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

Posted on December 4, 2010 by The Mathematician

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.

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1,177 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Yee says:

March 5, 2013 at 11:29 pm

The Cool Dude,
0/0, 1/(1/0), 0^2/0^ involve dividing by 0,
therefore there are undefined.

But 0^0, 0, 1/i, i^3 don’t,
there are no reasons to undefine them.

Dividing by 0 need not discussing here.
The Cool Dude says:

March 6, 2013 at 6:18 am

Yee,
0^0 and 0/0 are the same.
Yee says:

March 6, 2013 at 6:30 pm

The Cool Dude,

It’s a bad idea to say they are the same.
Yee says:

March 6, 2013 at 9:19 pm

The Cool Dude,
If 0^0 is regarded as 0/0,
we can’t get anything.
If 0^0 is defined as 1,
we can get some properties,
which bring some conveniences.
What do you think is better?
Noel Evans says:

March 7, 2013 at 11:31 am

If 0/0 is intended to represent a division, then this is not a permissible form since
the axiom introducing the multipliative inverses excludes zero; that is, the axiom
introduces multiplicative inverses for all NON-ZERO numbers x. 1/0 would
represent a multiplicative inverse for zero, which is disallowed by the axiom.
Thus defining division “a dividied by b” as a times the reciprocal of b would not
be possible for b=0.

If 0/0 is intended to represent a fraction, then this is disallowed by the definition
of the set F of fractions: F = { a/b | a and b are whole numbers and b is not zero }
a/0 for any a, much less 0/0 is disallowed by the definition of fractions.

So to say that o^0 = 0/0 is employing symbolism 0/0 which is not allowed by
either reciprocals or fractions.

Have a blesssed day! 🙂
The Cool Dude says:

March 7, 2013 at 3:10 pm

Yee,
Defining 0^0 as 0/0 doesn’t hinder any progress.
Although it isn’t immediately defined as as 1, it can be defined as anything, depending on circumstance.
If, for instance, using calculus, infinite sums, etc, 0^0 may be defined using a limit, which would make it 1.
An infinite number of potential results allows for infinite potential.
Yee says:

March 7, 2013 at 9:09 pm

The Cool Dude,
Continuity should not be overemphasized.
Limit is a part of math, but not all.

lim x^y
(x,y)->(0,0)

is different from 0^0.

lim x^y
(x,y)->(0,0)
depends on circumstance,
but 0^0 has only one reasonable definition 1.
The Cool Dude says:

March 8, 2013 at 4:44 pm

Yee,
0^0 can not be defined as any one value using any method at all, including limits, but if a particular equation were to evaluate to 0^0 at some key point, a second conditional could further define 0^0, in that particular circumstance.

In the same way that the point (4,2) is the solution to the intersection of the equations:
y=±√(x) and y=(7+e^(x-4))/4,
while the y value is 2, despite ±√(4) being a set of two numbers,

(0,9) is the solution to the intersection of the equations:
y=x^(-|x|) and y=x^4+9
While the y value is 9, despite 0^0 being a set of many numbers.

Noel Evans,
Many mathematical rules were set up specifically to avoid zero because at the time of creation, they did not know what to do with it. The purpose of discussing these scenarios is to “fill in the gap” so to speak.
Yee says:

March 8, 2013 at 8:58 pm

The Cool Dude,
You are ignoring the truth.
All circumstances show 0^0 is best defined uniquely as 1 except limit.
I have mentioned above.
The Cool Dude says:

March 9, 2013 at 12:18 am

Yee,
Perhaps all circumstances you use, but there are other uses for equations that have values of 0^0 at certain points.

Equations which equate directly to 0^0 at a certain point generally can be transformed to show a different equation, where the statement displayed is true for all values.
Yee says:

March 9, 2013 at 6:35 am

No equations are only paradoxes.
All circumstances show 0^0 is best defined uniquely as 1 except limit.
Yee says:

March 9, 2013 at 8:11 am

The Cool Dude,
The equations you listed mean nothing.
Yee says:

March 9, 2013 at 8:18 pm

The Cool Dude

The equations you listed are only paradoxes.
All circumstances show 0^0 is best defined uniquely as 1 except limit.
The Cool Dude says:

March 9, 2013 at 8:39 pm

Yee,
A definition with any exceptions is wrong.
0*n is exclusively 0 for any number or set n which does not contain 1/0.
0^m is equal to 0^0 only when m is 0.
Thus, 0^0 is exclusively 0^(0*n) for any number or set n which does not contain 1/0.
1^n, for any number or set n which does not contain 1/0, has infinite potential values.
The best definition is the definition which fits all cases with as much accuracy as possible, and which is as simple as possible.

