Which statement can be a wrong but tempting answer in the proof of (\sqrt{3})?
Answer and explanation
Correct answer: From (p^2=3q^2), (p=3q)
The tempting mistake is to treat a square equation as though taking square roots preserved the same linear relationship. From \(p^2=3q^2\), it is not valid to conclude directly that \(p=3q\). Taking square roots formally would suggest a factor of \(\sqrt{3}\), not 3, so that proposed step has no logical basis.
The correct argument uses divisibility. Since \(3\mid p^2\) and 3 is prime, it follows that \(3\mid p\). Write \(p=3r\) and substitute into the equation. Then \(9r^2=3q^2\), so \(q^2=3r^2\), which gives \(3\mid q\) as well. Thus both numerator and denominator share 3. Option A is the attractive but wrong statement.
Frequently asked questions
What is the correct answer to this question?
From (p^2=3q^2), (p=3q)
Why is this the correct answer?
The tempting mistake is to treat a square equation as though taking square roots preserved the same linear relationship. From \(p^2=3q^2\), it is not valid to conclude directly that \(p=3q\). Taking square roots formally would suggest a factor of \(\sqrt{3}\), not 3, so that proposed step has no logical basis.
The correct argument uses divisibility. Since \(3\mid p^2\) and 3 is prime, it follows that \(3\mid p\). Write \(p=3r\) and substitute into the equation. Then \(9r^2=3q^2\), so \(q^2=3r^2\), which gives \(3\mid q\) as well. Thus both numerator and denominator share 3. Option A is the attractive but wrong statement.
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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