Is 0^0 Equal to 1? An Explanation of Mathematical Concepts and Terminology

[coolul007]

Moderator's note: this was split from here[/color]

Here's a nice problem: Prove that

(Convention: 0^0=1.)

By definition the following is true it is NOT 1:
0^0=0

 

[jostpuur]

By definition the following is true it is NOT 1:
0^0=0

Did you prove my equation for all 0 < j < k already? So that you can now complain about notational issue with 0=j and j=k cases?

But seriously... I just said that I define 0^0 to be 1. Therefore 0^0=1, by definition. OK? What do you think about dialogue like this:

Person A: "I have defined f so that it is f(x)=x^2"

Person B: "I see. I think I'm going to define f so that f(x)=x^3... What a minute! Did you just say that f(x)=x^2? That's wrong! By (my) definition f(x)=x^3, and x^2\neq x^3".

Person A: ""

 

[micromass]

By definition the following is true it is NOT 1:
0^0=0

If there were a definition for 0^0then it would be 0^0=1. We often make that definition to make things work, for example, the binomial theorem:

(1+0)^0=\sum_{k=0}^0{\binom{0}{k}1^k0^{0-k}}=\binom{0}{0}1^00^0=1

Or set theory:

|\emptyset^\emptyset|=0^0=1

So if there is a definition, then it would be

0^0=1

So the OP's convention is certainly justified...

 

[coolul007]

The problem is that you are dividing by zero when 0^0 =1. It implies that (0/0)^0 =1, sorry for the lack of TEX.

 

[micromass]

Setting 0^0=1 is merely a convention which is handy a lot of times. There are obvious downsides to this convention (like the one you mention, and the continuity of 0^x for example), but still the convention is handy enough. You should just be aware of it's problems.

 

[Hurkyl]

The problem is that you are dividing by zero when 0^0 =1.

No, you are raising 0 to the zeroeth power when you write 0^0.

There are lots of reasons why leaving 0^0 is undefined -- one of those rationales is indeed that some exponent laws cannot be extended to cover 0^0 because it would result in a division by zero error.

Don't confuse the two things. (or overlook the details of the latter thing)


For the problem you originally responded to, there is a very good reason to define 0^0=1, and so he adopted the convention for stating his problem and explicitly stated he was doing so. He did everything right.

 

[coolul007]

It seems that the justification of 0^0 = 1, is justified by another definition of o! = 1. 0! = 1, is narrowly justified for manipulating factorials. Out of that context it has no relevance. I do appreciate the voice of reason, but to rely on 0^0 =1 for a proof seems very dangerous.

 

[JJacquelin]

For French readers : "Zéro puissance zéro"
http://www.scribd.com/JJacquelin/documents
.

 

[D H]

It seems that the justification of 0^0 = 1, is justified by another definition of o! = 1. 0! = 1, is narrowly justified for manipulating factorials. Out of that context it has no relevance. I do appreciate the voice of reason, but to rely on 0^0 =1 for a proof seems very dangerous.

jostpuur is not relying on 0^0=1 as a part of the proof. He is using 0^0=1 as a convention so as to reduce the size of the problem statement. If he didn't use that convention he would have to have written his challenge as four separate statements, one special case for j=0, k>0, another special case for j=k, k>0, yet another special case for j=k=0, and finally a general case for 0<j<k. Employing the convention 0^0=1 enabled jostpuur to eliminate those special cases.

 

[spamiam]

For French readers : "Zéro puissance zéro"
http://www.scribd.com/JJacquelin/documents
.

Thanks for the link, JJacquelin! I liked your treatment of the dilemma!

 

[uart]

1=a^n/a^n
a^(n-n)
a^0

Hi lolarogers. That's the standard demonstration that a^0 = 1 for a \neq 0. Notice however that if a=0 then your very first line starts out with 0/0. Hence the reason why a^0 is not well defined for a=0.

