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:

April 25, 2012 at 2:51 am

lim(x→0) f(x)^g(x) = (lim(x→0) f(x))^(lim(x→0) g(x))
This is incorrect at a discontinuous point.
Ron says:

April 25, 2012 at 3:11 am

Lee,

Once the limits were evaluated on the right side of the equation, it was no longer a limit equation. It was a straight forward 0^0= something. In this case, it equaled the limit as x->0 of f(x)^g(x), which is a proven value by the formal definition of limits. It is irrelevant that the limit doesn’t equal the function because we aren’t evaluating the limit of 0^0, nor are we even interested in the value of the function. We are evaluating the limit of a function where the proven value of the limit itself specifically and definitively equals 0^0.
Ron says:

April 25, 2012 at 12:28 pm

Actually, I meant to write it like this:

lim(x→0+) f(x)^g(x) = (lim(x→0+) f(x))^(lim(x→0+) g(x))

That corrects any discontinuity issues, and still demonstrates an obvious example where 0^0 does not equal 1
The Cool Dude says:

April 25, 2012 at 2:16 pm

Don,
You can’t disprove one value of 0^0 by proving another, because 0^0, is in fact, all values.

The question is what value is the most practical, and this is, in fact, of definition, and general consensus is that 0^0 should equal one.
Yee says:

April 25, 2012 at 7:46 pm

Ron,
lim(x→0+) f(x)^g(x) = (lim(x→0+) f(x))^(lim(x→0+) g(x))
is invalid when
lim(x→0+) f(x) = lim(x→0+) g(x) = 0
Yee says:

April 25, 2012 at 9:09 pm

If define 0^0 as a specific number,
lim(x→0+) f(x)^g(x) does not necessarily equal the number.
lim(x→0+) f(x)^g(x) = (lim(x→0+) f(x))^(lim(x→0+) g(x))
will not hold.

If define 0^0 as indeterminate,
Something indeterminate does not equal anything.
lim(x→0+) f(x)^g(x) = (lim(x→0+) f(x))^(lim(x→0+) g(x))
will not hold,either.

So lim(x→0+) f(x)^g(x) = (lim(x→0+) f(x))^(lim(x→0+) g(x))
will not hold in any case.
Ron says:

April 25, 2012 at 9:48 pm

“…because 0^0, is in fact, all values.”

Which means that it is indeterminate. Having all values means that it can never be defined as only one value. 1∈ℝ does not mean {1}=ℝ or ℝ={1} (the set of 1 equals the set of real numbers, or the set of real numbers equals the set of 1 ) by any stretch of the imagination.

To go back to a previous argument, which clearly is invalid:

x=x; which means {ℝ}={ℝ}
0=0; which means {0}={0}
does not lead to:
x=0; which means {ℝ}={0} Which is absurd and clearly false.

“general consensus is” Is an appeal to authority logical fallacy. Besides, I have yet to see anyone list off ten mathematicians, of significant standing, that agree that it must be defined as 1 across the board. Theorems and the consequence of definitions are derived from axioms, nothing else. Definitions can be created to define new ideas, but the consequences of those definitions can NEVER contradict the axioms. In this particular case, the definition in question is not that of 0^0, but the definition of exponents.

If the argument is that there is a consensus, it is entirely reasonable to expect that argument be supported by facts and references that actually demonstrate a real consensus. Not that a consensus matters…There was a time when the consensus was that the Earth was flat.
Ron says:

April 25, 2012 at 10:39 pm

Yee,

In math as it is in science, It is not enough to just claim something is invalid. It must be demonstrated with proof, contradiction, or evidence. Without that basic standard, message boards would devolve to:

1+1=2!

response: No it doesn’t.

a=a!

response: No it doesn’t

Which can obviously get old real fast for everybody.

