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**LuisRodg****Real Member**- Registered: 2007-10-23
- Posts: 322

I'd like to raise this question and see if there is an explicit answer or if this is just one of those topics that is not well established and yields different answers depending on who you speak to.

Basically all of this started for some reason during my Axiomatic Set Theory class, the topic just came and we asked our Professor who is a Set Theorist/Logician (Foundations) and he told us that 0^0 = 1. He gave a quick argument using Power Series:

He mentioned he also had another argument from a logic point of view but didnt show it.

So now yesterday in our Abstract Algebra, our Professor who is a Algebraic Geometer mentioned during the lecture for some reason that 0^0 is undefined to which we all told him the discussion we had in Axiomatic Set Theory. Well, he couldnt believe that our other Professor had said it was equal to 1 since he says that it is obviously undefined.

So in this thread I basically ask, which one of my Professors is right? Or is this one of topic in which no one is right and you assume whatever you want?

Regardless, I would like to see your thoughts

*Last edited by LuisRodg (2009-02-05 12:39:00)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I have always been of the opinion that 0[sup]0[/sup] should be defined to be 1, but some people (like **mathsyperson**) insist that that 0[sup]0[/sup] is undefined.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I just don't see how any value it could take would be logical.

We can all agree that:

x^0 = 1, for x≠ 0

0^x = 0, for x≠ 0

Whatever value 0^0 has, it will be breaking one of those patterns, so nothing would make sense.

So my feeling is that it's best to just write it off as a weird case and not assign any value to it.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

To ease power series notation, 0^0 is taken to be 1 in complex analysis.

I agree with both Jane and mathsyperson, in a way. It is perfectly fine to leave it undefined, but it is also perfectly fine to define it which ever way you like when it is useful. Usefulness is the key.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**lorddoomicus****Member**- Registered: 2008-12-01
- Posts: 4

For what it's worth, both the UNIX 'bc' program, and the basic on the C-64 interpret 0^0 as 1.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Ricky wrote:

I agree with both Jane and mathsyperson, in a way. It is perfectly fine to leave it undefined, but it is also perfectly fine to define it which ever way you like when it is useful. Usefulness is the key.

Im not convinced by Mathsys argument. He claims theres a pattern but I dont see any.

He is also claiming that for which is false.

I would need a much better argument to be convinced that 0[sup]0[/sup] should be left undefined.

*Last edited by JaneFairfax (2009-02-06 00:31:03)*

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**Daniel123****Member**- Registered: 2007-05-23
- Posts: 663

I've seen a nice combinatorial 'proof' that 0^0 = 1.

"The cardinal number n^m is the size of the set of functions from a set of size m into a set of size n. If m is positive and n is zero, then there are no such functions, because there are no elements in the latter set to map those of the former set into. Thus 0^m = 0 when m is positive. However, if both sets are empty (have size 0), then there is exactly one such function the empty function."

The only function between the empty set and itself is the identity function.

*Last edited by Daniel123 (2009-02-05 22:29:54)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oof...

Point taken, that was a silly mistake.

In that case, fair enough. 0^0 =1 is intuitive.

I'm not clear on what this topic is about though.

Are we discussing what 0^0 **is** or what it **should be**?

Those two things might have entirely different answers.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

No, Mathsy, its not a question of intuition. Its a question of usefulness (as Ricky has said).

This is, in detail, what you are suggesting above.

Hence there is no way to define

so that both and are continuous from the right at . Well, rather than being cynical and leaving to be undefined at this point, why not choose which of the two functions to make continuous (from the right) at 0?Now

is already undefined for half of the real line. Choosing to make it continous from the right at 0 by defining to be 0 would serve very little purpose indeed. On the other hand, defining would make continuous over the entire real line, whichAnd indeed, it turns out that in many mathematical applications, having

is very useful even to the point of seeming natural. That is why defining has advantages over leaving it undefined.Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 301

Doesn't 0^0 = 0/0?

I suspect that this is one of the proofs with a fundamental flaw (like the proof 1=2) but where is the flaw? Is it as simple as dividing by 0?

To me this says: 0^0 is undefined because 0/0 is undefined.

What is complex analysis? Is it simply analyzing complex functions the Reinman Hypothesis? Or does it mean something else?

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

bossk171 wrote:

To me this says: 0^0 is undefined because 0/0 is undefined.

Then to you is also undefined because is undefined.

**But **

So do you now see the flaw (in both proofs)?

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**bossk171****Member**- Registered: 2007-07-16
- Posts: 301

JaneFairfax wrote:

So do you now see the flaw (in both proofs)?

I see that there is a flaw, but I'm not sure what. If I had to guess, I'd say it's the divide by zero flaw, but which step specifically is the no-no? Is it 0^0 = 0^(1-1) step? I'm not sure why, but I think it is.

And surely 1 does equal x/x so what is the actually flaw? Like I said, I get that there is one, I just can't tell what it is *specifically*.

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Not that one. Here is the flawed step:

bossk171 wrote:

Compare that with this:

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm not clear on what this topic is about though.

Are we discussing what 0^0 is or what it should be?

Those two things might have entirely different answers.

Looks like we got a Platonist

In the case where properties break down, I believe the Platonist has to recognize that we will never know what something *is*. That is, as long as an argument can be made which contradicts what you think it should be, there is no way we can claim one as mathematically superior to the other. You'll always be left with with a feeling out doubt.

What is complex analysis? Is it simply analyzing complex functions the Reinman Hypothesis? Or does it mean something else?

