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:

December 18, 2012 at 6:04 pm

Your inference has nothing to do with 0^0.
The Cool Dude says:

December 22, 2012 at 8:50 am

My inference is extremely important to our attempt at defining 0^0. If I can perform an operation on 0^0, and then the same operation on 1, and they have different results, then defining 0^0 as 1 is wrong.

The idea is simple.
1^(1/i2π)=e
0^(0/i2π)=0^0
If 1=0^0
Then e=0^0.
It can’t be one or another. Defining 0^0 as 1 is inappropriate because the same operation has two different results on each object. You can only define it as both, or neither.
Yee says:

December 22, 2012 at 9:15 am

The Cool Dude,
If 0^0 is not defined as 1,
e^0=e^(i2π)
e^(0/i2π)=e^(i2π/i2π)
e^0=e^1
1=e
This is identical to you inference.
This inference does not depend on defining 0^0=1.
The Cool Dude says:

December 23, 2012 at 6:35 pm

Yee,
e^(i2π/i2π) has multiple solutions. e^1 is one of them. e^0 is also one of them.
(-2)^2=2^2
(-2)^(2/2)=2^(2/2)
(-2)^1=2^1
-2=2
This is thematically identical to your inference.
This inference is not thematically identical to mine.

The difference between 1 and 0^0 is multiple solutions. 1 has multiple solutions when taking imaginary roots. 0^0 does not. They are different.
Yee says:

December 23, 2012 at 10:30 pm

If you make 1 multiple values,
it is obviously a paradox.
A paradox proves nothing.
The Cool Dude says:

December 25, 2012 at 12:55 pm

1 does not have multiple values, it has multiple methods of expression.
1 does, however, have multiple values when taking imaginary roots, just like any other non-zero number.
This is because 1=e^(i*2*arcsin(0)), for any value of arcsin(0).
0^0 does not have multiple solutions when taking imaginary roots, because it can not be expressed in any other way than 0^0.
Taking an imaginary root of 0^0 does nothing.
Taking the imaginary root of 1 has an infinite number of exponentially periodic values.
Yee says:

December 25, 2012 at 4:12 pm

It is only a paradox.
The Cool Dude says:

December 26, 2012 at 3:23 pm

Are square roots a paradox?
Are inverse trigonometric functions a paradox?
Are logarithms a paradox?

√(1)=1,-1.
1^2=1
-1^2=1

arcsin(0)=0,π,2π…
sin(0)=0
sin(π)=0
sin(2π)=0

ln(-e)=1+iπ,1+3iπ,1+5iπ…
e^(1+iπ)=-e
e^(1+3iπ)=-e
e^(1+5iπ)=-e

1^(1/i2π)=e^0, e^1, e^2…
e^0i2π=1
e^1i2π=1
e^2i2π=1
Yee says:

December 27, 2012 at 10:13 am

We don’t need to discuss this paradox.
You can start a new topic.
The Cool Dude says:

December 28, 2012 at 3:58 am

The idea of multiple solutions in mathematics holds both extreme importance, and is completely logically consistent with its self. Calling it a paradox without any explanation doesn’t actually mean anything. 0^0=1 just as much as 1/0=∞, which is to say, only if 0=1/∞.
Yee says:

December 29, 2012 at 12:39 am

It has nothing to do with 0^0.
Please don’t mix them.
Gaurav says:

January 9, 2013 at 10:39 am

I have did the solution of 0^0
Sol: 0^0
Suppose 2^0=1
2^1-1=2^1/2^1
=2/2
=1
Similarly,
0^0=0^1-1
=0^1/0^1
=0/0

Therefore 0/0 is undefined.So the value 0^0 is undefined or meaningless
Yee says:

January 11, 2013 at 3:17 am

Gaurav,
If 0^0=0^(1-1)=0^1/0^1=0/0
then 0=0^1=0^(2-1)=0^2/0^1=0/0
mostafa says:

