Infinity Minus InfinityDate: 10/23/98 at 03:43:46 From: Joe Kuster Subject: Infinity minus Infinity Hi, I have been reading through your infinity related questions, and I can't seem to find exactly what I am looking for. I see that you have noted that there are two possible versions of infinity, countable and uncountable. Assuming that we are dealing with two instances of uncountable infinity, what would you get if you subtracted infinity from infinity? Would it not be the same as x minus x? I have argued, without a reason that I can document, that it would be zero. I don't know the reason, but it just seems right to me, or more correct than anything else. I understand that infinity is not a number, but an abstract idea. Any insight you could provide would be helpful. Thank you, Joe Kuster Date: 10/23/98 at 07:51:53 From: Doctor Jerry Subject: Re: Infinity minus Infinity Hi Joe, There are two situations that are sometimes referred to as "infinity minus infinity," one involving cardinal or ordinal numbers, and the other involving limits of functions. I'm guessing that you mean the second. Here's an example: find the limit of sqrt(x+1) - sqrt(x) as x becomes positively infinite (that is, as x grows beyond all bounds). This limit is categorized as an "infinity minus infinity" limit. Such limits are called indeterminate because, on further examination, they can turn out to be just about anything. One must change their form: sqrt(x+1) - sqrt(x) = 1/[sqrt(x+1)+sqrt(x)] and now it's obvious that the limit is 0. So in this case "infinity minus infinity" is 0. However, consider the limit of x^2 - 2^x as x->oo. This is also an "infinity minus infinity" situation. This is resolved by using the well known fact that x^2/2^x ->0 as x->oo. Here's the argument: x^2 - 2^x = 2^x(x^2/2^x - 1) The second factor goes to -1 and the first becomes infinite. So, the limit goes to -oo. As I mentioned, "infinity minus infinity" is an indeterminate form. As for subtracting infinite cardinal numbers like c-aleph_0, where c is the cardinal of the reals and aleph_0 is the cardinal of the positive integers, I presume that one does it by solving an addition problem: c-aleph_0 = b, where c = aleph_0 + b Then b would have to be c. Adding cardinals u and v is done by looking at the cardinal of the disjoint union of sets with cardinalities u and v. This is too big a subject to go into here. See the CRC Encyclopedia of Mathematics on Cardinal Numbers (the page is only intermittently available): http://www.astro.virginia.edu/~eww6n/math/CardinalNumber.html "Cardinal Number In informal usage, a cardinal number is a number used in counting (a Counting Number), such as 1, 2, 3, .... Formally, a cardinal number is a type of number defined in such a way that any method of counting Sets using it gives the same result. (This is not true for the Ordinal Numbers.) In fact, the cardinal numbers are obtained by collecting all Ordinal Numbers which are obtainable by counting a given set. A set has Aleph-0 members if it can be put into a One-to-One correspondence with the finite Ordinal Numbers. Two sets are said to have the same cardinal number if all the elements in the sets can be paired off One-to-One. An Inaccessible Cardinal cannot be expressed in terms of a smaller number of smaller cardinals. See also Aleph, Aleph-0, Aleph-1, Cantor-Dedekind Axiom, Cantor Diagonal Slash, Continuum, Continuum Hypothesis, Equipollent, Inaccessible Cardinals Axiom, Infinity, Ordinal Number, Power Set, Surreal Number, Uncountable Set" - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/