## Can any real number....

be represented in the form m/n with m and n integers?...by using continued fractions?...

if so what would happen if we take sqrt(-3) and expnad it in continued fractions?...

can be the continued fraction method be applied to complex numbers?
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 Recognitions: Homework Help Science Advisor No, in fact most (for a given meaning of the word most) real numbers can't be represented by any algorithm of quotients.
 Recognitions: Gold Member Science Advisor Staff Emeritus OTOH, I'm pretty sure there's a continued fraction for any real number.

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## Can any real number....

 Quote by Hurkyl OTOH, I'm pretty sure there's a continued fraction for any real number.
"OTOH"? Sorry I could of misread some proof, but there are only a countable number of algorithms of quotients to define real numbers are there not? Hmm, sounds silly when I say it actually, I'll look it up again tomorrow.

Anyway, still no to the "be represented in the form m/n with m and n integers?..."
 Recognitions: Gold Member Science Advisor Staff Emeritus I didn't claim there was an algorithm for coming up with the coefficients of the continued fraction.
 Recognitions: Homework Help Science Advisor Oh yeah didn't think about it that way, cool.
 Recognitions: Gold Member Science Advisor Staff Emeritus Those numbers that can be represented in the form m/n are, by definition, the "rational numbers". It is well known that "almost all" (all except a countable number) real numbers are NOT rational. Having started talking about real numbers, it makes no sense at all to then ask about √(-3) which is NOT a real number!
 Recognitions: Homework Help Science Advisor of course we need a definition of "represent". obviously every real number is a least upper bound, i.e. a limit, of rational numbers.
 Recognitions: Gold Member Of course, the square root of any positive integer can be expressed as a continued fraction. $$\sqrt(3) =1;\overline{1,2}$$
 Recognitions: Gold Member Science Advisor Staff Emeritus But I seriously doubt (definitely can not prove one way or the other) that e or pi can be written as a continued fraction!

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 Quote by HallsofIvy But I seriously doubt (definitely can not prove one way or the other) that e or pi can be written as a continued fraction!
Pi = http://www.research.att.com/cgi-bin/...i?Anum=A001203

e = http://www.research.att.com/cgi-bin/...i?Anum=A003417

Actually Euler proved that e was irrational by proving it had a simple infinite continued fraction.
 Recognitions: Homework Help Science Advisor Pi and e (and every other real number) sure do have representations as continued fractions as Zurtex points out! In fact, the familiar 22/7 and 355/113 approximations to pi are from truncating it's continued fraction expansion, 22/7=3+1/7 and 335/113=3+1/(7+1/(15+1/1)).
 Recognitions: Gold Member shmoe: Pi and e (and every other real number) sure do have representations as continued fractions as Zurtex points out! They do have representations as continued fractions, but that does not mean that such representations are regular or known beyond the decimal approximation. For example Pi does not seem to have any regular pattern as a continued fraction. Pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, 4, 2, 6, 6, 99, 1, 2, 2, 6, 3, 5, 1, 1, 6, 8, 1, 7, 1, 2, 3, 7, 1, 2, 1, 1, 12, 1, 1, 1, 3, 1, 1, 8, 1, 1, 2, 1, 6, 1, 1, 5, 2, 2, 3, 1, 2, 4, 4, 16, 1, 161, 45, 1, 22, 1, 2, 2, 1, 4, 1, 2, ...] See: http://www.mcs.surrey.ac.uk/Personal...RO.html#sqrtcf Yet from the same article Pi does have a pattern in the form 4/Pi, where the numerator is generally not 1. (Article explains that these things are a mystery.)
 Blog Entries: 2 The question is: Can ALL real numbers represented by continued fractions. I found a link that claims that this is true: http://www.maa.org/editorial/mathgam..._03_15_04.html (search for 'any real' to find the claim) and I found this one: http://www-math.mit.edu/phase2/UJM/vol1/COLLIN~1.PDF (see page 14, or search for 'any real') and another one: http://www.ams.org/bull/pre-1996-dat...30-1/niven.pdf (see page 1 upper part, or search for 'all real' )

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 Quote by Edgardo The question is: Can ALL real numbers represented by continued fractions.
I hope so because we are about to prove in one my classes in a lecture or 2
 Suppose r is a real number. Then let n=int(r). Let x1=1/(r-n) and d1=int(x1) then let x2=1/(x1-d1) and d2=int(x2) . In general let x[n]=1/(x[n-1]-d[n-1]) and d[n]=int(x[n]). Then [n;d1,d2 ... d[n]... ] is a continued fraction representation of r.
 Recognitions: Gold Member I don't doubt what chronon has to say. However, this question of representation has not been gone into. Suppose the number r is not known in terms of its digits? Take for example, I once read in Hardy and Wright, "Number Theory," that a prime generator exists, but they can only determine the digits after they have found all the primes. Thus there can not be an endless continued fraction formed since we would have to know all the primes first. Maybe this is just splitting hairs, and maybe not.

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