Infinite Of Infinates, whats the solution?

CRGreathouse

I noticed that the same page has one English paper, a crackpot piece asserting that [itex]0.999\ldots\neq1.[/itex]

csprof2000

I'd say, simply and without appeal to calculus, that nothing is continuous... space, time, temperature, are discrete, and therefore there is no paradox in walking to the door or in adjusting the knob. You could enumerate the different temperatures, if you were quick enough.

Why not?

matticus

the same reason you cannot enumerate all of the real numbers, no matter how fast you are. there is a one-to-one correspondence between a time interval and an interval of real numbers, and every interval of real numbers is uncountable.

csprof2000

I think you misunderstood my point, matticus.

The real numbers are not "real" in the general sense. There is not a one-to-one correspondence between mathematics and nature. Math is an approximation... it's a language for describing nature. Mathematics is true in its context. But there's no reason to expect it to apply *exactly* to the real world.

You said:
"there is a one-to-one correspondence between a time interval and an interval of real numbers"

Prove that, and you'll be the most famous mathematician who ever walked the face of the earth.

csprof2000

Furthermore, just for kicks, tell me about a real number you can't enumerate. Describe it for me. Give me an example... tell me what its value is.

matticus

i'll tell you what. enumerate them, and i'll show you one you missed :)

csprof2000

Simple.

Using the decimal expansion of every number, start by filling the ones place with 0-9, then the 10s and 10ths place with 0-9, etc. You'll never reach the end, but you wouldn't reach it if you enumerated the natural numbers either.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9
00.0, 01.0, 02.0, ..., 10.0, 11.0, ... etc.

Perhaps a more disturbing result of my above enumeration scheme is that it produces only a countable infinity of different numbers. (unless I missed a few...)

matticus

your enumeration scheme produces all the decimals, which is not countable, but has the cardinality aleph-1. To be countable there must be a bijection (1-1 correspondence) between the set and the set of naturals. Cantor's diagonalization process shows that no such bijection exists.

Hurkyl

While R is indeed uncountable, its cardinality is aleph-1 if and only if you adopt the continuum hypothesis.

csprof2000

But it is impossible to give an algorithm which produces all real numbers. The simple reason: there are only countably many algorithms, and as such, there are more real numbers than algorithms. There must exist some real numbers for which there is no algorithm which can tell you the digits... some examples of numbers for which there are algorithms to give you the decimal digits:

integers:
Put the integer x into a variable k.
Let n = 0.

Divide k by 10.
Put a zero into the place corresponding to the power 10^-n.
Put the remainder after division into the place corresponding to the power 10^n.
Increase n by 1.
Put the quotient back into k, and repeat (forever, or until k = 0).


rationals:
see Long Division


special irrationals (roots, etc.):

For square roots, for instance...

1.) Let k = 0, delta = 1, steps = 0
2.) if k^2 < n, add delta to k and go to step 1
3.) if k^2 = n, return this value of k
4.) if k^2 > n, subtract one from k
5.) divide delta by 10
6.) If steps > max, return the approximation k
7.) go to step 1, add 1 to steps



The interesting thing to notice is that even if you take the natural numbers, the rationals, and the "special irrationals" (such as roots, logs, exponentials, sines and cosines, special constants, etc.) then you have a countably infinite set. How so?

Well, consider an ordered 5-tuple of rational numbers.

(a, b, c, d, e)

a can be any rational number
b can be any power to raise a to... assuming it's valid for that a
c can be either 0 or 1. if it's zero, then evaluate the 5-tuple as b*exp(a). Otherwise, don't.
d can be either 0 or 1. if it's zero, then evaluate the 5-tuple as b*log(a). Otherwise, don't.
e can be either 0 or 1. if it's zero, then evaluate the 5-tuple as b*sin(a). otherwise, don't.

I could get really, really technical, and make it an ordered 50-tuple, and wipe out about any numbers you could imagine. Do you see the rub?

There are as many rational numbers as there are ordered 50-tuples of rational numbers. So I can't possibly compute (or even "define", if by define you mean give the value of) the majority of the real numbers. Which also means you will never see one or use one (directly).

HallsofIvy

What you are doing here will only give positive integers won't it?

If, however, you meant .0, .1, .2, .3, .4, .5, .6, .7, .8, .9, .10, .11, .12, .13. etc. that will give only terminating decimals, a small subset of the rational numbers (specifically those that can be written as fractions with only powers of 2 and 5 in the denominator) and so countable.