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Prove that √2 and √3 are irrational number |
| Answer» \tSuppose\xa0{tex}\\sqrt2{/tex}\xa0is a rational number. That is ,\xa0{tex}\\sqrt2{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z.\xa0We can assume the fraction is in lowest fraction, That is p and q shares no common factors.\tThen {tex}\\sqrt2q=p{/tex}\xa0\tSquaring both side we get,\xa0\t{tex}2q^2=p^2{/tex}\tSo\xa0{tex}p^2{/tex}\xa0is a multiple of 2,\tlet\'s assume\xa0{tex}p=2m{/tex}\xa0\tThen,\xa0{tex}2q^2=\\left(2m\\right)^2{/tex}\xa0\t{tex}2q^2=4m^2{/tex}\tOr {tex}q^2=2m^2{/tex}\tSo {tex}q^2{/tex}\xa0is a multiple of 2,\t{tex}\\therefore{/tex} q is multiple of 2\tThus p and q shares a common factor.This is contradiction.\t{tex}\\Rightarrow {/tex}{tex}\\sqrt { 2 }{/tex}\xa0is an irrational number.\t\t\tLet us assume that\xa0{tex}\\sqrt 3{/tex}\xa0be a rational number.\t{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \\neq{/tex}0\tSquaring both sides, we have\t{tex}\\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}\tor,\xa0{tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)\ta2\xa0is divisible by 3.\tHence a is divisible by 3..........(ii)\tLet a = 3c ( where c is any integer)\tsquaring on both sides we get\t(3c)2\xa0= 3b2\t9c2\xa0= 3b2\tb2\xa0= 3c2\tso b2\xa0is divisible by 3\thence, b is divisible by 3..........(iii)\tFrom equation(ii) and (iii), we have\t3 is a factor of a and b which is contradicting the fact that a and b are co-primes.\tThus, our assumption that\xa0{tex}\\sqrt 3{/tex} is rational number is wrong.\tHence,\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.\t | |