 

[D H]

Hi lolarogers. That's the standard demonstration that a^0 = 1 for a \neq 0. Notice however that if a=0 then your very first line starts out with 0/0. Hence the reason why a^0 is not well defined for a=0.

One of many reasons why 0^0 is not well defined.

Let y=a/ln(x) where a is some constant and let x approach 0 from above. Note that y also approaches 0 as x approaches 0. Thus we could think of 0^0 being the limit of x^y as x approaches 0. However, as defined, x^y is identically equal to a for all positive x.

In short, 0^0 can anything I want it to be. Which in turn means it can't be defined, at least in terms of limits.

 

[shunzhiaaa]

0^0=1,is right without a doubt, "by definition", in some specifical territory.

正如#12所说的一样，"0^0 can anything I want it to be." ,but we have to define or look for a specifical rule, and we'd better choose a suitable value. For example, let 0^0=x, x^b=(0^0)^b=0^(0^b)=0^0=x, where b is arbitrary number, x=1 is a singular value, so you could define 0^0=1. But as #11, 0^0 is inanition, so we can ignore that rule. Only we ignore some known rules, 0^0=1 maybe perfect.

 

[coolul007]

0^0 = 1, is wrong without a doubt as it violates the rules of exponents. We have definitions regarding algebraic statements that do not conform to the rules of algebra. log(0), 1/0, etc. a^0, a not equal to 0, can be derived from the laws of exponents. 0^0 cannot, as any statement preceding the conclusion is governed by one of the exceptions, mainly that one cannot divide by zero.

 

[micromass]

0^0 = 1, is wrong without a doubt as it violates the rules of exponents.

Maybe it's just that the rules of exponents don't apply to 0^0?

I'm fine with leaving 0^0 undefined, I think it's the best way. But in many occasions, it can be handy to define it as 1, for example, when working with sets, binomial theorems or Taylor expansions.

 

[D H]

0^0 = 1, is wrong without a doubt as it violates the rules of exponents.

This is your opinion, and it is inconsistent with the opinions of those whose opinions matter: Professional mathematicians. In the opinion of many (most!) mathematicians, defining 0^0 to be 1 for almost all places where one would run into 0^0 makes an immense amount of sense.

Defining 0^0 as 1 is arguably an abuse of notation, but an incredibly useful one. Think of how ugly the binomial theorem, power series, and set operations would be if we couldn't just use this convention.

There is also nothing wrong with defining 0^0 to be 1 even though it is an indeterminate form when examined from the concept of limits. The value of a function at some point and the limit of the function at that point do not necessarily have to agree with one another.

 

[Hurkyl]

But in many occasions, it can be handy to define it as 1, for example, when working with sets, binomial theorems or Taylor expansions.

Or in general, when limiting one's attention to integer exponents.

 

[coolul007]

Are those the same mathematicians that thought the Earth was flat. It is easy to bend the rules to make things work. It would be nice for a lot of proofs, trisection of angle, etc. But it is these rules that keep proofs honest. If we can't prove it unless we change the rules, what's the use? I guess we all have to go back to the pi = 4 thread and agree. After all, he bent the rules and demonstrated pi = 4. With special relativity pi can equal anything. But back here on Earth the rules should be the rules.

 

[Hurkyl]

If we can't prove it unless we change the rules, what's the use?

We're not changing the rules; we're changing the notation.

 

[micromass]

Are those the same mathematicians that thought the Earth was flat.

I don't want to be a smart-*** here. But people in the middle ages never really believed the Earth was flat. This is a myth that was created in the 18th century.

In fact, it were mathematicians like Erastothenes who first proved that the Earth was round and gave a method for calculating it's radius.

It is easy to bend the rules to make things work.

It's not bending the rules, it's giving a new definition. If we define 0^0=1 then some rules won't work anymore, and we need to be aware of that. And if rules still hold for 0^0=1, then we need to prove that.

