Proof of irrationality of square root 2 and square root 3

Correct answer: A. यह केवल सन्निकट मान है; \(\sqrt{3}\) का दशमलव प्रसार असांत और अनावर्ती है।

Explanation: A calculator shows only a rounded approximation. Since \(1^2<3<2^2\), \(\sqrt{3}\) lies between 1 and 2, but its decimal expansion is non-terminating and non-repeating, so it is irrational. Exam tip: never treat a finite display as the exact decimal expansion.

Correct answer: B. \(a\) and \(b\) are not coprime, so the original assumption is false

Explanation: If both \(a\) and \(b\) are even, each is divisible by 2, so they cannot be coprime. This contradicts the original assumption; hence \(\sqrt{2}\) is irrational. Exam tip: always use the lowest-terms condition to identify the contradiction.

Correct answer: C. \(a\) and \(b\) have a common factor other than 1

Explanation: If 3 divides both \(a\) and \(b\), then 3 is their common factor. This contradicts the initial condition that \(a/b\) is in lowest terms and \(a,b\) are coprime. Exam tip: link the contradiction to coprimality.

Correct answer: A. n is divisible by 3

Explanation: Since 3 is prime, an integer leaves remainder 0, 1, or 2 on division by 3. The squares of remainders 1 and 2 both leave remainder 1. Thus, for n² to leave remainder 0, n must leave remainder 0 and be divisible by 3. Exam tip: this result is central to the proof that √3 is irrational.

Correct answer: C. Both (a) and (b) are even

Explanation: The direct answer is C, both a and b are even. In the contradiction proof, assume
\(\sqrt{2}=a/b\), where a and b are integers, b is not zero, and the fraction is in lowest form, so a and b are coprime. Squaring gives \(2b^2=a^2\). Therefore \(a^2\) is even, and an integer whose square is even must itself be even, so write \(a=2k\). Substitution gives \(b^2=2k^2\), so b is also even. Thus both a and b have the common factor 2, contradicting their being coprime. Option A is only an intermediate result. Option B is another intermediate result, not the complete contradiction. Option C states the complete contradiction and is correct. Option D, \(\sqrt{2}>0\), is true but does not conflict with the lowest-form assumption. Memory cue: the contradiction is not merely “one number is even”; it is “both numerator and denominator are even.”

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

Explanation: Only (p) being divisible by (3) is not the final contradiction. Later (q) must also be proved divisible by (3).

Correct answer: A. The fraction cannot be in lowest form

Explanation: The governing concept is proof by contradiction together with the meaning of lowest terms. Assume √2 = a/b, where a and b are integers, b is nonzero, and gcd(a,b)=1. Squaring gives a² = 2b². Thus a² is even, so a is even; write a=2k. Substitution gives 4k²=2b², hence b²=2k², so b is also even. Therefore 2 is a common factor of a and b, which contradicts gcd(a,b)=1. The fraction cannot therefore have been in lowest form, and the original rational assumption fails. Option A states this exact contradiction. The result does not imply that the denominator is zero, that √2 is an integer, or that a and b are equal.

Correct answer: B. Area \(2\,\text{cm}^2\), side \(\sqrt{2}\,\text{cm}\)

Explanation: For a square, \(s^2=A\). When \(A=2\), the side is \(s=\sqrt{2}\), which is irrational. Hence, a rational area need not give a rational side. In exams, first take the square root of the area.

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

Explanation: From \\(p^2=3q^2\\), \\(3\\mid p^2\\); since 3 is prime, \\(3\\mid p\\). Put \\(p=3k\\) to get \\(q^2=3k^2\\), so \\(3\\mid q\\) too. This contradicts coprimality. Exam tip: if a prime divides a square, it divides the number itself.

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

Explanation: From \(3q^2=p^2\), \(p\) must be divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: check divisibility of both terms.

Correct answer: A. 3, \(a^2\) को विभाजित करता है; इसलिए 3, \(a\) को विभाजित करता है।

Explanation: From \(a^2=3b^2\), 3 divides \(a^2\). Since 3 is prime, it must divide \(a\). On putting \(a=3k\), one also obtains that 3 divides \(b\), contradicting coprimality. Exam tip: if a prime \(p\mid n^2\), then \(p\mid n\).

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

Explanation: Prime factor (3) appears in (p^2) only if it appears in (p). The equation forces this divisibility.

Correct answer: A. मान लें \(\sqrt{2}=\frac{a}{b}\), जहाँ \(a,b\) सह-अभाज्य हैं \(\rightarrow a^2=2b^2\rightarrow a\) सम \(\rightarrow b\) सम \(\rightarrow\) सह-अभाज्य होने के विरुद्ध

Explanation: Option A contains every essential step of the proof by contradiction. Assume \(\sqrt{2}=\frac{a}{b}\) in lowest terms. Then \(a^2=2b^2\), so \(a^2\), and hence \(a\), is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is also even. Thus both \(a\) and \(b\) are divisible by 2, contradicting that they are coprime. Option D is wrong because an infinite decimal alone does not prove irrationality; rational numbers may have infinite recurring decimals. Exam tip: always begin this proof by taking the fraction in lowest terms.

