Proof of irrationality of square root 2 and square root 3

Correct answer: A. Because \(p^2=3q^2\) shows that both \(p\), and then \(q\), are divisible by 3.

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: look for a common factor in both terms.

Correct answer: B. Assuming (\sqrt{3}=\frac{p}{q})

Explanation: First (\sqrt{3}) is assumed rational and written as (\frac{p}{q}). Then squaring gives (p^2=3q^2).

Correct answer: A. यदि \(a^2\) 3 से विभाज्य है, तो \(a\) भी 3 से विभाज्य है।

Explanation: Assume \(\sqrt{3}=a/b\) for integers \(a,b\). Then \(a^2=3b^2\), so \(3\mid a^2\), which implies \(3\mid a\). Substitution then gives \(3\mid b\), creating the contradiction. Exam tip: remember that for a prime \(p\), \(p\mid a^2\Rightarrow p\mid a\).

Correct answer: B. Their only common factor is (1)

Explanation: Coprime numbers have no common factor except (1). Both becoming divisible by (3) clashes with this.

Correct answer: A. Assume rational then square then contradiction of both even

Explanation: The direct answer is option A. To prove that \(\sqrt{2}\) is irrational, begin by assuming the opposite: suppose \(\sqrt{2}\) is rational. Write it in lowest form as \(\sqrt{2}=p/q\), where \(p,q\) are integers, \(q\neq0\), and they have no common factor. Squaring gives \(2=p^2/q^2\), so \(p^2=2q^2\). Thus \(p^2\) is even, so \(p\) is even; write \(p=2k\). Substitution gives \(q^2=2k^2\), so \(q\) is also even. This contradicts the lowest-form condition. Therefore \(\sqrt{2}\) is irrational. Option A gives this correct short order. Option B is wrong because drawing a picture does not prove irrationality. Option C is wrong because there is no reason to assume the square root is zero. Option D is wrong because a decimal approximation or guess cannot be a proof. Memory cue: assume rational, square, show both even, contradict lowest form.

Correct answer: A. Because the eventual result that both \(p\) and \(q\) are divisible by 3 contradicts this condition

Explanation: From \(3q^2=p^2\), \(p\) must be divisible by 3. Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: always assume the fraction is in lowest terms.

Correct answer: A. अनंत अनावर्ती दशमलव प्रसार वाली संख्या अपरिमेय होती है।

Explanation: A rational number has either a terminating or a non-terminating recurring decimal expansion. For example, \(\frac{1}{3}=0.333\ldots\) repeats, whereas \(\sqrt{2}=1.414\ldots\) is non-repeating and irrational. Exam tip: look carefully for the word “recurring.”

Correct answer: B. \(p\) सम होने पर \(p=2k\) रखने से \(q^2=2k^2\) मिलता है, इसलिए \(q\) भी सम है।

Explanation: Write \(p=2k\) because \(p\) is even. Then \(4k^2=2q^2\), so \(q^2=2k^2\) and \(q\) is also even. This contradicts coprimality. Exam tip: an even square has an even base.

Correct answer: A. Odd

Explanation: For integers, evenness of the square is linked with evenness of the number. If the square is not even, the number is not even.

Correct answer: B. (3)

Explanation: Prime factor (3) appears in the square only when it appears in the number. This supports the proof of (\sqrt{3}).

Correct answer: A. यदि \(\sqrt{3}=\frac{p}{q}\) सरलतम रूप में हो, तो \(p\) और \(q\) दोनों 3 से विभाज्य सिद्ध होते हैं।

Explanation: In proof by contradiction, assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), first \(p\), then \(q\), is divisible by 3, contradicting lowest terms. Exam tip: common divisibility by the same prime signals the contradiction.

Correct answer: A. 1.732 is only an approximation of \(\sqrt{3}\), not its exact value.

Explanation: 1.732 is not the exact value of \(\sqrt{3}\); it is only an approximation. Check: \(1.732^2=2.999824\), not 3. The decimal expansion of \(\sqrt{3}\) is non-terminating and non-repeating. In exams, do not treat an approximate value as an exact value.

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

Explanation: From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), making \(q\) divisible by 3 too. This contradicts coprimality. Exam tip: use prime divisibility of squares carefully.

