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In a proof by contradiction that \(\sqrt{3}\) is irrational, what fact produces the contradiction after assuming \(\sqrt{3}=p/q\), where \(p\) and \(q\) are coprime integers?यदि
the proof of
the irrationality of
the square root of 3 begins by assuming
the fraction
the fraction
?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(p\) and \(q\) are both divisible by 3, so they are not coprime\(p\) और \(q\) दोनों 3 से विभाज्य हैं, इसलिए वे सह-अभाज्य नहीं हैं
Assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), \(p\) is divisible by 3, and then \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: state “lowest terms” first.मान लें \(\sqrt{3}=p/q\), जहाँ \(p,q\) सह-अभाज्य हैं। \(p^2=3q^2\) से \(p\), और फिर \(q\), 3 से विभाज्य मिलता है। यह सह-अभाज्य होने का विरोधाभास है। परीक्षा में यह शर्त अवश्य लिखें।
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