my question is about recursion of continued fraction

Question

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

I cover a lot of this in a similar question here. The short version is that the "finite" versions of the fraction -- what we would call a convergent -- converge only to one of the values. This is what we would say the infinite fraction "equals." What that "$\cdots$" represents matters a lot, especially in how the infinite process you do converges to a given value. — PrincessEev, Mar 01 '20 at 03:50
I would find it a lot easier to read your question if you had used MathJax. — Lubin, Mar 01 '20 at 05:04
so the correct answer is, a rational number has a finite continued fraction? — Intellectual_, Mar 01 '20 at 05:25
Yes a rational number has a finite continued fraction. I don't know how that corresponds to your question. — Ross Millikan, Mar 01 '20 at 05:29

score 1 · Accepted Answer · answered Mar 01 '20 at 05:25

1

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

answered Mar 01 '20 at 05:25

Gerry Myerson

179,216

my question is about recursion of continued fraction

1 Answers1