If (\sqrt{2}=\frac{p}{q}) is assumed in lowest form and (p^2=2q^2) is obtained, which sequence is most logical to reach the contradiction?
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Answer and explanation
Correct answer: (p^2) even, then (p) even, then (p=2k), then (q) even
Step 1: From (p^2=2q^2), (p^2) is even, so (p) is even. Step 2: Putting (p=2k) gives (q^2=2k^2), so (q) is even. Step 3: Both being even contradicts coprimality of the lowest-form fraction.
Frequently asked questions
What is the correct answer to this question?
(p^2) even, then (p) even, then (p=2k), then (q) even
Why is this the correct answer?
Step 1: From (p^2=2q^2), (p^2) is even, so (p) is even. Step 2: Putting (p=2k) gives (q^2=2k^2), so (q) is even. Step 3: Both being even contradicts coprimality of the lowest-form fraction.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Proof of irrationality of √2, √3, √5.
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