
Re: What is infinity minus one?
Posted:
Oct 12, 2004 5:14 PM


Tim Brauch <RnEeMwOs.pVoEst@tbrauch.cNOoSPAMm> wrote: > Tracy Poff <pofft@gmx.net> wrote in news://2t0o2cF1qqiliU1@uniberlin.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 welldefined (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, floatingpoint 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.
David Cantrell