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

Explanation: In lowest terms, \(p\) and \(q\) cannot have a common factor. From \(3\mid p^2\), we get \(3\mid p\); substitution then gives \(3\mid q\). Thus 3 is a common factor. Exam tip: use the prime divisibility rule for squares.

Correct answer: A. It gives a genuine contradiction when both become even

Explanation: A rational number can be represented by many equivalent fractions, so merely writing √2=a/b is not enough for a contradiction. We choose the representation in lowest terms, meaning gcd(a,b)=1. From a²=2b², parity arguments show that a is even and then b is even. This conclusion means that 2 divides both numbers, so their greatest common divisor is at least 2. That directly conflicts with gcd(a,b)=1, producing the genuine contradiction required by the proof. Hence option A is correct. Without the coprime assumption, a pair such as 2/4 could be even in both entries without contradiction because it was not reduced. The other options do not follow from the equations.

Correct answer: A. A number can be irrational even if its square is rational.

Explanation: Although \((\sqrt{2})^2=2\) is rational, \(\sqrt{2}\) is irrational. Thus, a rational square does not prove that the original number is rational. Exam tip: always test the converse of a statement separately.

Correct answer: A. Both are even

Explanation: From a=2r, a is a multiple of 2, so a is even. From b^2=2r^2, b^2 is even. If the square of an integer is even, then the integer itself must be even; hence b is even. Therefore, both a and b are even. They need not both be zero, since non-zero even integers are also possible. Exam tip: In the irrationality proof, this result contradicts the assumption that a and b are coprime.

Correct answer: B. Both are divisible by 3

Explanation: From p = 3k, p is divisible by 3. Also, q² = 3k² shows that q² is divisible by 3. Since 3 is prime, if it divides the square of an integer, it must divide that integer itself. Hence q is also divisible by 3. Therefore, both p and q are divisible by 3. Option A has no basis because divisibility by 2 is not implied. Exam tip: In contradiction proofs of irrationality, showing that two assumed coprime integers share a prime factor produces the contradiction.

Correct answer: A. If a number is odd, its square is odd, so if the square is even, the number is even

Explanation: The relevant parity theorem is: an odd integer has an odd square. Its contrapositive is that if an integer’s square is even, then the integer itself must be even. In the proof, a²=2b² is even, so this rule establishes that a is even. After substituting a=2k, the resulting equation similarly shows that b² is even, and hence b is even. Therefore option A gives the correct logical direction. Option B reverses the parity result and is false because an even integer has an even square. An even square need not be zero, and its parity says nothing about a denominator being zero, so options C and D are also invalid.

Correct answer: A. If (3\mid p^2), then (3\mid p)

Explanation: The direct answer is A: if 3 divides p squared, then 3 divides p. The reason is the prime-factor property. A prime number appearing in the factorisation of a square must already appear in the factorisation of its base. For example, if p had no factor 3, then p squared would also have no factor 3. Option A is correct and is the key step in proving the irrationality of the square root of 3. Option B is wrong because divisibility by 3 does not imply divisibility by 2. Option C is wrong because 3 dividing p does not mean p equals zero; p could be 3, 6, 9, or another multiple of 3. Option D is wrong because the divisibility statement says nothing about p and q being equal. In the full proof, 3 divides p, then substitution shows 3 divides q, contradicting lowest form. Memory cue: prime divides a square, so it divides the original number.

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

Explanation: From \(a^2=3b^2\), \(a^2\), and hence \(a\), is divisible by 3. Putting \(a=3k\) gives \(b^2=3k^2\), so \(b\) is also divisible by 3. This contradicts coprimality. Exam tip: state the common factor explicitly.

Correct answer: A. The step is incomplete; first (p=3k) must be taken

Explanation: First (p) is proved divisible by (3) from (p^2). Then putting (p=3k) gives the conclusion for (q).

Correct answer: A. In the proof, a/b should be taken in lowest form with gcd(a,b)=1

Explanation: Every rational number can be written as a/b with integer a and non-zero integer b, but the irrationality proof needs a special, reduced representation. We choose a/b in lowest terms, expressed by gcd(a,b)=1. Assuming √2=a/b then gives a²=2b². The parity argument forces a and b both to be even, which contradicts the reduced-form condition because they would share the factor 2. Thus option A states the proof-ready form precisely. A zero denominator is not allowed in a fraction, and requiring both entries to be even would destroy the contradiction rather than create it. Equal numerator and denominator is also irrelevant and would describe only a restricted fraction, not every rational number.

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

Explanation: If \(\sqrt{n}=p/q\), then \(nq^2=p^2\). Since every prime exponent in a square is even, \(n\) must be a perfect square. Thus \(\sqrt{2}\) is irrational. Exam tip: first check whether the integer is a perfect square.

Correct answer: A. Both use rational assumption, squaring, and then contradiction through a prime factor

Explanation: Both proofs have the same structure, but the prime factors are (2) and (3) respectively. Mention this in comparison.