Saved Bookmarks
| 1. |
Prove that (5-2√3)2 is irrational number? |
|
Answer» We will prove it in a same way as we prove that\xa0{tex} \\sqrt{3}{/tex}\xa0is an irrational number..Let us assume that\xa0{tex}({5-2\\sqrt{3}})^2{/tex}\xa0is a rational number.\xa0Then {tex}( {5-2\\sqrt{3}})^2={p\\over q}{/tex} where p and q are co-prime.{tex} =>{/tex}{tex}25-2×5×2{\\sqrt 3}+(2{\\sqrt 3})^2={p\\over q}{/tex} [by using\xa0{tex}(a-b)^2=a^2-2ab+b^2{/tex}{tex}25-20{\\sqrt 3} +4×3={p\\over q}{/tex}{tex}25-20{\\sqrt 3}+12={p\\over q}{/tex}{tex}37-20{\\sqrt 3}={p\\over q}{/tex}{tex}37+{p\\over q}=20{\\sqrt3 }{/tex}{tex}{37\\over 20} +{p\\over 20q}={\\sqrt 3}{/tex} Clearly L.H.S. is a sum of two rational number and therefore L.H.S is rational.So\xa0{tex}{\\sqrt 3}{/tex}\xa0is a rational number.But we know that\xa0{tex}{\\sqrt3}{/tex}\xa0is an irrational number.So our assumption is wrong.Hence\xa0{tex}({5-2\\sqrt{3}})^2{/tex}is an irrational number. prove (5-2 root3)2\xa0=\xa025-20root 3+12=p/q consider it as rational firstbut solving it we get that root 3 could be represented as rational but it is irrational number thus our assumption that (5-2 root 3)2was wrong and hence it is irrational |
|