0^0=0/0, a set of all numbers n that solve for n*0=0, though can be considered to be one particular value, should a system require it.
CS Prof says:

March 10, 2013 at 6:48 pm

I’ve spent the last couple of weeks reading all of the comments in this thread. Cool Dude, I would like to commend you for your efforts to explain what’s really going on.

Yee, as I’ve read your comments over the last 18 months, I became quite irratated with your resistance to listen to any reasonable explanation and your repeated protestations of invalidity of certain operations, which Cool Dude repeatedly pointed out, are true regardless of circumstance.

I think I finally realized what the disconnect is. Cool Dude has been rightly discussing the issue using every number system in the toolbox. Yee’s discussion is limited to valid operations for real numbers. Only in real number systems is it invalid to divide by zero. And this has been Yee’s entire argument that 0^0 is not equal to 0/0 because it is not valid to divide by zero. Yee’s reason for this is that no valid real number solution can be assigned to the equation 0*U = 5. Since no real number can be assigned to solve the equation, then the operation required to generate the solution, which is to divide by zero is considered invalid.

Yee has used this line of reasoning over and over in this thread. The ln(0) is invalid. 0/0 is invalid. 1/0 is invalid. All equations that use any of this forms is invalid, because non of these forms falls within the realm of reals.

Yee then has dismissed limits and continuity, which of course gives a trivial way to describe the fact that 0^0 (and of course 0/0) are both indeterminate forms. 0^n is 1 for all real and complex values of n except 0 and n^0 is 1 for all real and complex values except 0. L’Hopital’s rule can be used to show that both forms can converge to any real value. Yet both are “overemphesized”.

But what Yee has consistently refused to do is to explain why the discussion is limited to real number systems and operations with no limits. While it is perfectly fine to teach Math by disallowing the square root of a negative number initially, there does become a point where large stretches of Mathematics cannot be explained without the use of the non real value of the sqrt(-1). Cool Dude’s used of U (and 1/0) above is an extension of that concept. It gives a clear and concise explanation of why the forms are indeterminate while validating the ability to use forms such as 0/0 and 1/0 in a correct manner.

So to summarize: 0^0 is equal to 0/0 when non real number systems are allowed. Both are indeterminate forms that contains at minimum all real and complex numbers. Since 1 is one of those values, it can be used as the value of 0^0 in certain limited situations. But neither form is defined to a single value.

Yee, please do not dismiss non real forms or operations (specifically 1/0, 0/0, ln(0), and dividing by zero) without giving a clear explanation of why you a limiting your discussion to a number system that excludes these forms. Quite a while ago Cool Dude asked you to explain what your rules were. You never have. Without understanding why you have imposed limits to this discussion, your responses really do not make a lot of sense.
Yee says:

March 10, 2013 at 6:59 pm

The Cool Dude,
0^0 is not 0/0.
0/0 is a set,
but 0^0 is not a set.
1/0 is not a number which need not discussing here.
0^0=1 fits all cases except limit.
Yee says:

March 11, 2013 at 2:09 am

CS Prof,
1/0, 0/0, ln(0), and dividing by zero need not discussing here.
They are not reasonble at all.
I don’t need to reply these digressive topics.
0^0 is different from 0/0.

I did not dismiss continuity,
but it should not be overemphasized.
Continuity should not be taken for granted.
If the limit does not exist,
function value can also be defined.
Yee says:

March 11, 2013 at 2:41 am

Show you 1 is indeterminate:
Let 1=x.
0*1=0
Therefore, 1 is the solution of 0*x=0
x can be any number.
Therefore 1 is indeterminate.

The Cool Dude’s arguments are like this:
List some equations, get many solutions, and conclude it’s indeterminate.
What do I have to reply?
The Cool Dude says:

March 11, 2013 at 7:46 pm

Yee,
I think it would be best for all of us if we remember three important facts:
1) An operation can equal many numbers
2) A number can equal many operations
3) Two different numbers are not the same ever.
CS Prof says:

March 11, 2013 at 9:44 pm

Yee,

Let’s take them one at a time:

0^0 is not 0/0. CSP: Except that it is.

