If (\sqrt{3}=\frac{u}{v}) is in lowest form, why is getting (3\mid u) and (3\mid v) a decisive contradiction?
Answer and explanation
Correct answer: Because (\gcd(u,v)\ge3) will hold
A fraction in lowest form has no common factor greater than 1 between its numerator and denominator. Therefore, if \(\sqrt{3}=u/v\) is assumed to be in lowest form, then \(\gcd(u,v)=1\). The proof begins with \(u^2=3v^2\). Since 3 divides \(u^2\) and is prime, 3 divides \(u\). Substituting this fact back into the equation then proves that 3 divides \(v\) too.
Consequently, both \(u\) and \(v\) are divisible by 3. Their greatest common divisor must therefore be at least 3, so \(\gcd(u,v)\ge3\). This directly contradicts \(\gcd(u,v)=1\), the lowest-form condition. The issue is not that \(v=0\), that \(u=v\), or that the root becomes an integer. Hence option A is decisive.
Frequently asked questions
What is the correct answer to this question?
Because (\gcd(u,v)\ge3) will hold
Why is this the correct answer?
A fraction in lowest form has no common factor greater than 1 between its numerator and denominator. Therefore, if \(\sqrt{3}=u/v\) is assumed to be in lowest form, then \(\gcd(u,v)=1\). The proof begins with \(u^2=3v^2\). Since 3 divides \(u^2\) and is prime, 3 divides \(u\). Substituting this fact back into the equation then proves that 3 divides \(v\) too.
Consequently, both \(u\) and \(v\) are divisible by 3. Their greatest common divisor must therefore be at least 3, so \(\gcd(u,v)\ge3\). This directly contradicts \(\gcd(u,v)=1\), the lowest-form condition. The issue is not that \(v=0\), that \(u=v\), or that the root becomes an integer. Hence option A is decisive.
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.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
