While proving the irrationality of \(\sqrt{3}\), a student assumes \(\sqrt{3}=\frac{a}{b}\), where \(a\) and \(b\) are coprime. After obtaining \(a^2=3b^2\), the student states that both \(a\) and \(b\) are divisible by 3. Which reasoning is necessary to justify this conclusion?
Answer and explanation
Correct answer: First use \(3\mid a^2\) to write \(a=3k\), then substitute to get \(b^2=3k^2\) and prove \(3\mid b\).
From \(a^2=3b^2\), \(3\mid a^2\), so prime-factor reasoning gives \(3\mid a\). Put \(a=3k\) to obtain \(b^2=3k^2\), hence \(3\mid b\), contradicting coprimality. Exam tip: always show the substitution step.
Frequently asked questions
What is the correct answer to this question?
First use \(3\mid a^2\) to write \(a=3k\), then substitute to get \(b^2=3k^2\) and prove \(3\mid b\).
Why is this the correct answer?
From \(a^2=3b^2\), \(3\mid a^2\), so prime-factor reasoning gives \(3\mid a\). Put \(a=3k\) to obtain \(b^2=3k^2\), hence \(3\mid b\), contradicting coprimality. Exam tip: always show the substitution step.
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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