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.

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

1,178 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Yee says:

October 23, 2011 at 8:06 pm

micheal,
1.
How can you say 0^0 = 0*0*…*0 (0 times) and of couse equal 0.
Why does it of course equal 0?
When it comes to other numbers,
a^0=a*a*…*a(0 times)
Do you think it of course equals 0?

Consider it in another way,
a^3=1*a*a*a
a^2=1*a*a
a^1=1*a
a^0=1
There is no exception for a=0.
Therefore 0^0=1
Dina Chandran says:

October 29, 2011 at 11:31 pm

x°=1

x°= x(²-²) or any same power of x ,
=x²*x-² or
=x²/x²=1
ie,
.´. x°=1
Yee says:

October 30, 2011 at 8:14 pm

Dina Chandran,
You can not explain 0^0=1 in this way.
Gia Shwan says:

November 2, 2011 at 6:08 pm

You know..
I am not a mathematician, I am not a 10th grade student, I’m not perfect in math. I am just a 15 year old boy trying to find answers.
I just stumbled upon this question when I thought this for a second:

x^0=1
0^x=0

0^0=?

In my opinion, you can calculate all you want, you can show the longest and most confusing formulas you can think of.
But the answer to 0^0 depends on how you think of the question.

It can either be both, or it can be none of the 2.

0^0 = 0/1
or
0^0 = –

The answer cannot be only one of the two.
Yee says:

November 2, 2011 at 7:26 pm

Gia Shawn,
0^x=0 is valid only for x>0,
but invalid for x<0.
It is unreasonable to define 0^0=0.
Buf if you define 0^0=1,
x^0=1 is valid for any x.
It is reasonable.
You should say it depends on how one thinks.
Yee says:

November 2, 2011 at 9:02 pm

You shouldn’t say it depends on how one thinks.
Gia Shwan says:

November 3, 2011 at 6:42 pm

Yee,
“0^x=0 is valid only for x>0,
but invalid for x<0.
It is unreasonable to define 0^0=0.
Buf if you define 0^0=1,
x^0=1 is valid for any x."

the first 2 lines don't prove anything. the last line also doesn't prove anything.
"0^x=0 is invalid for x<0"
0 is not less then 0.
Yee says:

November 3, 2011 at 6:58 pm

Gia Shawn,
There is nothing to be proved.
It’s a matter of rationality.
Chad says:

November 4, 2011 at 2:50 pm

0 ^ 0 does not exist.

Type it into your calculator.

0.000000000000000000001 ^ 0 = 1

(lim[x–>0] X) ^ 0 = 1

but 0 ^ 0 DNE.
Yee says:

November 4, 2011 at 10:46 pm

Chad,
That the limit does not exist does not mean
the function is undefined at this point.
You shouldn’t say 0^0 does not exist.
Yee says:

November 4, 2011 at 11:02 pm

Chad,
Can I say 0! does not exist?
Can I say 0 does not exist?
Whay can you say 0^0 does not exist?
Yee says:

November 4, 2011 at 11:03 pm

Why can you say 0^0 does not exist?
peace keeper says:

November 8, 2011 at 11:46 am

why cant it just be 1? its easier that way.
Yee says:

November 8, 2011 at 6:35 pm

peace keeper,
Of course it can be 1,
and it’s reasonable.
brandon poore says:

November 17, 2011 at 4:40 pm

what is 0 – 0 and what is 0
patty says:

November 20, 2011 at 4:12 pm

x^0=x^1 * x^-1=x/x

0^0=0/0
0^0=????

Ex: 6/2= 3 (multiply 3 to the denominator to equal 1)

a=(-inf,0)U(0,inf); a/0= DNF or undefined
(no number can be multiplied to the denominator equal the numerator)

so would 0/0= all real #’s??( the denominator already equals numerator, but any number past zero would still have the denominator numerator)

Ex: 0/0 =Z
0=0(Z)
Z= all real #’s
patty says:

November 20, 2011 at 8:51 pm

this is a better way of looking at it

a/b=x => a-bx=0(this is what division does)

Ex: 6/2=x => 6-2(3)=0

6/0=x => 6-0(x)=0 (because 6 will always equal 0 in this case{false statement} x= undefined)

0/0=x => 0-0(x)=0 because 0 will always equal 0 in this case {true statement} x= all real #’s)
Yee says:

November 21, 2011 at 5:52 am

Patty,
If you think 0^0=0^(1-1)=0^1/0^1=0/0
then 0=0^1=0^(2-1)=0^2/0^1=0/0
Therefore 0^0 is irrelevant to 0/0.
0^0=1
but 0/0 is indeterminate.
patty says:

November 22, 2011 at 12:05 am

Yee,

i see your logic for disproving my theory except for “0^1=0^(2-1)”(bear with me)

its like saying that [ x=x^2/x ] is the same thing as [x^2=x^2] (are they?)

if x=0 then, correspondingly, [0=0/0] is the same thing as [0=0]

Now if we say [ x=x^1/x^0 ] is the same as [x^1=x^1] ( variable A=0^0)
then(x=0); [ 0=0/(A)] is the same as [0=0] ( A= indeterminate)

also, would u please elaborate on why you think 0^0=1? ( ? for anybody)

i know that any #^0 is 1, however 0 is unlike other #( 0 is the only defined #(maybe) that can take part in an undefined expression)

also, what does 1^infinity=???( why is it not 1?)
patty says:

November 22, 2011 at 12:35 am

another question: could we rewrite “0^2/0^1″as “0(0/0)?” [can we factor out zeros?]
Yee says:

November 22, 2011 at 5:34 am

patty,

0^(-0)=1/0^0
(0^0)^2=0^(0*2)

Let binomial theorem be valid to power 0,
(1-1)^0=C(0,0)*1^0*(-1)^0=1

Let polynomial a[0]+a[1]*x+a[2]*x^2+…
be denoted as
Σa[n]*x^n
n

These are some reasons to define 0^0 as 1.

