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In a proof by contradiction for the irrationality of \(\sqrt{3}\), which conclusion is necessary to proceed after obtaining \(a^2=3b^2\)?\(\sqrt{3}\) की अपरिमेयता के विरोधाभास-प्रमाण में \(a^2=3b^2\) मिलने पर प्रमाण को आगे बढ़ाने के लिए कौन-सा निष्कर्ष आवश्यक है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(3\mid a\)\(3\mid a\)
Since 3 is prime, \(3\mid a^2\) implies \(3\mid a\). Put \(a=3k\); then \(b^2=3k^2\), so \(3\mid b\), contradicting coprimality. Exam tip: use prime divisibility carefully.चूँकि 3 अभाज्य है, \(3\mid a^2\) से \(3\mid a\) मिलता है। \(a=3k\) रखने पर \(b^2=3k^2\), अतः \(3\mid b\) भी होगा, जो सह-अभाज्यता के विरुद्ध है। परीक्षा टिप: अभाज्य गुणनखंड देखें।
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