When f(x)=e^(-1/x), lim(x→0+) f(x)=0
When g(x)=x, lim(x→0+) g(x) = 0

lim(x→0+) f(x)^g(x) = (lim(x→0+) f(x))^(lim(x→0+) g(x)) by the power law of limits
which gives us:
lim(x→0+) f(x)^g(x) =(0)^(0) = 0^0

Of course, f(x)^g(x) expands out to:
f(x)^g(x) = (e^(-1/x))^x
lim(x→0+) (e^(-1/x))^x = lim(x→0+) e^(-x/x) = e^(-1)

therefore,
lim(x→0+) f(x)^g(x)=1/e Which also equals 0^0
Yee says:

April 25, 2012 at 10:41 pm

Ron,
The value of 0^0 depends on definition.
It does not “in fact” equal any value.
0^0 = 1 does not lead to any contradiction.
Ron says:

April 25, 2012 at 11:38 pm

Yee, you are incorrect to say “If define 0^0 as indeterminate, Something indeterminate does not equal anything.”

Something that does not equal anything is called the empty set, or undefined. Indeterminate does not mean empty set or undefined. Indeterminate means that it is case specific in its value. In one case, it might be 1, in another case it might be 2, in another case it could equal 1/2, and so on.

You originally asked for an example of it not equaling 1…Hence, I showed you one case where it equaled 1/e.

If you replace the e with n in the exact set of equations, we would get 0^0={1/n, n∈ℝ and n>1 } Which means it can equal anything between 0 and 1. This example can never equal one or zero, but other examples of f(x) where the limit->0+ also can be substituted to show that 0^0 can also equal both zero and one. ie f(x)=3x, or f(x)=0
Yee says:

April 26, 2012 at 1:57 am

Ron,
If a=b,b=c
is a=c correct?
The Cool Dude says:

April 26, 2012 at 4:11 pm

Ron,
X=X is all solutions. The only value that is not X would nullify the basic ability of intelligent beings in all real and hypothetical planes of existence to give a solution.

The only things that are strictly true, that I thoroughly demonstrated, are these:
All values of X support X^0=1
All values of N support 0^0=N

Any other argument that you can propose, like for instance, what number BEST fits 0^0=N, is of pure opinion, and I have no opinion at all. I am neutral.
Yee says:

April 26, 2012 at 7:29 pm

The solution of X^0=1 and 0^0=N depends on the definition of 0^0.
Ron says:

April 27, 2012 at 1:01 pm

Actually Yee, the definitions of x^0 and 0^0=N depend on the definition of exponents and how that definition logically relates to the axioms of mathematics.

0^0 is not specifically defined by definition in mathematics.
Yee says:

April 27, 2012 at 8:26 pm

Ron,
Defining 0^0=1 is good and logical.
The reason most mathematicians define it as inderterminate is continuity.
They don’t define a function value at a point where limit does not exist.
Continuity is overemphasized and the function of 0^0=1 in other fields is despised.
This is not a good choice
Yee says:

April 29, 2012 at 8:03 am

It’s ridiculous to overemphasize continuity and despise other fields.
Do other mathematical fields disappear and only analysis remain?
Defining 0^0=1 is the only reasonable choice.
Ron says:

April 29, 2012 at 8:00 pm

“The reason most mathematicians define it as inderterminate is…”

Which naturally means that the consensus is that it is indeterminate…

“Do other mathematical fields disappear and only analysis remain?”

Logic is the basis of all rationalism systems to include mathematics. As for what disappears?? Nothing, unless we start irrationally defining specific cases to always equal something when it is clear that they don’t.

Does defining 0^0=1 save the binomial theorem?? Not at all! The reason is simple: It would destroy the definition of exponents which would, in turn, destroy the binomial theorem. Destroying the definition of exponents would be huge in mathematics. It could never be viewed as “reasonable.”

The binomial theorem is easy to save. We just need to be more precise in how we define a binomial, i.e. (x+0) shouldn’t be viewed as a binomial at all. it should be viewed as the monomial x. Interestingly enough, this solution doesn’t break anything in mathematics. This is the only “reasonable” choice.
Yee says:

April 29, 2012 at 11:11 pm

Ron,
This consensus overemphasizes analysis and despise other fields.

For (x+y)^n,
If define 0^0=1,
x,y can be any complex number and n can be nonnegative integer.

If not,
x,y can only be nonzero complex number and n can only be positive integer.

