If (a) is assumed odd in (a^2=2b^2), what contradiction appears?
Answer and explanation
Correct answer: \(a^2\) is odd, whereas \(2b^2\) is even
If \(a\) is odd, then its square \(a^2\) is also odd. On the other hand, \(2b^2\) is even for every integer value of \(b\), since it has 2 as a factor. Thus, the equation would make the same number both odd and even, which is the contradiction. Option B is not valid because the equation does not prove that \(b^2\) is odd. Exam tip: the square of an odd integer is odd, and the square of an even integer is even.
Frequently asked questions
What is the correct answer to this question?
\(a^2\) is odd, whereas \(2b^2\) is even
Why is this the correct answer?
If \(a\) is odd, then its square \(a^2\) is also odd. On the other hand, \(2b^2\) is even for every integer value of \(b\), since it has 2 as a factor. Thus, the equation would make the same number both odd and even, which is the contradiction. Option B is not valid because the equation does not prove that \(b^2\) is odd. Exam tip: the square of an odd integer is odd, and the square of an even integer is even.
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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