In the proof of (\sqrt{3}), what is the role of (n\neq0), and where does the final contradiction come from?
Answer and explanation
Correct answer: (n\neq0) keeps the fraction defined, contradiction comes from (\gcd(m,n)=1)
The condition \(n\ne0\) is necessary because a fraction with denominator zero is not defined. When proving irrationality, we assume \(\sqrt{3}=m/n\) with integers \(m,n\), where \(n\ne0\), and choose the fraction in lowest form. Squaring gives \(m^2=3n^2\), which leads to divisibility by 3.
The divisibility argument first shows that 3 divides \(m\). Writing \(m=3r\) and substituting gives \(n^2=3r^2\), so 3 also divides \(n\). Thus both integers have a common factor 3, contradicting \(\gcd(m,n)=1\). The nonzero-denominator condition does not imply \(m=n\), \(n=0\), or \(\sqrt{3}=3\). Therefore option A is correct.
Frequently asked questions
What is the correct answer to this question?
(n\neq0) keeps the fraction defined, contradiction comes from (\gcd(m,n)=1)
Why is this the correct answer?
The condition \(n\ne0\) is necessary because a fraction with denominator zero is not defined. When proving irrationality, we assume \(\sqrt{3}=m/n\) with integers \(m,n\), where \(n\ne0\), and choose the fraction in lowest form. Squaring gives \(m^2=3n^2\), which leads to divisibility by 3.
The divisibility argument first shows that 3 divides \(m\). Writing \(m=3r\) and substituting gives \(n^2=3r^2\), so 3 also divides \(n\). Thus both integers have a common factor 3, contradicting \(\gcd(m,n)=1\). The nonzero-denominator condition does not imply \(m=n\), \(n=0\), or \(\sqrt{3}=3\). Therefore option A is correct.
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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