Proof of irrationality of square root 2 and square root 3

Correct answer: A. Conclusion of divisibility by (3)

Explanation: First (a) is found divisible by (3), and then (b) is also found divisible by (3). This creates the contradiction.

Correct answer: A. Both get contradiction from coprime fraction assumption

Explanation: Both proofs start with rational assumption and a coprime fraction. Finally a common factor gives contradiction.

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

Explanation: Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). On writing \(p=3k\), the proof further shows that \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime. Exam tip: use this prime-factor property carefully.

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

Explanation: The direct answer is B: assume that \\(\\sqrt{3}\\) is rational. To prove irrationality by contradiction, we temporarily assume the opposite of what we want to prove. Thus write \\(\\sqrt{3}=p/q\\), where p and q are integers, q is not zero, and the fraction is in lowest terms. Squaring gives \\(3q^2=p^2\\). This eventually leads to the conclusion that both p and q are divisible by 3, contradicting the assumption that p/q was in lowest terms. Therefore \\(\\sqrt{3}\\) is irrational. Option A, saying it is a perfect square, is not the starting assumption and is not the required opposite statement. Option B is correct because rationality is assumed temporarily. Option C, saying it is zero, is false since its square is 3. Option D, saying it is a natural number, is also false and is not the standard contradiction assumption. Memory cue: in an irrationality proof by contradiction, first assume rational, then derive an impossibility.

Correct answer: B. Both \(m\) and \(n\) are divisible by 3

Explanation: From \(3n^2=m^2\), \(m^2\), and hence \(m\), is divisible by 3. Putting \(m=3k\) shows that \(n\) is also divisible by 3, contradicting coprimality. Exam tip: identify the common factor causing the contradiction.

Correct answer: A. मान लेते हैं कि \(\sqrt{2}=\frac{p}{q}\) है, जहाँ \(p\) और \(q\) सह-अभाज्य हैं; फिर विरोधाभास प्राप्त करते हैं कि दोनों सम हैं।

Explanation: In proof by contradiction, assume \(\sqrt{2}=p/q\) in lowest terms. Squaring gives \(p^2=2q^2\), so \(p\), and then \(q\), must both be even. This contradicts coprimality. Exam tip: always state that the fraction is in lowest terms.

Correct answer: A. 3 divides both p and q

Explanation: From

p^2=3q^2
,

p^2
is divisible by 3, so

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: use prime divisibility of squares.

Correct answer: A. \(1.414\) is only an approximate decimal value of \(\sqrt{2}\)

Explanation: \(1.414\) is a terminating decimal and hence rational, but it is not exactly equal to \(\sqrt{2}\). Check: \(1.414^2=1.999396\ne2\). Thus it is only an approximation. In exams, distinguish “approximately equal” from “equal.”

Correct answer: C. even

Explanation: In \(a^2=2b^2\), \(a^2\) is 2 multiplied by the integer \(b^2\). Hence \(a^2\) is divisible by 2, so it is even. The relation does not necessarily show that \(a^2\) is prime or zero. Exam tip: A number expressible as \(2\times\text{an integer}\) is even.

Correct answer: A. To establish rationality, both the numerator and denominator must be integers; here \(\sqrt{12}\) is not an integer.

Explanation: The equality is correct because \(\sqrt{12}=2\sqrt{3}\) and \(\frac{6}{2\sqrt{3}}=\sqrt{3}\). However, a rational number must be expressed as \(a/b\) with integers \(a,b\); the denominator here is irrational. Check the denominator in such claims.

Correct answer: A. (a) is even

Explanation: If the square of an integer is even then the integer is also even. This is the key fact in the proof of (\sqrt{2}).

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

Explanation: If prime factor (3) divides the square then it also divides the number. This is used in the proof of (\sqrt{3}).

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

Explanation: From \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\); substituting back shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: use the prime-factor property of a square carefully.

Correct answer: A. This is not possible because \(\sqrt{2}\) is irrational.

