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? Please pardon my lack of knowledge. 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.
>>>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.
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?
> 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'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.