January 15, 2013 at 11:17 am

e^0=e^(i2π)
e^(0/i2π)=e^(i2π/i2π)
e^0=e^1
1=e !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
but it is not correct conclusion
in second line we cant write e^(0/i2π)=e^(i2π/i2π) because 0 is not equal i2π so u cant conclude that 0/i2π is equal i2π/i2π its an incorrwct conclusion
in compex numbers if we have x^a=x^b then you cant absolutely conclude that a=b , this result is correct just in real numbers not in complex
mostafa says:

January 15, 2013 at 11:40 am

if o^o =x then ln (0^0) = ln x
then o*ln 0 =ln x
ln 0 in not defined like 1/0
but ln 0 ≅ -1/0
so 0*ln 0 ≅ 0*(-1/0)
so so we can say 0^0 ≅ 0/0
and then we can conclude the conclusion that i had wrote above
i mean this conclusion:

0^0= 0
We know :
(eq1): 0*x=0 or (x=0^0) ⇒
⇒∀xϵℝ we know ” 0*x=0 ” is logical ⇒ 0 can shows xϵℝ
In the other word 0/0 ≅ ∀x|xϵℝ ⇒ ∀x|xϵℝ ⊆ 0^0
⇒ ℝ ⊆ 0^0
Now
We rewrite eq1 in this way:
(eq2) y*0=0 like above ∀y|yϵℝ ⊆ 0^0 so
x*y*0 =0 ≅ ℝ*ℝ ⇒ x*y*0=0 ≅ ℝ^2 ⇒ ℝ^2 ⊆ 0^0
We can continue this process so we obtain:
ℝ^n ⊆ 0^0 , n ϵ N+{0}
Now again we know:
In eq1 we can have x|xϵ ℂ
So
We can conclude:
ℂ^n ⊆ 0^0 , n ϵ N+{0} and also 0^0 ≅ ℂ
in the other word it is a collection of number that any number can be 0^0
for example in physics if we wanna conclude the mass of light from below formula
m= mₒ/(1-v^2/c^2)^1/2 ( for any velocity)
velocity of light = c and mₒ for light =0
so m for light = 0/((1-c^2/c^2)^1/2=0/0
so we cant know what is the mass of the light here because for any m it is right
but from the other formula E=m*c^2
E for light = hf so m=hf/c^2
Yee says:

January 16, 2013 at 12:06 am

mostafa,

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

0^0 is irrelevant to 0/0.
Don’t mix them.
The Cool Dude says:

January 16, 2013 at 6:21 am

It’s true that you can’t take 0=i2π from e^i2π=e^0, because imaginary roots have multiple solutions, which makes the conclusion of 0=i2π exactly as valid as taking 2=-2 from -2^2=2^2.

The reason my conclusion works is because although 1 has many solutions when taking the imaginary root, 0^0 does not. This is because the angle in the complex plane of 0^0 is ln(0)*0
Yee says:

January 16, 2013 at 6:35 pm

The Cool Dude,

In complex variables,
e^a=e^b does not imply a=b.
And this topic is irrelevant to 0^0.

ln(a^b)=b*ln(a) is invalid when a=0.
0^0 is not ln(0)*0
The Cool Dude says:

January 17, 2013 at 6:36 am

I don’t see why it is invalid for 0^0. They seem to have quite a lot in common.
Yee says:

January 17, 2013 at 9:27 am

The Cool Dude,
Since ln(0) is undefined,
ln(a^b)=b*ln(a) is invalid when a=0.
The Cool Dude says:

January 17, 2013 at 6:35 pm

If ln(a^b)=b*ln(a) is invalid for 0^0, then what is it?
Similarly, what would (0^0)*(0/0) be? is it different from both of them?
Is 1^(1/0) also a completely different mathematical entity, for which is required a completely different set of rules and exceptions?
Is even 0*(0^0) different from either 0^0 or 0/0?

