Which option is a wrong start in the proof of \(\sqrt{2}\)?
Answer and explanation
Correct answer: Assuming \(\sqrt{2}=\frac{m}{0}\)
To prove that \(\sqrt{2}\) is irrational by contradiction, we first assume it is rational and write \(\sqrt{2}=\frac{m}{n}\), where \(m,n\) are coprime and \(n\ne0\). The expression \(\frac{m}{0}\) is undefined, so option C is an invalid start. In option A, the denominator is non-zero, so it is a valid assumption. Exam tip: whenever a rational number is written as \(\frac{p}{q}\), check that \(q\ne0\).
Frequently asked questions
What is the correct answer to this question?
Assuming \(\sqrt{2}=\frac{m}{0}\)
Why is this the correct answer?
To prove that \(\sqrt{2}\) is irrational by contradiction, we first assume it is rational and write \(\sqrt{2}=\frac{m}{n}\), where \(m,n\) are coprime and \(n\ne0\). The expression \(\frac{m}{0}\) is undefined, so option C is an invalid start. In option A, the denominator is non-zero, so it is a valid assumption. Exam tip: whenever a rational number is written as \(\frac{p}{q}\), check that \(q\ne0\).
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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