For polynomial a[0]+a[1]*x^1+…+a[n]*x^n
If define 0^0=1,
It can be written as
n
Σ a[k]*x^k
k=0

If not,
It has to be written as:
a[0] if n = 0

a[0] +

n
Σ a[k]*x^k
k=1

if n >= 1

The rationality and convenience is obvious.
Ron says:

April 30, 2012 at 1:51 pm

“Other fields” external to mathematics are irrelevant to mathematics. They are more than welcome to pretend it equals one for their own sake or poorly written functions and algorithms.

As for the binomial theorem, The only time there is a problem is when the domain of one of the monomials ={0}, which isn’t a problem at all because the equation simplifies to something much nicer than a polynomial…it simplifies to either x^n or y^n.
Ron says:

April 30, 2012 at 2:05 pm

“The rationality and convenience is obvious.” is false because it is far more convenient and rational to realize that (x+0)^n= simplifies to (x)^n where the binomial theorem is irrelevant.
Yee says:

April 30, 2012 at 7:34 pm

Ron,
0^0 = 1 is good and convenient in all mathematical fields,
except the limit of 0^0 form is different by case.
Therefore the best choice is defining 0^0 = 1.
These filed are no less than part of mathematics.

(x+y)^n
=
n
Σ C(n,k)*x^(n-k)*y^k
k=0

To write bionomial theorem in this form and make it valid to the largest domain.
0^0=1 is a must.

Besides, the combinatorial interpretation of 0^0 is the number of empty tuples of elements from the empty set.
There is exactly one empty tuple.
The Cool Dude says:

April 30, 2012 at 8:28 pm

I think that it is far more convenient and rational to realize that I have a proof where the binomial theorem is irrelevant.
Ron says:

April 30, 2012 at 11:05 pm

There is only one single, solitary example where this might be true: “To write bionomial theorem in this form and make it valid to the largest domain. 0^0=1 is a must.”

And that is (x+0)^n. This is the ONLY case where 0^0=1 can be said to matter, and yet….

(x+0)=x…. Do we REALLY need the binomial theorem for just ‘x’?? Of course not.

As for how we write the binomial theorem…It’s not like paper is expensive. Or ink for that matter. We can easily follow the definition with: where the domain of x,y do not equal {0}

Yee, you keep insisting that “0^0 = 1 is good and convenient in all mathematical fields” but could you elaborate?
Yee says:

May 1, 2012 at 2:29 am

Ron,
(x+y)^n
=
n
Σ C(n,k)*x^(n-k)*y^k
k=0
This is the best form for binomial theorem.
Should we expand (x+0)^n with binomial theorem?
It is the user’s business.
If a user insist to use binomial theorem,
It had better be okay.
It is good to make it valid to the largest domain,
it is bad to have some unnecessary constraints.
What do you think not elaborate?
Since it is useful in these examples,
should analysis be overemphasized and despise these example?
Ron says:

May 3, 2012 at 2:57 am

“It is the user’s business.”

Not if we define a binomial as (x+y), where the domain of x,y ≠{0}

The beauty of that definition is that it isn’t wrong, it doesn’t contradict the rest of math in anyway, nor does it hamstring “the user”(lol) in any critical way. (x+0) is easily x, and (0+y) is easily y.

Unlike trying to define 0^0=1, which is clearly wrong because the domain of 0^0 ≠{1}(definition of equality)

Who is this lazy user anyway?? Perhaps he/she is a computer software engineer who is too lazy to throw in a couple of ‘if’ statements into their code? Why should we break math for them? Besides, some IEEE standards already allow them to pretend it equals one without any additional effort. Java standards expect it to equal one as well for, in my humble opinion, coding laziness. Running through a binomial theorem loop to solve (x+0)^12000000 is absurd, and more importantly, inefficient. Try doing a billion of those calculations, and my code will take seconds, while theirs will take years.

To that end, software and electrical engineers should insist it is indeterminate as well, just to force better coding!

What is reasonable?? Something that is NOT wrong and solves the problem without creating others, or something that IS wrong and solves the problem in some cases but creates problems in others??

Yee, you say “it is bad to have some unnecessary constraints,” but are they actually unnecessary? After all, allowing (x+0) to be called a binomial gives us situations of 0^0 where we already know that it is indeterminate and has some people wanting to define it as 1 for the sake of this simple mathematical laziness.

