Proving 0/0 ≠ 1: Does Math Really Work This Way?

[Smurf]

Ok, after reading this, this, and the beginning of this I've realized 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.

 

[Hurkyl]

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.

 

[Smurf]

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

 

[Hurkyl]

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.

 

[Smurf]

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

 

[Icebreaker]

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

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

 

[Pyrrhus]

Ice, the factorial of 0 is defined as 1.

 

[Icebreaker]

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

 

[Smurf]

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

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?

 

[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)

 

[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!

 

[Hurkyl]

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)

 

[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 [itex]\frac{0}{0} = 1[/tex][/itex]

 

[Hurkyl]

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

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)

 

[quetzalcoatl9]

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)

wait a minute, i thought that's the whole point of L'Hopital's rule, evaluating indeterminate things of the form 0/0

 

[master_coda]

wait a minute, i thought that's the whole point of L'Hopital's rule, evaluating indeterminate things of the form 0/0

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.

 

[quetzalcoatl9]

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.

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.

 

[Hurkyl]

The equation I quoted is wrong. There is no theorem that let's you conclude that.

 

[vsage]

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 [itex]\frac{0}{0} = 1[/tex][/itex]

$<br /> What if f(x) = 2x? By your logic you'll still get the answer to 0/0 I think.<br /> <br /> Saying that 0/0 = 1 (and by me saying by your logic also = 2), implies that the "=" comparison does not hold the property of transitivity and therefore can't be an equivalence relation because clearly 1 != 2. I think then that using "=" the way you did would then make no sense (much like a lot of what I just said). I'm not sure if that's the definition of "undefined" though.$

 

[master_coda]

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.

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

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 $\lim_{x\rightarrow0}g(x)\neq 0$.

 

[honestrosewater]

Smurf,
If you're still there, I would ask here to resolve the following problem.
Let a denote any number. Say you have the following rules:
1) 0 * a = 0
2) a * 1/a = 1
(Surely, you recognize these.) Let a = 0. Then (2) says:
3) 0 * 1/0 = 1
If you don't see the problem already, let a = 1/0. Then (1) says:
4) 0 * 1/0 = 0
So does 1 = 0? Or does (2) not apply to 0? Or what?
There really isn't anything to argue about, the axioms I've seen always just say that 1 does not equal 0 and 1/0 is undefined, and no one gets to argue with the axioms. If she doesn't like them, she can always make up her own.

 

[quetzalcoatl9]

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.

you're right, this does make sense to me. i concede.

 

[HallsofIvy]

Yes, L'Hopital's rule works for fractions where the numerator and denominator both go to 0. That does not mean that you can write $$\frac{\lim_{x \rightarrow 0} f(x)}{\lim_{x \rightarrow 0} g(x)} = \frac{0}{0}$$ and think of 0/0 as anything other than a short hand way of saying "both numerator and denominator go to 0". "0/0" is not a number.

 

[chronon]

$$\frac{\lim_{x \rightarrow 0} f(x)}{\lim_{x \rightarrow 0} g(x)} = \lim_{x \rightarrow 0} \frac{f(x)}{g(x)}$$
is the incorrect step, at least if you are using it to go from the RHS being defined to the LHS being defined. Suppose you put g(x)=f(x) and choose f(x) which diverges as x->0. Then the RHS is defined, but the limits on the LHS are not.

 

[Night Owl]

Apart from what has already been said about writing $\frac{0}{0}$, you made another important mistake that nobody has mentioned so far, quetzalcoatl9. It deals with the following, which we've already asserted is an incorrect statement (because $\lim_{x \rightarrow 0} g(x) = 0$):

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

The proper use of L'Hospital's rule would have been as follows:

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

Then, provided that $\frac{\lim_{x \rightarrow 0} f'(x)}{\lim_{x \rightarrow 0} g'(x)}$ is defined, you could express the whole operation as follows, if you wanted to:

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

Don't anybody hesitate to correct me if I said something incorrect. I'm the first that would want to know!

 

[pallidin]

The problem here is that Zero is not a quantity. Zero is the absence of quantity.
Therefore, zero cannot be added, subtracted, multiplied, divided, factored, or anything else because has no quantified existence.
Therefore, equations which have, say, x/0 is pointless and inherently invalid.

 

[Hurkyl]

Well, seeing how we have no problem adding, subtracting, or multiplying with zeroes, and dividing with zeroes when they're not on the denominator, it should be clear your reasoning is flawed.

And zero is most certainly quantitative, even in the sense of natural language. For example, it can be an answer to a question of "how many".


But, this is all irrelevant: this is a math subforum, and things follow from definitions, not opinion.

 

[pallidin]

Well, seeing how we have no problem adding, subtracting, or multiplying with zeroes, and dividing with zeroes when they're not on the denominator, it should be clear your reasoning is flawed.

And zero is most certainly quantitative, even in the sense of natural language. For example, it can be an answer to a question of "how many".


But, this is all irrelevant: this is a math subforum, and things follow from definitions, not opinion.

Hmmm. I thought that .999~ is the true representation of "1"
Is that wrong?

 

[Hurkyl]

1.000~ and 0.999~ are both decimal representations of the real number "one". (decimals aren't the only way to represent real numbers, of course)

 

[Moo Of Doom]

Hmmm. I thought that .999~ is the true representation of "1"
Is that wrong?

While .999... = 1, one representation is not more true than the other. I'd say .999... is "a" true representation of 1, but not "the" true representation of 1.