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?

Joe says:

April 2, 2013 at 10:43 am

The cleverest student wrote out the problem as a limit and then used logarithms to simplify. Should it not be the natural logarithm instead of the common logarithm? It would make more sense to me.
Yee says:

April 2, 2013 at 7:12 pm

Joe,
Limit from only one path means little.
There are many other cases whose limit is not 1.
The Wonderer says:

April 4, 2013 at 6:12 am

Limit is one or zero depending on the function. I wouldn’t use it to much to prove this problem.
The Cool Dude says:

April 4, 2013 at 9:57 pm

Yee,
“You said 0! can also be a set.
Well, is 0!=1 a factual truth?”
It is true in that 1 is a single of the infinite results produced by the set “0!”.

“You are just overemphasizing continuity.”
Didn’t even use limits or an argument based on continuity once to prove that 0^0 is a complete set of all numbers, so I am actually completely ignoring continuity.

“The most intuitive and natural understanding of power zero
is identity element of multiplication.”
Zero has an infinite number of what were previously, and accurately named “local multiplicative identities”, so if the most intuitive and natural understanding of x^0 is the identity element of multiplication for x, then the most intuitive and natural understanding of 0^0 is all numbers x*0=0, where x*0 only produces 0 for all numbers and sets (or sets of sets, sets of sets of sets, etc) x that do not contain 1/0.
Yee says:

April 5, 2013 at 5:34 am

The Cool Dude,
How to define 0! and 0^0 depends on what you expect.
It is good to define 0^0 as a single value 1.

Listing certain equations and get many solutions about 0^0 is not difficult,
you did not prove anything.

Solutions of x*0=0 is the definition of 0/0, not 0^0.
The Cool Dude says:

April 8, 2013 at 3:32 pm

Yee,
“what you expect” is an argument of continuity.
Yee says:

April 8, 2013 at 6:57 pm

The Cool Dude,
What I expect refer to more fields of mathematics,
not only continuity.
The Cool Dude says:

April 9, 2013 at 6:19 am

Yee,
You base what you expect on the function x^0.
It is an argument that the expected result from x^0 can be expanded to x=0.
It is an argument of continuity.
Yee says:

April 9, 2013 at 10:42 am

The Cool Dude,
This is not an argument of continuity.
Those who expect continuity tend not to define it.
Whether I expect continuity is not the point.
The Cool Dude says:

April 9, 2013 at 4:11 pm

Yee,
“What to expect” is a statement of continuity.
You argue 0^0=1 because x^0=1 for all other x.
That is exactly what an argument of continuity looks like.
Yee says:

April 9, 2013 at 7:46 pm

The Cool Dude,
Expecting x^0=1 is not a consideration of continuity at all,
because 0^x is not continuous.
x^0=1 is important in some fields as I mentioned.
The Cool Dude says:

April 11, 2013 at 8:45 pm

Yee,
0^x technically IS continuous, if you consider 0^0 as a complete set of all real numbers.

If your definition of continuity is that the limit from any particular side is equal to at least one of the values of the function at that point, all standard functions can be proven continuous, including multiplication, exponentiation, trigonometric equations, etcetera.

Since the idea of continuity has no rules for when considering values that are sets, and there is literally no other possible way for a function to approach a point that is a set of all numbers, I think this extension of the definition of continuity fits in nicely, and allows all functions to be continuous at all points, both in terms of absolute value, and angle in the complex plane.
Yee says:

April 12, 2013 at 7:33 am

The Cool Dude,
I don’t overemphasize continuity.

lim x^y is a set,
(x,y)->(0,0)

so x^y is discontinuous at (0,0).
But it is good to 0^0=1.
The Cool Dude says:

April 12, 2013 at 5:28 pm

Yee,
If lim x^y is a set
But the limit is still equal to the actual value at that point
How is it discontinuous.
In what way is a function supposed to approach a set continuously

“Good” is an opinion, and one opinion is just as valid as another until backed with facts.