0/0 is a set, True

but 0^0 is not a set. CSP: It is because it is equivalent to 0/0

1/0 is not a number which need not discussing here. CSP: It is because it is ln(0) which is part of the result of taking the ln(0^0)

0^0=1 fits all cases except limit. CSP: This is actually not a relevant fact. We all agree that 0^0 can have the value of one. The disagreement is about the infinite planes of other values that it can also have.

1/0, 0/0, ln(0), and dividing by zero need not discussing here. CSP: Each have been shown to be relevant to the discussion.

They are not reasonble at all. CSP: I believe that you think that because none are reals that they are not reasonable. Dismissing them is exactly the same as dismissing the sqrt(-1) because it is not a real value.

I don’t need to reply these digressive topics. CSP: Actually you do. In over 18 months, these digressive topics have shown in multiple ways that 0^0 = 0/0 (which of course is simple algebra) and that both values are indeterminate sets. You say that they are digressive, but you never say why they are. You say that there are invalid operations, but never show why they are not valid. Cool Dude specifically showed equations where the indeterminate forms never had an inverse operation done yet produces the result of all complex numbers as that value of 0^0. You have not even attempted to explain why the form or operations are invalid, while Cool Dude went out of his way to show why the operations were valid even though the forms do not exist in the realm of complex numbers.

0^0 is different from 0/0. CSP: Prove it. I can prove that it is (actually I can simply repeat the steps of the “Clever Student” up to the step x/x. None of the operations are invalid. The form 0/0 isn’t invalid. What would be invalid is to do an operation on 0/0 as if it were a singular value. As you have agreed that 0/0 is indeterminate, so is 0^0 power. You cannot simply state that it is not so. You have to show through a series of operations that they do not represent the same thing, no matter what that thing is. I haven’t seen this proof from you in 18 months of this thread.

I did not dismiss continuity,
but it should not be overemphasized. CSP: Of course you did. You do every time you say that the value of 0^0 is 1, only 1, and exactly 1. Limits and continuity show that this is not true. You do the same every time you dismiss 0^0 being equal to 0/0. Limits can show this to be the case.

This discussion would have been done a long time ago with the simple statement that “0^0 is an indeterminate form that for convenience mathematians typically use the value of 1.” In fact that’s what the original article did. But the dogged insistance that any tool that can be used to show the indeterminate nature of 0^0 is in fact invalid has dragged this thread well past 800 comments.

Continuity should not be taken for granted. CSP: It seems you are doing exactly that. x^0 = 1 for all real x that is not 0. Therefore 0^0 is also one. That takes continuity for granted. But how do you reconcile it with 0^x = 0 for all real x that is not zero?

If the limit does not exist, function value can also be defined. CSP: Which is precisely can be used with L’Hopitals rule to show that 0^0 = 0/0 is indeterminate. It’s trivial to create two functions that converge to any value when divided into one another.

You have to prove your assertions or disprove Cool Dude’s. So let’s take your example:

Show you 1 is indeterminate:
Let 1=x.
0*1=0
Therefore, 1 is the solution of 0*x=0
x can be any number.
Therefore 1 is indeterminate.

CSP: The fallacy here are the two statements 1=x and x can be any number. Both cannot be true. Now if you had said that 0*y=0 and y can be any number, I would have been fine with that. But you have fallen into the association fallacy right here:

y can be any number.
1 is one of those numbers
x is 1
therefore 1 can be any number.

It fails because you mapped a single value of a set onto the entire set. In fact that’s the entire issue here. 1 is one of the infinite possible values that 0^0 = 0/0 can be. But it isn’t the sole valid value. a=b and b=c means a=c is not valid when b is a set.

The key in Cool Dude’s equations is that he never does an invalid operation that would do the set mapping on any of his indeterminate forms. If he took 0/0 and multiplied it by 0, then the same fallacy would arise. But go back and take a look. You will see those inverse operations are never done on the indeterminate forms.

So you cannot dismiss the other forms or the operations without proving that they are invalid. 1/0 is a form that cannot have real operations done to it because it is not a real. However, it is not an invalid form.
Yee says:

March 11, 2013 at 10:19 pm

What is CSP?
Yee says:

March 12, 2013 at 12:27 am

CS Prof,
You just repeat The Cool Dude’s digressive arguments.
I say again, I don’t need to reply them.

1/0, 0/0, ln(0), and dividing by zero are dismissed not only by me.
Why not say you dismiss 0^0. Dismissing 0^0 is exactly the same as dismissing the sqrt(-1)?

If you must discuss continuity, please prove discontinuous functions don’t exist.
If not, I don’t need to discuss continuity.

0^x=0 is valid only for x>0, not for all real x.

