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Which prime-divisibility property is crucial in a proof by contradiction that \(\sqrt{3}\) is irrational?\(\sqrt{3}\) की अपरिमेयता के विरोधाभास-प्रमाण में कौन-सी अभाज्य विभाज्यता-विशेषता निर्णायक होती है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: If \(3\mid k^2\), then \(3\mid k\).यदि \(3\mid k^2\), तो \(3\mid k\)।
Let \(\sqrt{3}=a/b\) in lowest terms. Then \(a^2=3b^2\), so \(3\mid a^2\); since 3 is prime, \(3\mid a\). Option C is unnecessarily strong because \(9\mid a\) need not follow. Exam tip: remember that if prime \(p\mid k^2\), then \(p\mid k\).मान लें \(\sqrt{3}=a/b\), जहाँ \(a,b\) सहाभाज्य हैं। तब \(a^2=3b^2\), इसलिए \(3\mid a^2\) और 3 अभाज्य होने से \(3\mid a\)। विकल्प C में \(9\mid a\) कहना जरूरी नहीं है। परीक्षा टिप: \(p\mid k^2\Rightarrow p\mid k\) नियम याद रखें।
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