## More positive integers than here are positive integers?

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VegiZombie
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### More positive integers than here are positive integers?

In two books I have recently read (I cannot currently rememebr the exact titles but if required I may be able to look it up), th proof given that you can't fit all decimals between 0 and 1 into Hilbert's hotel (the hotel with an infinite nuber of rooms) is that if you take the first digit from the first number and change it by one and make it the first digit of your new number, the second number from the second digit and change it by one and make it the second digit of your new number and carry on for every room in the hotel, then the new number you've made is not in any of the rooms as it can't be in room x as the xth digit as different.
I can see this making sense, although that dosn't mean much with infinity and I'm not sure if they used slightly kooky mathematics and that's why the next thing won't work.
What I then considered is why you can't do the same with integers, with the 1st digit left of the decimal being changed by one and being the first digit left of the decimal in your new number and repeat once again, then that number too is not in any of the rooms. However there's the same number of rooms as there are infinite integers, that being the principal of Hilbrt's hotel.
Is anyone able to help?

lightvector
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### Re: More positive integers than here are positive integers?

Because a "number" with infinitely many digits to the left of the decimal point is not an integer.

Talith
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### Re: More positive integers than here are positive integers?

There are two problems. The first is that no matter how you decide to order the natural numbers, if you try to build your 'unlisted integer', you will find that you get to a point where the kth member of your list doesn't have k digits in it because all natural numbers have a finite representation (try to prove this). The other more obvious problem (and the one where any argument will ultimately fail) is that your 'unlisted integer' isn't an integer at all because by necessity of its definition, it has an infinite number of digits which contradicts the fact that all integers have a finite representation.

What makes real numbers special is that every real number has an infinite representation (we can always put an infinite trail of zeros at the end if it happens to be a terminating rational number) and so the pitfalls that you would encounter in the case of integers won't appear for the reals.

ConMan
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### Re: More positive integers than here are positive integers?

Just to clarify something, the set of all finite strings of digits (i.e. positive integers) is smaller than the set of all infinite strings of digits (which, if you stick 0. in front of them, give you the real numbers between 0 and 1). Similarly, the set of all numbers with terminating decimal expansions - i.e. all numbers which can be written in a finite number of digits, including the decimal point - is a subset of the rational numbers and so has the same cardinality as the set of positive integers.

Also, the set of all sets of finitely many integers is the same cardinality as the set of integers, whereas the set of all sets of (not necessarily finite) sets of integers is not. The bijection is a little bit messy, but you can generate it by assigning each set a "size" equal to the sum of the absolute value of all its elements plus the number of elements in it, and listing them in order of size (so the only size 0 is the empty set, the only size 1 is {0}, the only size 2 is {1}, the size 3s are {0, 1} and {2}, and so forth).
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VegiZombie
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### Re: More positive integers than here are positive integers?

Thank you very much for your help. I get mixed up quite easily with infinities.
Also, sorry for the typo in the title and any others I may not have noticed.
I'd be interested in attempting to proof, however my mathematical knowledge is rather lacking, would you be able to give me an idea as to how I could go about proving that there's a finite representation of all natural numbers. Intuitively it is so but apart from intuition often being wrong, even if it wasn't it's not quite a formal proof.
Might a proof by contradiction work by assuming that it is not so and then showing that processes like multiplication and division, would not work as they involve a sequence of steps applied to each digit in order or is it possible for there to be an integer that it's not possible to be divided/multiplied (I don't include dividing other by it due to 0).
If anything I've said is stupid or unclear, please point it out. Thanks again.

skullturf
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### Re: More positive integers than here are positive integers?

The details may depend on exactly how you formulate your axioms and definitions, but basically, the reason each natural number has a finite representation is that that's just how the natural numbers are defined.

By definition, something is a natural number if and only if it occupies a specific position in the list

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...

Each number in the list is written with only finitely many digits.

Talith
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### Re: More positive integers than here are positive integers?

Oh sorry, when I said try to prove this, I meant try to prove that for all total orderings of the natural numbers there exists a k such that the kth natural number in the ordering has less than k digits in decimal notation (hint, how many natural numbers have one digit?).

VegiZombie
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### Re: More positive integers than here are positive integers?

So prove that k has less than log(base 10, k) digits?

EDIT: What I meant and wasn't at all clear is isn't that just a convention based on our representation of numbers rather than an intrinsic property of numbers themselves. Surely a system is theoretically feasible with every number having a single character representation, even if it might be a rather poor one.

skeptical scientist
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### Re: More positive integers than here are positive integers?

Certainly 1 has a finite base-b representation. Assuming n has a finite base-b representation, so does n+1: the usual base-b addition algorithm is guaranteed to terminate, since you can't have more carries than the number of digits of n. By induction, every natural number has a finite base-b representation.
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Osha
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### Re: More positive integers than here are positive integers?

VegiZombie wrote:Surely a system is theoretically feasible with every number having a single character representation, even if it might be a rather poor one.

Interestingly enough, nope! http://scienceblogs.com/goodmath/2009/0 ... ber_in.php
Unless you include infinitely sized characters, but that would be pretty silly

But a system like you suggest for all the numbers we're familiar with is pretty easy to construct... take standard valid mathematical notation that represents a number, such as 345, or Pi + 7e; draw a box around it, and call that a single digit ;P

VegiZombie
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### Re: More positive integers than here are positive integers?

Sorry, that was a mistake on my part, I meant all integers.