Which option gives the most correct full logical chain in the proof of \(\sqrt{2}\)?
Answer and explanation
Correct answer: मान लें \(\sqrt{2}=\frac{a}{b}\), जहाँ \(a,b\) सह-अभाज्य हैं \(\rightarrow a^2=2b^2\rightarrow a\) सम \(\rightarrow b\) सम \(\rightarrow\) सह-अभाज्य होने के विरुद्ध
Option A contains every essential step of the proof by contradiction. Assume \(\sqrt{2}=\frac{a}{b}\) in lowest terms. Then \(a^2=2b^2\), so \(a^2\), and hence \(a\), is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is also even. Thus both \(a\) and \(b\) are divisible by 2, contradicting that they are coprime. Option D is wrong because an infinite decimal alone does not prove irrationality; rational numbers may have infinite recurring decimals. Exam tip: always begin this proof by taking the fraction in lowest terms.
Frequently asked questions
What is the correct answer to this question?
मान लें \(\sqrt{2}=\frac{a}{b}\), जहाँ \(a,b\) सह-अभाज्य हैं \(\rightarrow a^2=2b^2\rightarrow a\) सम \(\rightarrow b\) सम \(\rightarrow\) सह-अभाज्य होने के विरुद्ध
Why is this the correct answer?
Option A contains every essential step of the proof by contradiction. Assume \(\sqrt{2}=\frac{a}{b}\) in lowest terms. Then \(a^2=2b^2\), so \(a^2\), and hence \(a\), is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is also even. Thus both \(a\) and \(b\) are divisible by 2, contradicting that they are coprime. Option D is wrong because an infinite decimal alone does not prove irrationality; rational numbers may have infinite recurring decimals. Exam tip: always begin this proof by taking the fraction in lowest terms.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Proof of irrationality of square root 2 and square root 3.
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