यदि \(\sqrt{3}\) के प्रमाण में अंत में (a,b) दोनों (3) से विभाज्य मिलते हैं, तो विरोधाभास किस शुरुआती शर्त से है?
If in the proof of \(\sqrt{3}\), both (a,b) are found divisible by (3) at the end, the contradiction is with which initial condition?
Explanation opens after your attempt
Correct Answer
C. (\gcd(a,b)=1)
Concept
In lowest form, (\gcd(a,b)=1) should hold. If both are divisible by (3), (\gcd(a,b)\ge3).
Why this answer is correct
The correct answer is C. (\gcd(a,b)=1). In lowest form, (\gcd(a,b)=1) should hold. If both are divisible by (3), (\gcd(a,b)\ge3).
Exam Tip
सरलतम रूप में (\gcd(a,b)=1) होना चाहिए। दोनों (3) से विभाज्य होने पर (\gcd(a,b)\ge3) होगा।
Login to save your score, XP, coins and progress.