If you want to talk about operations.
What operation is a must? What are not? And why?
CS Prof says:

March 12, 2013 at 10:01 am

CSP (CS Prof) is me. I have not yet figured out how to quote your comments so that I can respond to them. So I prefaced my responses with my handle.

You still need to prove that 0^0 is not equal to 0/0. You are welcome to do this by contradiction, by starting with the assumption that the two are equal, do a set of valid operations on them, then show that they produce different results. I am happy to stipulate that neither are reals, so certain operations (such as multiplying by 0) are invalid to do. But you cannot hide behind 0/0 is an invalid form because you cannot divide by zero.

I reiterate the items that have been shown here:

1. 0^0 is equal to 0/0. See “Clever Student above.” and leave off the last replacement of 0/0 = 1.

2. Both are indeterminate forms that can be valued to any complex number.

3. These forms can have the value of 1. However, they can attain an infinite number of other values too as each are equivalent sets with an infinite number of members.

4. Neither form is a real number. So certain operations on these forms (such as multipling by zero) is invalid. However, they can still be used in equations as long as these invalid operations are not performed on these forms.

Yee, you are welcome to refute any or all of these assertions. However, you cannot simply dismiss the forms or operations as invalid without justification. These non real/complex forms are required because they are the only forms that have the ability to solve equations that have no real or complex solutions. Cool Dude described the ln(0) as an example. It is relevant because natual log is an operation that is valid to do on the form 0^0 by definition, i.e. ln(0^0) = 0 * ln(0). This is by definition. Now of course ln(0) is one of these non real/complex forms, specifically 1/0. It is the solution to the equation:

0 * x = 1

Clearly there is no real or complex value that can solve this equation. And while it is clear that the solution is a non real/complex form, it is not acceptable to call it invalid solely because it is non real/complex. It is perfectly valid to restrict the operations that can be normally done to real/complex values on these forms. However again, you cannot simply call them invalid forms because they are not real/complex values. So let’s get back to it. Cool Dude’s reasoning is this:

1. Presume 0^0 is equal to 1.

2. Raise each form to the n^th power (0^0)^n = 1^n (note n is real). Now note that I did not assign a value to the left hand side, so the operation is valid.

3. Take the natural log of both sides: n*ln(0^0) = n*ln(1) -> n*0*ln(0) = n*ln(1)

4. Now on the left hand side, n*0 is a valid operation so 0*ln(0) = n * ln(1)

5. Note the left hand side ln(0) = 1/0 which is the solution to the equation 0*x = 1. Also note on the right hand side that ln(1) = 0. So by substitution: 1 = n * 0

6. Note again that n is a real number so n*0 = 0. Therefore 1 = 0

Since the result is invalid, the original premise is invalid. We all understand that this is impossible to do without the non real/complex form 1/0. But this form exists because there are no real/complex solutions that can multiply by zero and produce a non-zero result. And exponentiation by definition is a repetitive multipliciative operation. Finally no invalid operations (such as multiplying by a real number) was done to any of the non real/complex forms.

Note in step one you can replace 1 with any real/complex value and still get a similar impossible result at the end. What it shows is that 0^0 is not a single real/complex value. Therefore it is indeterminate.

So 1/0 is relevant. 1/0 is non real/complex. 1/0 can be used as long as no real operations are done on it. Only definitions of operations have been done. They are valid on non real/complex forms because they are definition.

Yee, you are welcome to disprove the login. However, simply stating that it is invalid without defining why it is so isn’t sufficient.
The Wonderer says:

March 12, 2013 at 12:51 pm

The first comment is adapted from your comments. CSP is a shortened part of his name. So when CSP is stated in his argument that is his rebuttal to your comment.
CS Prof says:

March 12, 2013 at 2:18 pm

CS Prof,
You just repeat The Cool Dude’s digressive arguments.
I say again, I don’t need to reply them.

CSP: Sure you do. You have to explain why they are digressive. They are the results of operations on the form in question.

1/0, 0/0, ln(0), and dividing by zero are dismissed not only by me.

CSP: Who else then? And why?

Why not say you dismiss 0^0. Dismissing 0^0 is exactly the same as dismissing the sqrt(-1)?

CSP: You tend to invalidate forms that are not real. sqrt(-1) isn’t a real form either.

If you must discuss continuity, please prove discontinuous functions don’t exist.
If not, I don’t need to discuss continuity.

CSP: This is a non sequitur.

0^x=0 is valid only for x>0, not for all real x.

