InterviewSolution
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InterviewSolution
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Show that \(\frac{1}{\sqrt2}\) is an irrational number. |
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Answer» Let’s assume on the contrary that \(\frac{1}{\sqrt2}\) is a rational number. Then, there exist positive integers a and b such that \(\frac{1}{\sqrt2}\) = \(\frac{a}{b}\) where, a and b, are co-primes ⇒ (\(\frac{1}{\sqrt2}\))2 = (\(\frac{a}{b}\))2 ⇒ \(\frac{1}{2}\) = \(\frac{a^2}{b^2}\) ⇒ 2a2 = b2 ⇒ 2 | b2 [∵ 2|2b2 and 2a2 = b2] ⇒ 2 | b ………… (ii) ⇒ b = 2c for some integer c. ⇒ b2 = 4c2 ⇒ 2a2 = 4c2 [∵ b2 = 2a2] ⇒ a2 = 2c2 ⇒ 2 | a2 ⇒ 2 | a ……… (i) From (i) and (ii), we can infer that 2 is a common factor of a and b. But, this contradicts the fact that a and b are co-primes. So, our assumption is incorrect. Hence, \(\frac{1}{\sqrt2}\) is an irrational number. |
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