Proof of irrationality of square root 2 and square root 3

Correct answer: A. \(a^2\) is odd, whereas \(2b^2\) is even

Explanation: If \(a\) is odd, then its square \(a^2\) is also odd. On the other hand, \(2b^2\) is even for every integer value of \(b\), since it has 2 as a factor. Thus, the equation would make the same number both odd and even, which is the contradiction. Option B is not valid because the equation does not prove that \(b^2\) is odd. Exam tip: the square of an odd integer is odd, and the square of an even integer is even.

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

Explanation: For a prime \(p\), \(p\mid n^2\Rightarrow p\mid n\). From \(a^2=2b^2\), 2 divides \(a\), and then \(b\). But \(p^2\mid n\) need not hold. Exam tip: state this lemma first.

Correct answer: A. \(\sqrt{12}=2\sqrt{3}\), and 2 is a non-zero rational number; therefore, the result is irrational.

Explanation: \(\sqrt{12}=2\sqrt3\). A non-zero rational times an irrational is irrational, so A works. B wrongly uses \(\sqrt{a+b}=\sqrt a+\sqrt b\). Exam tip: extract perfect-square factors first.

Correct answer: A. The side is \(\sqrt{2}\text{ cm}\) and is irrational; a rational area does not necessarily give a rational side.

Explanation: For side \(s\), \(s^2=2\), so \(s=\sqrt{2}\). If \(\sqrt{2}=a/b\) is in lowest terms, \(a^2=2b^2\) makes both \(a\) and \(b\) even, giving a contradiction. Exam tip: a rational area need not give a rational side.

Correct answer: A. \(p\) and \(q\) have no common prime factor

Explanation: In lowest terms, \(p\) and \(q\) are coprime. From \(p^2=3q^2\), we get \(3\mid p\), and then \(3\mid q\), creating the contradiction. Exam tip: show that the same prime divides both.

Correct answer: B. Since \(p^2\) is even, \(p\) is even; putting \(p=2k\) shows that \(q\) is also even

Explanation: Since \(p^2\) is even, \(p\) must be even. Put \(p=2k\): then \(q^2=2k^2\), so \(q\) is also even, contradicting coprimality. Exam tip: begin with the fraction in lowest terms.

Correct answer: A. यदि 3, \(p^2\) को विभाजित करता है, तो 3, \(p\) को भी विभाजित करता है।

Explanation: Let \(\sqrt{3}=p/q\) be in lowest terms. From \(p^2=3q^2\), \(3\mid p^2\) implies \(3\mid p\); substitution then gives \(3\mid q\), a contradiction. Taking \(p=3\) disproves the other claims. Exam tip: use this rule for primes.

Correct answer: A. Non-terminating and non-repeating

Explanation: Since \(\sqrt{2}\) is irrational, its decimal expansion is non-terminating and non-repeating. A rational number \(p/q\) always has a terminating or repeating decimal. Exam tip: a non-terminating decimal is irrational only when it is also non-repeating.

Correct answer: A. So that \(p\) and \(q\) are not both divisible by 3

Explanation: From \(p^2=3q^2\), 3 divides \(p\). On writing \(p=3k\), it also follows that 3 divides \(q\), contradicting lowest terms. Exam tip: identify this common-factor contradiction.

Correct answer: A. The square of every odd integer is always odd.

Explanation: 2b² is even, hence a² is even. An odd integer has an odd square, so a is even. Substituting a=2k makes b even too, contradicting coprimality. Exam tip: state the odd-square fact.

Correct answer: A. 3 divides both \(p\) and \(q\)

Explanation: From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisor property of a square.

Correct answer: A. Prime divisor property: If a prime divides the square of an integer, it also divides that integer

Explanation: Since 3 is prime and \(p^2=p\times p\), Euclid’s lemma gives \(3\mid p\) from \(3\mid p^2\). Applying the same idea to \(q\) contradicts coprimality. Exam tip: this inference specifically requires the divisor to be prime.

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

Explanation: From \(p^2=3q^2\), \(3\mid p^2\), so \(3\mid p\). Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: state the lowest-terms contradiction clearly.

Correct answer: A. Both \(p\) and \(q\) are even, so they cannot be coprime.

Explanation: From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Putting \(p=2k\) gives \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: if a square is even, its integer root is even.

Correct answer: A. At the beginning they should be taken coprime in lowest form

Explanation: Both even is not the assumption but the contradiction. At the start (\gcd(a,b)=1) should be assumed.

Correct answer: A. At the start (p,q) should be coprime

Explanation: Both being divisible by (3) is derived later as contradiction. It is not assumed at the beginning.

