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prove that 15+17root3 be an irrational Number. |
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Answer» Suppose\xa0{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are co-prime integers, {tex}b\\ne 0{/tex}Squaring both sides,\xa0{tex}\\Rightarrow 3 ={/tex}\xa0{tex}\\frac { a ^ { 2 } } { b ^ { 2 } }{/tex}Multiplying with {tex}b{/tex}\xa0on both sides,\xa0{tex}\\Rightarrow 3b ={/tex}\xa0{tex}\\frac { a ^ { 2 } } { b }{/tex}LHS = {tex}3\\times b{/tex}{tex}= Integer{/tex}RHS =\xa0{tex}\\frac { a ^ { 2 } } { b } = \\frac { \\text { Integer } } { \\text { Integer } }{/tex}{tex}= Rational\\ Number{/tex}{tex}\\Rightarrow \\mathrm { LHS } \\neq \\mathrm { RHS }{/tex}{tex}\\therefore{/tex}\xa0Our supposition is wrong.{tex}\\Rightarrow \\sqrt { 3 }{/tex}\xa0is irrational.Suppose\xa0{tex}15 + 17{/tex}{tex}\\sqrt 3{/tex}\xa0is a rational number.{tex}\\therefore 15 + 17 \\sqrt { 3 } = \\frac { a } { b }{/tex}, where {tex}a\\ and\\ b{/tex} are co-prime, {tex}b\\ne0{/tex}{tex}\\Rightarrow \\quad 17 \\sqrt { 3 } = \\frac { a } { b } - 15{/tex}{tex}\\sqrt { 3 } = \\frac { a - 15 b } { 17 b }{/tex}{tex}\\frac { a - 15 b } { 17 b }{/tex}\xa0is rational number,{tex}\\sqrt 3{/tex}\xa0is irrational.{tex}\\therefore \\quad \\sqrt { 3 } \\neq \\frac { a - 15 b } { 17 b }{/tex}{tex}\\therefore{/tex} Our supposition is wrong.{tex}\\Rightarrow \\quad 15 + 17 \\sqrt { 3 }{/tex}\xa0is irrational. thnx |
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