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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.
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Expert · Level 18 · number systems, irrational numbers, proof by contradiction, square roots, conjugates, class 9 mathematicsView options
Then \(1/x=\sqrt{3}-\sqrt{2}\) would be rational; adding them gives \(2\sqrt{3}\) rational, which is impossible.
\(x^2=5\), so \(x\) is rational.
Every square root is irrational, so \(x\) is irrational.
\(\sqrt{3}-\sqrt{2}\) is irrational, so the reciprocal of \(x\) must also be irrational.
Expert · Level 18 · number systems, irrational numbers, square roots, proof of irrationality, decimal approximationView options
She has treated an approximation as an exact value.
She has not written the decimal number as a fraction.
She has chosen the positive square root although a negative value could also be taken.
She has not checked whether 3 is a perfect square.
Expert · Level 18 · number systems, irrational numbers, proof by contradiction, square root 2, parity, coprime integersView options
\(p\) is even
\(q\) is odd
\(p\) is prime
\(p=q\)
Expert · Level 18 · number systems,irrational numbers,square root 3,proof of irrationality,decimal approximation,common misconceptionsView options
1.732 is only an approximation of \(\sqrt{3}\), not its exact value.
\(\frac{1732}{1000}\) is not a rational number.
Every terminating decimal is irrational.
The exact value of \(\sqrt{3}\) is 1.732.
Expert · Level 18 · number systems, irrational numbers, square roots, perfect squares, proof of irrationalityView options
\(n\) is a perfect square
\(n\) is an even number
\(n\) is a prime number
\(n\) is a composite number
Expert · Level 18 · number systems, irrational numbers, proof by contradiction, square root 3, coprime integersView options
\(3\mid p\)
\(q\mid p\)
\(p\) and \(q\) are consecutive integers
\(p\) and \(q\) are both prime
Expert · Level 18 · number systems, irrational numbers, proof by contradiction, square root 2, coprime integersView options
If \(p\) is even, write \(p=2k\); then \(q\) is also even, contradicting that \(p\) and \(q\) are coprime.
If \(p\) is even, then \(q\) is odd, so no contradiction arises.
From \(p^2=2q^2\), both \(p\) and \(q\) are proved to be prime numbers.
From \(p^2=2q^2\), we get \(p=q\), so \(\sqrt{2}=1\).
Expert · Level 18 · number-systems,irrationality,proof-method,Proof of irrationality of square root 2 and square root 3,Number Systems,Mathematics,Class 9 MCQView options
Approximate decimal value
Lowest-term fraction
Prime divisibility
Method of contradiction
Question 1ExpertLevel 18
If \(x=\sqrt{2}+\sqrt{3}\), which argument correctly disproves the assumption that \(x\) is rational?
Correct answer: A
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.
Rima says, “\(\sqrt{3}=1.732\); therefore, \(\sqrt{3}\) is rational.” What is the main error in her reasoning?
Correct answer: A
\(1.732\) 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 \(\approx\).
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 \(p^2=2q^2\)?
Correct answer: A
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.
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?
Correct answer: A
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.
Which condition guarantees that the square root of a natural number \(n\) is rational?
Correct answer: A
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.
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?
Correct answer: A
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.
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?
Correct answer: A
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.
In the proofs of √2 and √3, what should not be treated as the basis of proof?
Correct answer: A
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.
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