Tracy Poff <email@example.com> wrote: > David W. Cantrell wrote: > >>>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! > > But that is also defined on the set of complex numbers, isn't it?
Exactly. Just because a property holds in one system, we obviously can't say that it won't hold in other systems.
And of course, just because a property holds in one system, we obviously can't say that it _will_ hold in other systems either.
> Please pardon my lack of knowledge.
Of course! You're in the learning process.
> I'm reading several texts on number theory > at the moment and that reason seemed logical to me. > > Could you please say then, what is the reason why it is undefined? If > you've a link to a site that explains it I'd be happy to accept that as > well.
Why is oo - oo undefined in the extended reals?
The best explanation is, unfortunately, something which I suspect you're not ready for yet. (Or maybe I'm wrong about that. Have you learned about constructing the reals from the rationals, via Dedekind cuts or equivalence classes of Cauchy sequences? If so, then there is a similar construction of the extended reals, and it shows clearly why oo - oo is undefined in that system.)
In lieu of that sort of explanation, let me suggest that you look at the paragraph beginning with "The above statements", below equation (8), at .
> >>>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. > > Again, as you've said, that property *is* defined on the set of > complexes, which I took to be the reason it could work. One of the texts > on number theory that I am reading opens with a definition of the > properties of addition, multiplication, and inequality as they can be > used to describe an ordered integral domain. Specifically, this text > says that "there exists an element 1 such that a*1=1*a=a." > > So I took this to be the meaning of the number 1 as used. That is, the > number such that when multiplied by a number 'a' yields the product 'a.' > If this is not a valid way to understand the identity property of the > reals, please excuse my ignorance, and enlighten me.
It's perfectly valid. But my point was that it's not just a "property of the reals". It is valid in other systems, such as the extended reals, too.
> Further, I googled for "system of extended real numbers" and achieved > only one result, unrelated to this. So I will assume that this is not > the name I should be looking for when attempting to learn. What should I > search for in order to get more useful results?
Searching for "extended real" will give you many pertinent links.
> > 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. > > I think I must be missing some crucial piece of knowledge about this. > The only time I can clearly recall dealing with infinity as such was > when working with integrals.
I suspect that, if you had asked your teacher at that time, he would have said that you were not actually "dealing with infinity as such" even then, and that the symbol oo was being used merely as an abbreviation for "increases without bound".
> I'll admit that I am no authority on > calculus as I'm presently taking multivariable calc, which is by no > means a high-level course. So, if you can tell me in what particular way > arithmetic with infinity is used in floating-point calculations, I'll be > most grateful.
To avoid a lengthy answer, let me recommend that you look at the end of the web page cited earlier. The last sentence mentions floating-point arithmetic and gives a link to an excellent paper by Goldberg.