CSP: True. I made a mistake. The mistake you keep making though is saying that 0^x is invalid when x < 0. The result is not real. But just because it is not real does not automatically make it invalid.

If you want to talk about operations.
What operation is a must? What are not? And why?

CSP: Operations are premature at this point. The discussion needs to be about number systems first. They are used to describe math concepts that extend beyond existing systems. Some examples:

If all you know are natural numbers, then what is less than 1?

If all you know are whole numbers, then what is between 1 and 2?

If all you know are rational numbers, then what is the ratio between a circle's radius and circumference?

If all you know are real numbers, then what is the sqrt(-1)?

That's the problem here. The math concepts here have forms which extend beyond complex numbers. But the fact that these forms do not fit into one of these systems does not make it invalid, no more than 1/2 is invalid because it is not in the realm of integers.

0^x is still defined as 1/(0^(-x)) if x < 0 even though the result is a non complex form.
It is not invalid. It's just not a complex number.

Until there is agreement on this, the discussion goes nowhere. If we can agree on this, then we can move on to operations.
Yee says:

March 12, 2013 at 7:10 pm

CS Prof,
Whether 0^0 is equal to 0/0 is a matter of definition.
Definiton can not be proven.
Defining 0^0 as 0/0 is not good because we can’t benefit form it.
That is abuse of index laws.

ln(a^b)=b*ln(a) is invalid when a=0,
0^0 does not equal 0*ln(0)

Operating with 1/0 is ridiculous.
Yee says:

March 12, 2013 at 11:24 pm

What operations do we expect?
Do we benefit from the definition?
Does the definition make trouble?

These are what we should discuss here.
CS Prof says:

March 13, 2013 at 12:06 am

Whether 0^0 is equal to 0/0 is a matter of definition.

CSP: Let me correct you here. 0^0 is equal to 0/0 because of definition. Specifically the definition of exponentiation. You keep confusing the valid operations that are done with the results that are produced. Just because the result isn’t a real or complex number doesn’t mean that the operation that produced it is invalid.

0^0 = 0^(1 + -1) = 0^1 * 0^(-1) = 0/0

This is true because of the definitions of the operations, not the definitions of either 0^0 or 0/0.

Definiton can not be proven.

CSP: However, Yee definitions can be disproved. It has been done multiple times in both the original article and the hundreds of comments that have followed. I can define 1=2. I guarantee that you can disprove this.

Defining 0^0 as 0/0 is not good because we can’t benefit form it.
That is abuse of index laws.

CSP: It’s defined by the index laws. By the index laws x^0 -> x/x period end of discussion. Just because the result of 0/0 is a result that you have a challenge comprehending doesn’t invalidate the very laws that defines the equivalence of 0^0 and 0/0. To do otherwise is in fact an abuse of the index laws.

ln(a^b)=b*ln(a) is invalid when a=0,

CSP: You can repeat this as much as you like. You may even be right to say that non complex results are generated when a=0. However, you have given absolutely no reason why the result is invalid simply because the result is a complex number.

Cool Dude has done yeoman’s work here. I am here to break you of this misconception that an operation or law is invalid when the result happens to not be in the number system that you perceive as valid. Saying that ln(0) is invalid because it’s not a complex value is just about as useful as saying that 1/2 is invalid because it happens not to be in the realm of whole numbers. The result isn’t a whole number, but that doesn’t make it invalid. It makes it a value outside of that realm.

If you are going to continue this crusade, you will have to prove why forms that are outside of the complex plane are considered invalid. You can agree with me and say that they are non complex forms. You can also stipulate there are certain operations that cannot be done on them because they are not complex numbers, and therefore those operations produce provably inconsistent results. But I want to be clear that I will not accept you simply stating that they are invalid without a reasonable proof as to why they are.

0^0 does not equal 0*ln(0)

CSP: It has to, Yee. It’s the definition of exponentiation in terms of logarithms. Now I don’t disagree that ln(0) is a non complex form. This is the reason it is disallowed when results have to be in the complex plane, as there are no complex solutions. But once again you confuse the operational definition with the result.

Many times you have made the statement along the lines of “It is reasonable to define 0^0 = 1”. It is perfectly reasonable to also say that “ln(0) is undefined because there is no complex value that e can be raised to that produces a result of zero.” However, just because the result is a non complex form, does not make it invalid.

Operating with 1/0 is ridiculous.