Mathematicians do not define 0^0 as 1 for a reason: it is indeterminate.
Yee says:

May 4, 2012 at 1:16 am

Ron,
You insist 0^0 is indeterminate in advance.
Therefore, if it is useful in some examples, you deny it.
It is irrelevant to right and wrong.
It is irrelevant to true and false.
Convenience if a kind of mathematical beauty.
Pursuing convenience and mathematical beauty is reasonable.
In mathematical history,
0^0 = 1 had been generally accepted until continuity was overemphasized.
Mathematicians do not define 0^0 only because of continuity.
Continuity is overemphasized and others are despised.
It results in twisted mathematics.
Yee says:

May 4, 2012 at 1:19 am

Ron,
Do you think 0 exists?
Do you think imaginary numbers exist?
What is your reason to say 0^0 is indeterminate?
Yee says:

May 4, 2012 at 1:36 am

Ron,
Of course 0^0 = 1 does not contradict anything.
Discontinity is never contradiction.
If you think 0^0 = 1 contradicts something,
please give an example.
Yee says:

May 4, 2012 at 7:03 am

Convenience is a kind of mathematical beauty.
The Cool Dude says:

May 4, 2012 at 3:06 pm

The PURPOSE of math is to be convenient. If math was not convenient, it would not exist. If math becomes inconvenient, then whatever you are doing is probably wrong.
Yee says:

May 4, 2012 at 8:41 pm

Ron,
If 0^0 is indeterminate,
what is (x+y)^n (n is nonnegative integer) if you are not sure x,y≠0?

(x+y)^n =
indeterminate if x+y=0,n=0
0 if x+y=0,n>0
x^n if y=0,x≠0
y^n if x=0,y≠0

n
Σ C(n,k)*x^(n-k)*y^k
k=0

else.

See how troubling it is.
Yee says:

May 5, 2012 at 10:17 am

Write binomial theorem in a somewhat simpler form:

(x+y)^n
x,y are complex numbers,
n is positive number.
(x+y)^n=
x+y if n=1

x^n+y^n+

n-1
Σ C(n,k)*x^(n-k)*y^k
k=1
if n > 1

Does anyone write binomial theorem in this form?
Does anyone forbids x,y to be 0?
If not, 0^0 = 1 is a must.
Ron says:

May 5, 2012 at 1:00 pm

Yee, you asked, “If 0^0 is indeterminate, what is (x+y)^n (n is nonnegative integer) if you are not sure x,y≠0?”

The answer is, the binomial equation, where the results of that equation is another equation involving the variables x and y. i.e. x^2 + 2xy+ y^2. Since we don’t know what their values are, they must remain variables.

x^0 and y^0 fall out of the equation that is produced by the binomial theorem such that 0^0 is never numerically evaluated if it is later determined that x or y equals zero. In other words, not knowing the value of x or y is never a problem.

If we know x or y before hand, and one of them is zero, then we don’t need the binomial theorem. The result is either x^n or y^n

“Convenience is a kind of mathematical beauty.”

You mean like the 10 to 15 years we spent learning it? And, after all that effort learning it, how convenient would math be if it gave us wrong answers?

“See how troubling it is.”

You didn’t illustrate anything that was troubling. There is nothing wrong with a list of consequences, even if that list takes 3 lines on a sheet of paper or even if it takes three pages to describe. There are plenty of definitions in math that require more than one line to describe.
Yee says:

May 5, 2012 at 9:22 pm

Ron,
“how convenient would math be if it gave us wrong answers”
What do you mean by saying it wrong?
What’s wrong with 0^0=1?
Ron says:

May 5, 2012 at 11:07 pm

“What’s wrong with 0^0=1?”

It doesn’t only equal one. The definition of equality tells us if a=b then b=a. 1≠0^0 because 0^0 has a range of values.
Yee says:

May 6, 2012 at 12:27 am

Ron,
If define 0^0 = 1.
we don’t need to be worried about what values the variables will equal later.
It is convenient.
Yee says:

May 6, 2012 at 3:54 am

Ron,
Can I say 0! does not equal 1?
Can I say 0 does not equal a value?
Can I say imaginary numbers don’t exist?
Yee says:

May 6, 2012 at 4:51 am

Ron,
Why does 0^0 has a range of values?
Yee says:

May 6, 2012 at 5:09 am

If you insist limit of 0^0 form has different values,
so 0^0 must have different values,
you are overemphsizing limit.
Ron says:

May 6, 2012 at 10:05 am

“you are overemphsizing limit”

How so? limits are the basis of the entirety of calculus. This is not a trivial thing.
Ron says:

May 6, 2012 at 10:17 am

“If you insist limit of 0^0 form has different values,
so 0^0 must have different values,”

I’m not talking about the limit of 0^0. I never have. That wouldn’t prove anything.

