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

 

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

<br /> <br /> What if f(x) = 2x? By your logic you&#039;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 &quot;=&quot; comparison does not hold the property of transitivity and therefore can&#039;t be an equivalence relation because clearly 1 != 2. I think then that using &quot;=&quot; the way you did would then make no sense (much like a lot of what I just said). I&#039;m not sure if that&#039;s the definition of &quot;undefined&quot; 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&#039;(x)}{g&#039;(x)} = \lim_{x \rightarrow 0} \frac{f(x)}{g(x)}

Then, provided that \frac{\lim_{x \rightarrow 0} f&#039;(x)}{\lim_{x \rightarrow 0} g&#039;(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&#039;(x)}{g&#039;(x)} = \frac{\lim_{x \rightarrow 0} f&#039;(x)}{\lim_{x \rightarrow 0} g&#039;(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.

 

[AndersHermansson]

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.

It is a common misperception that 0/0 is undefined. It is merely indeterminate.

Consider that expression:
\frac{0}{0}=a

is equivalent to:
0=a\cdot 0

which is true for any number a (it is not undefined). Hence 0/0 is indeterminate.

 

[Hurkyl]

It's not a misperception if it's right.

0/0 is infix notation for the application of the binary function '/' to the argument (0, 0) -- a pair that is outside of the domain of '/'. When one writes a string of symbols that denotes the evaluation of a function someplace outside of its domain, one says that string of symbols is undefined.

 

[AndersHermansson]

It's not a misperception if it's right.

0/0 is infix notation for the application of the binary function '/' to the argument (0, 0) -- a pair that is outside of the domain of '/'. When one writes a string of symbols that denotes the evaluation of a function someplace outside of its domain, one says that string of symbols is undefined.

But the statement is clearly true for all a. How can an undefined statement be true? No that doesn't add up my friend!

 

[Hurkyl]

0 = a * 0 is true for all a. 0 / 0 = a is not, because it's undefined.


Leaving formalism behind and just speaking heuristically for a moment, a/b is supposed to be the unique number such that b * (a/b) = a. There is no x that is the unique solution to 0 * x = 0. Thus, there is no x that could satisfy 0/0 = x.

 

[AndersHermansson]

Leaving formalism behind and just speaking heuristically for a moment, a/b is supposed to be the unique number such that b * (a/b) = a. There is no x that is the unique solution to 0 * x = 0. Thus, there is no x that could satisfy 0/0 = x.

Which is why i would call it indeterminate.

An expression in mathematics which does not have meaning and so which is not assigned an interpretation. For example, division by zero is undefined in the field of real numbers.

Since a/b=c/d iff a*d=c*b, the expression 0/0=a isn't meaningless, hence I wouldn't call it undefined.

I guess it blows down to what you want undefined to mean, exactly. I get your point about uniqueness. Only I wouldn't call it undefined, since something undefined is just something that lacks definition. And this particular expression can be defined this way and not cause any trouble ...

 

[Zurtex]

If it causes no particular trouble and 0/0 = a, for some a we don't know (one might some undefined a but nevermind). Then how can we be sure 0/0 = 0/0? And if equality isn't so simple then surely it does cause trouble!

 

[shmoe]

Since a/b=c/d iff a*d=c*b,

Have you ever looked into a rigorous construction of the rationals from the integers? You take all pairs of integers (a,b) where b is non-zero, (a,b) intended to represent a/b, and define the equivalence relation you have above, (a,b)~(c,d) iff ad=bc. Key here is b is by definition non-zero. If you allowed b to be zero you'll run into many problems, so (a,0) is not allowed to represent a fraction for any a and we leave it undefined.

If you want to allow 0/0=a for all a (which would happen if we allowed (0,0) above) then -1=0/0=1, in fact everything ends up equal. Or you would have to admit that your = relation is no longer transitive, which will put you in a world of hurt.

 

[CTS]

http://mathworld.wolfram.com/Indeterminate.html

 

[Hurkyl]

For example, a limit of the form 0/0 ...

I think this is the biggest reason for confusion -- people hear "indeterminate form" for this kind of limit form, but make the mistake of thinking that the label is referring to this arithmetic expression, rather than to the form of a limit.

 

[JonF]

0/0 turns up pretty regularly in almost all levels of math. I was helping a trig student verify some trig identities and they were asked to show that: sin(x)/tan(x) = cos(x). But that isn’t true because sin(x)/tan(x) isn’t defined at 0 and cos(x) is.

 

[master_coda]

I guess it blows down to what you want undefined to mean, exactly. I get your point about uniqueness. Only I wouldn't call it undefined, since something undefined is just something that lacks definition. And this particular expression can be defined this way and not cause any trouble ...

Something isn't undefined because it can't be defined; it's undefined because it's not defined. Since 0/0 has no defined value, it is therefore undefined. The fact that defining it doesn't break as many rules as defining 1/0 would doesn't really change that.

 

[AndersHermansson]

Ok, thanks for all the interresting responses. I will delve deeper into this!

 

[AndersHermansson]

I think this is the biggest reason for confusion -- people hear "indeterminate form" for this kind of limit form, but make the mistake of thinking that the label is referring to this arithmetic expression, rather than to the form of a limit.

Yes! It seems I have to think about this some more. Thank you =)