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Prove root 3 is an irrational |
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Answer» let root 3 is rational number. root 3=p/q [where,p and q are integers,q is not equal to 0 and p and q are co-primes]. root 3q=p squaring both sides, 3q2 =p2...............[1] 3 divides p2 3 divides p.............[2] p2=3q2 p2=3m Put the value of p2 in eqn [1]. p2=3q2 [3m]2=3q2 9m2=3q2 3m2=q2 q2=3m2 3 divides q2 3 divides q [3] By eqn [2] and [3].... 3 is the common factor of both p and q . This is contradiction that and q are co-primes. Our assumption is wrong. Hence , root 3 is irrational number. \xa0let us assume that\xa0{tex}\\sqrt 3{/tex}\xa0be a rational number.{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \\neq{/tex}0Squaring both sides, we have{tex}\\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}or,\xa0{tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)a2\xa0is divisible by 3.Hence a is divisible by 3..........(ii)Let a = 3c ( where c is any integer)squaring on both sides we get(3c)2\xa0= 3b29c2\xa0= 3b2b2\xa0= 3c2so b2\xa0is divisible by 3hence, b is divisible by 3..........(iii)From equation(ii) and (iii), we have3 is a factor of a and b which is contradicting the fact that a and b are co-primes.Thus, our assumption that\xa0{tex}\\sqrt 3{/tex} is rational number is wrong.Hence,\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number. |
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