# Why does 0/0 = undefined?

Hello, i was just wondering....why does 0/0 = undefined. To me, it works mathematicaly.

12/[ ] = 4
fill in the blank...its 3
check the answer...3*4 = 12 woohoo!
12/0 = 12 0*12 = 12 whoops that doesnt work!
12/0 = 0 0*0 = 12 whoops, doesnt work either!
I can see why that is undefined.

0/0 = 0 0*0 = 0 Seems to work...
But there is a problem, because in that case any number could work
0/0 = any# (any#)*0 = 0 Is THIS why its undefined??

I also want to know why 0^0 = 1
I know that any # to the 0 power = 1, but its just weird when u think of it in different ways.
0^0 = 0^x/0^x = 0/0 = undefined

And one more question, why does 0! = 1?

Thanks
Armo

Galileo
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ArmoSkater87 said:
But there is a problem, because in that case any number could work
0/0 = any# (any#)*0 = 0 Is THIS why its undefined??
Yes, you can't assign a number to it in any logical way.
Multiplication by zero is not an invertible operation.
Division is the inverse operation of multiplying, so division by zero
is not possible (even for 0/0).

ArmoSkater87 said:
I also want to know why 0^0 = 1
I know that any # to the 0 power = 1, but its just weird when u think of it in different ways.
0^0 = 0^x/0^x = 0/0 = undefined
0^0 is also undefined, because of what you just showed.

ArmoSkater87 said:
And one more question, why does 0! = 1?
Because it's convenient to define 0!=1
Lots of situations which yield equations which use the factorial n!
are valid in the special case where n=0 is to be considered when we take 0!=1.

The most important example is the binomial coefficient:
$${n \choose k}:=\frac{n!}{(n-k)!k!}$$
which is the number of ways to pick k objects out of a group of n objects.
If k is equal to zero, we should have 1, which is the case if 0!=1.
The binomial coefficient occurs so often, it would be a crutch not to define
0!=1

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matt grime
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Something they don't tell you at school is that the real numbers are a mathematical construction, a lot of work went into getting a good description of them, clearly the idea of decimal expansions doesn't do it as the whole slew of 0.99... not 1 posts shows.

So, the inverse of an element is only defined for non-zero numbers. For some reason this annoys non-mathematicians, though why is a mystery since any attempt at a definition would require us to redefine multiplication. Something you've pretty much nailed.

arildno
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I think it is unfortunate that the standard way of teaching maths in elementary school emphasizes that "subtraction" is something completely different and independent from "addition", and that "division" is something completely distinct from "multiplication".

I believe that a possible reason for "ordinary" persons failure to accept that "division by zero" is inherently "unacceptable" (that is, as long as we demand the "usual" properties of add./mult.)
might be thoughts along this line:
"Hey, I can add whatever two numbers I want together, same with subtraction and multiplication.
So why the f*'k isn't this allowable with division?
It's not like division is less fundamental than the other three ways, is it?"

Just my thoughts, anyway..

matt grime
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I think you may have got it there, arildno, though how you get round that problem seems unobvious.

Thanks all for your responses, i think i understand why 0/0 is undefined and why 0!=1 But i still dont understand the case of 0^0, because it DOES = 1, it isnt undefined.

0^0=1 so that a^0=1 is satisfied (there really isn't any reason to make a special exception for 0^0).

Things like 0^0=0^1*0^-1=0/0 seem like a problem, but of course 0^-1 isn't defined, so that fact that expressions involving it give weird results isn't really a problem. And similarly 0^x for any x<0 is undefined.

lets look at 0^0 in this way:

we know x^0=1 for all x belonging to R

but when you say 2^3, it means 2 is multiplied by itself 3 times.
when you say 2^0, similarly it should mean 2 multiplied by itself 0 times. i.e no operation being performed on 2. now obviously you cant interpret this answer as 1. how did the mathematicians then define x^0=1? i think matt came close to explaining this?

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arildno
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kdkdkd said:
lets look at 0^0 in this way:

we know x^0=1 for all x belonging to R

but when you say 2^3, it means 2 is multiplied by itself 3 times.
when you say 2^0, similarly it should mean 2 multiplied by itself 0 times. i.e no operation being performed on 2. now obviously you cant interpret this answer as 1. how did the mathematicians then define x^0=1?

