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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.)
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