# 1/oo = 0 ?

1. Aug 7, 2007

### Nick666

Well? Is it equal to zero ? If there are threads with this subject, redirect me to them please.

2. Aug 7, 2007

### Kummer

Please look in the philosophy forum.
(Because this is not a topic mathematicians discuss, just philosophers.)

Last edited: Aug 7, 2007
3. Aug 7, 2007

### mgb_phys

To an engineer or physicist yes.
We aren't as squeamish as mathematicians when it comes to needing an answer.

4. Aug 7, 2007

5. Aug 7, 2007

### homology

Or, heck, check out nonstandard analysis... But I assume you're (Nick666) talking about calculus and perhaps a limit that comes up? The idea is that 1/big is small, and 1/bigger is smaller, and so I can always choose an x to make 1/x as small as you'd like it (or as close to zero).

Cheers,

Kevin

6. Aug 7, 2007

### Moridin

Here is the correct mathematical notation:

$$\lim_{x\to\infty} \frac{1}{x} = 0$$

1 over infinity is not a valid computation. Actually, it should be that "the limit of 1 over x as x goes to infinity is equal to zero".

7. Aug 7, 2007

### matt grime

Oh dear, so many misconceptions here.

1/oo is a perfectly good symbol. In the extended complex plane it is 0. As it would be in the extended reals - you do not need limits at all to answer that. However, the symbol 1/oo does not have a canonical meaning - I can think of no symbol in mathematics that has a canonical meaning. It's not even true that there is a unique meaning for the symbol 1, or 0 for that matter, is there, so why should there be such a meaning here?

8. Aug 8, 2007

### jostpuur

Nick666, do you know yourself what you mean with the infinity? Is there a definition you are using?

9. Aug 8, 2007

### Nick666

Let oo be 999... :) . ( oh, can 999... be infinity ?)

Last edited: Aug 8, 2007
10. Aug 8, 2007

### HallsofIvy

Now you're just pulling our chain!

11. Aug 8, 2007

### jostpuur

If you write 999..., I'm afraid I'll have to ask again, that do you know yourself what you mean by that?

For example, a number 123 is $1\cdot 10^2 + 2\cdot 10^1 + 3\cdot 10^0$. In general natural numbers can be written as $\sum_{k=0}^N a_k 10^k$, where for all k $a_k\in\{0,1,2,\ldots,9\}$. Your number starts like $9\cdot 10^{?} + \cdots$, and what do you have up there in the exponent?

Writing ...999 would make more sense, because it would be $\sum_{k=0}^{\infty} 9\cdot 10^k$, but I don't know what this means either, because the sum doesn't converge towards any natural number.

It seems your problem is, that you don't know what you mean with the infinity. If you are interested in the basics of analysis, I think Moridin's answer has the point. $\infty$ is a symbol, that usually means that there is some kind of limiting process. The symbol doesn't have an independent meaning there, but it gets meaning in expressions like $\lim_{n\to\infty}$ and $\sum_{k=0}^{\infty}$.

12. Aug 8, 2007

### Nick666

999... As in infinitely many 9`s .

And 1/"that sum you wrote" = ?

13. Aug 8, 2007

### Nick666

And another question about the sum you wrote. Isnt every element of that sum a natural number? (9, 90, 900, 9000 etc) Or let me put it another way. 10^k, when k ->oo , isnt that a natural number ? I mean, if we multiply 10 by 10 by 10........ and so on, shouldnt we get a natural number?

Last edited: Aug 8, 2007
14. Aug 8, 2007

### CompuChip

Yes it is.
So is every partial sum (cutting off the summation after a finite number of terms).
But the sum itself isn't.

15. Aug 8, 2007

### Nick666

See my above edited post.

But if we add a bunch of natural numbers, no matter how many, isnt it logic that we should also get a natural number ? (or maybe this is why I got low grades at math haha)

16. Aug 8, 2007

### jostpuur

No at all! $\lim_{k\to\infty} 10^k$ is not a natural number.

17. Aug 8, 2007

As you cannot compute $\infty[/tex] (division by zero is undefined), how would it be possible to compute something that involves it without using limits? I'm not sure I understand. 18. Aug 8, 2007 ### matt grime It's just a symbol. One that is used in the context of limits in analysis, and one that is not "computed' (whatever that means) in terms of limits in other contexts. 19. Aug 8, 2007 ### Moridin 20. Aug 8, 2007 ### Nick666 I still dont understand how, if you add a natural number to a natural number and another natural number and so on,you dont get a natural number. If you add 1 apple and 1 apple and 1 apple and so on, dont you get an infinite number.... of apples ??? 21. Aug 8, 2007 ### daveb A natural number is an integer. Infinity is not an integer. 22. Aug 8, 2007 ### CRGreathouse If you add any finite number of integers, you get an integer. If you add an infinite number of integers you could get a natural number or an undefined result. In the 'extended integers' you could get infinity or -infinity in addition to those two. 23. Aug 8, 2007 ### Kummer I really hope the moderators move it to the Philosophy forum. My comments are not to address this "question" but to show you a beautiful phenomona that happens with infinite things. Now, $$1$$ is rational. $$1+\frac{1}{2^2}$$ is rational. $$1+\frac{1}{2^2}+\frac{1}{3^2}$$ is rational. ... And so one. You are assuming that since the finite sums are rational, then infinitely many of them are rational. But that is not true since, $$1+\frac{1}{2^2}+\frac{1}{3^3}+... = \frac{\pi ^2}{6}$$. Not a rational. ~~~~~ Here is another example, Consider the numbers, $$\{ 1 , 2 , 3, \}$$ This set has a maximum value , namely, 3. In fact given any finite set (non-empty) it has a maximum number. And if this set happens to be a set of rational numbers then its maximum is also a rational number. But look what happens with infinite sets. Say we have, $$\{ \mbox{ all rational numbers } < \sqrt{2} \}$$. What is the maximum for that set? It is not a rational number because we can always choose a larger one getting closer to $$\sqrt{2}$$. In fact its maximum is $$\sqrt{2}$$ it is not longer a rational number! Eventhough the set was filled with only rational numbers. So part of mathematics studies infinite things. Whether infinitely many real numbers (that is called real analysis) or infinite sets (that is called set theory) and so one. Because the finite things cause no problem. It is the infinite concepts that are fun and sometimes strange. Note: Have you ever wondered why professional mathematicians never argue (or even discuss) these topics? Because as I said it has little to do with math (perhaps even nothing). Last edited: Aug 8, 2007 24. Aug 8, 2007 ### morphism 1, 2, 3, ..., [itex]\omega$!

25. Aug 9, 2007

### Gib Z

How do you expect him to know what professional mathematicians discuss lol