It is a fact that if you define 0^0=1 to satisfy the continuity, and consistency of the function x^0, you are actually making far MORE functions discontinuous and inconsistent. If defined to support consistency, it should be defined situation by situation, not just one value for all situations. If you want consistency, that is the way to go.
LarryD says:

April 13, 2013 at 1:10 am

I think that most of you are going round in circles. If you want 0^0 to behave in certain ways then there has to be another algebra with different rules that satisfy appropriately. There is nothing wrong in making another ‘algebra’ (there are many to date) to ensure certain crtiteria but it must be done Logically.
In the basic Algebra zero is an UNDEFINED assumption (such as in Basic Logic with ‘&’ and ‘not’ and all others are defined using just those two). Zero is already used in more than one way, as a starting point and as a ‘place’ marker and is not a number. If something is not a number then cannot have defined limits in Calculus or anything else. This ambiguity was set long before modern maths came into being but it seems we are stuck with it. Sumerians didn’t like these problems so they didn’t use zero as a number but its counterpart was a ‘place’ maker.
Yee says:

April 13, 2013 at 2:01 am

The Cool Dude,
Since

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

is not a single value,
x^y is discontinuous at (0,0) whatever value 0^0 is defined.

Defining 0^0=1 to satisfy x^0=1 is irrelevant to continuity.
There is nothing inconsistent in defining value at discontinuous points.
The Cool Dude says:

April 13, 2013 at 11:17 am

Yee,
Continuity is an idea which has no definitions for points that are sets!
You can not say for certain if it is discontinuous even if it has many different limits from many different sides, because there are no rules of continuity for when the point is a set. If your definition of continuity were correct, then NO points that are sets would be continuous. Ignoring a problem instead of trying to solving it is one thing a mathematician should never do.

You define 0^0=1 to satisfy consistency of power zero, but you are thereby ignoring the consistency of all the other aspects of it. To make your idea work you throw out index rules, logarithms, the complex plane, consistency from directions that aren’t considered in a simple 2-D plane, and the only way for exponential equations that have an infinite area of results to express all their solutions. It may patch up logical holes in the mathematical problems you consider, but it tears a senseless void in every other area which does not comply to what you think is good, of which there are infinitely many.
Yee says:

April 14, 2013 at 9:50 am

The Cool Dude,
Defining 0^0=1 satisfies all aspects of mathematics.
Discontinuity is not wrong, this problem need not solving.
You are repeating the aspect of overemphasis of continuity
and ignoring all other aspects.
The Cool Dude says:

April 15, 2013 at 12:35 pm

Yee,
“Defining 0^0=1 satisfies all aspects of mathematics.”
Doesn’t satisfy index laws.
Doesn’t satisfy logarithms.
Doesn’t satisfy complex exponents.
And it definitely doesn’t satisfy continuity.

“Discontinuity is not wrong, this problem need not solving.”
0^0 equaling only 1 is literally wrong mathematically.
If it’s wrong mathematically, then all you have left is continuity.
0^0=1 does not satisfy continuity.

“You are repeating the aspect of overemphasis of continuity
and ignoring all other aspects.”
Your entire argument is based on consistency, which is the same thing as continuity. My entire argument is based on satisfying mathematical laws, all of which point directly to 0^0 being a complete set of all numbers.

Give an actual reason for why 0^0 equals only 1 is good, present an example.
Yee says:

April 16, 2013 at 2:49 am

The Cool Dude,

a^(m-n)=a^m/a^n is invalid when a=0,
There are no such index laws to satisfy.

a^b=b*ln(a) is invalid when a=0.
There are no such logarithms to satisfy.

Complex exponents does not discuss base 0,
There are no such complex exponents to satisfy.

I have listed some reasons above.
Yee says:

April 16, 2013 at 4:16 am

When the limit does not equal 1,
if 0^0 is undefined,
It is discontinuous.

When the limit equals 1,
if 0^0 is undefined,
It is still discontinuous.

Satisfying continuity by not defining 0^0 is impossible.
The Cool Dude says:

April 16, 2013 at 9:41 pm

Yee,
If a^0=a/a is invalid when a=0,
then a^0=1 is invalid for a=0.
a^0 was only ever 1 because of the original premise of a^0=a/a.
Multiplication and division come before exponentiation, because exponentiation is DEFINED by multiplication and division.
Your argument defeats its self.