What I am talking about are limits of functions that = 0^0 and also equal something other than 1. That is a very different thing. a=b, b=c, therefore a=c.
Yee says:

May 6, 2012 at 11:23 am

Ron,
Limit is not the only viewpoint of mathematics.
It is unreasonable to discuss 0^0 only in limit.
Except limit, you have no reason to say 0^0 equals other values.
Take many fields of mathematics into consideration,
defining 0^0=1 is reasonable.
The Cool Dude says:

May 6, 2012 at 11:30 am

If an object has multiple values, then it is up to the user to decide what value fits best. It is simply a matter of practicality to choose one value. If we are all agreeing to accept the possibility that 0^0 is all values, then it becomes a matter of opinion of how to write it, and technically can not be said to be wrong.

The question “What is best” has no constraints, and therefor, is simply arbitrary. We could instead answer the question of how to best determine the value, but then we are asking another question of opinion. Turtles all the way down.

This being said, we should stop measuring imaginary things, like opinions, and instead measure real things, like 0^0.

I say 0^0=1.
You say 0^0=8.
My friend Dave says 0^0=1/0
We are all correct.
It is a fact that we are all correct. Asking who of us is the most correct is ridiculous. We are all equally correct.

P.S. Just a quick aside, 0^0=U because 0U=0, because 0U=0/0, and 0/0 has a possible solution of 0. (0*0=0)
Ned says:

May 6, 2012 at 7:36 pm

I defy anyone here to cite a specific text, at the level of calculus or above, that claims 0^0 = 1. Moreover, every calc book lists 0^0 together with 1^infinity, 0*infinity, 0/0 and infinity/infinity as indeterminate forms.

Please note that I am asking f0r a specific textbook, not a vague claim that you once saw one.
Yee says:

May 6, 2012 at 8:03 pm

The Cool Dude,
What is the reason that 0^0 can equal many values except limit?
If we distinguish 0^0 from
lim x^y
x->0,y->0
0^0 only equal 1 in all cases.
Therefore defining 0^0=1 is reasonable.
Yee says:

May 6, 2012 at 8:10 pm

Ned,
Calculus mainly discusses limit.
Limit of 0^0 form is indeterminate.
Because the majority of mathematicians overemphasize limit,
it may be difficult to find a book that claims 0^0=1.
However it is informally used in some cases.
Binomial theorem is an examples.
Ron says:

May 7, 2012 at 3:56 am

“it may be difficult to find a book that claims 0^0=1”

In fact, probably impossible since mathematicians do not define it as 1.
Ron says:

May 7, 2012 at 4:12 am

“I say 0^0=1.
You say 0^0=8.
My friend Dave says 0^0=1/0
We are all correct.”

Actually, none of those are correct with the exception of 1 on occasion. It is not the person that defines what it is, it is the mathematical circumstance that defines it at that moment. The correct overall value is indeterminate. It is not “defined” as anything since the definition of exponents already covers the situation.

Also, no one has been able to show that it equals all values (certainly not 1/0 or 8 without doing invalid division by zero proofs or invalid multiplying both sides of an equation by zero proofs.). At best, it’s range has been demonstrated as [0,1] with variations of lim (x->0+) f(x)^g(x)=0^0
Yee says:

May 7, 2012 at 9:56 am

Ron,
If one wants to say 0^0 equal a value,
he must have some reasons.
What is the reason?
Limit is not a good reason.
What’s your definition of exponent?
Yee says:

May 7, 2012 at 9:59 am

Ron,
Books may not claim 0^0=1.
However, 0^0=1 is used hiddenly in some cases.
Binomial theorem is an example.
This is the result of overemphasis of limit.

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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