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Suppose \(r=\sqrt{2}+\sqrt{3}\) is a rational number. Which conclusion from this assumption proves by contradiction that \(\sqrt{2}+\sqrt{3}\) is irrational?मान लीजिए कि \(r=\sqrt{2}+\sqrt{3}\) एक परिमेय संख्या है। इस मान्यता से प्राप्त कौन-सा निष्कर्ष विरोधाभास सिद्ध करता है कि \(\sqrt{2}+\sqrt{3}\) अपरिमेय है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(r^2=5+2\sqrt{6}\), so \(\sqrt{6}\) would be rational, which is impossible\(r^2=5+2\sqrt{6}\), अतः \(\sqrt{6}\) परिमेय होगा, जो असंभव है
If \(r\) were rational, \(r^2\) would also be rational. From \(r^2=5+2\sqrt6\), \(\sqrt6=(r^2-5)/2\) would be rational, impossible because 6 is not a perfect square. Exam tip: retain the cross term \(2\sqrt6\).परिमेय \(r\) के लिए \(r^2\) भी परिमेय है। \(r^2=5+2\sqrt6\) से \(\sqrt6=(r^2-5)/2\) परिमेय होगा, जबकि 6 पूर्ण वर्ग नहीं है। टिप: वर्ग करते समय \(2\sqrt6\) न छोड़ें।
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