Undefined is not a definition. It is literally the lack of a definition. That’s why it’s called undefined.

Continuity has no established factual definition for sets of points, so if 0^0 is a set of all numbers, which it is, then you can not actually say if it is or is not continuous, because continuity does not rely on other mathematical concepts, it is simply a method of observation. If you define continuity as being the limit from all sides of a point are equal to the value at that point, and if that point is a complete set of all numbers, and the limit from all sides spans all numbers, then it is continuous. If you define continuity as being the limit from all sides of a point are equal, AND equal to the value at that point, then it is not continuous. The only difference between these two definitions is that one doesn’t allow for sets, and the other does. I have never actually heard the second definition of continuity until I made it up right now, because it is far too specific for anyone practical to want to use, and mathematicians tend to ignore sets until ignoring them causes problems. This being said, the second definition needlessly excludes all sets from being continuous, which is a ridiculous oversight, in my opinion, so the first definition is the most appropriate.
Yee says:

April 17, 2013 at 4:44 am

The Cool Dude,

a^0=a/a is an unreasonable definition.
Continuity should not be overemphasized.
You are only repeating the same arguments.
So do I.
The Cool Dude says:

April 17, 2013 at 2:35 pm

Yee,
That is the point!
You argue 0^0=1 because
“The most intuitive and natural understanding of power zero
is identity element of multiplication.”,
but a^0=1 is an idea which comes from the observation that a/a=a^0, and a/a tends to be 1, which means a^0=1 RELIES on the fact that a^0=a/a, and is only true when a/a=1!

If you say a^0=a/a is consistent at a=0, then you are agreeing with my point that 0^0=0/0,
If you say a^0=a/a is inconsistent at a=0, then defining 0^0=1 has no premise, and you remove all support and value from your argument!
Yee says:

April 18, 2013 at 12:41 am

The Cool Dude,
“a^0=1 is an idea which comes from the observation that a/a=a^0″
No.
Is a^1 an idea which comes from the observation that a^2/a=a^1?
a^0 is the identity element of multiplication, which is irrelevant to division.
You are repeating the argument that confuses a^0 and a/a.
The Cool Dude says:

April 18, 2013 at 9:33 pm

Yee,
“No.”
Yes.

“Is a^1 an idea which comes from the observation that a^2/a=a^1?”
My argument is that exponentiation is BASED on multiplication and division, which is literally as factual as it gets. It will never get more factual than that. Further, the conclusion of x^0=1 would have NEVER been reached on its own, without considering simpler functions as multiplication and division. x^0=1 is based on the fact that x^0=x/x, and x/x tends to equal 1. You can not tack on definitions to concepts which are already fully described and defined by other things, as that would defeat the point of the original definition, and invalidate the entire concept. For that reason, you can not claim that x^0=1 is independent of the original definition of exponentiation, because that would be tacking on a definition to exponentiation which is already covered in its original definition.
Your argument has no relation to mine. Instead of stepping down in terms of complexity, and stepping through a medium which is the only thing that DEFINES that concept, as I did, you instead took a short step up in complexity, through an idea that does NOT define the concept in any way shape or form. Your argument has no relation, and is enormously fallacious.

“a^0 is the identity element of multiplication, which is irrelevant to division.”
Exponentiation is defined by multiplication and division. It is literally always relevant to division, ESPECIALLY when the exponent of that number is an integer. Exponentiation will never be independent of multiplication and division. Multiplication and division are functionally more important than exponentiation, in the same way that addition and subtraction are functionally more important than multiplication or division. If you for some reason are under the impression that exponentiation is capable of existing independently of multiplication and division, you are equally as wrong as if you were to say multiplication is independent of addition and subtraction. If you believe that exponentiation is independent of division, but not multiplication, you are equally as wrong as if you were to say that multiplication is independent of subtraction, but not addition. Multiplication is in fact equally dependent on both multiplication and addition. If you think that multiplication is independent of addition, I have no response to that.

