Which option gives the correct complete logical chain for the proof of \(\sqrt{2}\)?
Answer and explanation
Correct answer: Assume rational \(\rightarrow\) \(m^2=2n^2\) \(\rightarrow\) (m) even \(\rightarrow\) (n) even \(\rightarrow\) contradiction
The proof begins by assuming the opposite of what must be shown: suppose \(\sqrt{2}\) is rational and write it as \(m/n\) in lowest form, with \(n\neq0\). Squaring gives \(m^2=2n^2\). This equation shows that \(m^2\), and therefore \(m\), is even. Writing \(m=2k\) and substituting back shows that \(n^2\), and therefore \(n\), is also even.
Thus both \(m\) and \(n\) have 2 as a common factor. That contradicts the assumption that \(m/n\) was in lowest form, meaning the two integers were coprime. This contradiction proves that the original assumption was false. Hence option A gives the correct logical chain; the other options omit the essential algebra and contradiction.
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What is the correct answer to this question?
Assume rational \(\rightarrow\) \(m^2=2n^2\) \(\rightarrow\) (m) even \(\rightarrow\) (n) even \(\rightarrow\) contradiction
Why is this the correct answer?
The proof begins by assuming the opposite of what must be shown: suppose \(\sqrt{2}\) is rational and write it as \(m/n\) in lowest form, with \(n\neq0\). Squaring gives \(m^2=2n^2\). This equation shows that \(m^2\), and therefore \(m\), is even. Writing \(m=2k\) and substituting back shows that \(n^2\), and therefore \(n\), is also even.
Thus both \(m\) and \(n\) have 2 as a common factor. That contradicts the assumption that \(m/n\) was in lowest form, meaning the two integers were coprime. This contradiction proves that the original assumption was false. Hence option A gives the correct logical chain; the other options omit the essential algebra and contradiction.
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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