CSP: Why? It allows for a definition for a non complex form that exists outside of the realm of complex number. It is a form that facilitates explaining what happens with 0^0. It extends the number system into a space where other non complex forms such as 0^0 and 0/0 can be explained.
Yee says:

March 13, 2013 at 7:08 am

CS Prof,

1.”0^0 is equal to 0/0 because of definition.”
This is not wrong, but this is not good.
Defining 0^0=1 is better, as I mentioned above.

2.”0^0 = 0^(1 + -1) = 0^1 * 0^(-1) = 0/0″
0 = 0^1 = 0^(2 + -1) = 0^2 * 0^(-1) = 0/0
Is this true?

3.”ln(a^b)=b*ln(a) is invalid when a=0
You can repeat this as much as you like.”
Yes, I will.

4.”Operating with 1/0 is ridiculous.
Why?”
I don’t need to answer why it is ridiculous.
You can start a new topic to discuss 1/0.
Yee says:

March 13, 2013 at 9:57 am

About dividing by 0:
Division is the inverse operation of multiplication.
Multiplying by 0 is irreversible,
so dividing by 0 can not be done.
Maurizio Paolini says:

March 13, 2013 at 1:27 pm

Mathematics is a game that has rules… rules can be changed, but everyone
playing the game *must* agree to the rules, otherwise they play a different game.
It is customary among mathematicians to assume that functions assume a single value. They have a domain (a set on which they are defined).
if x is in the domain of f, then f(x) takes one and only one value. In x is not
in the domain of f, then f(x) simply is not defined.
For this reason, for example, sqrt(4) is 2 and not {-2,2}.
Another example: a/b = x if and only if the equation ax=b has one and only one solution (existence and uniqueness), which excludes 0/0 (more than one
solution) and 1/0 (no solutions at all).
Binary operators are just functions defined on (subsets) of some AxA with
A e.g. the real numbers or the complex numbers.
Why functions are forced to take only one value? Because of convenience: in this way we can easily compose functions, composing multivalued functions would force to also define the value of functions when computed on sets, and that cause problems.
Furthermore, including or not some specific value in the domain of a function is a matter of convenience: do the usual properties of the that function still hold true in the extended definition?
This is the reason why 0/0 is not defined e.g. to be 1…
The whole point of the discussion is whether or not is “convenient” to define 0^0 to be 1.
(I definitely vote for YES)
The Cool Dude says:

March 13, 2013 at 3:23 pm

Yee,
“Defining 0^0=1 is better, as I mentioned above.”
Is an opinion. An opinion which can be proved is either correct or incorrect, making it a prediction. An opinion which can’t be proved is pointless. Predicting a result is only necessary when the result needs to be used immediately, or to show where to look to prove the prediction in a timely manner. This conversation has no current material application, and thus, is not bound by Time.

Once again, 0/0 is a set, which means
0 = 0^1 = 0^(2 + -1) = 0^2 * 0^(-1) = 0/0
This IS technically true.

“I don’t need to answer why it is ridiculous.”
If you want me to stop talking about it you do.

“Multiplying by 0 is irreversible,
so dividing by 0 can not be done.”
This is circular logic.
Multiplying by zero is assumed irreversible because dividing by zero is assumed impossible, and dividing by zero is assumed impossible because it does not result in a real number, which is only a valid argument for as long as √(-1) is assumed impossible.
CS Prof says:

March 13, 2013 at 4:50 pm

1.”0^0 is equal to 0/0 because of definition.”
This is not wrong, but this is not good.
Defining 0^0=1 is better, as I mentioned above.

CSP: At least we agree that 0^0 is equal to 0/0 because of definition is not wrong.

2.”0^0 = 0^(1 + -1) = 0^1 * 0^(-1) = 0/0″
0 = 0^1 = 0^(2 + -1) = 0^2 * 0^(-1) = 0/0
Is this true?

CSP: Perfectly true. And exactly the point. Both 0^0 and 0/0 (which are equivalent) are both sets that contain at minimum all complex values.

3.”ln(a^b)=b*ln(a) is invalid when a=0
You can repeat this as much as you like.”
Yes, I will.

CSP: The reason is that the result is a non complex form. That fact does not invalidate the operation. But 0^0 (and 0/0) are also non complex forms. There’s no way to discuss them without allowing other non complex forms into the discussion.

4.”Operating with 1/0 is ridiculous.
Why?”
I don’t need to answer why it is ridiculous.
You can start a new topic to discuss 1/0.

