Can any real number....

eljose

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?

Zurtex

No, in fact most (for a given meaning of the word most) real numbers can't be represented by any algorithm of quotients.

Hurkyl

OTOH, I'm pretty sure there's a continued fraction for any real number.

Zurtex

"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?..."

Hurkyl

I didn't claim there was an algorithm for coming up with the coefficients of the continued fraction.

Zurtex

Oh yeah didn't think about it that way, cool.

HallsofIvy

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!

mathwonk

of course we need a definition of "represent". obviously every real number is a least upper bound, i.e. a limit, of rational numbers.

robert Ihnot

Of course, the square root of any positive integer can be expressed as a continued fraction. [tex]\sqrt(3) =1;\overline{1,2}[/tex]

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!

Zurtex

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.

shmoe

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)).

robert Ihnot

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.)

Edgardo

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' )

Zurtex

I hope so because we are about to prove in one my classes in a lecture or 2

chronon

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.

robert Ihnot

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.