my question is about recursion of continued fraction

52 Views Asked by At

1 is not equals 2, then whats wrong with the following... 1=2/(3-1) and if we replace (can we or cannot?) 1 on right hand side by 1=2/(3-1) that is 1=2/(3-2/(3-1)) and if we continue replacing 1 on right hand side that is 1=2/(3-2/(3-2/(3-2/(3-2/(3-2/(3-... ---(A)

similarly 2=2/(3-2) 2=2/(3-2/(3-2/(3-2/(3-2/(3-2/(3-... ---(B)

this says (A)=(B) or 1 = 2 kindly explain what is not mathematical here

1

There are 1 best solutions below

0
On BEST ANSWER

$${2\over3-{2\over3-2_{\ddots}}}$$ has to be interpreted as the limit of $$2,{2\over3-2},{2\over3-{2\over3-2}},\dots$$ and not as the limit of $$1,{2\over3-1},{2\over3-{2\over3-1}},\dots$$