# 0/0 (continued)

1. Jun 7, 2005

### Smurf

Ok, after reading this, this, and the beginning of this I've realised there seems to be a consensus that 0/0= indeterminate.

Now, today my friend who's a math person has come up with this 'theory' that 0/0=1. She presented two arguments to support her theory, first she said that 0!/0! = 1, then she said that anything divided by itself = 1, (x/x=1),

I think she's full of crap, but I'm at a bit of a loss, so can you guys help me prove her wrong (or help me understand why she's right). I already tried arguing my basic understanding of the issue, but she won't give in.

2. Jun 7, 2005

### Hurkyl

Staff Emeritus
It's worse than indeterminate -- it's undefined.

It boggles me why people insist on using x/x = 1 to "prove" 0/0 = 1, but they never accept 0/x = 0 to "prove" 0/0 = 0.

Try asking her to actually mathematically prove it. I imagine she won't even know where to begin.

3. Jun 7, 2005

### Smurf

I already did and she gave me the 0!/0!=1 thingy.

4. Jun 7, 2005

### Hurkyl

Staff Emeritus
Then tell her she has failed, because she wasn't able to come up with a proof. She won't believe you, but it's still true.

5. Jun 7, 2005

### Smurf

Hmmm, ok. Is there a way to show her that she hasn't provided proof?

6. Jun 7, 2005

### Icebreaker

But what does 0! has to do with anything? n! does not equal to n.

Btw, what's the latex command for !=?

7. Jun 7, 2005

### Pyrrhus

Ice, the factorial of 0 is defined as 1.

8. Jun 7, 2005

### Icebreaker

I know. But if you were to show that 3/4 = 0.75, you wouldn't go with 3!/4!.

9. Jun 7, 2005

### Smurf

That's what I said and she yammered off some gibberish. Neither of us know the theory behind factorals. Otherwise I would be able to show her where she's wrong. What is 0! anyways? 0*1? 0*-1? And how can it be defined as 1?

Last edited: Jun 7, 2005
10. Jun 7, 2005

### vsage

I don't understand how this person reached 0/0 = 1 because 0!/0! = 1. Factorial signs can hardly be divided out like that! Did you ask her to look at a graph of 1/x and inspect it to the left and right of x = 0? It should alread be clear to her that depending on which side you look at, the graph approaches different values. What's more, placing 1/x = 1 at x = 0 would throw off the symmetry of a graph perfectly symmetric about the line y = x (or y = -x)

11. Jun 7, 2005

### quetzalcoatl9

being ignorant in such matter, i'll suggest the following:

let
$$f(x) = x^n, x \epsilon R$$ and $$g(x) = x^n, x \epsilon R$$

then,

$$\frac{f(x)}{g(x)} = \frac{x^n}{x^n} = x^{n-n} = x^0 = 1$$

so then $$\frac{f(x)}{g(x)} = 1$$ for all the reals, including zero!

12. Jun 7, 2005

### Hurkyl

Staff Emeritus
If you mean to say that f and g are functions, then you're wrong. f(x) / g(x) is not defined at x = 0. (Precisely because 0/0 is undefined)

One might say that f(x) / g(x) has a removable singularity at x = 0.

Or, if you meant to say that f(x) and g(x) are elements of a ring of polynomials, and division is being done in the field of rational functions, it would be correct to say that f(x) / g(x) = 1. (And that would be true "at x = 0", at least the way that phrase is usually given meaning)

13. Jun 7, 2005

### quetzalcoatl9

ok then, how 'bout this.

let
$$f(x) = x, x \epsilon R$$ and $$g(x) = x, x \epsilon R$$

then
$$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} x = 0$$ and $$\lim_{x \rightarrow 0} g(x) = \lim_{x \rightarrow 0} x = 0$$

so then

$$\frac{\lim_{x \rightarrow 0} f(x)}{\lim_{x \rightarrow 0} g(x)} = \frac{0}{0}$$

but by L'Hopital's:

$$\frac{\lim_{x \rightarrow 0} f(x)}{\lim_{x \rightarrow 0} g(x)} = \lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f(x)'}{g(x)'} = \lim_{x \rightarrow 0} \frac{x'}{x'} = \lim_{x \rightarrow 0} \frac{1}{1} = 1$$

therefore $\frac{0}{0} = 1[/tex] 14. Jun 7, 2005 ### Hurkyl Staff Emeritus That's incorrect. The relevant limit theorem specifically states that the denominator cannot be zero when making that equality. (Just like people forget that the identity x/x = 1 specifically states x cannot be zero) 15. Jun 7, 2005 ### quetzalcoatl9 wait a minute, i thought that's the whole point of L'Hopital's rule, evaluating indeterminate things of the form 0/0 16. Jun 7, 2005 ### master_coda It is. But there's a difference between saying that a limit is of the form 0/0 and saying that the limit equals 0/0. A limit is a number (or not defined) and 0/0 is not a number, so how could a limit be equal to it? However, it can look like 0/0 if you attempt to evaluate it in a naive way. Also, you made the mistake of thinking that if the limit of a function is defined at a point, then the function itself must be. 17. Jun 7, 2005 ### quetzalcoatl9 i really didn't assume anything...i tried to be as thorough about it as possible, showing each step clearly. so tell me: which part of it is wrong? because your explanation is a bit 'wishy-washy' (no offense). it's not that i don't believe you, but i want an actually mathematical argument against it. every professor that i showed the L'hopital thing to, did some hand-waving but could never mathematically explain where the flaw was. 18. Jun 7, 2005 ### Hurkyl Staff Emeritus The equation I quoted is wrong. There is no theorem that lets you conclude that. 19. Jun 7, 2005 ### vsage What if f(x) = 2x? By your logic you'll still get the answer to 0/0 I think. Saying that 0/0 = 1 (and by me saying by your logic also = 2), implies that the "=" comparison does not hold the property of transitivity and therefore can't be an equivalence relation because clearly 1 != 2. I think then that using "=" the way you did would then make no sense (much like a lot of what I just said). I'm not sure if that's the definition of "undefined" though. Last edited by a moderator: Jun 7, 2005 20. Jun 7, 2005 ### master_coda Hurkyl already explained what you did wrong; you're trying to use the limit theorem that states $$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow0}f(x)}{\lim_{x\rightarrow0}g(x)}$$ but you're ignoring the fact that this theorem requires that [itex]\lim_{x\rightarrow0}g(x)\neq 0$.

Last edited: Jun 7, 2005