Bumpyride said:
Another way to look at it is to consider both zero and infinity as being limits defined in terms of each other where infinity is 1/0 and zero is one divided by infinity. It's interesting that people seem have more trouble with infinity than with zero, though.
Ironic, since you're wrong. 1/0 is not infinity, it's undefined.
You can prove it in so many ways. For example, consider f(x) = 1/(1/x) and g(x) = 1/(-1/x) As x-> inf, both f(x) and g(x) approach 1/0, only f(x) approaches to 0+ and g(x) approaches to 0-. Yet, if you simply the sequences, you get f(x) = 1,2,3,...inf and g(x) = -1,-2,-3,...-inf
Hence, the limit does not exist, and therefore 1/0 is undefined. If you assume otherwise, you are faced with a contradiction.
You can also try to use the relation you stated (1/0 = inf, or 0 * inf = 1, etc). If you use this relation, algebra breaks down.
For example, assume 1/0 exists and call it X. 1/0 = X. Or 1 = 0 * X. But we know that zero times anything is zero.
If you made a special exception for 0 times infinity, you are left with nonsense. Let 0 * X = 1 be written as (X-X)*X = 1. Now, we have X^2 - X^2 = 1. So infinity squared minus infinity squared = 1? Umm, ok.
But what about 2/0? Clearly we have X^2 - X^2 = 2, so infinity squared minus infinity squared = 2!?! What you end up with is complete nonsense. No sensible definition of 1/0 exists for algebra.
Now, in analysis there are all sorts of ways to deal with infinity and different kinds of infinity that can be consistent, but none of them assert that 1/0 = infinity.