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InterviewSolution
| 1. |
Prove 3+√(5 is irrational |
| Answer» Let us assume, to the contrary, that is\xa0{tex}3 + \\sqrt { 5 }{/tex}\xa0rational.That is, we can find coprime integers a and b\xa0{tex}( b \\neq 0 ){/tex}\xa0such that{tex}3 + \\sqrt { 5 } = \\frac { a } { b } \\text { Therefore, } \\frac { a } { b } - 3 = \\sqrt { 5 }{/tex}{tex}\\Rightarrow \\frac { a - 3 b } { b } = \\sqrt { 5 }{/tex}{tex}\\Rightarrow \\frac { a - 3 b } { b } = \\sqrt { 5 } \\Rightarrow \\frac { a } { b } - \\frac { 3 } { 2 }{/tex}Since a and b are integers,We get\xa0{tex}\\frac { a } { b } - \\frac { 3 } { 2 }{/tex}\xa0is rational, also so\xa0{tex}\\sqrt { 5 }{/tex}\xa0is rational.But this contradicts the fact that\xa0{tex}\\sqrt { 5 }{/tex}\xa0is irrational.This contradiction\xa0arose because of our incorrect\xa0assumption that\xa0{tex}3 + \\sqrt { 5 }{/tex}\xa0is rational.So, we conclude that\xa0{tex}3 + \\sqrt { 5 }{/tex}\xa0is irrational. | |