Integers with Infinitely Many Digits

[whc.03grady]

@Trinopus

(Is it even…or odd? Is it prime…or composite? You can’t answer those questions.)

I was set to tell her that it’s not an integer since an integer has to be even or odd, but it seems to me that 10000…, if it exists, is even, and is not prime. 12345… I’m less sure about.

She isn’t quite six, so I’m gonna have to dial down the answer. I think I’ll cash it out in Thudlow Boink’s terms, and it’ll give me a good chance to talk about place value: in 12345…, what place value does the ‘2’ represent? Since there’s no good answer (as far as I can tell), she’ll probably accept that. I just hope she doesn’t bring up …54321.

Thanks all for the interesting discussion so far. Don’t feel a need to end it since I’ve chimed in (and please don’t if what I propose to propose is glaringly wrong).

[Buck_Godot]

@Thudlow_Boink

Let’s think about repeating decimals (rational or irrational) for a minute. What do they mean? Take a relatively simple one, like 0.6363636363… (the decimal representation of 7/11). The value of this repeating decimal is the sum of the infinite series 610[sup]-1[/sup] + 310[sup]-2[/sup] + 610[sup]-3[/sup] + 310[sup]-4[/sup] + 610[sup]-5[/sup] + 310[sup]-6[/sup] + …

In other words, you can specify exactly what each individual digit represents: the first 6 after the decimal point represents 6 tenths; the next digit represents 3 hundredths; the next digit represents 6 thousandths; and so on.

In a “number” like 12345… or 10000…, what does each individual digit represent? That 2 in 12345… means 2 of what? Ask your daughter that.

Continuing in this vein, I’ve been taught that when you see … in an informal mathematical setting what you really have is an infinite series. So from this point of view,

1234567…

is the following series

n[sub]1[/sub]=1
n[sub]i[/sub]=i+n[sub]i-1[/sub]*10[sup]f(i)[/sup]

where f(i) is the number of digits in i.

Now continuing this on towards infinity its clear that if fails to converge. So unless you want to invoke some fancy definitions of convergence where 1+2+3+4+…= -1/12

You are going to have to just leave it as being infinity or undefined. In any case not an integer.

[Saint_Cad]

One reason the idea of integers having infinite digits won’t work. We know the cardinality of integers is countable and the number of reals (0, 1) is uncountable. If integers could have infinite digits then they could be put into 1-1 correspondence with an uncountable set making Cantor go crazy(er).

[ZenBeam]

@Saint_Cad

One reason the idea of integers having infinite digits won’t work. We know the cardinality of integers is countable and the number of reals (0, 1) is uncountable. If integers could have infinite digits then they could be put into 1-1 correspondence with an uncountable set making Cantor go crazy(er).

You could turn that around and say that the proofs that the number of integers is smaller than the number of reals are flawed, because they all assume (explicitly or implicitly) that there are no integers with an infinite number of digits.

[md2000]

The problem is infinity.

I suppose there’s a branch of math that treats infinity as a number (something a bit more esoteric than I’ve gotten into) , but basically, infinity is… infinite. It has no upper bound. As a result of “no upper bound”, you can’t for example, subtract 1 from it. It’s still infinite, by definition.
Mathematically, any changes or alterations to infinity are meaningless.

If your daughter is bored, maybe she can deduce the rules for math and infinity, based on the fact that it has no upper bound. I’m sure someone has already.
infinity (times/minus/plus/divided by) X = infinity (where X not infinity)
infinity/infinity = 1 (?)
Infinity x 0 = 0 (?) (for sufficiently large values of infinity )

Infinity-1=infintity. (one less than infinite is still infinite)
Infinity x 2 = infinity (thus 11111… x 2 =22222…
11111… x10 = 111111…

The problem is you are going at it the wrong way. Writing it 111111… implies there is a left-hand end-point. There isn’t, if it is infinite. More correctly it could be written …1111111.0
( …11111.0 which equals …111110.0 is more correct)
both are infinity so are equal, neither is larger than the other, so therefore neither is smaller than the other, therefore they are equal.

[Saint_Cad]

@ZenBeam

You could turn that around and say that the proofs that the number of integers is smaller than the number of reals are flawed, because they all assume (explicitly or implicitly) that there are no integers with an infinite number of digits.

