On Fri, 19 Jul 2013 22:10:11 +0200, Orchid Win7 v1 <voi### [at] devnull> wrote:

> On 19/07/2013 07:54 PM, Nekar Xenos wrote:
>> I don't know.
>>
>> Maybe I should start with something simpler.
>>
>> Infinity + Infinity = ?
>>
>> Is there any other answer than 2(Infinity)?
>>
>> If so please explain.
>>
>> -Unlike Warp I don't know the answer and really want to know :)
>
> "Infinity" is rather vague.
>
> Let's talk specifics. Suppose we have the set of all natural numbers  
> (i.e., all positive whole numbers). There are infinity of these.
>
> This set is commonly denoted by the latter N. I'm going to refer to the  
> size of this set as being A0.
>
> Now, suppose we compare N, the set of all *positive* integers, to Z, the  
> set of *all* integers, negative and positive. Clearly there are twice as  
> many of these. For every item in N, there are two corresponding items in  
> Z, one negative and one positive. So the size of Z is A0 + A0.
>
> But wait...
>
> I can map every odd number in N to a negative number in Z, and every  
> even number in N to a positive number in Z. In this way, every element  
> of N is mapped to *one* element of Z, and every element of Z is mapped  
> to *one* element of N.
>
> The existence of this mapping _proves_ that the sets N and Z are  
> ACTUALLY THE SAME SIZE! O_O
>
> In other words, A0 + A0 = A0, exactly unchanged.
>
>
>
> This seems utterly bizarre. But it's a consequence of infinitely large  
> quantities. To understand, try doing the same trick to prove that the  
> set {1 .. 5} is the same size as the set {-5 .. +5}. If you try this,  
> you'll discover the following:
>
>    {1 .. 5}   {-5 .. +5}
>       1    <->    -1
>       2    <->    +1
>       3    <->    -2
>       4    <->    +2
>       5    <->    -3
>                   +3
>                   -4
>    Oh bugger!     +4
>                   -5
>                   +5
>
> It doesn't work. You can't match them up. Because - AS IS FRIGGING  
> OBVIOUS - the sets aren't the same size. Basically you run out of  
> elements of the left set before you've matched up all the elements of  
> the right set.
>
> The ONLY reason this works with infinite sets is that, because they're  
> infinite, you will never "run out" of stuff to match up. Because of  
> this, even "doubling" the size of an infinite set actually leaves its  
> size unchanged - bizarre as that is!
>
>
>
> If adding a set to itself once does nothing, doing this twice also does  
> nothing. That is,
>
>    A0 + A0 + A0 = A0
>
> In general,
>
>    x * A0 = A0
>
> This is clearly true for any finite x. But what if x ISN'T FINITE? What  
> happens if we do this:
>
>    A0 * A0 = ???
>
> Remember that A0 is the size of N, the set of all integers. So if we  
> construct the set of all PAIRS of integers, like this...
>
>     (1, 1)
>     (1, 2)
>     (1, 3)
>       .
>       .
>       .
>    infinity
>     (2, 1)
>     (2, 2)
>     (2, 3)
>       .
>       .
>       .
>    infinity
>     (3, 1)
>     (3, 2)
>     (3, 3)
>       .
>       .
>       .
>    infinity
>
> Surely, the size of this set is CLEARLY equal to A0 * A0.
>
> But wait...
>
> If I take both numbers, add zeros to pad them out to the same length,  
> and then interleave their digits, I get this:
>
>       ...           ...
>       ...           ...
>    (123, 456) <-> 142536
>    (123, 457) <-> 142537
>    (123, 458) <-> 142538
>       ...           ...
>       ...           ...
>
> Every number on the right matches exactly one pair on the left. Every  
> pair matches on number on the right. THE SETS HAVE THE SAME SIZE! We  
> just proved that A0 * A0 = A0.
>
>
>
> Just when it's starting to look like A0 is A0 no matter what you do to  
> it, something interesting happens.
>
> The size of N is A0. If we construct all possible subsets of N, then for  
> each element of N, a given subset of N either DOES or DOES NOT contain  
> that element. That's 2 possibilities, independently for each element of  
> N. So that's a total of 2 ^ A0 possible subsets.
>
> It turns out that (2 ^ A0) > A0.
>
> Yes sir, there are actually DIFFERENT SIZES of infinity. Let us write
>
>    A1 = 2 ^ A0
>
>    A1 > A0
>
> But, by the same algorithm, 2 ^ A1 > A1, so we actually have
>
>    A(n+1) = 2 ^ A(n)
>
>    A(n+1) > A(n)
>
> In summary, infinity comes in different sizes. (Now you see why the  
> original question "what is infinity plus infinity?" is vague; there's  
> one than one infinity you may ask about.)
>
> In general,
>
>    A(x) + A(y) = A(y)   iff y >= x
>
> That is, if you add two infinities, the result is the size of whichever  
> is the largest infinity. (Thus, when you add a particular infinity to  
> itself, nothing happens.)

This is all very interesting. :)

I was thinking of adding any particular infinity to itself.

-- 
-Nekar Xenos-

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