InterviewSolution
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InterviewSolution
| 1. |
Prove that 7-2*3 is an irrational number. |
| Answer» \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.Let us assume that 7 - 2{tex}\\sqrt 3{/tex}\xa0is a rational number7 -\xa02{tex}\\sqrt 3{/tex}\xa0=\xa0{tex}\\frac { p } { q }{/tex} (q\xa0{tex}\\neq{/tex}0 and p and q are co-primes){tex}\\style{font-family:Arial}{\\begin{array}{l}7-2\\sqrt3=\\frac pq\\\\-2\\sqrt3=\\frac pq-7\\\\2\\sqrt3=7-\\frac pq\\\\2\\sqrt3=\\frac{7q-p}q\\\\\\sqrt3=\\frac{7q-p}{2q}\\end{array}}{/tex}Here 7q-p\xa0and 2q both are integers, hence\xa0{tex}\\sqrt 3{/tex} is a rational number.But this contradicts the fact that\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.This contradict is due to our assumption that\xa0{tex}\\style{font-family:Arial}{7-2\\sqrt3}{/tex}\xa0is rational.Hence, 7 -\xa02{tex}\\sqrt3{/tex}\xa0is irrational. | |