Making the two unrelated rises an infinite number of questions about their combined nature, and the nature of themselves without reference to the other.
Where does the ambiguity stop?
Yee says:

January 18, 2013 at 5:26 am

The Cool Dude,

ln(a^b)=b*ln(a) is invalid for a=0.
0/0 is undefined, so (0^0)*(0/0) is undefined.
1/0 is undefined, so 1^(1/0) is undefined.
0*(0^0) depends on the definition of 0^0,
and 1 is the best definition of 0^0.
The Cool Dude says:

January 18, 2013 at 10:34 pm

What does being undefined actually mean? Could we define undefined for a moment, so we know what we’re actually working with here?

Is √(-1) undefined?
√(-1)^2=-1, and that seems like a definition to me, so it’s defined now, right?

But surely 1/0 is still undefined.
Unless (1/0)^-1=0, which it does.
If that isn’t a definition, I don’t know what is.
Either (1/0)^-1 is somehow magically not zero, or √(-1) is somehow undefined, despite it being completely defined in at least an infinite number of ways.

But then, if I say 0^n=0/0, what is n?
If 0/0≠0^0, then n can’t be 0, right?
But if n is any positive or negative number, it would just be 0 or 1/0, but 0/0≠0, and if 0/0=1/0, then that means that (0/0)^-1=0, but that isn’t true, because (0/0)^-1=0/0.

So what then? Is n imaginary or complex? Is n ALSO undefined? I can’t get behind that, why can’t we just define n already, what’s the problem?
=) says:

January 22, 2013 at 4:47 pm

Peace & Love

(º±º)
Yee says:

January 23, 2013 at 11:21 pm

The Cool Dude,
0! can also be undefined.
Alexander says:

January 24, 2013 at 8:50 pm

Unsubscribing to this question
The Cool Dude says:

January 24, 2013 at 10:54 pm

Yee,
Undefined things are meant to be defined. If one is unable to define a thing, or a set of things, one can define objects in that set as related to one another. Refusing to define one thing in a family of objects makes defining other things in that family pointless to define.

An obvious initial definition might be that 0*n=0, for any n. More appropriately, one would pronounce the “=” symbol, not as equals, or is equal to, but rather, “has a possible solution of”.

One could easily define 1/0 as some imaginary number, V. In the same way that we can not solve √(-1), but we can still relate it to other concepts, like ln(-1), or e^(iπ), we can allow 1/0 to be V, and use the substitution to simplify other concepts.

Being that 0*-1=0, and 0^-1=V, it would be easy to say that 0^0=V^0.
In the same way, it would be easy to say that 0/0=V/V.

Shifting the definition of 0*n=0, one may receive n=0/0, for any n*0=0. This being so, if one desired, one could easily define ln(0) as V, by saying that ln(0/0)=ln(n), while n is a complete set of all real and imaginary but finite numbers, and the natural logarithm having an infinite range over both the real and imaginary planes, ln(n)=n, and that ln(0/0)=ln(0)-ln(0), or 0*ln(0)=n. Since n=0*V, and 0*ln(0)=n, then ln(0) must be V.
Additionally, since V*-1=V for the same reason that 0*-1=0, one could say that -1*ln(0)=ln(0), or that ln(V)=ln(0), making V a solution to ln(X)=X. Making that connection would follow that e^V=V.

However, something was missed in the conclusion of e^V=V. Since ln(V) AND ln(0)=V, and V=e^V, then e^V must also equal zero. If e^V=0, then this means that e^(0*V)=0^0, or that, e^(0/0)=0^0, and to finish, that ln(0^0) must equal 0/0, which is equivalent to 0*V, or 0*ln(0).

