यदि \(\sqrt{2}\) के प्रमाण में अंत में (x,y) दोनों सम मिलते हैं, तो विरोधाभास किस शुरुआती शर्त से है?
If in the proof of \(\sqrt{2}\), both (x,y) are found even at the end, the contradiction is with which initial condition?
Explanation opens after your attempt
Correct Answer
A. (\gcd(x,y)=1)
Concept
In lowest form, (\gcd(x,y)=1) should hold. If both are even, (\gcd(x,y)\ge2).
Why this answer is correct
The correct answer is A. (\gcd(x,y)=1). In lowest form, (\gcd(x,y)=1) should hold. If both are even, (\gcd(x,y)\ge2).
Exam Tip
सरलतम रूप में (\gcd(x,y)=1) होना चाहिए। दोनों सम होने पर (\gcd(x,y)\ge2) हो जाता है।
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