Proof of irrationality of square root 2 and square root 3

Correct answer: C. Because \(b^2=2r^2\) shows that \(b^2\) is even

Explanation: Substituting \(a=2r\) gives \(a^2=4r^2\). Using this in \(a^2=2b^2\), we get \(4r^2=2b^2\), or \(b^2=2r^2\). Thus, \(b^2\) is even. If the square of an integer is even, the integer itself must be even; hence \(b\) is even. Being negative is not a reason for \(b\) to be even. Exam tip: this result is used to obtain a contradiction in irrationality proofs.

Correct answer: A. Because substitution gives \(q^2=3k^2\), so \(q^2\) is divisible by 3

Explanation: Given \(p=3k\), substitute it into \(p^2=3q^2\): \((3k)^2=3q^2\), so \(9k^2=3q^2\). Dividing by 3 gives \(q^2=3k^2\); hence \(q^2\) is divisible by 3. Since 3 is prime, if the square of an integer is divisible by 3, then the integer itself is divisible by 3. Therefore, \(q\) is divisible by 3. Exam tip: for a prime \(r\), \(r\mid n^2\) implies \(r\mid n\).

Correct answer: A. (\sqrt{2}) uses evenness by (2) and (\sqrt{3}) uses divisibility by (3)

Explanation: In (\sqrt{2}), (2) is the key factor, and in (\sqrt{3}), (3) is the key factor. Identify the difference by the factor.

Correct answer: A. They should be assumed coprime in lowest form

Explanation: At the start the fraction is in lowest form so (a) and (b) are assumed coprime. Both even is the final contradiction.

Correct answer: B. At the start (p) and (q) are assumed coprime

Explanation: At the start (\frac{p}{q}) is in lowest form. Both divisible by (3) is derived later as a contradiction.

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

Explanation: \(\sqrt{2}\) and \(\sqrt{3}\) are irrational because 2 and 3 are not perfect squares. But \(\sqrt{4}=2\) and \(\sqrt{9}=3\) are rational. Exam tip: the square root of a perfect-square integer is an integer.

Correct answer: A. Since \(3\mid p^2\), \(3\mid p\); putting \(p=3k\) shows that \(3\mid q\) as well.

Explanation: As \(3\mid p^2\) and 3 is prime, \(3\mid p\). Put \(p=3k\): \(9k^2=3q^2\), so \(q^2=3k^2\) and \(3\mid q\). Thus both share 3, contradicting coprimality. Exam tip: state the prime-divisor rule clearly.

Correct answer: C. √3 is irrational

Explanation: Assume for contradiction that √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. Squaring gives p² = 3q². This implies that 3 divides p², and therefore 3 divides p; writing p = 3k then gives q² = 3k², so 3 also divides q. Thus p and q have the common factor 3, contradicting the assumption that p/q was in lowest terms. The contradiction does not make √3 an integer, rational, or zero. Instead, it disproves the rational assumption. Therefore √3 is irrational, and option C is the correct final conclusion.

Correct answer: B. The statement is incorrect; if \(5+\sqrt{2}\) were rational, subtracting 5 would make \(\sqrt{2}\) rational too.

Explanation: \(\sqrt{2}\) is irrational. If \(5+\sqrt{2}\) were rational, subtracting the rational number 5 would make \(\sqrt{2}\) rational, a contradiction. Hence the sum is irrational. Exam tip: isolate the irrational term by adding or subtracting a rational number.

Correct answer: C. (p^2=3q^2), (p=3k), (q^2=3k^2)

Explanation: In the proof of (\sqrt{3}), the chain uses divisibility by (3). It shows (q) is also divisible by (3).

Correct answer: A. (a) and (b) coprime and (b\neq0)

Explanation: A rational number is written in lowest form as a ratio of coprime integers. The denominator cannot be zero.

Correct answer: B. p and q are coprime and q ≠ 0

Explanation: When a number is assumed to be rational, it can be represented as p/q with p and q integers, q ≠ 0, and the fraction reduced to lowest terms. Lowest terms means that p and q are coprime, so they have no common divisor greater than 1. This condition is essential because the proof later derives that both are divisible by 3, producing the contradiction. Option A describes the result obtained later, not the initial assumption. Options C and D are impossible because a denominator cannot be zero and p/q would not be a valid reduced representation if both were zero. Therefore option B correctly states the required starting condition.

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

Explanation: Writing the fraction in lowest terms makes \(p\) and \(q\) coprime. From \(p^2=3q^2\), 3 divides \(p\), and then \(q\), creating a contradiction. Exam tip: state “lowest terms” explicitly.