This all being a very nice way to tie together many numerous ideas relating to the concept of the imaginary number V, through acceptance of one disputable definition. Not accepting these definitions is a matter of personal choice, but accepting an idea which answers much less than another, another which is much simpler, and has much more mathematical support, makes math a messy room.
Yee says:

January 25, 2013 at 3:44 am

The Cool Dude,
Relating 0^0 to these undefined things is only a paradox.
0*1=0,
1 can be any number.
Therefore 1 must be undefined.
The Cool Dude says:

January 25, 2013 at 10:07 pm

Yee,
a=b and b=c therefor a=c relies on b not being a set.
0/0 is a set.
There is only a paradox if we mistakenly assume 0/0 is not a set.

Furthermore, I retain a complete inability to see any reason whatsoever as for why ln(x^x)=x*ln(x) is invalid for x=0. Having a complete lack of realness has not stopped there from being an entire branch of definitive mathematics surrounding √(-1), so why should it be any different for imaginary numbers that lack finite properties.
Yee says:

January 27, 2013 at 9:31 pm

The Cool Dude,
Because ln(0) is undefined,
ln(a^b)=b*ln(a) is invalid when a=0.
No more explanation is needed.
The Cool Dude says:

January 28, 2013 at 4:15 pm

Yee,
Isn’t 0^0 also undefined? How does it make sense to attempt to define 0^0 if we can not define other things in the process, or relate it to other ideas?
Noel Evans says:

January 28, 2013 at 6:39 pm

As the Mathematician said, it is convenient, nice, elegant, etc. to define 0^0 to be 1 in
certain situations. In others some may prefer to say it is undefined. So why not make
the definition context sensitive? Define it how one wishes in the context they are dealing with. It’s nice to be able to say (ax+b)^0=1 without having to make an exception of “if x is not -b/a.” And as mentioned the definition of the binomial coefficients is nicer when exceptions are not needed. And power series are “nicer” when we don’t have to make an exception for the ax^0 = a term. But when dealing with situations where defining 0^0 to be 1 causes problems, make it undefined in
those contexts. Just make it clear how 0^0 is to be treated in the discussion that follows.

As for me, I feel that it should be defined or left undefined way before calculus and complex numbers arrive on the scene. After these arrive then make it context sensitive there too. Are there not other context sensitive concepts that are defined in math? For example, is 3.5 the quotient of 7/2 or is it 3?

Some definitions are “extended” as the context is “extended.” The definition of “even
vs odd” is different when extended to include fractions. So is “prime vs composite.”
And lcm and gcd can be extended to include fractions:
gcd(a/b, c/d) = gcd(a/b)/gcd(c/d)
lcm(a,b, c/c) = lcm(a/b)/gcd(c/d).
But these definitions are not the same as for integers.
Hopefully, the extended definitions will be “compatible” with the original definitions.

Mathematicians will probably never completely agree on many concepts and definitions in mathematics, but the math is still useful and can usually be followed if
we pay attention to the context of the discussion.
Yee says:

January 28, 2013 at 11:17 pm

Don’t overemphasize continuity,
and it is good to define 0^0=1.
The Cool Dude says:

January 30, 2013 at 6:36 am

Math isn’t about good,
it is about correct.
Noel Evans says:

January 30, 2013 at 5:53 pm

Math isn’t about good, bad, ugly, correct, logical. It’s a language and it’s about
communication. At times it is good, at times bad, at times consistent, at times
inconvenient, at times non existent, at times logical, at times convenient, at
times misleading, at times awkward, etc. It was created (and is still in the
process of being created) over thousands of years by folks who usually didn’t
have the ability to communicate with each other. So we have a hodgepodge of
terminology, definitions, symbolism, etc. that is not nearly as consistent and
logical as most people are led to believe. And that is probably why so many
people have difficulty with it.

A simple illustration: Going from 4/8 to 1/2 is commonly called “reducing”
(or “simplifying”) a fraction. What is the process of going from 1/2 to 4/8
called; that is, what is the common term for this? “Antireducing?”, “Enlarging?”,
“Fattening?”, …

And both of these are OPERATIONS we perform on fractions. What are the SYMBOLS for these operations?

