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| 1. |
Prove that √2 + √3 are irrational number |
| Answer» Let us assume that\xa0{tex}\\sqrt 2 + \\sqrt 3{/tex}\xa0is a rational numberLet {tex}\\sqrt2+\\sqrt3=\\frac{\\mathrm a}{\\mathrm b}{/tex} Where a and b are co-prime positive integersOn squaring both sides, we get{tex}(\\sqrt2+\\sqrt3)^2=\\frac{\\mathrm a^2}{\\mathrm b^2}{/tex}{tex}2+3+2\\sqrt6=\\frac{\\mathrm a^2}{\\mathrm b^2}{/tex}{tex}5 + 2\\sqrt 6 =\\frac{a^2}{b^2}{/tex}\xa0{tex}2\\sqrt6=\\frac{\\mathrm a^2}{\\mathrm b^2}-5{/tex}{tex}2\\sqrt6=\\frac{\\mathrm a^2-5\\mathrm b^2}{\\mathrm b^2}{/tex}{tex}\\sqrt6=\\frac{\\mathrm a^2-5\\mathrm b^2}{2\\mathrm b^2}{/tex}Now\xa0{tex}\\frac{\\mathrm a^2-5\\mathrm b^2}{2\\mathrm b^2}{/tex}\xa0is a rational number.This shows that\xa0{tex}\\sqrt 6{/tex}\xa0is a rational number.But this contradicts the fact that\xa0{tex}\\sqrt 6{/tex}\xa0is an irrational number.This contradiction has raised because we assume that\xa0{tex}\\left( {\\sqrt 2 + \\sqrt 3 } \\right){/tex}\xa0is a rational number.Hence, our assumption is wrong and\xa0{tex}\\left( {\\sqrt 2 + \\sqrt 3 } \\right){/tex}\xa0is an irrational number. | |