Tracy Poff <firstname.lastname@example.org> wrote in news://2t0o2cF1qqiliU1@uni-berlin.de:
> Is this because a + (-a) = 0 is defined on the set of reals, and > infinity is nonreal? > > This question was posed to my calc teacher in high school when > discussing indeterminate forms, and she replied (in essence) that it > was a magical thing that didn't have a reason, while I disagreed. > > Similarly, if my reasoning is correct, this would mean that 1*infinity > is undefined, since that is defined as the multiplicative identity > property of the reals. Is this correct? > > Tracy Poff >
I think what you have is a pretty good reason. I never thought about it like that, but it is a nice, simple explanation. The way I learned to look at it was to draw a contradiction of some sorts thus showing it is not well-defined (which is essentially the same as undefined).
Here is the way I get a contradiction for oo - oo. Define:
A1 := 1 + 2 + 3 + 4 + 5 + 6 +... --> oo
A2 := 2 + 4 + 6 + 8 + 10 + 12 +... --> oo
A3 := 3 + 4 + 5 + 6 + 7 + 8 + ... --> oo
A1 - A2 = 1 + 3 + 5 + 7 + ... --> oo
Thus oo - oo = oo
A1 - A3 = 1 + 2 = 3.
Thus 3 = oo - oo = oo or 3 = oo #
Therefore oo - oo is not defined.
You can actually make oo - oo be any real number you want by starting your series A3 at n+1.
The reason 1*infinity is undefined, I would agree, is that infinity is not really a number (not in the reals). You simply cannot multiply a number by something that is not a number in any mathematical fashion that will make sense all of the time.
3*lemon does not make much sense mathematically.
There are times when you can multiply a number times a non-number and have significant results, like a scalar times a matrix or a number times a set. But in these cases, the multiplication has been very clearly defined.