Correct answer: A. Because (3) is a prime factor

Explanation: The correct answer is option A: because 3 is a prime factor. Start with the fact that if a prime number divides a square, it must divide the number whose square was taken. Here, the proof gives that 3 divides p². The prime-factor rule therefore says that 3 divides p, so p can be written as p=3k for some integer k. This is not because p is zero, equal to q, or always even. Option A is correct because 3 is prime. Option B is wrong because p=0 is neither given nor suitable in the rational representation. Option C is wrong because p=q is not obtained and is unnecessary. Option D is wrong because p need not be even; divisibility by 3 has nothing to do with being even. Remember: for a prime r, if r divides n², then r divides n.

Correct answer: B. The square root of a rational number need not be rational; \(\sqrt{3}\) is irrational

Explanation: The square root of a rational number is not always rational. If \(\sqrt{3}=p/q\) in lowest form, then \(p^2=3q^2\), so 3 divides \(p\) and then \(q\), a contradiction. Exam tip: the square root of a non-perfect-square integer is irrational.

Correct answer: A. \(\frac{3+\sqrt{2}}{5}\)

Explanation: If \(\frac{3+\sqrt{2}}{5}\) were rational, multiplying by 5 and subtracting 3 would make \(\sqrt{2}\) rational, a contradiction. B, C and D equal 2, 1 and 0. Exam tip: use closure of rational numbers to test such expressions.

Correct answer: C. \(p\) and \(q\) are both divisible by 3, so they are not coprime

Explanation: Assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), \(p\) is divisible by 3, and then \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: state “lowest terms” first.

Correct answer: A. \(q\) is also 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. If a prime \(r\) divides \(n^2\), then \(r\) also divides \(n\).

Explanation: Since 3 is prime, \(3\mid a^2\) gives \(3\mid a\). Set \(a=3k\); then \(b^2=3k^2\), so \(3\mid b\), contradicting lowest terms. Exam tip: state the prime-divisor rule clearly.

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

Explanation: Assuming \(\sqrt{3}=a/b\) gives \(a^2=3b^2\). Thus \(a^2\), hence \(a\), is divisible by 3; putting \(a=3k\) shows that \(b\) is also divisible by 3. This contradicts coprimality. Exam tip: if a square is divisible by 3, its root number is divisible by 3.

Correct answer: A. √2 is rational

Explanation: The proof uses contradiction. It begins by assuming that √2 is rational and can be written as a/b in lowest terms. The equation a² = 2b² then forces a to be even. Substituting a = 2r gives b² = 2r², so b is also even. This means that a and b share the factor 2, contradicting the claim that a/b was in lowest terms. Consequently, the assumption that √2 is rational is false, and the correct conclusion is that √2 is irrational. Positivity and reality are not disproved: √2 is a positive real number. Thus only option A identifies the statement rejected by the proof.

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

Explanation: The direct answer is Option B: the assumption that √3 is rational. In a contradiction proof, we temporarily assume the opposite of what we want to prove. Suppose √3 = p/q, where p and q are coprime integers and q is not zero. Squaring gives p² = 3q². Therefore p² is divisible by 3, so p is divisible by 3; write p = 3r. Substitution gives q² = 3r², so q is also divisible by 3. This contradicts the statement that p and q have no common factor. Thus the temporary assumption that √3 is rational must be rejected, and √3 is irrational. Option A and Option D merely state positivity, which is true but is not the rejected assumption. Option C says it is real; that is also true and is not contradicted. Memory cue: reject the starting assumption that creates the impossible common factor.

Correct answer: A. The equation first shows that 3 divides \(a\), not \(b\)

Explanation: From \(a^2=3b^2\), 3 divides \(a^2\), so since 3 is prime, it divides \(a\) first. Put \(a=3k\); then 3 also divides \(b\), contradicting coprimality. Exam tip: track the numerator first.

Correct answer: B. Irrational because the rational assumption makes both p and q divisible by 3

Explanation: Assume √3 = p/q in lowest terms, with q ≠ 0. Squaring gives p² = 3q². Since 3 divides p², the prime-divisibility property implies that 3 divides p. Let p = 3k; substitution gives q² = 3k², so 3 divides q as well. This creates a common factor 3 in p and q, contradicting the lowest-terms assumption. Therefore the initial assumption that √3 is rational is false, and √3 is irrational. Option C merely repeats an intermediate equation and does not establish rationality. Options A and D confuse the contradiction with an unrelated conclusion. Hence option B gives both the correct conclusion and its reason.

Correct answer: A. Both having 2 as a common factor

Explanation: Coprime numbers have 1 as their only common factor, so their HCF is 1. If both a and b are divisible by 2, then 2 is a common factor and they cannot be coprime. In contrast, one number may be even and the other odd, such as 2 and 3, which are coprime. Exam tip: For coprime numbers, check that no common factor other than 1 divides both numbers.