In the proof of \(\sqrt{3}\), which conclusion is rejected by the rational assumption?
Answer and explanation
Correct answer: \(\sqrt{3}\) is rational
In proof by contradiction, assume that \(\sqrt{3}=p/q\), where \(p\) and \(q\) are coprime integers. On squaring, \(3q^2=p^2\), which shows that \(p\), and then \(q\), are both divisible by 3. This contradicts their being coprime. Hence the assumption that \(\sqrt{3}\) is rational is rejected. \(\sqrt{3}\) is still positive and real; those facts are not rejected by the contradiction. Exam tip: when a contradiction is obtained, reject the initial assumption used in the proof.
Frequently asked questions
What is the correct answer to this question?
\(\sqrt{3}\) is rational
Why is this the correct answer?
In proof by contradiction, assume that \(\sqrt{3}=p/q\), where \(p\) and \(q\) are coprime integers. On squaring, \(3q^2=p^2\), which shows that \(p\), and then \(q\), are both divisible by 3. This contradicts their being coprime. Hence the assumption that \(\sqrt{3}\) is rational is rejected. \(\sqrt{3}\) is still positive and real; those facts are not rejected by the contradiction. Exam tip: when a contradiction is obtained, reject the initial assumption used in the proof.
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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