For the record, and future reference, a^2/a always has at least one possible solution of a, no matter what a actually is, but a does not rely on a^2/a because that doesn’t even make any sense.
Additionally, zero has an infinite number of multiplicative identities, so your point is once again invalid.
Yee says:

April 19, 2013 at 12:22 am

The Cool Dude,
x^0 is identity element of multiplication.
Identity element does not come from inverse operation.
The Cool Dude says:

April 19, 2013 at 6:41 am

Yee,
Division and multiplication are no different than addition and subtraction.
This does not stop your point from being fallacious.
a^0=1 is an idea which can ONLY come from the preexisting idea that a^0=a/a, and a/a=1, because there is no other place the idea of a^0=1 can come from.
If you say a^0=a/a is invalid for zero, then a^0=1 for a=0 has no support.
If you say a^0=a/a is valid for zero, then a^0=1 for a=0 is provably wrong.
Yee says:

April 19, 2013 at 7:40 pm

The Cool Dude,
No!
Identity element does not come from inverse operation,
no matter what operation it is.
The Cool Dude says:

April 19, 2013 at 10:45 pm

Yee,
Yes!
Exponentiation is defined only by multiplication, and division is the same thing as multiplication of the reciprocal, in the same way that subtraction is the same thing as addition of the negative.
There are no features of exponentiation which are independent of multiplication and division. In fact, there are no features of exponentiation that are dependent on anything other than multiplication and division.
Yee says:

April 20, 2013 at 2:31 am

The Cool Dude,
How do you define the identity element of multiplication?
The Cool Dude says:

April 20, 2013 at 2:15 pm

Yee,
I don’t view the number system as if there is anything that is both absolute and perfectly consistent. There are some things which tend to be absolute, and there are some things that tend to be perfectly consistent, but one should not title them so, because many similar statements can be made that match the original statement, but with subtle details, and choosing the wrong details is where all the most detrimental mistakes come from.
To say that there is one number that, when multiplied by x, is x, is technically correct, in the sense that there is in fact a number that does that, however, to say that there is only one implies that no other number does that in any other situation, which is not true. For an infinite range of x or n in “x*n=x”, there are an infinite number of values, x, which satisfy n being multiple things. Although 1 is the only solution that *seems* to satisfy all x, it is not so. If n were a complete set, then x*n would have no choice but to also be a complete set, and since complete sets contain all values, one value in it would be x, and it would satisfy the solution.
Even if you were to change the equation from “x*n is x”, to “x*n is only x”, there are still an infinite quantity of values for x that have multiple solutions for n. There are even an infinite quantity of values for x where x is a finite set, so that n can be multiple values while still not being a set.
The only way for 1=n to be special in the equation “x*n=x” is for n to not be a set, x to not be a set, and x isn’t any value in “0^±1”, which divides the range of the equation infinitely, an infinite number of times over.

To answer your question, I do not.
There is no one solution.
Yee says:

April 21, 2013 at 2:54 am

The Cool Dude,
The identity element of multiplication has only on value.
The Cool Dude says:

April 21, 2013 at 12:33 pm

Yee,
The multiplicative identity for some number x is equal to x/x, but x/x only has one value, of 1, when x is not a set, or zero. If x is a set, or zero, x/x has multiple values, and so the multiplicative identity for that number can be many things.

1 is only important when you restrict your values to single, non-zero, finite values.
Mathematics is far more complex than that.
Chris says:

April 21, 2013 at 12:51 pm

0 raised to the 0 is: “not 0”
0 raised to the 0 raised to the 0 is: “not not 0”
This fluctuates back and forth. Eventually the cycle repeats, and fluctuates between two numbers.
Chris says:

April 21, 2013 at 12:55 pm

I meant to say “vascillates”, not “fluctuates’.
Chris says:

April 21, 2013 at 1:00 pm

To clarify, eventually it vascillates between the _last two_ numbers. Sorry for any confusion.
Yee says:

April 22, 2013 at 1:11 am

The Cool Dude,
No!
The identity element of multiplication has only one value 1.
Although e*x=x and x*e=x has many solutions when x=0,
the identity element of multiplication is only 1.
x^0 is the identity element of multiplication no matter what x equals.
The Cool Dude says:

April 22, 2013 at 12:53 pm

Yee,
Yes!
x^0 is only presumed 1 because x^0=x/x, and x/x tends to be 1, but when x is zero, or a set, x/x can be many things. Sets having multiple multiplicative identities can be demonstrated at your request.
If you don’t assume x^0=x/x, then x^0=1 has no reasoning
If you do assume x^0=x/x, then the exclusivity of 0^0=1 can be shown nonexistent.

