Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
Proof of irrationality of square root 2 and square root 3
√2 और √3 की अपरिमेयता का प्रमाण
In Class 9 Mathematics, this Number Systems topic explains how to prove that √2 and √3 are irrational numbers. Students use proof by contradiction: they assume a square root can be written as a fraction in lowest terms, then apply prime-factor and divisibility properties to show that the assumption leads to an impossibility. The lesson strengthens understanding of rational and irrational numbers, factors, parity, and the logic of mathematical proof, while helping learners present each step clearly and accurately.
Practice questions
01 A student claims that \(\sqrt{3}\) is rational and writes it as \(\frac{p}{q}\) in lowest terms, where \(p,q\) are coprime. If \(p^2=3q^2\), what is the error in this claim?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: B. Both \(p\) and \(q\) become divisible by 3, contradicting that they are coprime.
Explanation: Since \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\): \(9k^2=3q^2\), hence \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: if a prime divides a square, it divides the number.
02 A student claims that \(\sqrt{2}+\sqrt{3}\) is a rational number. Which argument correctly identifies the error in the claim?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. यदि \(\sqrt{2}+\sqrt{3}=r\), जहाँ \(r\) परिमेय है, तो \(\sqrt{3}=r-\sqrt{2}\) होगा; इसलिए \(\sqrt{3}\) परिमेय है।
Explanation: Assume \(\sqrt{2}+\sqrt{3}=r\), with \(r\) rational. Option A is not a valid conclusion because rational minus irrational can still be irrational. Squaring gives \(r^2=5+2\sqrt{6}\), so \(\sqrt{6}=(r^2-5)/2\) would be rational, a contradiction. Exam tip: square a sum of surds to isolate the remaining radical.
04 If a student assumes that \(\sqrt{2}+\sqrt{3}\) is a rational number \(r\), which conclusion correctly proves a contradiction in this claim?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: C. \(\sqrt{2}=\dfrac{r^2-1}{2r}\), so \(\sqrt{2}\) would be rational
Explanation: Assume \(r=\sqrt{2}+\sqrt{3}\). Then \(\sqrt{3}=r-\sqrt{2}\); on squaring, \(3=r^2+2-2r\sqrt{2}\). Hence \(\sqrt{2}=(r^2-1)/(2r)\) would be rational, a contradiction. Exam tip: isolate the radical carefully after squaring.
05 A student says that \(\sqrt{3}\) is irrational because its decimal expansion \(1.732\ldots\) continues endlessly. What is the main error in this argument?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. Not every non-terminating decimal is irrational; it may be recurring.
Explanation: A non-terminating decimal alone does not prove irrationality: \(\frac{1}{3}=0.333\ldots\) goes on forever but is rational. An irrational number has a non-terminating, non-recurring decimal expansion. Exam tip: distinguish recurring from non-recurring decimals.
06 In the proof by contradiction that \(\sqrt{3}\) is irrational, which property is used to conclude \(3\mid p\) from \(3\mid p^2\)?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: C. If \(3\mid p^2\), then \(3\mid p\).
Explanation: Every prime factor in \(p^2\) has an even exponent. Thus, if \(3\mid p^2\), the factor 3 must already occur in \(p\), so \(3\mid p\). It need not imply \(9\mid p\). Exam tip: use prime-factor exponents in irrationality proofs.
07 Why is it necessary to take the fraction \(\frac{p}{q}\) in lowest terms in the contradiction proof that \(\sqrt{2}\) is irrational?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. To ensure that \(p\) and \(q\) are not both even
Explanation: In lowest terms, \(p\) and \(q\) have no common factor. From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even, giving a contradiction. Exam tip: link lowest terms with coprime numerator and denominator.
08 A student assumes that \(\sqrt{3}=\frac{p}{q}\), where \(p\) and \(q\) are coprime. On squaring, the student gets \(p^2=3q^2\) and concludes that \(p\) is divisible by 3. What is the next essential step to complete the proof?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. Show that \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime.
Explanation: From \(p^2=3q^2\), let \(p=3k\). Substitution gives \(9k^2=3q^2\), so \(q^2=3k^2\) and \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: establish divisibility of both numerator and denominator.
