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Which initial assumption is made to prove the irrationality of \(\sqrt{2}\) by the method of contradiction?विरोधाभास विधि से \(\sqrt{2}\) की अपरिमेयता सिद्ध करने के लिए कौन-सी प्रारम्भिक मान्यता ली जाती है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\sqrt{2}=\frac{p}{q}\), where \(p,q\) are coprime integers and \(q\ne0\)\(\sqrt{2}=\frac{p}{q}\), जहाँ \(p,q\) सह-अभाज्य पूर्णांक हैं और \(q\ne0\)
Assume \(\sqrt{2}=p/q\) with coprime integers. Squaring gives \(2q^2=p^2\); this eventually makes both integers even, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.मानते हैं कि \(\sqrt{2}=p/q\), जहाँ \(p,q\) सह-अभाज्य हैं। वर्ग करने पर \(2q^2=p^2\) मिलता है; आगे दोनों सम निकलते हैं, जो सह-अभाज्यता के विरुद्ध है। परीक्षा में ‘सरलतम रूप’ अवश्य लिखें।
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