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| 1. |
prove that √2\xa0is irrational\xa0 |
| Answer» Let us assume\xa0{tex}\\sqrt{2}{/tex}\xa0be a rational number and its simplest form be\xa0{tex}\\frac{a}{b}{/tex}, a and b are coprime positive integers and b\xa0{tex}\\ne{/tex}\xa00.So,\xa0{tex}\\sqrt{2}{/tex}\xa0=\xa0{tex}\\frac{a}{b}{/tex}{tex}\\Rightarrow{/tex}\xa0a2 = 2b2\xa0Thus, a2 is a multiple of 2{tex}\\Rightarrow{/tex} a is a multiple of 2Let a = 2m for some integer m{tex}\\therefore{/tex}\xa0b2 = 2m2Thus, b2 is a multiple of 2{tex}\\Rightarrow{/tex}\xa0b is a multiple of 2Hence 2 is a common factor of a and b.This contradicts the fact that a and b are coprimesHence\xa0{tex}\\sqrt{2}{/tex}\xa0is an irrational number. | |