of u. Consequently, 3 ∣ u² implies 3 ∣ u. In the proof of √3, an equation such as u²=3v² first shows that 3 divides u², and this rule then gives 3 ∣ u. Option A states exactly this valid implication. Option B uses the wrong prime, option C wrongly requires u to be zero, and option D introduces an unrelated equality.

Correct answer: C. It contradicts the condition that \(p\) and \(q\) are coprime; hence \(\sqrt{3}\) is irrational

Explanation: If 3 divides
\(p^2\), then 3 must divide
\(p\). Put
\(p=3k\); this gives
\(q^2=3k^2\), so 3 also divides
\(q\). This contradicts coprimality. Exam tip: apply this prime-divisibility rule in irrationality proofs.

Correct answer: A. It is incomplete; first set u = 3t

Explanation: The governing idea is that the divisibility argument must be carried out in two justified stages. From u²=3v², the right-hand side is divisible by 3, so 3 divides u². Because 3 is prime, 3 divides u; therefore write u=3t for an integer t. Substituting gives 9t²=3v², and dividing by 3 gives v²=3t². Only now is v² visibly divisible by 3, allowing the prime-square rule to prove 3 divides v. Thus v is not obtained directly from the original equation without explanation. Option A correctly identifies the missing intermediate step. The other choices either accept an unsupported conclusion or discuss unrelated denominator and decimal issues.

Correct answer: A. The claim is false; if \(3\sqrt{2}\) were rational, dividing by 3 would make \(\sqrt{2}\) rational too.

Explanation: The claim is false. If \(3\sqrt{2}\) were rational, division by the non-zero rational number 3 would make \(\sqrt{2}\) rational, a contradiction. Exam tip: divide by a non-zero rational factor to test such claims.

Correct answer: C. (\sqrt{3}=\frac{u}{v}), where (\gcd(u,v)=1) and (v\neq0)

Explanation: The direct answer is C: square root of 3 equals u/v, where u and v are coprime integers and v is not zero. To prove irrationality by contradiction, assume the number is rational. By the definition of a rational number, it can be written as a quotient of integers with a nonzero denominator. We may reduce that fraction to lowest form, so gcd(u,v)=1. Squaring gives u squared equals 3v squared. Then 3 divides u squared, so the prime property gives 3 divides u; substitution then gives 3 divides v. This contradicts coprimality. Option A is impossible because division by zero is undefined. Option B is not the required general form; a rational number need not be an integer sum. Option C is correct and contains every needed condition. Option D is too restrictive and does not represent every rational assumption. Memory cue: rational number = coprime integer fraction with nonzero denominator.

Correct answer: A. Since 3 divides \(p^2\) and 3 is prime, 3 also divides \(p\)

Explanation: From \(p^2=3q^2\), 3 divides \(p^2\). A prime dividing a square must divide its base, so \(3\mid p\). On putting \(p=3k\), one also gets \(3\mid q\), contradicting lowest terms. Exam tip: explicitly state the prime-divisibility rule.