Can any real number be represented by a continued fraction?

[eljose]

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?

 

[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, 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?..."

 

[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. $$\sqrt(3) =1;\overline{1,2}$$

 

[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]

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/access.cgi/as/njas/sequences/eisA.cgi?Anum=A001203

e = http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?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/R.Knott/Fibonacci/cfINTRO.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/mathgames/mathgames_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-data/199430-1/niven.pdf
(see page 1 upper part, or search for 'all real' )

 

[Zurtex]

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

 

[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.

 

[Hurkyl]

Thus there can not be an endless continued fraction formed since we would have to know all the primes first.

 

[Zurtex]

I think it's just one of the odd things about real numbers, there are an uncountable number of real numbers but only a countable number of real numbers you can uniquely define :shy:

 

[Hurkyl]

I think it's just one of the odd things about real numbers, there are an uncountable number of real numbers but only a countable number of real numbers you can uniquely define

There are relevant technicalities... the statement "a countable number of real numbers you can uniquely define" is made in a higher order logic than the statement "there are an uncountable number of real numbers", so the two aren't really comparable.

In fact, I'm quite sure it's logically consistent for each real number to be uniquely specifiable.

 

[Zurtex]

Ugh yeah, I've got my wording all horribly wrong, I do mean something I just don't know how to word it Stupid brain, I can think entirely in mathematical realms and concepts but when it comes to the English language I am lost.

 

[chronon]

Take for example, I once read in Hardy and Wright, "Number Theory," that a prime generator exists,

I take this to mean that there is a real number corresponding to the continued fraction [0;2,3,5,7,11,13,...]

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.

It is certainly possible to generate the digits one by one for example 0.4323320871859...
I don't really see why you think of this real number as being less well defined than any other irrational such as pi or e.

 

[HallsofIvy]

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

Be sure to report on it here!

 

[Zurtex]

Be sure to report on it here!

It might not be for a few weeks yet but I will ask the lecturer tomorrow to show me the proof, I've visited his website but unfortunately he hasn't posted any of the documents there.

 

[shmoe]

I take this to mean that there is a real number corresponding to the continued fraction [0;2,3,5,7,11,13,...]

While this is certainly true, the number robert is referring to from Hardy&Wright is likely

$$\alpha=\sum_{m=1}^{\infty}p_{m}10^{-2^{m}}$$

Where $$p_m$$ is the mth prime. With the right multiplications by powers of 10 and use of the floor function you can pull out any prime you like from this decimal. Of course this is useless for generating primes, but this doesn't mean that $$\alpha$$ isn't well defined, just like the continued fraction made from primes above.


Zurtex, if you're impatient for a proof that any real number can be represented by a continued fraction, take a look at chronon's recursive formula for generating the continued fraction of a real number. What remains is to show that this construction converges to what you started with. You can try to fill in the convergence issues yourself if you're feeling brave, or check out the details in Hardy&Wright or any good text dealing with continued fractions.