That's pretty much it. The property of being differentiable as a complex function is much stronger than being differentiable in the real sense, and complex analysis is figuring out how much stronger it is and precisely what it gives us to play with.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Google says 0^0 is 1.

Yahoo says 0^0 is 1.

Windows Live Search says 0^0 is 1.

I says 0^0 is 1.

Linux FTW

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

simron wrote:

Google says 0^0 is 1.

Yahoo says 0^0 is 1.

Windows Live Search says 0^0 is 1.

I says 0^0 is 1.

If that's the case, I have a used car to sell you. I have a bunch of people that will tell you it's a great deal.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**bossk171****Member**- Registered: 2007-07-16
- Posts: 301

simron wrote:

Google says 0^0 is 1.

Yahoo says 0^0 is 1.

Windows Live Search says 0^0 is 1.

I says 0^0 is 1.

Excel says it's undefined.

TI-83 says it's undefined.

I say I still don't know enough about it to make an educated decision.

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Ricky wrote:

If that's the case, I have a used car to sell you. I have a bunch of people that will tell you it's a great deal.

Sorry, I'm too young to drive. My feet don't even reach the pedals.

Actually, I've done a little thinking and changed my mind on it. 0^n is 0, so I'm inclined to believe that 0^0 is equal to 0 as well. Think about this: the definition of a^b is axaxaxa... b times. If it's raised to the 0 power, then a^0 is just this: -> <-. 0 is defined to be the absence of something, so 0^0 is 0. At least in my opinion.

*Last edited by simron (2009-02-08 07:22:57)*

Linux FTW

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

simron wrote:

Think about this: the definition of a^b is axaxaxa... b times. If it's raised to the 0 power, then a^0 is just this: -> <-. 0 is defined to be the absence of something, so 0^0 is 0. At least in my opinion.

In my opinion, this is Alice-in-Wonderland nonsense. Sorry.

Moreover, you appear to be confusing empty sum with empty product . When you multiply no numbers together, you get the empty product, not the empty sum. The empty sum is 0, but the empty product is 1, not 0.

*Last edited by JaneFairfax (2009-02-08 10:56:05)*

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Good point Jane...

*head explodes*

I give up.

Linux FTW

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

Here's an argument that 0^0 could be 0 or 1.

This argument may not be valid (I came up with it, and this is the first place I am placing it in for criticism), but it claims that 0^0 is both 1 and 0 in ordinary arithmetic!

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

0^0 can be defined nicely immediately after introducing the field axioms which give us multiplication and the multiplicative identity 1.

There are a number of formulas in mathematics that are much nicer if 0^0 is defined to be 1.

You have probably seen several of them, so I won't go into them.

I don't see any reason to go up to calculus to try to make 0^0 work. The indeterminant form

0^0 in calculus is a **form** that limits may tend to BUT they never actually become 0^0. Hence using these types of limits seems inappropriate.

Let's go back and define exponentiation in a field (we will use the real number field). The usual definition is something like: ... x^3=x*x*x, x^2=x*x, x^1=x and following this pattern we would get x^0 to be a blank space. Hence x^0 is usually defined separately as x^0=1 if x is not equal to zero. And then 0^0 is usually said to be one or undefined.

To do it a little differently, let's define exponentiation as follows:

..., x^3=1*x*x*x. x^2=1*x*x. x^1=1*x. so following this pattern x^0=1. We don't need a separate definition for x^0. And this definition automatically makes 0^0=1 since it doesn't matter what the x is. Just drop the last x off of the right side of the definition for x^1=1*x.

This definition has been around a while, but it is still not widely known.

0*x=0 comes from a similar definition: ...3*x=0+x+x+x, 2*x=0+x+x, 1*x=0+x, 0*x=0.

This kind of definition works for any group. If e is the identity for the group and # is the symbol for the operation and x is an element of the group then we have

... x^3=e#x#x#x, x^2=e#x#x, x^1=e#x, x^0=e.

And this could be done with coefficients (like 3x, 2x, 1x, 0x above but in groups repeating the operation with a fixed element is usually represented by exponentiation).

This also works for union and intersection in topology where the identity for union is the null set and the identity for intersection is the whole space X. This eliminates the need for vacuous arguments for unions and intersections over a null family of sets.

How we define things in mathematics is extremely important. Also the symbolism we use can "make or break" a concept.

*Last edited by noelevans (2012-07-21 11:29:51)*

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,845

Hi noelevans

That definition x^0 is applicable only when x<>0. Why? When you divide two powers of x you subtract the exponents and vice versa, so x^0=x^(1-1)=x/x. Now, when you want to see what 0^0 is, yo would get that it is the same as 0/0, which is indeterminate i.e. undefined.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

I'm afraid there is a fallacy in the argument that 0^0 = 0/0.

Let y be the reciprocal of x. Then xy=1. Now x^0 = x^(1-1) = x^1*x^(-1) = x*y = 1 (not x/y Oops)

This works fine for non-zero x. But when x=0 the y in the formula does not exist since zero has no reciprocal. Hence the quantities x^(-1) and y in the third and fourth expressions do not exist. Hence we cannot conclude that 0^0 = 0/0.

It is not valid to apply the law of exponents x^(1-1) = x^1*x^(-1) when x=0.

So 0^0 is not equal to 0/0 via this argument. We are then free to define 0^0 unless we can find some valid reason that it is not permissible; that is, some valid reason that it must remain undefined. So if we want to claim that 0^0 is undefined we need to find a valid proof of this.

*Last edited by noelevans (2012-07-22 12:05:31)*

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,845

Hi

Do you agree that x^0=x^(1-1)?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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