Tim Brauch <RnEeMwOs.pVoEst@tbrauch.cNOoSPAMm> wrote: > Tracy Poff <email@example.com> wrote in news://2t0o2cF1qqiliU1@uni-berlin.de: > > > Is this because a + (-a) = 0 is defined on the set of reals, and > > infinity is nonreal?
That's certainly not quite the reason. By that very reasoning Sqrt(-1) + (-Sqrt(-1)) should be undefined too. But after all, we don't want to deal strictly with reals all the time!
> > 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.
Again, by reasoning such as that, 1*Sqrt(-1) should also be undefined. Of course, it _is_ undefined in the reals. But 1 is also the multiplicative identity both in the system of complex numbers and in the system of extended real numbers, thereby allowing us to claim 1*Sqrt(-1) = Sqrt(-1) and 1*infinity = infinity in appropriate systems.
> I think what you have is a pretty good reason. I never thought about it > like that, but it is a nice, simple explanation.
And of course I disagree!
> 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 > > Look at: > > 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.
Your idea is not bad. But using series of _integers_, it's hard to see how you'd "make oo - oo be any real number". It would be hard to get pi, for example.
> The reason 1*infinity is undefined, I would agree, is that infinity is > not really a number (not in the reals).
Well, _of course_, since the reals don't include infinity, it's not defined _there_. (But so what? The same argument would allow us to say that Sqrt(-1) is undefined.) Infinity _is_ an element of the extended reals (and whether we choose to call it a "number" or not is essentially irrelevant), and so 1*infinity is defined there.
> 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.
The only products which are normally undefined in the extended reals are those of 0 and +-infinity. For nonzero x, you can indeed simply multiply:
x * infinity = infinity if x > 0, -infinity if x < 0.
[snip] > But overall, I would say your reasoning is pretty good and catches the > essential parts of why arithmetic with infinity is not a good idea.
If you think it's a bad idea, then perhaps you should contemplate why the most used (in terms of number of computations performed per day) number system in the world, floating-point arithmetic, incorporates arithmetic with infinity (and does so according to an internationally accepted standard). You might also ask yourself why, if it's a bad idea, all computer algebra systems (or at least all known to me) implement arithmetic with infinity.