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| 1. |
Show that 7-sq. root 5 is irrational |
| Answer» Let us assume that {tex}7 - \\sqrt 5{/tex}\xa0is a rational number.Then, there must exist positive co primes a and b such that{tex}\\begin{array}{l}7 - \\sqrt 5=\\frac{\\mathrm a}{\\mathrm b}\\\\\\end{array}{/tex}{tex}-\\sqrt5=\\frac{\\mathrm a}{\\mathrm b}-7{/tex}{tex}\\begin{array}{l}\\sqrt5=7-\\frac{\\mathrm a}{\\mathrm b}\\\\\\end{array}{/tex}{tex}\\begin{array}{l}\\sqrt 5=\\frac{7\\mathrm b-\\mathrm a}{\\mathrm a}\\\\\\end{array}{/tex}The right side\xa0{tex}\\begin{array}{l}\\frac{7\\mathrm b-\\mathrm a}{2\\mathrm a}\\\\\\end{array}{/tex}\xa0is a rational numbers so\xa0{tex}\\sqrt5{/tex} is a rational numberThis contradicts the fact that {tex}\\sqrt 5{/tex}\xa0is an irrational numberHence our assumption is incorrect and {tex}7 - \\sqrt 5{/tex}\xa0is an irrational number. | |