Infinity / 0

1. Mar 6, 2005

B-Con

I've heard, and agree, that infinity (I) divided by 0 is impossible, yet somewhere I saw that I/0 * 0/I = 1

is this correct? It seems to make sense, somewhat....

2. Mar 6, 2005

Zurtex

First you need to define what infinity is and then you need to define what a/0 is for all numbers you want a defined for. Before you do that you can say nothing about your problem.

3. Mar 6, 2005

Hurkyl

Staff Emeritus
I can't think of any example of an infinite object used by mathematicians for which that is true.

Also, realize that the / in your equation is not ordinary division, because ordinary division only operates on finite objects and does not allow division by zero.

4. Mar 8, 2005

npgreat

Someone who has newly learnt division and multiplication is making fun.

5. Mar 8, 2005

cronxeh

9.9999999999999999 * 10^99999 looks kinda big enough to be infinity, but so does
9.9999999999999999 * 10^999999999

They would be both considered 'infinity' in some context. Division by zero is impossible since it is undefined, and not 'infinity'. Undefined means that there is no value for such a thing, and the limit of such a division is infinity

so inf/0 * 0/inf = 1 is false

6. Mar 8, 2005

Ba

I seem to remember something like that with the directrix and foci of an elipse where the foci were a^2/c and the distance between foci bieng 2c so as the foci moved inwards to the center then the directrix moved out to infinity. So a circle could be thought of as an elipse with a directrix infinitly far away. 2a was the distance of the minor axis.

7. Mar 8, 2005

strid

why not just 10^100000 ?

waste of keyboard :tongue:

8. Mar 8, 2005

nnnnnnnn

inf/0 * 0/inf = 1

Assuming you can cancel the 0's out, you get inf/inf which is not defined under normal circumstances so this problem being equal to 1 is most probably not true.

9. Mar 8, 2005

nnnnnnnn

or 99999999^999999999

10. Mar 9, 2005

Zurtex

Or much much better yet:

A(99999999^999999999,99999999^999999999)

Where A(m,n) is the Ackermann Function

11. Mar 9, 2005

HallsofIvy

Staff Emeritus
Yes, you start with a circle that has one "focus", the center, and then let one of the foci move away- you get ellipses with greater and greater eccentricity. When the focus "goes to infinity", the eccentricity goes to 1, the "other" end of the ellipse goes to infinity and you have a parabola!

If you keep going ("to infinity and BEYOND!") the eccentricity becomes greater than 1: a hyperbola- and the other end of the 'ellipse' now appears on the other side of the world!

Yes, I've used that in classes myself- it's cute but it isn't mathematics: don't try to make a precise mathematical statement out of an analogy.

12. Mar 9, 2005

BobG

No, but you could say that the limit of a/b*b/a as 'a' approaches infinity and as 'b' approaches 0 equals 1, no matter how close each gets to infinity or zero (just as long as neither quite reaches its destination).

13. Apr 1, 2005

leon1127

from series, we know that inf/inf could be converge. The key is the rate of growing to inf
here is an example
Limit[n^2/n!, n>inf], this is a case that inf/inf, but the answer actually converge to zero.
because the rate of growing of numerator and denominator are different.
So first of all, if we assume we can reduce the zeros, we get inf/inf, but since these two infinities may be growing in different rate, it might actually converge to a real number(it might not too!)

if i have any mistake, please tell me

14. Apr 14, 2005

Tom Something

i wouldn't trust "I/0 * 0/I = 1", and here's why:

first, as mentioned in someone else's reply, "/" only refers to finite values. I was first thinking there could be an exception, for instance, if you have two formulae, and plugging in x for both of them gives you "I/0" in one and "O/I" in another. You _could_ do the division of the formulae before plugging in x and simplifying, and that could cancel out the problem and even lead to answers that aren't 1. But the question that came to mind from there is this: what would happen at that point on the graph?

0/0 on a graph often leads to a "hole", a single point on the graph that is undefined, but every number infinitely close to that point _is_ defined.

That's why I don't think I/0 x 0/I exists in practical math.

15. Apr 14, 2005

Hurkyl

Staff Emeritus
Minor nitpick -- in the reals, for any number x, the only number infinitely close to x is x itself.