2^3=2*2*2=1*2*2*2
2^2=2*2=1*2*2
2^1=2=1*2
2^0=1

arildno
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A more formal answer is the following:

We wish to work with functions with various properties.
One such class of functions could be:
For every choice of x and y, we want a function F which satisfy:
F(x+y)=F(x)F(y)

To give an "example" of such a function, consider 2^x
By the rules you've learnt about exponents, you have 2^(x+y)=2^x*2^y
(Right?)
Hence, 2^x "satisfy" the defining property to be an F (Similarly, 3^x is such a function)

Let's now study F(x+0), where x is an arbitrary number.
By the defining property, we have:
F(x+0)=F(x)*F(0)
But since x+0=x, we have:
F(x)=F(x)*F(0), valid for any choice of x!

This equation has really only two basic classes of solutions:
1. If F(x)=0 for all x, the equation is obviously satisfied for every x
2. If F(0)=1, the equation is trivially true (F(x)=F(x)).

Here is my proof that 0! equals one.

By _definition_ of factorial (regardless for what we chose of n) we have: n*(n-1)(n-2)(n-3)...1=n!
This is equal to n*(n-1)*(n-1-1)(n-1-1-1)...1=n! However, this is also equal to
n*[(n-1)!]=n!
thus
1!=1*(0!)
thus
1=1*(0!)
thus
(0!)=1

Dare I say qed? (if I'm off, I expect those more skilled at mathematics than I to correct me!)

Also, 0/0 is undefined for numerous reasons. Most people don't know what undefined really means. Multiplication for numbers is DEFINED in a certain way. I.e, there is a mathematical definition. Division is defined the same way. However, the definition for division does not include the number zero. So, saying anything divided by zero is equivalent to saying: The dog sfdjhsdf the cat.

These notes are pretty AMAZING at describing what you need: http://www.math.ucla.edu/~tao/131ah.1.03w/week1.pdf [Broken]

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Zurtex
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master_coda said:
0^0=1 so that a^0=1 is satisfied (there really isn't any reason to make a special exception for 0^0).

Things like 0^0=0^1*0^-1=0/0 seem like a problem, but of course 0^-1 isn't defined, so that fact that expressions involving it give weird results isn't really a problem. And similarly 0^x for any x<0 is undefined.
How else would you define $a^0$ other than $a^1 * a^{-1}$?

master_coda said:
0^0=1 so that a^0=1 is satisfied (there really isn't any reason to make a special exception for 0^0).
There's a very good reason to make a special exception:

$$\frac {x^a}{x^b} = x^{a-b}$$

$$0^0 = 0^{x - x} = \frac{0^x}{0^x}$$

which is undefined. Thus 0^0 is undefined.

arildno
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Zurtex said:
How else would you define $a^0$ other than $a^1 * a^{-1}$?
1.
Let Exp(x) be a function defined as follows:
$$Exp(x)=1+\sum_{i=1}^{\infty}\frac{x^{i}}{i!}$$

2.
It may be proven that Exp(x) has a lot of fascinating properties; in particular it is positive and invertible.
Let the inverse be called Log(x).

3.
Define the function:
$$E_{a}(x)=Exp(xLog(a))$$
Here, a is some positive number
We introduce a more common notation, by writing:
$$a^{x}=E_{a}(x)$$

4. There is no particular reason to define $$a^{0}$$ in any special manner; it's just equal to Exp(0)=1.

Note that 0^0 cannot be defined in this manner, since 0 is not a positive number (the function $$0^{x}$$ cannot be defined by this procedure).

Neither is there any other logical way to assign some value to 0^0; it remains undefined.

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HallsofIvy
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If a is any positive number, a0= 1 so it would seem natural to define 00= 1. (We've seen that given on this board as a proof that 00= 1.)

However, if a is any positive number, then 0a= 0 so it would seem just as natural to define 00= 0. That is why it is not correct to say, without some proviso, that 00= 1 (or any other number).

By the way, we often say that terms such as 0/0 or 00 are "undetermined" rather than "undefined" because the problem is not that there is no way to define it but, rather, too many ways.

Galileo
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In some cases it is conventional to define $0^0=1$.
For example, take the Taylor series of a function about some point a:

$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$

If we enter x=a:

$$f(a)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}0^n=f(a)0^0$$
So, in this case we obvioulsy want $0^0=1$.

