Question 1118033
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If sqrt(5)= 2+ 1/a, show that a= 4+1/a
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If {{{sqrt(5)}}} = 2 + {{{1/a}}},  then  {{{1/a}}} = {{{sqrt(5)-2}}}     (1)


It implies   a = {{{1/(sqrt(5)-2)}}}    (2)


Rationalize the denominator in the right side of (2):


a = {{{1/(sqrt(5)-2)}}} = {{{1/(sqrt(5)-2)}}}.{{{(sqrt(5)+2)/(sqrt(5)+2)}}} = {{{(sqrt(5)+2)/((sqrt(5))^2 -2^2)}}} = {{{(sqrt(5)+2)/(5-4)}}} = {{{sqrt(5)+2}}}.   (3)


Now in the formula  a = 4 + {{{1/a}}}    (*)


    the left  side is  a = {{{sqrt(5)+2}}}    (according to (3))


    the right side is  4 + {{{1/a}}} =  (now replace {{{1/a}}} by {{{sqrt(5)-2}}}, based on (1), to get )  = 4 + {{{(sqrt(5)-2)}}} = {{{sqrt(5)+2}}}.


Thus the both sides of (*) are the same.


The statement is proved.
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