CSP: Why would I do that when it is totally relevant here?
Yee says:

March 13, 2013 at 7:17 pm

The Cool Dude,

1.
Defintion is irrelevant to right or wrong, ture or false,
correct or incorrect.
Trying to prove a definiton is wrong.

2.
If 0=0/0,
0 must also be undefined.

3.
Dividing by 0 is ridiculous because multiplying by 0 is irreversible.
Multiplication is defined prior to division.
It’s not circular logic.
Yee says:

March 13, 2013 at 7:23 pm

CS Prof

1.
You still don’t understand what we should discuss here.
We don’t discuss what definition is right,
but what definition is good.
Defining 0!=0 is also not wrong.

2.
If 0=0/0 is true,
0 must be indeterminate, it must be a set.
CS Prof says:

March 14, 2013 at 4:19 am

Yee,

You may have decided that this discussion is about what is a good definition. But the original question asked “What is the value of 0^0?” The whole point of the nearly 900 comments so far is that the value of the form cannot be defined. It is indeterminate. It is specifically a set that contains the entire complex plane at a minimum. Its value is indeterminate because there is no single specific value in the complex plane that can completely represent it.

As to your second point, specifically because the forms are these sets, when they are mapped into the complex plane, they can attain the value of any single element in that plane. But it’s a one way street. 0^0 can be mapped to 0, or 1, or e, or pi, or -6+5.2i. But that does not mean that any of these values completely represent 0^0. These mappings, in my opinion, are like a shadow of a 3D object that is passing perpindicular through a 2D plane. There is a projection of the object onto the plane, but the plane doesn’t have the capacity to completely capture the object, because the object is outside of the domain of the plane. And just because there is a point on the plane that is a part of the projection, that point does not, in fact cannot, represent the entire 3D object that is causing the projection.

Continuing the analogy, the very operations that Yee continually invalidates are precisely the ones that take points in the plane and “lifts” them perpindicular to it. So those results also become 3D forms that are no longer completely representable on the 2D plane. It is analogous of taking a 2D point, doing an operation, and having a result that disappears from the plane.

I agree with Maurizio about needing to define what are the rules of the game. Yee has had this discussion with feet firmly planted in the 2D plane. Any form or operation that projects off that plane is invalid or ridiculous. The problem is that since the original form starts off the plane, there is absolutely no way to describe it completely while solely using operations that keep you on that plane. All that can be done then is to show projections of the shadow on the plane that gives results that defy logic within the domain of the plane.

To finish up, just because an operation produces a result that is no longer on the plane does not mean that the operation is invalid, nor that the resulting form is still not subject to the operation’s rules in the new space. This is the reason that there have been multiple posts about the translation of 0^0 = 0 * ln(0). In the 2D plane of complex numbers, this is invalid. It is invalid because the ln(0) is a result that no longer exists on the plane. However it is a form that exists on in the 3D space (as 1/0), and certain operations, definitions and rules still apply to it (such as it being the solution to the equation 0*U = 1 which of course no complex value on the plane can satisfy). Since it is a fully formed 3D form, it can be used to describe other 3D forms.

I know that The Cool Dude has said on multiple occasions that the fault isn’t the defining of 0^0 as 1, it is the pinning of it to that single value to the exclusion of all others. Since 0^0 projects to every value on the plane, it is possible to use it to represent each of the values. The problem lays in the simplification that it is only that single value when clearly it is not.

So in short: 0^0 projects (does not equal) to 0. 0^0 projects to 1. 0^0 projects to e. However 0,1, and e are distinct points on the 2D plane of complex numbers. They do not equal each other, nor do any of them equal 0^0 even though it is possible to project 0^0 onto each of them.

There is no need to “define” 0^0. Call it indeterminate. Project it onto whatever complex value you need it to be for the context you happen to be working in at the time. That’s why in the original piece the Mathematician stated that the value was 1 for the binomial theorem. But please recognize that no matter what context that it is used in the complex plane, that in fact it is a form that exists outside of the plane and there is no true way to completely define it within that plane.

I hope this helps to move the discussion along.
The Wonderer says:

March 14, 2013 at 8:41 am

Yee, A=B and B=C. A doesnt have to equal C if B is a set. A and C dont have to be a set though.
The Cool Dude says:

March 14, 2013 at 5:18 pm

Maurizio Paolini,
The title of the discussion is literally “What does 0^0 (zero raised to the zeroth power) equal?”

