In the proof of (√2), if (m) is assumed odd, what problem arises from (m²=2n²)?
Answer and explanation
Correct answer: (m²) should be odd but the equation gives even
The relevant parity rule is that the square of an odd integer is odd. If m is assumed odd, then m² must be odd. However, the equation m² = 2n² has an even right-hand side because it is two times an integer square. It therefore forces m² to be even. An integer cannot be both odd and even, so the assumption that m is odd creates the contradiction. This parity observation supports the usual proof: m² is even, hence m is even, and later the equation shows that n is even too. Option A accurately states the problem. The other options are unrelated conclusions and do not follow from the equation.
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What is the correct answer to this question?
(m²) should be odd but the equation gives even
Why is this the correct answer?
The relevant parity rule is that the square of an odd integer is odd. If m is assumed odd, then m² must be odd. However, the equation m² = 2n² has an even right-hand side because it is two times an integer square. It therefore forces m² to be even. An integer cannot be both odd and even, so the assumption that m is odd creates the contradiction. This parity observation supports the usual proof: m² is even, hence m is even, and later the equation shows that n is even too. Option A accurately states the problem. The other options are unrelated conclusions and do not follow from the equation.
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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