Correct answer: A. यदि किसी पूर्णांक का वर्ग सम है, तो वह पूर्णांक भी सम होता है।

Explanation: Assume \(\sqrt{2}=p/q\), where \(p\) and \(q\) are coprime. Then \(p^2=2q^2\), so \(p^2\) is even and hence \(p\) is even. This also makes \(q\) even, contradicting coprimality. Exam tip: remember that an even square implies an even integer.

Correct answer: A. \(p=3k\) रखने पर \(q\) भी 3 से विभाज्य सिद्ध होता है।

Explanation: Since 3 divides \(p\), write \(p=3k\). Substituting in \(p^2=3q^2\) gives \(9k^2=3q^2\), so \(q^2=3k^2\); hence 3 also divides \(q\). This contradicts lowest form. Exam tip: track prime factors after squaring.

Correct answer: B. So contradiction with coprime assumption can be shown

Explanation: In the standard contradiction proof, assume that \(\sqrt{3}=p/q\), where \(p\) and \(q\) are integers and the fraction is in lowest form. Lowest form means that \(p\) and \(q\) have no common factor. Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), and therefore 3 divides \(p\). Writing \(p=3k\) then shows that 3 also divides \(q\).

That conclusion contradicts the original statement that \(p/q\) was in lowest form. The contradiction proves that \(\sqrt{3}\) cannot be rational. Thus option B is correct. Without the coprime assumption, finding a common factor would not create a contradiction, so taking lowest form is essential. The supplied explanation is accurate.

Correct answer: B. 1.73 is only a rational approximation; the decimal expansion of \(\sqrt{3}\) is non-terminating and non-repeating.

Explanation: 1.73 is only \(\sqrt{3}\) rounded to two decimal places, not its exact value. Since \(\sqrt{3}\approx1.732\ldots\) is non-terminating and non-repeating, it is irrational. Exam tip: never infer rationality from a rounded decimal.

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

Explanation: Since \(1^2<2<2^2\), 2 is not a perfect square, so \(\sqrt{2}\) is irrational. In contrast, \(\sqrt{4}=2\) is rational. Exam tip: first check whether the radicand is a perfect square.

Correct answer: A. In lowest fraction form

Explanation: To prove that a square root such as \(\sqrt{2}\) or \(\sqrt{3}\) is irrational, the proof begins by assuming the opposite: that the number is rational. Every rational number can be written as a fraction \(\frac{m}{n}\), where m and n are integers, n is nonzero, and the fraction is in lowest terms. The lowest-terms condition means m and n have no common factor.

This form is essential because the later argument shows that both m and n must be divisible by the same number, usually 2 for \(\sqrt{2}\) or 3 for \(\sqrt{3}\). That contradicts their being coprime. A decimal or percentage form does not provide this useful coprime condition. Therefore option A, the simplest fraction form, is correct.

Correct answer: A. निष्कर्ष सही है, पर कथन पूर्ण प्रमाण नहीं है; विरोधाभास द्वारा उचित तर्क देना आवश्यक है।

Explanation: \(\sqrt{3}\) is indeed irrational, but merely saying that 3 is not a perfect square is not a complete formal proof. Assume \(\sqrt{3}=p/q\); then \(p^2=3q^2\), so both \(p\) and \(q\) are divisible by 3, a contradiction. Exam tip: state that \(p,q\) are coprime.

Correct answer: A. Rational number

Explanation: A number expressible as \(\frac{p}{q}\), with integers \(p,q\) and \(q\neq0\), is rational. Coprime \(p,q\) indicate lowest terms; not every rational number is whole or natural. Exam tip: the denominator can never be zero.

Correct answer: A. To show they cannot be written as lowest rational fractions

Explanation: An irrational number cannot be written as a ratio of two integers. Both proofs establish this fact.

Correct answer: A. \(a^2\) is even

Explanation: In \(a^2=2b^2\), the right-hand side is a multiple of \(2\), so \(a^2\) is even. The next step is to conclude that \(a\) is also even. We cannot conclude at this stage that \(b\) is odd. Exam tip: In such proofs, first compare the parity of both sides of the equation.

Correct answer: A. \(q\) is also even

Explanation: If \(p=2k\), then \(4k^2=2q^2\), so \(q^2=2k^2\) and \(q\) is even too. Thus both numerator and denominator have factor 2, contradicting coprimality. Exam tip: use the fact that an even square has an even root.

Correct answer: B. (\sqrt{2}) is rational

Explanation: In contradiction method we first assume the opposite that (\sqrt{2}) is rational. Later this assumption is shown false.