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Zorodius said:
There's a very good reason to make a special exception:

$$\frac {x^a}{x^b} = x^{a-b}$$

$$0^0 = 0^{x - x} = \frac{0^x}{0^x}$$

which is undefined. Thus 0^0 is undefined.
You'll note that I did mention things like 0^x * 0^-x and I pointed out that they aren't a good reason to not define 0^0, since 0^-x isn't defined anyway and so any "bad" results you can derive using it are not significant.

However I would agree that there isn't a truly fundamental reason that we should define 0^0=1. You can make a valid argument that 0^0 should be zero since 0^a=0.

But whenever 0^0 actually pops up in an equation, you can usually avoid using a special case by just defining 0^0=1. So in practice 0^0 is almost always just defined as equal to one, or left undefined if defining it isn't useful for the particular work you are doing.

I apologize if any of this is redundant...

if x/y=z then z boxes each containing y balls will fill a box with x balls.

for example:
8/4=2 means 2 boxes each containing 4 balls will fill a box with 8 balls.

0/1=0 because 0 boxes each containing 1 ball will fill a(n empty) box with 0 balls.

5/0=z means z boxes each containing 0 balls will fill a box with 5 balls. Think about it. How many boxes each containing 0 balls will fill a box with 5 balls?

0/0=z means z boxes each containing 0 balls will fill a box with 0 balls. Think about it. How many boxes each containing 0 balls will fill a box with 0 balls?
***
As for 0^0=1, m^n for natural numbers m and n is the number of functions from a set of n elements to a set of m elements.

SO 0^0 is the number of functions from the empty set to the empty set. There is one function: the empty function.
***
3!=4!/4
2!=3!/3
1!=2!/2
therefore
0!=1!/1=1

The natural extension of factorials to the reals is the gamma function: http://mathworld.wolfram.com/GammaFunction.html

This formulation also implies 0! = 1 and this is the only convention that I'm aware of. The key point is that AFAIK there are no reasons to define it otherwise - no equally valid argument can be made for 0!=0.
The problem with the forms 0/0 and 00 is that different functions suggest different results. For example, for any constant c, the limit c*x/x is of the form 0/0 as x -> 0 and there is no obvious reason to pick one of them. It has been suggested that since f(x)=0x is of no mathematical interest, 00 should be set to one, as many interesting functions would suggest that.

Wow, that all makes sense to me, thanks all. I understand why 0!=1 and why 0/0=undef. As for 0^0...its basicly dependent on the situation and sometimes convenience, right?

Ethereal
Doesn't it make anyone uncomfortable to know that the value of 0^0 can be either 1 or 0 depending on the situation? Can anyone point out other similar situations in mathematics whereby the same mathematical expression can yield a different answer dependent on the context? Otherwise it seems to me that sooner or later, heretics might claim 1+1=3.

lol, thats a great point. It bothers me very much actually, which is why i posted that question here. My common sense told me that it should be 0, then I saw that 0^0 = 0^x/0^x = 0/0 = undefined. This is what bothered me, because supposidly 0^0 = 1, at least it does where ever ive seen.

ArmoSkater87 said:
lol, thats a great point. It bothers me very much actually, which is why i posted that question here. My common sense told me that it should be 0, then I saw that 0^0 = 0^x/0^x = 0/0 = undefined. This is what bothered me, because supposidly 0^0 = 1, at least it does where ever ive seen.
0^0 does not equal one. Someone might decide to replace 0^0 with one for the sake of a certain application, but 0^0 is an indeterminate form. There is no single value you can assign to it and definitively say, "this is exactly what 0^0 means".

If you have a good calculator and put 0^0 into it, you'll usually get a domain error.

ethereal said:
Can anyone point out other similar situations in mathematics whereby the same mathematical expression can yield a different answer dependent on the context?
Multi-valued functions give multiple outputs for a single input. The inverse trigonometric functions are one good example: If I know that the tangent of some angle is one, that angle could be 45 degrees, 225 degrees, -135 degrees, 405 degrees, 585 degrees, and so on. I'd have to look at the situation to decide which angle or angles is the "right" one.

And, of course, there are the other indeterminate forms

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arildno
$$1^{\infty},\infty-\infty,\frac{\infty}{\infty},0*\infty,\infty^{0}$$