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Which property is used decisively in the proof by contradiction that \(\sqrt{3}\) is irrational?\(\sqrt{3}\) की अपरिमेयता के विरोधाभास-प्रमाण में कौन-सा गुण निर्णायक रूप से प्रयुक्त होता है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: यदि \(3\mid n^2\), तो \(3\mid n\)।यदि \(3\mid n^2\), तो \(3\mid n\)।
From \(p^2=3q^2\), \(3\mid p^2\); since 3 is prime, \(3\mid p\). Put \(p=3k\) to obtain \(3\mid q\), a contradiction. C fails at \(n=3\). Tip: use the prime-divisor rule.\(p^2=3q^2\) से \(3\mid p^2\), अतः अभाज्य 3 के कारण \(3\mid p\)। \(p=3k\) रखने पर \(3\mid q\) भी आता है, जो विरोधाभास है। C गलत है: \(n=3\)। टिप: अभाज्य-विभाजक गुण लगाएँ।
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