It would be nice for a lot of proofs, trisection of angle, etc. But it is these rules that keep proofs honest.

No, it's the proof of these rules that keep things honest. If you can't prove a rule for 0^0, then that rule doesn't hold there. We're not doing something dishonest here...

 

[coolul007]

I was being a bit facetious with the flat earth, so I will state what I really meant, Anytime you need to bring "consensus" into an argument it is a failed argument. Let's all vote on it, still doesn't make it so. My objection is that if we bend that rule and not bend the others to conform as the 1 = 2 "proof" then why have rules. I am a purist when it comes to the foundation and rules of number systems etc. I know they get "bent/modified" as we go from reals to complex numbers. As a purist I believe that 3 - 2 should really be written as 3 + (-2), this makes rules clear and concise. This eliminates the operation of "Subtraction" and its problems. etc. etc. etc.

 

[micromass]

I was being a bit facetious with the flat earth, so I will state what I really meant, Anytime you need to bring "consensus" into an argument it is a failed argument.

Not at all. We need consensus. If there was no consensus on what + is, then how could you prove 1+1=2?? Mathematicians need to agree on what the symbols mean and what with what axioms they work. So in any argument, we need consensus.
Of course, you can make up your own axioms and definitions. There's nothing wrong with that! But then you need to state clearly that you have done so. And it's also the question whether mathematicians are interested in a work that doesn't follow the consensus.

Let's all vote on it, still doesn't make it so. My objection is that if we bend that rule and not bend the others to conform as the 1 = 2 "proof" then why have rules.

We are not bending any rules. We just define the rules differently. If I want to have a number system where 1/0=9, then I can make it (I'll have to accept the consequence that it'll be useless however). If I want a number system where 0^0=1, then it can be done and it turns out to be a very useful one! (but some rules don't hold anymore...)

I am a purist when it comes to the foundation and rules of number systems etc.

Trust me, so am I.

I know they get "bent/modified" as we go from reals to complex numbers.

Uuh, in the complex numbers, we simply define a new number system. There are no rules being bent here, the real numbers will still behave like you want them to.

As a purist I believe that 3 - 2 should really be written as 3 + (-2), this makes rules clear and concise. This eliminates the operation of "Subtraction" and its problems. etc. etc. etc.

Yes, but that would make all texts unreadable.
And, if you want to be a purist, then you should also want to write 2 as \{\emptyset,\{\emptyset\}\}...

 

[coolul007]

Here is the rule being "broken" in the complex numbers sqrt((-1)^3) in the reals it would remain sqrt(-1) in the complex it is -sqrt(-1).

I think 2 existed well before set theory.

But I have stated the reasons and arguments, and of course true to most intelligent people, minds never get changed.

 

[micromass]

Here is the rule being "broken" in the complex numbers sqrt((-1)^3) in the reals it would remain sqrt(-1) in the complex it is -sqrt(-1).

sqrt(-1) doesn't exist in the real numbers. The sqrt function is only defined on positive real numbers.
And in fact, I'd like to say that the sqrt(-1) isn't even defined for complex numbers, but some books tend to agree and some disagree with that...

I think 2 existed well before set theory.

It is only set theory that makes the mathematics correct and rigorous. So a true purist would write everything down in set theory notation...

But I have stated the reasons and arguments, and of course true to most intelligent people, minds never get changed.

Well, if that is what you believe...

 

[JJacquelin]

Hi !
In set theory, 0^0=1
In basic algebra, 0^0=1 requires a conventional statement.
In general analysis, 0^0 is equivalent to exp(0/0)
As shown in : http://www.scribd.com/JJacquelin/documents

 

[hillzagold]

Ok, taking zero to be what I think it is, and taking exponents to be what I think they are, can someone explain how they equate to 1? It's not like 1/0, where we give up and call it undefined, there must be a specific reason 0^0 is 1, and not -1 or 0 or 2 or e or infinity.