It is literally about equality, and not definition.
Yee says:

March 14, 2013 at 7:25 pm

The Cool Dude,
What 0^0 equals depends on definition.
Yee says:

March 14, 2013 at 7:32 pm

The Wonderer,
Why do we regard 0^0 as a set?
Maurizio Paolini says:

March 14, 2013 at 7:54 pm

Cool Dude: No! It is about definition. Mathematics is mostly a construction.
Yee says:

March 15, 2013 at 12:21 am

CS Prof,

“the value of the form cannot be defined”

This is wrong.
It can be defined.
What properties it has depends on how it is defined.
You can not criticize it by the properties that need proving depending on the definition.
CS Prof says:

March 15, 2013 at 7:43 am

“the value of the form cannot be defined”

This is wrong.
It can be defined.
What properties it has depends on how it is defined.
You can not criticize it by the properties that need proving depending on the definition.

CSP: I find it interesting that I spend the time to write a very clear description of what is happening, and you pull a single snippet out of it, declare it wrong, without explanation.

It cannot be defined as any single complex value because any definition of it can be disproven. When you set 0^0 to any specific value, which is in fact a definition, it can be shown that 0^0 can be in fact another distinct value. That is proof by contradiction. Which means it cannot be the original value, even if you define it to be so.
You’ve done it several times yourself in your comments.

Yee, your premise is false because it is clear that the definition of 0^0 (which once again is equal to 0/0) is a set, not a single value. That is the correct definition. And by defining it to be so, it cannot solely be defined as a single element of that infinite set of values that it represents. At best, which we have all agreed to, is that you can use a specific context dependent value of the set when you need to. But here’s the important point that I’ve been trying to make clear: Just because you use the value 1, which is in the set today, does not invalidate using the value -9-3.2i from the set tomorrow in a different context. Both are valid, and 0^0 is not equal to either value. That is why in my previous post I changed the verb equals to projects.

The fact that 0^0 is a set precludes defining it as a single value.
The Cool Dude says:

March 15, 2013 at 4:54 pm

Yee,
Maurizio Paolini,
There is not a single correct mathematical definition in existence that can not be proven.

All operations in math need only one line to define what they do, and all operations, hold addition, are defined by the previously established operations. Addition is not defined by previous operations, because addition is the first operation, and is defined by the number line, which is the premise of all math, and hence, is the first “given” in any math question. Any other property of any operation outside of the single line needed to establish its basic property can be proven using that one, basic line.

There are zero correct definitions, outside of only the most extremely basic, that can not be proven on their own, and there are an uncountable number of correct and unique definitions in mathematics. Given this, the percent chance of defining something you can’t prove being wrong is just shy of a guarantee.
Yee says:

March 15, 2013 at 7:39 pm

CS Prof,
“the definition of 0^0 (which once again is equal to 0/0) is a set, not a single value.”

This is a bad definition which makes trouble.
Defining 0^0=1 is better.
The Cool Dude says:

March 15, 2013 at 10:46 pm

Yee,
A definition that directly conflicts facts is always a bad definition.
I have proved in 847 different ways that 0^0=0/0. Both are sets, and defining either as any possible value can prove them to be sets of all values possible in the realm of mathematics.
Your definition is in direct conflict with the most basic laws of mathematics.
0^0=1 is a good definition in certain very specific situations.
0^0=1 is a terrible definition in all other situations.
Yee says:

March 15, 2013 at 11:59 pm

The Cool Dude,
“There is not a single correct mathematical definition in existence that can not be proven.”

No definition can be proven.
On the contrary, operations need proving according to definitons.
Definitons decide what operations are true.
Yee says:

March 16, 2013 at 12:15 am

Definitions determine what operations are true.
Yee says:

March 16, 2013 at 10:47 am

The Cool Dude,
Dividing by 0 is really terrible.
Defining 0^0=0/0 is really terrible.
The Cool Dude says:

March 16, 2013 at 11:24 am

Yee,
0^0=1 is not a definition.
It is an answer.
It is a wrong answer, because 0^0 is a set.

“Definitions determine what operations are true.”
This is also wrong.
Operations are simply a method to compress an already existent idea.

I can create some operation “#”, so that
a#b=((+a)*a)^a…
Continuing through hyper operations until there have been “b” operations applied, where a#1=0+a.
This operation can not be proven, because I made it up.
I made it up to express an already existent idea.

You can define a#b to mean whatever you want it to, because it is nothing more than an abbreviation. The idea I showed of a#b, will always exist. Changing the definition of the # operation doesn’t make the idea go away. Changing the definition of a#b only means that the original idea of a#b can’t be expressed with the # symbol.

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

1,177 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

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