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| 1. |
prove that 1\\√3 is irrational |
| Answer» {tex}\\frac { 1 } { \\sqrt { 3 } } = \\frac { 1 } { \\sqrt { 3 } } \\times \\frac { \\sqrt { 3 } } { \\sqrt { 3 } } = \\frac { \\sqrt { 3 } } { 3 } {/tex} .............(i)If possible, let {tex}\\frac { 1 } { \\sqrt { 3 } }{/tex} be rationalThen, from (i) it follows that {tex}\\frac { \\sqrt { 3 } } { 3 } {/tex} is rational.Let {tex}\\frac { \\sqrt { 3 } } { 3 } = \\frac { a } { b }{/tex} where a and b are non-zero intergers having no common factor other than 1.Now, {tex}\\frac { \\sqrt { 3 } } { 3 } = \\frac { a } { b }{/tex}{tex} \\Rightarrow \\sqrt { 3 } = \\frac { 3 a } { b }{/tex} ...............(ii)But 3a and b are non-zero integers{tex}\\therefore \\frac { 3 \\mathrm { a } } { \\mathrm { b } }{/tex} is rational. So, {tex}\\sqrt { 3 }{/tex} is rational.This contradicts the fact that {tex}\\sqrt { 3 }{/tex} is irrationalThe Contradiction arises by assuming that {tex}\\frac { 1 } { \\sqrt { 3 } }{/tex} is rational.Hence {tex}\\frac { 1 } { \\sqrt { 3 } }{/tex} is irrational. | |