Correct answer: A. On putting \(p=3k\), \(q\) is also proved divisible by 3, contradicting the coprimality of \(p\) and \(q\).

Explanation: Putting \(p=3k\) gives \(9k^2=3q^2\), so \(q^2=3k^2\) and hence \(q\) is also divisible by 3. This contradicts coprimality. In exams, track prime factors in squares carefully.

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

Explanation: The key idea is that an irrationality proof begins by temporarily assuming the opposite of what we want to prove. Here, we assume that \(\sqrt{3}\) is rational. A rational number can be written as \(p/q\), where \(p\) and \(q\) are integers, \(q\ne0\), and the fraction is in lowest terms. The contradiction finally shows that this assumption cannot be true.

Thus, the rejected assumption is “\(\sqrt{3}\) is rational,” which is option C. The proof does not reject the fact that \(\sqrt{3}\) is real; it is certainly a real number. Also, \(\sqrt{3}>0\) and \(q\ne0\) are valid facts or conditions used in the argument, not the assumption being disproved. Therefore the supplied answer is correct.

Correct answer: A. (\frac{a}{b}) can be reduced to (\frac{r}{s}), so the initial lowest form was impossible

Explanation: The direct answer is A. In a proof by contradiction, we first suppose that \\(\\sqrt{2}=a/b\\), where a and b have no common factor and b is not zero. Squaring gives \\(a^2=2b^2\\). This shows that a is even, so write \\(a=2r\\). Substitution gives \\(4r^2=2b^2\\), hence \\(b^2=2r^2\\), so b is also even and can be written \\(b=2s\\). Therefore both a and b have the common factor 2. But that contradicts the original statement that the fraction was already in lowest terms. Equivalently, \\(a/b=(2r)/(2s)=r/s\\), so it could be reduced. Option A expresses this precise conclusion. B is wrong because the argument proves that \\(\\sqrt{2}\\) is not an integer. C is wrong because b is a denominator and must not be zero; the proof does not show b=0. D is wrong because nothing proves a and b are equal. Exam cue: a common factor obtained after assuming lowest form is the contradiction.

Correct answer: A. irrational

Explanation: The side is \(s=\sqrt{2}\) cm because \(s^2=2\). Thus, the perimeter is \(4s=4\sqrt{2}\) cm, which is irrational. \(2\sqrt{2}\) represents only two sides. Exam tip: the square root of a positive integer is rational only when the integer is a perfect square.

Correct answer: C. Non-terminating non-recurring decimal expansion

Explanation: \(\sqrt{3}\) is irrational, so its decimal expansion is non-terminating and non-recurring. Terminating or recurring decimals are always rational. Exam tip: use this decimal property to identify irrational numbers.

Correct answer: A. To ensure that the fraction is in lowest terms and \(p\) and \(q\) have no common factor

Explanation: If \(p^2=3q^2\), then 3 divides \(p\); substituting \(p=3k\) shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: state that the fraction is in lowest terms.

Correct answer: A. If \(1+\sqrt{2}\) were rational, subtracting 1 would make \(\sqrt{2}\) rational, which is impossible.

Explanation: Assume \(1+\sqrt{2}\) is rational. Rational numbers are closed under subtraction, so \((1+\sqrt{2})-1=\sqrt{2}\) would be rational. This contradicts the known irrationality of \(\sqrt{2}\). Exam tip: adding or subtracting a rational number cannot make an irrational number rational.

Correct answer: A. a is divisible by 3

Explanation: From a² = 3b², 3 divides a². Since 3 is prime, it must divide a. Then 3 also divides b, contradicting that a/b is in lowest terms. Exam tip: use the prime-divisibility rule for squares.

Correct answer: B. Because (q) must also be proved divisible by (3)

Explanation: The correct answer is option B: q must also be proved divisible by 3. To prove that √3 is irrational, assume √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest form, so gcd(p,q)=1. Squaring gives p²=3q². Therefore 3 divides p², and hence 3 divides p. Write p=3k. Substitution gives 9k²=3q², so q²=3k². Thus 3 also divides q. Now both p and q have the common factor 3, contradicting gcd(p,q)=1. Merely showing that p is divisible by 3 does not yet contradict coprimality, because p alone may have a factor that q does not have. Option A is wrong because p=0 is not the required contradiction and would not follow appropriately. Option B is correct because both numerator and denominator must share 3. Option C is wrong because p=q is neither required nor obtained. Option D is wrong because q=0 is forbidden from the beginning, not the final contradiction. Memory cue: in irrationality proofs, “both numerator and denominator share a factor” breaks lowest form.