Can any real number

1. Feb 9, 2005

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?

2. Feb 9, 2005

Zurtex

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

3. Feb 9, 2005

Hurkyl

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

4. Feb 9, 2005

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?..."

5. Feb 9, 2005

Hurkyl

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

6. Feb 9, 2005

Zurtex

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

7. Feb 10, 2005

HallsofIvy

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 &radic;(-3) which is NOT a real number!

8. Feb 10, 2005

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.

9. Feb 10, 2005

robert Ihnot

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

Last edited: Feb 10, 2005
10. Feb 11, 2005

HallsofIvy

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!

11. Feb 11, 2005

Zurtex

12. Feb 11, 2005

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)).

13. Feb 11, 2005

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.)

Last edited: Feb 11, 2005
14. Feb 11, 2005

Edgardo

15. Feb 13, 2005

Zurtex

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

16. Feb 13, 2005

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.

17. Feb 13, 2005

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.

18. Feb 13, 2005

Hurkyl

Staff Emeritus

19. Feb 13, 2005

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:

20. Feb 13, 2005

Hurkyl

Staff Emeritus
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.