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Infinity Minus Infinity

Date: 10/23/98 at 03:43:46
From: Joe Kuster
Subject: Infinity minus Infinity


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 

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 

   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):   

  "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 

   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   
Associated Topics:
High School Analysis

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