 

[JJacquelin]

Ok, taking zero to be what I think it is, and taking exponents to be what I think they are, can someone explain how they equate to 1? It's not like 1/0, where we give up and call it undefined, there must be a specific reason 0^0 is 1, and not -1 or 0 or 2 or e or infinity.

That is what is done in the paper referenced above
.

 

[coolul007]

and for those of us who don't read French?

 

[coolul007]

in algebraic terms, one cannot even do a series of operations that will legitimately get me to a step where the result is 0^0.

1. ?
2. 0^0

but I can get to any other number to the zero power through legitimate algebraic operations.

 

[Hurkyl]

in algebraic terms, one cannot even do a series of operations that will legitimately get me to a step where the result is 0^0.

How about:

• Start with 0
• Raise the current result to the 0-th power

 

[Hurkyl]

Hi !
In set theory, 0^0=1

I understand this, if you really mean "^ is exponentiation of cardinal numbers", rather than "The real number exponential 0^0 has a value and is 1".

In basic algebra, 0^0=1 requires a conventional statement.

Oddly, I would think exactly the opposite -- in basic algebra, we usually limit ourselves to natural number or integer exponents, and define exponentiation by repeated multiplication, in which case 0^0 = 1 follows directly from the definition.

In general analysis, 0^0 is equivalent to exp(0/0)

This makes no sense.

 

[micromass]

This makes no sense.

Judging by the scribd-file she gave earlier, the argument would be

x^y=e^{y\log(x)}=e^{y/t(x)}

where t(x)=\frac{1}{\log(x)}. Now if x goes to zero, t(x) goes to 0. If furthermore y goes to zero, then

0^0=e^{0/0}
​

Well, let's just say that this argument looks like an argument from a high-schooler who is confused about limits. Firstly, it ignore that the problem is multivariate (we have a function of x and y). Furthermore, you can't say that

\frac{1}{0^2}=+\infty

just because

\lim_{x\rightarrow +\infty}{\frac{1}{x^2}}=+\infty

there is a reason why there is a limit in front of the function...

 

[coolul007]

How about:

• Start with 0
• Raise the current result to the 0-th power

OK, e^x = 0, I can write it, but it's meaningless. I can write any absurd statement.

 

[Hurkyl]

OK, e^x = 0, I can write it, but it's meaningless. I can write any absurd statement.

And we can write well-defined statements too. 0⁰ is quite well-defined for many versions of exponentiation.



You've learned a correct fact -- that for real number exponentiation, 0⁰ is undefined -- but you are making the mistake of trying to apply the conclusion outside of the domain where the hypothesis is valid.

(moderator's hat on)
You've had your say. If you are just going to keep repeating your assertions repeating yourself without actually digesting what people are trying to teach you, then don't post anymore in the thread. OTOH, if you get to the point where you're ready to ask questions and try to understand, then please continue.[/color]
(moderator's hat off)


P.S. e^x=0 is a perfectly meaningful (and identically false) predicate in the variable x.

 

[coolul007]

And we can write well-defined statements too. 0⁰ is quite well-defined for many versions of exponentiation.



You've learned a correct fact -- that for real number exponentiation, 0⁰ is undefined -- but you are making the mistake of trying to apply the conclusion outside of the domain where the hypothesis is valid.

(moderator's hat on)
You've had your say. If you are just going to keep repeating your assertions repeating yourself without actually digesting what people are trying to teach you, then don't post anymore in the thread. OTOH, if you get to the point where you're ready to ask questions and try to understand, then please continue.[/color]
(moderator's hat off)


P.S. e^x=0 is a perfectly meaningful (and identically false) predicate in the variable x.

I'm to the point of leaving a forum/moderator that uses "power" to get his point across. Intellectual discourse is frowned upon, so adios, auswiedersehn.

 

[Ashwin_Kumar]

You can use taylor expansion to prove that 0⁰ is one.

(x+1)^{0}=1+\frac{f&#039;(a)}{1!}(x-a)+\frac{f&#039;&#039;(a)}{2!}(x-a)^{2}...

x-a= 1
f'(a)= 0
f''(a)=0
.
.
.
So we take x=-1, and we get
0⁰= 1+0+0...
= 1
As simple as that.

 

[Ashwin_Kumar]

The evidence micromass pointed out is sufficient. The idea that 0^{0}=0 doesn't make sense.

And,
\frac{1}{0^{2}} is equal to infinity, without the need for limits.

 

[micromass]

You can use taylor expansion to prove that 0⁰ is one.

You can never prove that 0⁰=1. You can only define it as being 1, but you can never prove it.

Your Taylor series argument is merely an indication why it is sometimes useful to set 0⁰=1, but it is no proof of it.

 

[micromass]

The evidence micromass pointed out is sufficient. The idea that 0^{0}=0 doesn't make sense.

It does make sense. But it just isn't useful. (Except in studying limits of functions such as 0^x).

And,
\frac{1}{0^{2}} is equal to infinity, without the need for limits.

Uuh, I was actually pointing out that this was not the case. You can not divide by zero.

 

[Ashwin_Kumar]

Oh- wait- no. Taylor expansion proves that 0⁰ is infinity.

f&#039;(a)=0^{-1}=\infty

So we get 0^{0}=\infty

 

[micromass]

Uuh, are you serious or just trolling??

Oh- wait- no. Taylor expansion proves that 0⁰ is infinity.

f&#039;(a)=0^{-1}=\infty

You can't divide by 0, and I have no clue what your f is...

 

[Hurkyl]

I'm to the point of leaving a forum/moderator that uses "power" to get his point across. Intellectual discourse is frowned upon, so adios, auswiedersehn.

And this is why I was closing the discussion. It isn't intellectual discourse to continually insist to others that mathematics should be practiced your way while simultaneously closing your mind to the how and why of mathematics is actually practiced. I put my moderator hat on not to get a point across, but because it's time to clean up.

 

[hillzagold]

Let me start by saying I don't speak French. Pointing to an untranslated French paper has zero meaning to me.

Now, if the idea idea of 0^0 is mystifying to the average person, at what point can one expect to learn how to explain it? Don't say to use google, because it will immediately jump to a confusing answer, and even independent study needs a structure that provides necessary prerequisite knowledge. Just tell me to learn X, then Y, then finally Z

 

[Ashwin_Kumar]

Using, different methods, you get different results for 0⁰
But micromass said you cannot divide a number by zero to get infinity(or am i mistaken). And way, you obviously can:

\frac{1}{0}=0^{-1}<br /> =(\frac{x}{\infty})^{-1}=\frac{\infty}{x}=\infty

 

[micromass]

Using, different methods, you get different results for 0⁰

You can never obtain different results for 0⁰, you can only define these results. There is no way you can write 0⁰=1 or 0⁰=infinity, with defining things that way. All we are doing here is setting a convention. You cannot prove anything about this.

But micromass said you cannot divide a number by zero to get infinity(or am i mistaken). And way, you obviously can:

\frac{1}{0}=0^{-1}<br /> =(\frac{x}{\infty})^{-1}=\frac{\infty}{x}=\infty

Sigh... Dividing by 0 is not allowed in the real numbers. Not now, not ever. And infinity is not a real number, so we can't work with that either.

 

[3.1415926535]

Using, different methods, you get different results for 0⁰
But micromass said you cannot divide a number by zero to get infinity(or am i mistaken). And way, you obviously can:

\frac{1}{0}=0^{-1}<br /> =(\frac{x}{\infty})^{-1}=\frac{\infty}{x}=\infty

No no no, you can't divide by 0. You can use limits like this
\lim_{x\rightarrow 0}{x^0}=<br /> \lim_{x\rightarrow 0}{x^{n-n}}=\lim_{x\rightarrow 0}{\frac{x^n}{x^n}}=\lim_{x\rightarrow 0}1=1
or like this
\lim_{x\rightarrow0}\left (e^{-\frac{1}{t}} \right )^t=\lim_{x\rightarrow0}e^{-\frac{t}{t}}=\lim_{x\rightarrow0}e^{-1}=\frac{1}{e}\approx 0.367879
As you can see 0^0 varies depending the limit, therefore it is an inditerminate form
Without limits however, mathematicians avoid writing 0^0 because it rests undefined

 

[timthereaper]

Okay, I've read all the posts and have tried to figure out the controversy of 0^0=1. From what I understand (correct me if I'm wrong), but it seems that 0^0 is actually indeterminate, but the general consensus is that defining 0^0=1 makes things easier in some ways, but it violates some rules somehow.

I'm no mathematician (I'm an engineer), so why is it convenient to define it this way? I guess I'm trying to figure out what problems arise if we didn't say that 0^0=1.

P.S. I'm not trying to beat a dead horse or rile things up, but I'm trying to learn more about this interesting topic.

 

[micromass]

Okay, I've read all the posts and have tried to figure out the controversy of 0^0=1. From what I understand (correct me if I'm wrong), but it seems that 0^0 is actually indeterminate, but the general consensus is that defining 0^0=1 makes things easier in some ways, but it violates some rules somehow.

That's a very good summary!

I'm no mathematician (I'm an engineer), so why is it convenient to define it this way? I guess I'm trying to figure out what problems arise if we didn't say that 0^0=1.

Well, for example in Taylor series. The Taylor series of e^x is

e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...

No problems so far, but if we want to write it more compactly, we get

e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}

However, when we evaluate this in 0 (thus if we want to evaluate e⁰), then we get 0⁰ in the first term. That is:

e^0=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+...

Now the conventions 0!=1 and 0⁰=1 are handy because they allow us to calculate e⁰=1 (which we already knew to be true).
If we didn't set 0⁰=1, then we would had to write

e^x=1+\sum_{k=1}^{+\infty}{\frac{x^k}{k!}}

which is less elegant. So you see, we only define 0⁰=1, because it is sometimes more elegant to do so. We won't create new mathematics with it, we won't run into problems with it, it just makes things nicer.

In a way, it's the same thing as setting 0!=1. This is just a handy convention that makes a lot of things easier. But setting 0!=2 would have been as good, but it would make the formula's uglier...

 

[3.1415926535]

That's a very good summary!



Well, for example in Taylor series. The Taylor series of e^x is

e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...

No problems so far, but if we want to write it more compactly, we get

e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}

However, when we evaluate this in 0 (thus if we want to evaluate e⁰), then we get 0⁰ in the first term. That is:

e^0=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+...

Now the conventions 0!=1 and 0⁰=1 are handy because they allow us to calculate e⁰=1 (which we already knew to be true).
If we didn't set 0⁰=1, then we would had to write

e^x=1+\sum_{k=1}^{+\infty}{\frac{x^k}{k!}}

which is less elegant. So you see, we only define 0⁰=1, because it is sometimes more elegant to do so. We won't create new mathematics with it, we won't run into problems with it, it just makes things nicer.

In a way, it's the same thing as setting 0!=1. This is just a handy convention that makes a lot of things easier. But setting 0!=2 would have been as good, but it would make the formula's uglier...

Same goes with the taylor expansion of \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

Generally, If we define 0^0\neq1 most of the Taylor-Mc Laurin expansions and some other power series would not be well defined since
f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n

Lastly the function f(x)=x^0 would be "deprived" of its continuity and differentiability

 

[micromass]

Lastly the function f(x)=x^0 would be "deprived" of its continuity and differentiability

That's not really an argument, since now the function

f(x)=0^x

get's deprived of its continuity and differentiability.