# Homework Help: How to proof 0\0=?

1. Oct 15, 2007

### hashimcom

how to proof 0\0=???

hi
im anew member..........
and i search about an appear proof to 0\0 = undefind
becuse i want to arrive to scince fact after proof
hashim​

2. Oct 15, 2007

### CompuChip

Hi, welcome to PF.

I think this discussion has already been held quite often, so might use the search function. In short: íf it existed, it would be a limit, as we cannot divide by zero. Moreover, íf this limit which we call 0/0 were defined, it would be unique.

But both
$$\lim_{x \to 0} \frac{0}{x} = 0$$
and
$$\lim_{x \to 0} \frac{x}{x} = 1$$
are good candidates, though they are not equal, and therefore there is no limit.

You can also view it as a multiplication problem, 0/0 would have to be the number you need to multiply 0 with to get 1 (which is the reason division by zero is not allowed for any number)

3. Oct 15, 2007

### Gib Z

It's simple to prove any real number divided by zero is undefined, because we define division as the inverse operation to multiplication. Division is defined as such:
$$\frac{a}{b} = c$$ If and only if $$a=bc$$, where a, b and c are real numbers.

Now 2 cases: For values of a not equal to 0, and when a=0.
When a is not equal to zero, then the definition states that $$\frac{a}{0}=c$$ if and only if $$a=0\cdot c = 0$$. But we already decided that a is not equal to 0. So the first case of a not being equal to 0 can not hold, it is not defined.

The second case of when a is also 0 becomes thus:
$$\frac{0}{0} = c$$ where c is some real number, if and only if $$0 = 0 \cdot c$$.

However, we can see that any value of 0 will satisfy that condition. Hence, if we were to evaluate 0/0, we could say it is equal to any value we like. We can't do that because then many inconsistencies come up when dealing with real numbers, so mathematicians like to leave it undefined.

4. Oct 15, 2007

### hashimcom

mr GIB Z
the proof is not clearlly.......
when i was studdy at universty the doctor proof it to our .........but
i search about it but i dot find it???

5. Oct 15, 2007

### Gib Z

What I did WAS Proof by contradiction.

There were 2 cases. For the first one, I assume a is not equal to 0 but the definition implies that a IS 0, so that is a contradiction.

In the second case, The definition shows that you can choose whatever you like for the answer, which is again a contradiction. Read through my post again.

6. Oct 16, 2007

### hashimcom

mr GIb Z

i dont understand proof???? please hellp me??
hashim

Edit: hashim, you have been warned several times to stop with the obnoxious formatting.

Last edited by a moderator: Oct 16, 2007
7. Oct 16, 2007

### Hurkyl

Staff Emeritus
0/0 is undefined because the definition of division says so. (I'm assuming you mean division of real numbers, division of rational numbers, or something 'ordinary' like that)

There is nothing to prove here; there can only be explanations of why people chose to use that definition.

8. Oct 16, 2007

### Numzie

So 0/0 can be proven to be defined but the definition states that it must be undefined?

9. Oct 16, 2007

### Hurkyl

Staff Emeritus
Incorrect.

0/0 cannot be proven to be defined, and the definition states that it must be undefined.

10. Oct 16, 2007

### Numzie

This because 0/0=1 but 0(0)=0?

11. Oct 16, 2007

### Hurkyl

Staff Emeritus
No. This is because domain of the division operator requires a nonzero denominator. Mathematically speaking, 0/0 is gibberish.

12. Oct 17, 2007

### Sleek

N/0 where N != 0 doesn't seem to have any solutions (by definition of division) as Gib Z proved. But 0/0 seems to have infinitely many solutions, and is thus indeterminate. I.e., in the case of 0/0, any number multiplied by 0 yields zero. Thus if division of two functions gives an indeterminate form, it can have a finite value depending on the functions, if it exists. For reference, one can take a look at L'Hospital's rule. Ofcourse, all the above uses concept of limits.

Regards,
Sleek.

13. Oct 17, 2007

### Gib Z

Ok say we have what you are talking about: $$\lim_{a\to 0, b\to 0} \frac{a}{b}$$.
This is actually the ratio of two infinitesimals, or differentials. And you know that can have a finite value, like you said. But that is different to the actual, 0/0, rather than this limit. Basically given values of a and b, arbitrarily small, there will still be a finite value that it evaluates to. ie In the definition of division i proposed, there WILL be a solution to that equation. However when b is EXACTLY 0, there no unique number will work, and assigning some principal value will ALWAYS lead to inconsistencies in arithmetic.

14. Oct 17, 2007

### CompuChip

OK, one final attempt to convince people:
There is also a theorem, I think, that says that
$$\lim_{(a, b) \to (0, 0)} f(a, b)$$ (1)
exists iff
$$\lim_{a \to 0} \left( \lim_{b \to 0} f(a, b) \right)$$ (2)
and
$$\lim_{b \to 0} \left( \lim_{a \to 0} f(a, b) \right)$$ (3)
both exist, in which case all the limits are equal.
Now take $f(a, b) = a / b$ and you will see that the limit in (2) does not exist.