Maybe we should concentrate on improving the language and not dwell so much
on some of the other aspects that we tend to “tend to infinty” on. We could
certainly profit from “agreeing to disagree” at times and trudge on as if we
knew what we were doing! Progress is made by the interaction of our minds as
we communicate with each other. Viva la math forums!

Have a very blessed day! 🙂

If you wish to explore math as a language you might check out the short book
entitled “Math: The Original Four Letter Word.” It can be freely downloaded.
Yee says:

January 30, 2013 at 8:16 pm

The Cool Dude,

http://en.wikipedia.org/wiki/Exponentiation

“The choice whether to define 00 is based on convenience, not on correctness.”
Noel Evans says:

January 31, 2013 at 8:24 pm

Amen! and hooray for convenience.

Products *{a,b,…} and sums +{a,b,…} over sets can be defined in a similar manner to
exponentiation. (Third case is for groups with identity e and operation #.)
For example:
*{a,b} = 1*a*b +{a,b} = 0+a+b #{a,b} = e#a#
*{a} = 1*a +{a} = 0+a #{a} = e#a
*{ } = 1 +{ } = 0 #{ } = e

Then we can easily define exponentiation from this. For example:
x^n = *{x,x,x,…,x} = 1*x*x*x*…*x where there are n x’s in the set.
and also
nx = +{x,x,x,…,x} = 0+x+x+x+…+x where there are n x’s in the set.
a^n = #{x,x,x,…,x} = e#a#a#a#…#a where there are n a’s in the set.

Similarly in topology using the null set as the identity for unions and the whole space
X as the identity for intersections, the union over the null family becomes the null set
and the intersection over the null family becomes the whole space X. This alleviates
the need for vacuous arguments to obtain the same results.

So in defining repeated “multiplication” or “addition” starting with the identity for
the operation is handy in a number of areas. In all of these the operation performed
on the elements of the null set produces the identity for that operation. In each case
raising to the zeroth power or multiplying by zero produces the identity for the
operation.

0^0=1 falls right in line with such definitions. Convenient!

Definitions are at times good, awkward, precise, concise, lousy, imprecise, etc. but
they are not logical statements that can be judged as true (correct) or false. They
simply provide a word, phrase or sentence to describe or name a statement.

Example: Given the set W={0, 1, 2, 3, …} to say that an element n of W is even is
to say n=2*k for some k in W.
Whether “n=2*k for some k in W” is true or not depends on what n is.

Higher levels of mathematics are built of layer upon layer of definitions which provide
the language for communicating the ideas at that level. Good definitions definitely aid
in our understanding of the ideas. Lousy definitions create quite a hassle. Proofs of
theorems very often require bringing in definitions of the concepts being combined in
the theorems.

We must take great care to make definitions that are precise, concise, suggestive of
what they deal with, easy to apply to discussions and proofs, incapable of being interpreted in more than the one way intended, etc. And looking back through the
history of mathematics we often see that definitions evolve and improve as time
passes.

Note to administrator: Is it possible to “preview” these comments? Sometimes the
lines really get “out of whack” and don’t look like the input due to spacing and unseen
characters in the lines?

Also is it possible to inject laTex into the comments?
Error: Unable to create directory uploads/2024/11. Is its parent directory writable by the server? The Physicist says:

February 1, 2013 at 8:13 am

@Noel
Yup! You can enter the LaTex math environment by typing “$latex” then your math then ending with “$”. So, $latex1 x^3$ gives you $x^3$ .
That extra “1” is there to screw up the compiling so you can see “behind the scenes”.
Noel Evans says:

February 1, 2013 at 10:17 pm

@The Physicist
Thanks for the reply!

Are you familiar with the following site? It is an excellent laTex editor which can be
used to generate the laTex code and then copy and paste it into comments. It allows
you to see the laTex and the output simultaneously.
“http://www.codecogs.com/latex/eqneditor.php”

Also I haven’t figured out how to “preview” my comments before officially posting them. Any suggestions? The MathIsFunForem has such a feature and it’s really
handy!

