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InterviewSolution
| 1. |
Prove that root 2+root 5 is irrational no. |
| Answer» Let us assume on the contrary that {tex}\\sqrt { 2 } + \\sqrt { 5 }{/tex} is a rational number.Then, there exist co-prime positive integers a and b such that{tex}\\sqrt { 2 } + \\sqrt { 5 } = \\frac { a } { b }{/tex}{tex}\\Rightarrow \\quad \\frac { a } { b } - \\sqrt { 2 } = \\sqrt { 5 }{/tex}{tex}\\Rightarrow \\quad \\left( \\frac { a } { b } - \\sqrt { 2 } \\right) ^ { 2 } = ( \\sqrt { 5 } ) ^ { 2 }{/tex}\xa0[Squaring both sides]{tex}\\Rightarrow \\quad \\frac { a ^ { 2 } } { b ^ { 2 } } - \\frac { 2 a } { b } \\sqrt { 2 } + 2 = 5{/tex}{tex}\\Rightarrow \\quad \\frac { a ^ { 2 } } { b ^ { 2 } } - 3 = \\frac { 2 a } { b } \\sqrt { 2 }{/tex}{tex}\\Rightarrow \\quad \\frac { a ^ { 2 } - 3 b ^ { 2 } } { 2 a b } = \\sqrt { 2 }{/tex}{tex}\\Rightarrow \\sqrt { 2 }{/tex}\xa0is a rational number .∵ a, b are integers .\xa0∴ {tex}\\frac { a ^ { 2 } - 3 b ^ { 2 } } { 2 a b }{/tex} is a rational numberThis contradicts the fact that\xa0{tex}\\sqrt{2}{/tex} is irrational.So, our assumption is wrong.Hence,\xa0{tex}\\sqrt { 2 } + \\sqrt { 5 }{/tex}\xa0is irrational. | |