# Why isn't zero divided by zero equal to one?

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Can somebody please explain to me why zero violates common sense? Every other number including infinity, divided by itself equals one, so why doesn't the same apply to zero as well? After all, since there are numbers which are less than zero, zero itself is logically a number too, and zero does go into itself once. It just seems totally counter-intuitive that 0/0 doesn't equal one.

And while we're talking about counter-intuitive math, can someone also explain why a negative times a negative equals a positive, even though a positive times a positive doesn't equal a negative.

Please don't use fancy terms that none but a math major could understand.

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Division is defined as the inverse of multiplication. That is,

$x/y = z \; \Rightarrow \; yz=x$

Now set x and y to 0. The product of y and any number is zero; in other words there is no unique solution to $0\cdot z = 0$.

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I don't know if this is the accepted answer, but it makes sense to me like this: NOTHING divided by anything at all, even itself, is still NOTHING. If you divide nothing up, even into no parts - there is nothing still nothing there.

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English guys, english. There has to be something to be nothing, and as I said, since there are numbers less than zero, zero must therefore be something and something divided by itself equals one.

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I saw some bogus proof on wikipedia claiming that if 0/0=1 than you could prove that 1=2, it was a lot of nonsense though, whoever wrote that proof seriously needs to go back to school.

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Can somebody please explain to me why zero violates common sense?
It doesn't. It's nothing, while all the other numbers represent something. Even negative numbers aren't nothing, they represent a deficit of something.
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But it is still a number divided by itself.

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It's still a number being divided by zero.

[imath]\frac{0}{0}[/imath] is basically "how many times can you put nothing into nothing?" As D H pointed out, the answer isn't only 1, it's "as many times as you want".

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But it is still a number divided by itself.
Zero's uniqueness overrides the rule of dividing a number by itself.

Imagine you have all the positive numbers in your right hand and all the negative numbers in your left. Zero sits between your two hands, representing the absence of either.

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But it is still a number divided by itself.

Then perhaps your inherent assumption that every number can be divided by itself and then equals one is wrong.

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Look at the graph of y=c/x where c is an arbitrary constant. Try to take the limit as x approaches 0.

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He's talking about 0/0, not 3/0. This case is somewhat different.

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zero isn't an arbitrary constant?

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

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$\lim_{x{\rightarrow}0}\frac{c}x$ does not exist. You can't divide by zero. I think my point stands.

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Dividing anything by zero is considered undefined.

If zero is any other number liek you claim it is, why would having zero in the numerator change things?

It's still undefined.

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For clarification:

$\lim_{x\to +0} \, c/x = \left\{ \begin{array}{rcl} \infty & : & c>0 \\ 0& : &c=0 \\ -\infty & : & c<0 \end{array} \right.$

$\lim_{x\to -0} \, c/x = \left\{ \begin{array}{rcl} -\infty &:& c>0 \\ 0&:&c=0 \\ \infty &:& c<0 \end{array} \right.$

=> $\lim_{x \to 0} \, c/x = \left\{\begin{array}{rcl} 0&:&c=0 \\ \text{non-existant} &:& c \neq 0 \end{array} \right.$

But imho limits have little to nothing to do with the question at hand which I consider purely algebraic, not analytic.

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Historically the debate in the smoked filed room was; Lim x/x; x-->0 equals 1 right up to the end, then conceptual debate. The second school of thought was 0 divided by anything is 0 but this also created debate due to the third school of thought. The third alternative fought for anything divided by zero is infinite, also created debate "Taste great, less filling, cost less." It was settled, on the second alternative. A few more brandies....

Part of what may have settled it is 0+0=0, 0-0=0, 0X0=0; so they didn't want to break up the set. 0X0=0 was transposed into division.

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My school of though is $\lim_{x\to 0} \frac{42x}{x} = 42 \Rightarrow \frac{0}{0}=42$.

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$\lim_{x{\rightarrow}0}\frac{c}x$ does not exist. You can't divide by zero. I think my point stands.

Zero is not an arbitrary constant; $\lim_{x\to 0}\frac 0 x$ does exist. Because $\frac 0 x \equiv 0 \;\forall x \ne 0$, $\lim_{x\to 0}\frac 0 x = 0$.

The same argument is not valid for some arbitrary non-zero constant. $\frac c x$ grows unbounded as $x \to 0$ if $c\ne 0$.

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• 2 weeks later...

Both 0/0 and infinity/infinity are undefined. The best way to understand this IMO is to replace the 0's or infinity's by a function that tends to 0 or infinity. By changing which functions you use, you can get any result you want.

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I guess that depends what field you are in.

Physicists have no problem making infinity / infinity equal a finite number.

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All it comes down to is that [imath]\frac{}{0}[/imath] doesn't mean anything.

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Well, a set of axioms must be established. I think everyone reading can agree that

Axiom 1:

0 = 1/∞

Therefore 0/0 identical to (1/∞)/(1/∞). Using (x/y)/(p/m)=(x/y)x(m/p), it must be true that 0/0 is the same as ∞/∞. This is very similar to the first question. Why isn't ∞/∞ simply one?

Well, ∞ = ∞+∞. Following on, if ∞/∞ was 1, then what is (∞+∞)/∞? Splitting the fraction, it becomes apparent that (∞+∞)/∞ = (∞/∞)+(∞/∞).

1 can never be equal to 2.

Edited by DeanK2
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Well, a set of axioms must be established. I think everyone reading can agree that

Axiom 1:

0 = 1/∞

Therefore 0/0 identical to (1/∞)/(1/∞). Using (x/y)/(p/m)=(x/y)x(m/p), it must be true that 0/0 is the same as ∞/∞. This is very similar to the first question. Why isn't ∞/∞ simply one?

Well, ∞ = ∞+∞. Following on, if ∞/∞ was 1, then what is (∞+∞)/∞? Splitting the fraction, it becomes apparent that (∞+∞)/∞ = (∞/∞)+(∞/∞).

1 can never be equal to 2.

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Well, I DON'T agree with your "axiom 1". In fact, since $\infty$ is NOT a real number, I have no idea what you could mean by $1/\infty$. Assuming that you are working in some extended that allows arithmetic with $\infty$, do you have any evidence that (x/y)/(p/m()= (x/y)(m/p) is true in that system?

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