Have a blessed day!
The Wonderer says:

February 4, 2013 at 3:01 pm

I asked my teacher and he said that 0^0 can equal different things depending on the context of the problem. He is a high school teacher, calculus teacher, and a mathematician. So he can see the points that were described in the last few comments and has said that most are right in the fact 0^0 power equal 0, 1, or undefined.
Yee says:

February 4, 2013 at 9:24 pm

The Wonderer,
Don’t overemphasize continuity,
and 1 is the only reasonable definition of 0^0.
The Wonderer says:

February 5, 2013 at 1:54 pm

I see where you are coming from Yee. I do believe it can equal one. I am not disagreeing with that statement. I understand that it is generally accepted to be 1. I also know I can’t be a really to helpful for why it is not right now. It just doesn’t seem logical to me. On the other hand it is logical because a^0 = 1. I am going to do some more research but I do believe you are correct.
Yee says:

February 6, 2013 at 1:38 am

The Wonderer,
It is irrelevant to right and wrong.
It is irrelevant to correct and incorrect.
It is irrelevant to true and false.
0^0 is undefined just because continuity is overemphasized.
Abhishek says:

February 13, 2013 at 12:54 pm

Well, I’m a simple pimple engineer and as per my understanding :

For calculating 0^0, let’s assume a function f(x,y) = x^y where x and y can assume any value in the real domain.

now,
as y approaches to 0 we observe that the function f(x,y) approaches to 1, but why not increase y and see the impact?
as y increases, the value of f(x,y) increases exponentially (as expected). So if y were to tend to 0, no matter what value of x might have been. it would have been considered as 1. Good.

Consider that we move x towards 0 with y set at a constant non zero value c.
as the value of x decreases, we see that as soon as the value of x decreases, f(x,y) decays faster.

Now let’s fluctuate both x and y such that x = y, and let’s start from x = y = 1
so, f(x,y) = 1;
for x = y = 0.5, f(x,y) = 0.7071
for x = y = 0.2, f(x,y) = 0.7247
for x = y = 0.1, f(x,y) = 0.7943
for x = y = 0.0009, f(x,y) = 0.9937080

Now if look at the statistics, f(x,y), when x = y increases as the value of x decays given that x1 as x decays towards zero.

BUT MY MAN, WHEN x = 0, and you say if I raise nothing by nothing I get something, THAT’S JUST WRONG LIL’ BROTHER

ANYTHING RAISED TO POWER ZERO IS ZERO, THAT’S JUST HARSH TRUTH, AND WE HAVE TO LIVE WITH IT:-)
Yee says:

February 13, 2013 at 11:38 pm

Abhishek,
Limit is only a part of math, not all.
If limit does not exist, function value still can be defined.
Overemphasizing continuity is not a good choice.
Defining 0^0=1 is the best choice.
The Wonderer says:

February 14, 2013 at 10:06 am

Yee pick a side here. You told me it is undefined and told Abhishek that it is 1.
The Cool Dude says:

February 14, 2013 at 3:39 pm

In Abhishek’s reply, his final conclusion was actually irrelevant to limits, where in fact, his final argument was circled around how limits do not matter.
Yee says:

February 14, 2013 at 5:52 pm

The Wonderer,
0^0 is undefined because continuity is overemphasized.
It is not a good definition.
0^0=1 is a good definition.
The Cool Dude says:

February 15, 2013 at 6:35 am

Yee,
Calling something undefined is not defining it, it is a title that establishes that it has no definition. Furthermore, if the literal truth is not a good definition, then no definition can apply, because no other statement can be more true.
The Wonderer says:

February 15, 2013 at 4:57 pm

So if a problem such as this, where there is no one correct answer, should have the answer of undefined. We can also say it depends on the situation to keep the binomial theory correct.

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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