09 If \(x=\sqrt{2}+\sqrt{3}\), which argument correctly disproves the assumption that \(x\) is rational?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. Then \(1/x=\sqrt{3}-\sqrt{2}\) would be rational; adding them gives \(2\sqrt{3}\) rational, which is impossible.
Explanation: Since \((\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})=1\), a rational \(x\) would have a rational reciprocal. Their sum would make \(2\sqrt{3}\) rational, a contradiction. Exam tip: first check the product of conjugates.
10 Rima says, “\(\sqrt{3}=1.732\); therefore, \(\sqrt{3}\) is rational.” What is the main error in her reasoning?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. She has treated an approximation as an exact value.
Explanation: \(1.732\) is only an approximation of \(\sqrt{3}\), not its exact value, since \(1.732^2=2.999824\), not 3. As 3 is not a perfect square, \(\sqrt{3}\) is irrational. Exam tip: always distinguish \(=\) from \(\approx\).
11 While proving the irrationality of \(\sqrt{2}\) by contradiction, if \(\sqrt{2}=\frac{p}{q}\) where \(p\) and \(q\) are coprime, which conclusion necessarily follows from \(p^2=2q^2\)?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(p\) is even
Explanation: Since \(p^2=2q^2\), \(p^2\) is even. The square of an integer is even only when the integer itself is even, so \(p\) is even. Substitution then makes \(q\) even too, contradicting coprimality. Exam tip: use parity of squares carefully.
12 A student claims that \(\sqrt{3}\) is rational because its decimal form is 1.732 and \(1.732=\frac{1732}{1000}\). What is the error in the student’s reasoning?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. 1.732 is only an approximation of \(\sqrt{3}\), not its exact value.
Explanation: 1.732 is a terminating decimal approximation. In fact, \(1.732^2=2.999824\), not 3. Thus \(\frac{1732}{1000}\) is rational, but it is not equal to \(\sqrt{3}\). Exam tip: distinguish an approximate decimal from an exact value.
13 Which condition guarantees that the square root of a natural number \(n\) is rational?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(n\) is a perfect square
Explanation: If \(n=k^2\) for an integer \(k\), then \(\sqrt{n}=k\), which is rational. Being even or composite alone is not enough; for example, \(\sqrt{6}\) is irrational. Exam tip: in prime factorisation, all exponents must be even for a perfect square.
14 In the standard proof by contradiction, suppose that \(\sqrt{3}=\frac{p}{q}\), where \(p\) and \(q\) are coprime integers. From \(p^2=3q^2\), which deduction is essential for obtaining the contradiction?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(3\mid p\)
Explanation: From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, if it divides the square of an integer, it divides that integer; hence \(3\mid p\). This later gives \(3\mid q\), contradicting coprimality. Exam tip: explicitly state the prime-divisor property.
15 Reema says that if \(\sqrt{2}=\frac{p}{q}\), where \(p\) and \(q\) are coprime integers, then \(p^2=2q^2\) only shows that \(p\) is even; therefore \(\sqrt{2}\) is not proved irrational. Which essential point is missing from Reema's argument?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. If \(p\) is even, write \(p=2k\); then \(q\) is also even, contradicting that \(p\) and \(q\) are coprime.
Explanation: If \(p=2k\), then \(4k^2=2q^2\), so \(q^2=2k^2\) and hence \(q\) is even too. Thus both have factor 2, contradicting coprimality. In exams, state this contradiction explicitly.
16 In the proofs of √2 and √3, what should not be treated as the basis of proof?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. Approximate decimal value
Explanation: A mathematical proof requires an exact chain of justified statements, not merely numerical evidence. An approximate decimal value can suggest that √2 or √3 is not an integer, but any finite approximation can be close to a rational number and therefore cannot establish irrationality. The rigorous proofs assume a lowest-term rational fraction and then use equations such as x² = 2y² or h² = 3k² together with prime divisibility to obtain a contradiction. Thus option A is correct: the approximate decimal is useful for intuition, not as the logical basis. Options B, C, and D are essential structural components of the standard proofs.
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy
