# Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Clever student:

I know!

$x^{0}$ =  $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$.

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$$0^{1+x-1}$$0^{1} \times 0^{x-1}$$0 \times 0^{x-1}$$0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$.

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since  $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$, we have

$0^{x} = 0$.

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$.

On the other hand, for real numbers y such that $y \ne 0$, we have that

$y^{0} = 1$.

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$, $y^{0}$ stays at $1$.

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$. In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$.

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$.

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$.

Look, we’ve just proved that $0^0 = 1$! But this is only for one possible definition of $y^x$. What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$.

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$, which seems to be recursive. The reason it is okay is because we are working here only with $z>0$, and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$. But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider the binomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I’ve used that $0^k = 0$ for k>0, and that  $\binom{x}{0} = 1$. Now, it so happens that the right hand side has the magical factor $0^0$. Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$.

If mathematicians were to use $0^0 = 0$, or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$.

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

This entry was posted in -- By the Mathematician, Math, Philosophical. Bookmark the permalink.

### 987 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

1. Yee says:

The Cool Dude,
At most, define the value of function at a point where the limit is not 1.
Discontinuity is not wrong.
It is still okay.

2. The Cool Dude says:

Yee,
If for whatever reason, 0^0 is to be defined as one number, it should only be done so per specific situation, and not for all general situations.
The reason for this is that it has been clearly shown to all participants and readers of this conversation that nailing down one specific value of it is extremely difficult, and one could argue for just about any value, for any number of reasons.

I argue for it being a set of all numbers, rather than no numbers at all, or any one number, because I believe that the fundamental laws of mathematics are always true, even if they conflict. If they conflict, then something that you are dealing with has multiple values. Given this, it can be shown that, if you have some two operations, f(a,b) and g(a,b), so that f(a,1)=a, and g(f(a,b),a)=f(a,b+1), that
f(a,0)=g(a,n), where g(g(a,n),a)=a,
because g(a,n)=f(a,0), and g(f(a,b),a)=f(a,b+1), and f(a,1)=a
What this means is that f(a,0) is any number that when operated in g with a, does nothing to a, which is to say, our functional identity. In essence, these are our index laws, and there is currently nothing that further defines these functions.

If, perhaps, g(a,b)=a+b, and f(a,b)=a*b, of which the function mechanics support exactly, by design, then it is clear that a+(a*0)=a, and so that a*0 is any number that satisfies this situation. If a is any general number, a*0=0, which we clearly know. However, if a is not a general number, as to perhaps, 1/0, then a*0=0/0, and then 0/0 is any number that when added to 1/0, is 1/0. It is clear that it is all numbers, because adding a number “c” to 1/0, results in
1/0+c=1/0+0*c/0=(1+0)/0=1/0
The same thing, so by consequence, 0/0 must be all numbers, because all numbers added to 1/0 are 1/0.

If, perhaps, g(a,b)=a*b, and f(a,b)=a^b, of which the function mechanics support exactly, by design, then it is clear that a*(a^0)=a, and so that a^0 is any number that satisfies this situation. If a is any general number, a^0=1, which we clearly know. However, if a is not a general number, as to perhaps, 0, then a^0=0^0, and 0^0 is any number that when multiplied to 0, is 0. It is clear that 0^0 is all numbers, because multiplying a number c to 0 results in
0*c=0
The same thing, so by consequence, 0^0 must be all numbers, because all numbers multiplied by 0 are 0.

This is just my position on the subject, however there are many other reasons for why 0^0 should be all numbers, but the fact of the matter is that there are no good arguments against this, because if you say any one of those laws do not work when a number is zero, for any reason, then even despite the fact that there is absolutely no reason to add such a restriction on the function, it still stands that claiming any other value after would be baseless, because the function has no further guidelines to work off of, and it again, might as well be anything.

3. The Cool Dude says:

0^0 can be nothing but anything.

4. Yee says:

The Cool Dude,
0! can also be all numbers.
The reasons you listed mean nothing.
The question is not what it can be,
but what it is good to be.

5. The Cool Dude says:

Yee,
0^0 has been conceded by numerous arguments as equivalent to a set of all numbers.
What it can be is anything.
What it is good to be is directly coherent to further qualifiers and limitations put on it, which are decided per situation.

Examples of common qualifiers and limitations:
Satisfaction of continuity
Satisfaction of logarithms
Satisfaction of power zero
Satisfaction of intersecting lines
Measure of finite areas
Creation of determinate sums
Finite range of infinite results

Power zero is not the only qualifier to be considered, and it is certainly not the best for all situations.
This will not degrade into an argument for which qualifier is best, which is, regardless of support for any stance, a question without an answer. Further, it is not the topic of conversation on this question to consider limitations and qualifiers.

What the topic question is not:
“What is it good for 0^0 to be?”
“What does 0^0 equal when it must equal its limit from a specific direction?”
“What does 0^0 equal when it equals x^(1/ln(x))?”
“What does 0^0 equal if x^0 is always 1?”
“what does 0^0 equal when finding the intersection of two lines when one of them equals 0^0 at a point?”
“What does 0^0 equal based on the area under the curve |x^x|*|ln(x)+1| from 0 to -∞?”
“What does 0^0 equal if the sum from 0 to ∞ of a^n is 1/(1-a) at a=0?”
“What does 0^0 equal if it is positive and less than five?”

What the topic question is:
“What does 0^0 equal?”
And the answer is a complete set of all values.

6. Yee says:

The Cool Dude,
You think defining 0^0 as a set satisfies some properties.
In fact, not at all.
Something indeterminate satisfies nothing.
What does 0^0 equal depends on definition.

7. The Cool Dude says:

Yee,
I don’t define 0^0.
I solve 0^0.
Your definition of “define” is wrong.

0^0 is not indeterminate.
0^0 is determined by the definition of exponentiation as a set of all values,
And knowing the exact value of something is the exact opposite of indeterminate.
Your definition of indeterminate is wrong.

0^0 does depend on definition, you are correct in that
But the definition which it depends on is not yours.
It depends on the definition of the operation of exponentiation.
Your determination of 0^0 is wrong.

The topic question is:
“What does 0^0 equal? Why do mathematicians and high school teachers disagree?”
The second half, “why do … disagree” has already been answered in the primary article, and can be observed by reading through this posts 957 responses. The remainder is “What does 0^0 equal?”
The qualifier “equal” does not imply user satisfaction, it implies the satisfaction of accurate replacement.
If you claim 0^0 equals 1, then you are equally correct as if I were to say 0^0 equals 7.
Your understanding of “equal” is wrong.

8. Yee says:

The Cool Dude,
It is you that don’t understand definition.
Definition is irrelevant to right or wrong.
Definition is irrelevant to correct or incorrect.
Definition is irrelevant to true or false.
Definition is relevant to good or bad.
Definition is relevant to reasonable or unreasonable.

The properties you think are satisfied depends on definition.

9. The Cool Dude says:

Yee,
This question is relevant to right and wrong.
This question is relevant to correct or incorrect.
This question is relevant to true or false.
This question is not relevant to good or bad.
This question is not relevant to reasonable or unreasonable.

In the primary post, the three students and the two teachers argue for which value of 0^0 is correct.
They attempt to answer the question “What does 0^0 equal”, and their dispute is the writers way of answering the question “Why do mathematicians and high school teachers disagree”.

(Proceeding, I will refer to any generic mathematical function of constants that result in a universal set as “Ω”, because I can)
The mathematician argues for which value is good, as you do, but that is not the question. If that were the question, and I am willing to entertain the idea for a time, then your reasoning behind why 0^0=1 is good can be easily encompassed in a much broader and simpler idea, which does not need qualifiers, which is:
Ω, if not acting as its true value, should simply act as any value that would satisfy the current situation.

This idea is far superior to any other, because not only does it make 0^0=1 when it is good, or convenient, as you and many others would have it, but on top of that, it includes different values for different situations which are “good”, and it is actually 100% correct, and coheres to the laws of mathematics. There is some indirect method to transform any function so that a variable can be solved to have a possible solution of Ω. If Ω is attained by some function directly, and a different, “standard” value for the same function is attained indirectly, then the “good” is that indirect value.
The reasoning behind this is because a potential solution to all equations is Ω, Ω is not a particularly notable solution, although it is literally always correct. Because of this, getting the value of Ω in the direct method, rather than through some indirect method, is strange at best. In actuality, there may be certain values in Ω that, on their own, do not satisfy the function, and a second, more relevant solution can be made by only including numbers that would satisfy the function on their own, and these values might be called “Primary solutions”.

0^0, on its own, has no primary solution, as all solutions can be shown to satisfy.
This being so, “defining” 0^0 as any one thing is sloppy and reckless, because it can be anything.

For 0^0, a good value is often 1, but not always.
Saying “0^0=1 is good” because it fits the most common situations is overlooking a much more accurate solution to solve for “What is good”.

10. Yee says:

The Cool Dude,

What about defining 0! as a set of all values?
Is it correct or incorrect?

It is easy to list equations which has many solutions.
Let 1=x
0*1=0
0*x=0
1 is the solutions of 0*x=0.
Therefore 1 is the set of all values.
Isn’t it a joke?

There are not any reasonable cases in which 0^0 use a value which is not 1.

11. The Cool Dude says:

Yee,
Exponentiation is one of the most basic functions.
If we can not settle the issue of 0^0, we should not even begin to think about factorials.

Your math shows a clear association fallacy.
I could show the same issue using squares and square roots, instead of multiplication by zero.
I have always been aware of the association fallacy, I have never made this mistake, though you may have misinterpreted something I have said.

There are plenty of reasonable cases in which 0^0 should use a value which is not 1.
x^(1/ln(x)) is e at all points, except for when x is zero, when it is 0^0. It doesn’t make any sense for the entire equation to be e, except at zero, and be 1.
If I want to solve:
Σ(-1,5) (x^(1/ln(x)))
For a determinate solution, it doesn’t make any sense to get 6e+1.
It makes sense to get 7e.
You can argue that 1/ln(0)≠0, but really you’re just splitting hairs.
Extremely small, infinitesimally relevant hairs.
And also you would be wrong.

Showing a proof against a value being a set using an association fallacy is easy.
Giving an actual mathematical reason for why 0^0 should not be a set of all numbers, for a random, unbound general solution, is impossible. Giving a good explanation for why 0^0=1 is better than all other methods of definition is arguably impossible, depending on ones definitions of the words “good” and “better”. Saying “There are not any reasonable cases” is clearly not true, which leaves you only with “0^0=1 is better”, which is a statement of pure, distilled opinion.

A statement of pure opinion has no value when dealing with a subject of such literal and absolute internal applicability.

12. Yee says:

The Cool Dude,

1.
Settle what issue?

2.
ln(0) is undefined, nor is 1/ln(0).
I don’t need to explain.

3.
Defining 0^0=1 is good and reasonable
because it is useful in some cases.
I have listed some cases above.
And you can not find a reasonable case in which a different value is used.

13. The Cool Dude says:

Yee,
1.
The issue of what 0^0 equals, which is the only point of this entire conversation, and is clearly displayed at the top of this page.

2.
ln(0) does not need to be defined, like 0^0, it needs to be solved. I have solved ln(0) to be equal to 1/0, based on the limit definition of logarithms, which holds true in all cases, because it is literally the only thing that defines the logarithmic function. 1/0 is not a real number, but calling it undefined is incorrect. It is specifically defined as the reciprocal of zero, as you can plainly see, it doesn’t need a number value to be used, because if that were the case, you may as well throw out the entire complex plane. Even if you reject my solving of ln(0) to 1/0 without reasonable argument, as you have done many other times, it is still true that ln(0) is infinite in nature, and the reciprocal of anything infinite in nature is still zero, so my point stands.

3.
Defining 0^0 is bad and unreasonable
BECAUSE it is only useful in SOME cases.
I literally just showed a reasonable case in which a different value is used, and I can, and many others have, found hundreds or thousands more.

You can not blindly dismiss everything I say simply because the conclusion is something that you reject. Responding to the topic questions requires one to decide if they approach this question as though it is a matter of what is true, or as though it is a matter of what is easy.
If it is about what is true, then 0^0=0/0 because of the sole definition of exponentiation, and 0/0=1 because of the sole definition of division, and thus expansively, is a set containing all other numbers due to the multiplicative definition of zero. This is what is true, because it assumes that all mathematical laws are always true no matter what, and that when faced with conflicting laws, an entity takes the stance of being multi-valued.
If it is about what is easy, then 0^0 equals what it makes sense to equal, because what makes sense is what is easy. Though what makes sense at first is not always the instant truth behind mathematics, what is true is that different values can make sense in different scenarios, and so it should be defined per situation. It makes sense for 1/ln(0) to be zero, and so 0^(1/ln(0)) makes sense to be 0^0, and it makes sense for x^(1/ln(x)) to equal e at all times, rather than all times but one, so in this scenario, it makes sense for 0^0 to equal e. You can not argue for 0^0=1 because of convenience, while arguing against 1/ln(0)=0 because it defies fact, when it also works with convenience, and 0^0=1 also defies fact. That is hypocritical, and defeats the point of your method of approaching the question, which is to define values that make equations easy. Regardless of which way you approach the question, defining 0^0 as a single value for all situations is inherently wrong.

14. Yee says:

The Cool Dude,

1.
There is no issue in defining 0^0=1.

2.
ln(0) and 1/0 are undefined.
This is globally accepted.
I don’t accept it.

3.
0^0=1 is used in all reasonable cases.
0^0=0/0 is unreasonable.
Does 0 equal 0^2/0^1 ?

15. The Cool Dude says:

Yee,
1.
Ignoring the issue and other points of view doesn’t make your point more valid, it just makes your argument more irrational. If there was no issue, then this question wouldn’t even exist, however, this question does exist, and one half of the question implies that there is in fact disagreement, and the following written article clearly shows the dispute.

2.
No it is not, as earlier, one of the moderators claimed that ln(0)=-∞.
Just saying something over and over doesn’t make it true, unless you have reasoning.
General consensus is a horrible reason. The entire human population once thought that
flies were spontaneously generated from meat, and at that time, the entire human population was wrong. Things that are undefined are meant to be defined. Early mathematicians left so many holes and exceptions in mathematical laws because they did not know how, or rather, simply did not care to deal with them, at the time.

3.
No it is not. I have given you reasonable cases. Ignoring them doesn’t make you right, it only invalidates your argument. 0^0=0/0 being unreasonable is a baseless accusation. In the factual mathematical sense, 0^0=0/0 is 100% true, and you have no argument against it, other than invalidating the sole definition of exponentiation at zero. I too can invalidate mathematical laws at zero, and like you, I would be quoting many old and famous mathematicians. I could say that x^0=1 is invalid at x=0, which it is, but I am not, because that would be a backless argument. The theme of excluding functions at zero was created because on those functions, zero caused strange things to occur, and at the time, they did not feel it mattered to establish the meaning behind it, and certainly, it still doesn’t actually matter. Repeating their arguments and using it to further define 0^0 is wrong, as you are using the argument to do exactly what it was designed to avoid doing.
In the sense of what is easy, or what SHOULD be right, rather than what is, claiming 0^0=1 should be right while ln(0)=-∞ should not be is hypocritical, because both claims are mathematically false, until you invalidate mathematical laws surrounding those functions at zero, and once you do invalidate mathematical laws surrounding those functions at zero, you are left only with what makes sense. e^-∞=0 makes sense, but is not technically correct. 0/0≠0^0=1 makes sense, but is not technically correct.
Honestly, e^-∞=0 makes more sense than 0^0≠0/0, because instead of invalidating mathematical laws, we are instead simply ignoring an infinitesimal difference.
And yes, for every single time you have asked it I have said, and will say once again, 0^2/0^1 is in fact a set, which contains zero. The factual or nonfactual nature of 0^2/0^1=0 depends on how you interpret the equals sign, in reference to sets.

If your only argument against my points is ignoring them, then this conversation is over, because I can have nothing more to add, and I don’t need to validate my point further.

16. Yee says:

The Cool Dude,

1.
I did not ignore any point of view.
There is nothing invalid in defining 0^0=1.

2.
You cannot make ln(0)=1/0 true by saying it over and over.

3.
0^0 does not equal 0/0.
0/0 is a set which contains 1,
0^0 is only 1.

4.
Falacies deserve to be ignored.
This discussion must have been over.

17. CS Prof says:

Yee,

I have to resist the urge to think that a robot is posting under your name. As Cool Dude has been pointing out for months, the argument is long over except for your insistence that you are correct and everyone else and every other argument is wrong. I feel this is now simply you wanting to have the last post in this thread. That would be fine if your arguments were valid. But since they are not, I will continue:

“I did not ignore any point of view.”

This is actually a true statement. You didn’t ignore others’ points of view. You dismissed them as invalid. The problem is that your invalidation has no rational basis. You have dismissed any line of thought that discusses non complex forms. The problem is that since 0^0 is a non complex form, there is no rational discussion about it without involving other non complex forms, which you have dismissed out of hand. As I have said before sometimes this discussion seems like trying to explain fractions to a math student who only understands whole numbers. Since 1 divided by 3 isn’t a whole number, it must be invalid. In that environment asking the question “What does 7 divided by 8 equal?” really isn’t a question that can be answered correctly due to the enforced limitations of the student.

Everywhere that you use the word invalid Yee, in order to have this discussion in a rational manner, you must extend into the concept of a non complex form. Now the form may not have a “value” per-se, but just because it is non-complex does not make it invalid any more than the fact that 7 divided by 8 does not produce a whole number does not invalidate either the operation of division, or the result.

I’m going to trudge on with that, even though I already know that you will insist that each and every non complex form produced by valid operations is in fact invalid.

“There is nothing invalid in defining 0^0=1.”
Yes there is. It’s not true. That makes it invalid. We’ve shown over and over again that using L’Hopital’s rule make it trivial to define continuous functions that approach any value on the complex plane as the function value approaches 0/0. I know you will dismiss continuity and limits at this point, but again your dismissing that fact does not invalidate it. 0^0 can be 1. It can also be any other value on the complex plane too. It is multivalued. So therefore it cannot have just one sole single solitary value. If you would simply say “There are certain situations where using the value of 0^0 as 1 is useful, but it can have other values too in different situations.” then we’d be finished here. But your insistence that it is 1, defined as 1, cannot be anything else other than 1, is simply invalid. And it’s now approaching two years in this thread explaining why that is.

“You cannot make ln(0)=1/0 true by saying it over and over.”
I believe this type of statement is just as attributable to “0^0=1″. Cool Dude layed out a proof of the value of the ln(0) based on definition of the natural log function. It’s in the comments. You can go read it. He didn’t say this was true, he proved it.

“0^0 does not equal 0/0.”

It must be the definition of the operations. Both are non-complex forms. But by definition it must be true. You have said over and over again that any operation that produces a non-complex form is invalid. However, you have no proof that it is invalid, only proof that the result is not a value in the complex plane.

“0/0 is a set which contains 1,
0^0 is only 1.”

Multiple times this has been disproved by example. You can say this until you are blue in the fact. Doesn’t make it any more true. You simply cannot explain away the trivial equation of 0^x = 0 lim x->0. There simply is no justification for the discontinuity at the 0^0 point that has a rational explanation.

“Falacies deserve to be ignored.
This discussion must have been over.”

And I think we are all well past the point of wishing that this thread was over. But it cannot be as long as you espouse that 0^0 has the sole, singular, value of 1 without any reasonable proof.

CSProf

18. The Cool Dude says:

Yee,
“This is globally accepted.”
That statement is ignoring every point of view anyone who has ever thought otherwise has had. You are being ignorant of millions of peoples different views. A large sum of individuals would say ln(0)=-∞, even thinking that everyone accepts ln(0) is undefined is ridiculous.

I don’t need to say ln(0)=1/0 over and over again, because I already proved it with mathematics, you have chosen to ignore it, so I will say it once again
e^x is defined EXCLUSIVELY by the limit equation:
lim(n→±∞),(1-x/n)^n=e^x
Given this, the inverse equation is
lim(n→±∞),((1-(x^(1/n)))*n)=ln(x)
For x=0,
ln(0)=lim(n→±∞),((1-(0^(1/n)))*n)
ln(0)=lim(n→±∞),((1-{0,1/0})*n)
ln(0)=lim(n→±∞),({1,1/0}*n)
ln(0)={∞,-∞,1/0}
Since lim(x→0) |1/x| is ∞, the difference between 1/0 and ∞ is that 1/0 is in multiple directions in the complex plane, and in this case, since the set contains both positive and negative infinity, along with 1/0, the answer as a set, {∞,-∞,1/0}, is virtually the same as just 1/0. I have given this argument before. You ignored it. I don’t need to say it over and over again to make it true, when it has already been proven by the sole definition of the natural logarithmic function.

Fallacies are to be corrected. Fallacies that exist will continue to cause problems indefinitely, so it is the responsibility of those who see them to point them out and ensure they are fixed, otherwise they will face the consequences of that fallacy influencing their progression as a logical entity.

You have no hard mathematical support to your claims. Your remaining support, of convenience, is overruled by the fact that there are more convenient solutions, which many people already use in certain areas of Calculus.

19. eliaspaul says:

1/infinity =0
therefore (1/infinity)/(1/infinity)=1/infinity*infinity/1=infinity/infinity
also since 1/infinity=0 (1/infinity)/(1/infinity)=0/0
so infinity/infinity=0/0
0^0=0/0
0/0=infinity/infinity=infinity^0=infinity^(-1)+1=(1/infinity)*infinity=0*infinity=0
so 0^0=0 also is possible

20. CS Prof says:

Eliaspaul wrote:

1/infinity =0
————————-
infinity isn’t a number. So the operation above is in fact invalid. The correct formulation is

lim n->infinity 1/n = 0

http://www.mathsisfun.com/calculus/limits-infinity.html

As such all of the operations you have listed after are also invalid.

On the other hand 0/0 is a valid operation. However, it does not produce a single complex value as a result. The easiest way to see it is using a modified limit equation from above:

lim n->0 Cn/n = C

where C is any complex number. Both Cn and n will approach zero. But the ratio of them is constant.

Finally 0^0 = 0/0 even though again both are non complex forms. As such 0^0 can have a single result anywhere in the complex plane and its “value” is the infinite set of all of those possibilities.

Hope this helps.

CSProf

21. Laurence Lu says:

I believe it is 0 to the power of 0 is undefined, and I have a rather different way of proving it. The reason that 10 to the power of 1 is ten, 10 to the power of -1 is 0.1 and that ten to the power of 0 is 1 is basically because the exponent of “1″ means “[insert number] times [nothing]“, the exponent of “2″ means “[insert number] times itself”, the exponent of “-1″ means “[insert number] divided by itself twice” and an exponent of “0″ means “[insert number] divided by itself”. So, wouldn’t 0 to the power of 2 be 0 because it’s (0)(0), and zero to the power of zero technically be 0/0, and since anything divided by zero (even zero itself) is undefined, 0 to the power of 0 is undefined? I guess it’s kind of simple, but really, technicalities prove this in the first place.

22. John T says:

It seems to me that it is not merely convenient, but the correct conclusion that 0^0=1, because of the definition of x^y for integers = 1*x*x…, as this is the most basic definition of powers, the one from which all other definitions are derived. We can then go on to define x^(y/z) from this definition to calculate rational powers, but we must refer to the first definition to do so (that is, x^(1/2) is defined as satisfying 1 = (x^(1/2))^2. I think that because calculus is convenient for estimating the value of powers where irrational numbers are involved it can be confused with a way to define powers, but to do so is a mistake. What we can only estimate does not cease to have a real value simply because we cannot calculate it, therefor the estimate may not be said to define the value which exists without the estimate.

What say you? Forgive my imperfect presentation of the argument, but perhaps you can still parse it to find the error.

23. John T says:

Ps, the equation above should read x=(x^(1/2))^2, but I have more thinking to do on my comment besides that correction. Perhaps you someone who might respond can see what I’m getting at, but if it’s just a jumble to you I can’t blame you.

24. Confused one says:

I am so confused!!!!
0^0= 0
or
0^0= 1????

25. Confused one says:

26. Big_J says:

Zero to the power of zero is undefined (or from a more philosophical stand point, it is all numbers at the same time).

If we use lim(x->0) x^0, it clearly equals 1. This only applies, however, if we approach the limit from a positive direction (i.e. 3, 2, 1, 0). If we approach 0 from a negative direction, then we get lim(x->0) x^0 = -1. So 0^0 equals both 1 and -1 at the same time.

If we graph the function of n^x for various numbers n, then we see as the numbers approach 0 from a positive direction, the function gets closer and closer to an L shape, where for any positive number x, 0^x would equal 0, while 0^0 would equal all positive numbers at the same time. If we approach from a negative direction, we find 0^0 would equal all negative numbers at the same time. Interestingly, both of these graphs seem to be the opposite of graphing n^x as n->infinity, where infinity^x equals 0 for all negative numbers x and infinity^0 equal all positive numbers at the same time.

I think this also shows why 0^x for all negative numbers x would be undefined.

I’d be very interested if anyone was able to put what I’m trying to demonstrate her into a function, and even more interested to see if I could understand said function.

27. Christi says:

I see one very big problem with saying 0^0 can be any number (or have multiple values). If we say that 0^0 = 7 and also 0^0 = 1, then by transitivity 7=1.

28. Big_J says:

A function cannot equal more than one number, which is why is has to be undefined. This is actually what makes it undefinable.

29. Izzath says:

Mathematician is the correct one… (y)

30. Mike Oliver says:

Actually there are two (or more) distinct operations being conflated as “exponentiation” here. The argument for 0^0=1 is quite convincing — if both 0′s are considered natural numbers, or put another way, if we’re doing natural-number-to-natural-number exponentiation. Even for real-number-to-natural-number exponentiation, the answer 1 (or 1.0, if you like) is unproblematic.

But real-number-to-real-number exponentiation is an intensionally *different* thing, and for it, none of the arguments presented for the value 1 (or 1.0) have any force. *That* 0^0 (or 0.0^0.0, if you like) is correctly left undefined.

None of this is really a problem. It’s quite common in mathematics to overload notation so it means different things (even rather subtly different, as in this case) in different contexts.

31. Joshua says:

In engineering we have a theorem that reads if you get 0^0 by accumulated round off and the calculation could not have been simplified or short-circuited (ignoring a hole but not a pole) then the calculation result is 1.

32. Following is a number-theoretic rationale for leaving 0^0 undefined. It makes no use of the concept of a limit or the set of real numbers.

There are precisely two binary functions on the set of natural numbers that satisfy the usual Laws of Exponents. One has 0^0 = 1, the other has 0^0 = 0. At all other points, they agree.

If e and e’ are binary functions on N (including 0) such that

(1) for all x in N, if x=/=0 then e(x,0) = 1 and e’(x,0) = 1

(2) for all x,y in N, e(x,y+1) = e(x,y) * x and e’(x,y+1) = e’(x,y) * x

then it can be shown that, irrespective of the values of e(0,0) and e’(0,0), we have

(1) for all x in N, if x=/0 then e’(x,0) = e(x,0) (trivial)

(2) for all x,y in N, if y=/=0 then e’(x,y) = e(x,y)

Therefore, every exponential function on the set of natural numbers is identical except for any value that may be assigned to 0^0. But there are only two such possibilities: 0^0 = 0 or 0^0 = 1.

If the Product of Powers Rule is to hold, then we must have

0^0 = 0^(0+0) = 0^0 * 0^0

or

0^0 = 0^0 * 0^0

Therefore, 0^0 = 0 or 1. To my knowledge, there is no purely number-theoretic justification for eliminating either possibility.

Given this ambiguity, the prudent course is to leave 0^0 undefined (like division by zero), especially in general purpose programming languages. Currently, most programming languages seem to have 0^0 = 1.

33. Idetrorce says:

very interesting, but I don’t agree with you

34. Hmm, I learned in high school that exponents are simply a form of multiplication with the exponent being defined as the number of times the affected number is multiplied by itself. By definition, the word zero has neither positive or negative value. Therefore zero to any power will always equal zero. This is known as the KISS principle.

35. Engineer Wakman Atal says:

it is all interesting but sorry i m not agree with you that 0^0 is unidentified

36. My last posting here was a bit of a false start. It turns out that there exists not just 1 or 2 “exponent-like” functions on N, but an infinite number of them! And they disagree only on the value of 0^0. Also, from any one of them, you can derive all the usual Laws of Exponents on N. So, while 0^0 can be seen as a natural number, it doesn’t matter which one! As such, 0^0 is ambiguous or “undefined.”

For formal proofs, etc., see my latest blog posting, “Oh, the ambiguity!” at http://dcproof.wordpress.com/

37. kyrbis says:

0^0=(i^2-i^2)^(i-i)=(i+i)(i-i)^i / (i+i)(i-i)^i= 1^i

e^pi*i=-1
.
.
.

0^0=-e^-pi= -0,04321391826