Let a and b be positive integers. Show that √2 always lies between 𝑎𝑏 𝑎𝑛𝑑 𝑎+2𝑏/ 𝑎+𝑏

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We do not know whether \frac{a}{b} < \frac{a + 2 b}{a + b} or ​ \frac{a}{b} > \frac{a + 2 b}{a + b} Therefore, to compare these two numbers, let us compute \frac{a}{b} - \frac{a + 2 b}{a + b} ​ We have, \frac{a}{b} - \frac{a + 2 b}{a + b} = \frac{a (a + b) - b (a + 2 b)}{b (a + b)} = \frac{a^{2} + ab - ab - 2 b^{2}}{b (a + b)} = \frac{a^{2} - 2 b^{2}}{b (a + b)} ∴ \frac{a}{b} - \frac{a + 2 b}{a + b} > 0 \frac{a^{2} - 2 b^{2}}{b (a + b)} > 0 a^{2} - 2 b > 0 a^{2} > 2 b a > \sqrt{2} b and \frac{a}{b} - \frac{a + 2 b}{a + b} < 0 \frac{a^{2} - 2 b^{2}}{b (a + b)} < 0 a^{2} - 2 b < 0 a^{2} < 2 b a < \sqrt{2} b Thus, \frac{a}{b} > \frac{a + 2 b}{a + b}, if a > \sqrt{2} b and \frac{a}{b} < \frac{a + 2 b}{a + b}, if a < \sqrt{2} b So, we have the following cases : Case 1 : When a > \sqrt{2} b In this cas, we have \frac{a}{b} > \frac{a + 2 b}{a + b} i . e ., \frac{a + 2 b}{a + b} < \frac{a}{b} We have to prove that \frac{a + 2 b}{a + b} < \sqrt{2} < \frac{a}{b} We have, a > \sqrt{2} b a^{2} > 2 b^{2} a^{2} + a^{2} > a^{2} + 2 b^{2} [adding a^{2} both sides] 2 a^{2} + 2 b^{2} > (a^{2} + 2 b^{2}) + 2 b^{2} [adding 2 b^{2} both sides] 2 a^{2} + 2 b^{2} + 4 ab > (a^{2} + 2 b^{2}) + 2 b^{2} + 4 ab [adding 4 ab both sides] 2 (a^{2} + b^{2} + 2 ab) > a^{2} + 4 ab + 4 b^{2} 2 {(a + b)}^{2} > {(a + 2 b)}^{2} \sqrt{2} (a + b) > a + 2 b \sqrt{2} > \frac{a + 2 b}{a + b} Again, a > \sqrt{2} b \frac{a}{b} > \sqrt{2} We get, \frac{a + 2 b}{a + b} < \sqrt{2} < \frac{a}{b} Case 2 : When a < \sqrt{2} b In this case, we have \frac{a}{b} < \frac{a + 2 b}{a + b} We have to show that \frac{a}{b} < \sqrt{2} < \frac{a + 2 b}{a + b} We have, a < \sqrt{2} b a^{2} < 2 b^{2} a^{2} + a^{2} < a^{2} + 2 b^{2} [adding a^{2} both sides] 2 a^{2} + 2 b^{2} < (a^{2} + 2 b^{2}) + 2 b^{2} [adding 2 b^{2} both sides] 2 a^{2} + 2 b^{2} + 4 ab < (a^{2} + 2 b^{2}) + 2 b^{2} + 4 ab [adding 4 ab both sides] 2 (a^{2} + b^{2} + 2 ab) < a^{2} + 4 ab + 4 b^{2} 2 {(a + b)}^{2} < {(a + 2 b)}^{2} \sqrt{2} (a + b) < a + 2 b \sqrt{2} < \frac{a + 2 b}{a + b} Again, a < \sqrt{2} b \frac{a}{b} < \sqrt{2} We get, \frac{a}{b} < \sqrt{2} < \frac{a + 2 b}{a + b} Hence, \sqrt{2} lies between \frac{a}{b} and \frac{a + 2 b}{a + b}

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