Umm . . . no. Cantor’s diagonal proof simply shows the reals (0,1) cannot be put into a 1-1 correspondence with the integers. It is a corollary of that that there are no integers with infinite digits.

[ZenBeam]

Cantor’s diagonal proof assumes you can enumerate the integers. Is that possible when you include all the “integers” with infinite number of digits?
I have other objections, but I don’t have time to properly write them up now.

[Learjeff]

@Jragon

Those can be numbers, but we don’t talk for them because for most practical purposes, they’re not particularly useful.

Great point.

@whc.03grady

I was set to tell her that it’s not an integer since an integer has to be even or odd

Good one: 12345… clearly loses one interesting property we associate with integers.

@Saint_Cad

One reason the idea of integers having infinite digits won’t work. We know the cardinality of integers is countable and the number of reals (0, 1) is uncountable. If integers could have infinite digits then they could be put into 1-1 correspondence with an uncountable set making Cantor go crazy(er).

I like that one!

@md2000

The problem is infinity.

You can say that again. And again. And again …

@ZenBeam

Cantor’s diagonal proof assumes you can enumerate the integers. Is that possible when you include all the “integers” with infinite number of digits?

Another great point!

I have nothing so profound, but one property of integers is that they’re a well-ordered set. I wouldn’t know how to begin ordering these other numbers. One could define an ordering, perhaps, but it would be rather arbitrary, rather than true by definition as is the case with integers.

[Chronos]

There would be a partial ordering, in that you could say (for instance) that …22222 > …11111 . But you’d have a hard time comparing …313131 to …222222 .

Alternately, you could impose a well-ordering by assigning greater significance to the digits closer to the decimal point. But it wouldn’t be the same well-ordering as is usually assigned to the integers.

[Jragon]

@Chronos

There would be a partial ordering, in that you could say (for instance) that …22222 > …11111 . But you’d have a hard time comparing …313131 to …222222 .

Alternately, you could impose a well-ordering by assigning greater significance to the digits closer to the decimal point. But it wouldn’t be the same well-ordering as is usually assigned to the integers.

Indeed, the p-adic numbers are not a total order (neither are the complex numbers, for that matter).

[Learjeff]

@Chronos

There would be a partial ordering, in that you could say (for instance) that …22222 > …11111 . But you’d have a hard time comparing …313131 to …222222 .

Alternately, you could impose a well-ordering by assigning greater significance to the digits closer to the decimal point. But it wouldn’t be the same well-ordering as is usually assigned to the integers.

You’d need different ordering definitions for different “styles” of these ill-defined “numbers”. Give that, I doubt the different styles fall into the same class of “number”.

For example, you could fully order 12345… and 11111… and 1000… with a simple enough algorithm. Likewise, you could fully order …54321 and …11111. But you couldn’t order the two styles, so they wouldn’t be comparable numbers (literally).

[Hari_Seldon]

@ftg

I was never happy with the way p-adics were handled.

…9999 + 1 = 0? Not to this Computer Scientist. There’s an overflow digit that has to be accounted for. Let’s use “:” to account for this. I.e., …9999 +1 = 1:0. The number before the “:” is overflow from operations on the “lower” digits. Addition and subtraction can be defined more reasonably, etc.

Multiplication has to be able to cascade these overflows. E.g., 1:0 * 1:0 is 1:0:0.

But at least some key properties of primes and such are preserved.

Division gets trickier. 1/1:0 is some type of infinitesimal. Which I am a big fan of but not too many others.

BTW: What’s the name for the opposite of an infinitesimal? “Infinitesimax”?

Another problem with p-adics is years ago back on the sci.math newsgroup there was a problematic individual with an obsession with them, so they got tainted as a crank topic.

If you can make the arithmetic work, be my guest. But the carries go on indefinitely and in these 10-adic numbers …99999 + 1 =0.

The real answer is that you can define any number system you like and give it any rules you like. But can you use it for anything? Are your rules consistent? The rules of p-adic (or even n-adic) arithmetic are consistent and have models. Same with power series. But while you can also extend power series by allowing a finite number of negative powers, you cannot define the algebraic operations if you allow infinitely many both negative and positive exponents. Similarly p-adics with infinitely many digits before the decimal point can have only finitely many after. If you do that you get the p-adic rationals.

In order to add two such numbers you start on the right, not the left end. But you need an end. Incidentally, in the p-adics, 1 > 11 > 111 > 1111 > …