Last edited: Oct 17, 2007
15. Oct 17, 2007

### Gib Z

I think you meant that the first limit does not exist, but either way that theorem works =]

16. Oct 20, 2007

### Owen Holden

It makes no sense to say 0/0 is defined as undefinable.

If we define division between the numbers x and y as: x/y =df (the z: x=y*z),
then neither 0/0 or x/0 are unique.
That is to say neither 0/0 nor 1/0 exist, even though they are defined!

There is no unique number that 0/0 is.
There is no unique number that 1/0 is.

Note: within hypercomplex numbers x/(a zero divisor) does not exist either.

17. Oct 20, 2007

### Hurkyl

Staff Emeritus
Whether or not that's true, it's not quite what I said. I said that / is defined so that 0/0 is an undefined expression.

That's not a definition. It fails two key semantic criteria: a function must have a value at each point in its domain, and that value must be unique.

But in the spirit of what you were trying to do, you can define a ternary relation, such as
"z is a quotient of x and y" iff x = y * z.​
In this case, the expressions
z is a quotient of 0 and 0​
and
z is a quotient of 1 and 0​
are, in fact, well-defined. (and, of course, the former is identically true and the latter is identically false)

18. Oct 20, 2007

### Owen Holden

You said "0/0 is undefined because the definition of division says so."
Presumably then, your defiition of 0/0 says that 0/0 is undefined??

What exactly does your definition of division say??

Quote:
If we define division between the numbers x and y as: x/y =df (the z: x=y*z),
then neither 0/0 or x/0 are unique.

What the hell are you talking about?

Where do you get your information?

19. Oct 20, 2007

### CompuChip

Here, for example, which corresponds to my definition by the way. Note the part where it says
or
This is also what I learned in my first year Mathematics course. What's your definition?

20. Oct 20, 2007

### Hurkyl

Staff Emeritus
Definition A real number x is said to be invertible iff there exists a real number y such that xy = 1.

Definition $\mathbb{R}^\times$ is the set of invertible real numbers.

Definition $(\cdot)^{-1}$ is the function whose domain and range is $\mathbb{R}^\times$ and satisfies the equation $x x^{-1} = 1$ for all invertible x.

Definition / is the binary function whose domain is $\mathbb{R} \times \mathbb{R}^\times$, whose range is $\mathbb{R}$ and is given by $x/y = xy^{-1}$.

Sorry for the verbosity, but the typical presentation really does go in stages like this. You could combine the definitions to get something that looks similar to yours, but with one key difference: it requires the denominator to be an invertible number.

Since 0 is not invertible, (1,0) and (0,0) are not in the domain of /, and so 1/0 and 0/0 are undefined expressions.

Last edited: Oct 20, 2007
21. Oct 20, 2007

### Gokul43201

Staff Emeritus
Actually, it is easier! The correct definition (in the reals) is:

$$\frac{a}{b} = c$$ If and only if $$a=bc$$

where a, b and c are real numbers, and b is nonzero.

Edit: Oops, didn't see page 2.

22. Oct 20, 2007

### Gib Z

Well, whenever a definition has one less restriction and that restriction comes about naturally, it seems like a more beautiful definition no :)? Thats what i tried to do, except in the case where a=0 the proof was quite shaky? Then again, when a definition rules out any possibility of dispute, I guess thats nice too =]

Many years of hard work and study, he is very talented I assure you. Normally I would say to someone if your answer doesn't agree with someone else's, check both answers over and try to see whose made the mistake. But when its Hurkyl, just check your answer again =]

23. Oct 22, 2007

### JonF

If you are talking about /0 as “divided by 0”: The division operator is defined to be the multiplicative inverse of the argument multiplied. The additive identity has no multiplicative inverse.

If you are talking about 0/0 as a number, look how fields of quotients are constructed.

If you want to know WHY we don’t define 0/0 there are plenty of good reasons. Many of the so called “proofs” in this thread illustrate them.

24. Oct 22, 2007

### drpizza

Another way to think about it is to attach physical meaning to it. 15 divided by 3 equals 5. This could mean 15 dozen eggs divided into 3 piles. How many in each pile?

Or, 15 divided by 30 = 1/2. 15dozen eggs, divided into 30 piles = 1/2 a dozen in each pile.

But, divide 15 dozen eggs up into zero piles??? To quote Hurkyl, "gibberish"

25. Oct 23, 2007

### CompuChip

Though, drpizza, the analogy doesn't extend (much) beyond integers.
E.g. $15 / (2/3)$ or $15/\sqrt{2}$ are perfectly well defined, though I'm having trouble imagining $\sqrt{2}$ piles (let alone dividing 15 eggs among them, though $\sqrt{8}$ eggs would be no problem).

But I'm probably just niggling