This isn’t to say that 0^0≠1,
But 0^0 is not exclusively 1
Which has been, and will continue to be my point.
Yee says:

April 22, 2013 at 11:38 pm

The Cool Dude,
I don’t assume x^0=x/x.
But I assume x^0 as the identity element of multiplication.
Then x^0 has reasoning.
Yee says:

April 23, 2013 at 12:45 am

x^0*x=x^1=x
x*x^0=x^1=x
Therefore x^0 is the identity element of multiplication.
This is a reasonable definition.
It is not good to relate x^0 to division.
The Cool Dude says:

April 23, 2013 at 10:35 pm

Yee,
This only shows that x^0 is some number that, when multiplied by x, is x, which I have already agreed to, but 1 is not the only number that does this. When x is zero, it can be anything, because the solution to x*0=0 can be anything.

Further, the only way to make the jump from x^0*x to x^1 without already assuming x^0 is a multiplicative identity, is to use index laws. You can not assume that x^0 is already a multiplicative identity, because then your statement would be circular in logic, because your point was that it must be the multiplicative identity because of that jump. At the same time, you have declared numerous times that index laws to not work when x is zero, which means you have made two statements which contradict each other. You could choose to redact your previous statements, and say yes, index laws do work at zero, but then we are left with the issue of x^0=x/x. On the other hand, you could also choose to redact your current statement, and say that, no, index laws still do not work at zero, but then you would be left with no current argument. Your remaining logical option is to say instead that, yes index laws work for multiplication, but no they do not work for division, but that brings up the very simple problem of replacement. If I take your statement and replace x with 1/x, then:
(x^0)*x=x^1
Turns into
(x^-1*0)/x=x^-1, or
(x^0)/x=x^-1
which clearly shows that in order for index laws to work with multiplication, they must also work with division. Finally, you could choose a stance somewhere in between, saying that, yes, index laws work for literally all multiplication and division, except for once, which is that 1/x=x^-1 is invalid for x=0. This, in my opinion, is a ridiculous exception to index laws, which would seem to serve no other purpose, and have no other reason to exist, other than to support this route of logical declaration, but opinions aside, it can once again be shown contradictory to your earlier statement by simply replacing the variable with its reciprocal. If x^-1=1/x is invalid for x=0, then x^1=x is also invalid for x=0, which directly contradicts the ability of (x^0)*x=x^1 to show (x^0) is a multiplicative identity.
Yee says:

April 23, 2013 at 11:39 pm

The Cool Dude,
Relating identity element to inverse operation is ridiculous.
You cannot replace x with 1/x when x=0.
And the identity element of multiplication is 1 although the operand 0 exists.
Yee says:

April 24, 2013 at 3:39 am

The Cool Dude,
You can not assume that x^0=x/x.
The Cool Dude says:

April 30, 2013 at 3:16 pm

Yee,
Variable replacement is a part of function mechanics, and has nothing to do with the actual value of the variables you’re replacing. If f(x)=g(x) for any domain R, then f(h(x))=g(h(x)) when h(x) has a range of R.

h(x) being non-real is irrelevant if your domain includes non-real values.
Yee says:

April 30, 2013 at 11:17 pm

The Cool Dude,
Replacing 0^0 with 1 is okay.
The Cool Dude says:

May 1, 2013 at 6:29 am

Yee,
In certain situations, no
In certain others, absolutely.
Yee says:

May 1, 2013 at 6:43 pm

The Cool Dude,
In all situations, yes.
The Cool Dude says:

May 2, 2013 at 6:36 am

Yee,
No.
It would only be okay to replace 0^0 with 1 in all situations if 0^0 was exclusively 1, or if all situations were limits and the limit was always 1, but multiple arguments for the prior have gone without refutation, and the latter is not the case for as long as I can ask for the actual value of an equation at a point.
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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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