Explanation: By Pythagoras’ theorem, \(d^2=1^2+1^2=2\), so \(d=\sqrt{2}\). It cannot be written as a ratio of integers. If both terms are even, the fraction is not in lowest form. Exam tip: write \(d^2\) first for a square’s diagonal.

Correct answer: A. \(b^2=2k^2\)

Explanation: Substituting \(a=2k\) into \(a^2=2b^2\) gives \((2k)^2=2b^2\), or \(4k^2=2b^2\). Dividing both sides by 2 gives \(b^2=2k^2\). The relation \(b^2=3k^2\) introduces an unjustified factor of 3, so it is incorrect. Exam tip: after substitution, expand the square first and then cancel common factors carefully.

Correct answer: B. \(q^2=3r^2\)

Explanation: Substitute \(p=3r\) into \(p^2=3q^2\). This gives \((3r)^2=3q^2\), or \(9r^2=3q^2\). Dividing both sides by 3 gives \(q^2=3r^2\). Hence, option B is correct. There is no basis for \(q^2=2r^2\). Exam tip: while substituting, remember that \((3r)^2=9r^2\).

Correct answer: C. (b) is even

Explanation: Since the right-hand side of (b^2=2k^2) is a multiple of 2, (b^2) is even. The square of an integer is even only when the integer itself is even; hence (b) is even. If (b) were odd, its square would also be odd, so option A is incorrect. Exam tip: in a contradiction proof of irrationality, this step helps show that both integers have the common factor 2.

Correct answer: D. q is divisible by 3

Explanation: Given q² = 3r², q² is divisible by 3. Since 3 is prime, if 3 divides the square of an integer, it must also divide the integer itself. Therefore, q is divisible by 3. Divisibility of q² by 3 does not imply that q is even or zero. Exam tip: For a prime p, use p | a² ⇒ p | a.

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, \(p\) is divisible by 3. Put \(p=3k\) to obtain that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: show this substitution clearly.

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

Explanation: From \(2q^2=p^2\), \(p^2\), and hence \(p\), is even. Put \(p=2k\); then \(q^2=2k^2\), so \(q\) is also even. This contradicts coprimality. Exam tip: identify the common factor 2 to state the contradiction.

Correct answer: A. Because it leads to an impossible situation

Explanation: The governing concept is proof by contradiction, also called reductio ad absurdum. To prove a statement such as the irrationality of a square root, we temporarily assume the opposite—for example, that the number is rational and can be written as a fraction in lowest terms. Algebraic reasoning then produces a result that conflicts with the original lowest-form or divisibility condition. Since a valid assumption cannot logically lead to an impossibility, the assumed opposite statement must be false. Therefore, option A is correct. Being small, containing a decimal, or lacking a diagram has no role in deciding whether the assumption is false.

Correct answer: B. Only perfect-square integers have integer square roots; \(3\) is not a perfect square.

Explanation: Being an integer is not enough; only perfect squares have integer square roots. If \(\sqrt{3}=p/q\), then \(p^2=3q^2\), so both \(p\) and \(q\) become divisible by 3, a contradiction. Exam tip: check perfect squares and prime factors.

Correct answer: B. If (p^2) is divisible by (3) then (p) is divisible by (3)

Explanation: If prime (3) divides the square, it divides the number too. This is the main step in the proof.

Correct answer: B. \(26^2=676\), whereas \(3\times15^2=675\)

Explanation: If \(\sqrt{3}=\frac{26}{15}\), squaring both sides would give \(26^2=3\times15^2\). But \(676\ne675\), so the claim is false. Exam tip: verify a claimed square root fraction by squaring it.

Correct answer: B. \(\sqrt{3}\) is not rational

Explanation: An irrational number cannot be written in the form \(p/q\), where \(p\) and \(q\) are integers and \(q\ne0\). Hence, proving that \(\sqrt{3}\) is irrational directly means that \(\sqrt{3}\) is not rational. It is neither an integer nor zero nor negative; in fact, \(\sqrt{3}>0\). Exam tip: remember that “irrational” means “not rational.”