Discontinuity is not a good reason not to define 0^0.

0 cannot be a dividor.
0^1=0^(2-1) is correct,
but 0^(2-1)=0^2/0^1 is incorrect.
patty says:

November 22, 2011 at 8:09 pm

Yee,

thanks for trying but i am still not convinced mostly because my math is not that high
for “(1-1)^0=C(0,0)*1^0*(-1)^0=1” what value does C(0,0) hold and why?

and i also don’t know what the significance of Σa[n]*x^n is?

the only way I will be able to believe that 0^0=1 is by you giving an algebraic equivalent
I thought that x^0=x/x was, but u proved it wrong
given x^y=x/x, solve for y in terms of x, i got y= log(x^(1-1))/log(x)= 0/log(x)

if x=1, y=o/o ; if x=0 , y= indeterminate because zero can’t go in logs
Yee says:

November 22, 2011 at 11:37 pm

patty,

C(0,0)=1
It’s combinational math.

b
Σ x[k]
k=a

=x[a]+x[a+1]+x[a+2]+…+x[b]

x^0=x/x
is valid when x does not equal 0.

1^y=1/1=1
y can be any number.

0^y=0/0
is not a correct equation.
It’s not worth discussing.
patty says:

November 23, 2011 at 1:28 am

Yee,
now i am starting to think that 0^0=1
if 0^0=1, then 0*log(0) =0 (sounds responsible)
//but what does log(0)=????(undefined, indeterminate, infinity??)
this website tries to prove that 0*log(0)=0:

http://arcsecond.wordpress.com/2009/03/19/0log0-0-for-real/

if their proof is correct, then we have make the assumption that 0^0=1 because log(1)=0

Thanks Yee
Yee says:

November 23, 2011 at 2:59 am

patty,
log(a^b)=b*log(a)
is invalid when a=0.
Yee says:

November 23, 2011 at 3:01 am

log(0^0)=log(1)=0
log(0^0) does not equal 0*log(0)
patty says:

November 23, 2011 at 7:48 pm

why does 0*log(0) not equal 0?? what would it equal then???
Yee says:

November 23, 2011 at 8:00 pm

log(0) is undefined.
Therefore 0*log(0) is undefined.
Yee says:

November 23, 2011 at 8:01 pm

0*log(0) can be discussed only in limit.
Its value depends.
patty says:

November 24, 2011 at 1:15 am

log(0)=undefined=0^a (a<0)
0*log(0)=
0^b*0^a=
0^(b+a)=
because b+a is indeterminate in this case,
0^(b+a)=indeterminate( could be either zero or undefined )
therefore, 0*log(0)= indeterminate
also indeterminate if log(0^0) equals 0*log(0)
Yee says:

November 24, 2011 at 4:08 am

log(0^0) does not equal 0*log(0)
patty says:

November 25, 2011 at 3:25 am

if 0^0=1, then log(0^0) MUST = 0*log(0)
if they don’t, then how could 0^0 have been created? otherwise,
it would be impossible to form 0^0 from any circumstance other than this, then it’s value should be indeterminate, especially if it can’t be derived from basic principles

“log(a^b)=b*log(a)
is invalid when a=0.”
I am unconvinced of this statement because of this example:
log(0^10)=10*log(0) [ log(0)= -infinity ; thought it was undefined, still confused but having it -inf makes more sense]
log(0)=10*log(0)
-inf=10*-inf
-inf=-inf
may not be the best example, but shows that they have same limits

what does 0*-inf=??(thinking its indeterminate)
(1-1)-inf ( zero identity)
-inf+inf (distribute inf )= indeterminate

possible proof that 0^0 does not equal 1

ln(x^x)=x
e^x=x^x
(1+1/x)^(x^2)=x^x ( e is (1+(1/x))^x)

0^0= (1+(1/0))^0
0^0= (1+undefined)^0
0^0= (undefined)^0
undefined to the zero power is not one, 0^0 is undefined
Yee says:

November 25, 2011 at 10:53 am

There is no MUST.
Yee says:

November 26, 2011 at 8:24 am

Directly define z^0=1 for any complex number.
patty says:

November 27, 2011 at 12:47 am

z^0=1
log(base z) z^0= log(base z) 1
log z^0/log z= log 1/ log z (log division rule)
log 0^0/log 0= log 1/ log 0 (plug in zero)
-log 0^0/inf = -0/inf (0^0 could equal 1, 10, or 100 or whatever cuz infinity is the denominator, so no matter what real # is on top, the equation will result with “0=0” )
-log 100/inf = -0/inf
-2/inf = -0/inf
0=0
Yee says:

November 27, 2011 at 4:20 am

When base is 0,
you can not discuss with log.
patty says:

November 27, 2011 at 2:30 pm

fine i’ll just let this be, but one last question

whats wrong with this proof:
e^ipi=-1
-e^ipi=1
(-1)*e^ipi=e^0
e^ipi*e^ipi=e^0
e^ipi+ipi=e^0
ipi+ipi=0 (this is false)
Pingback: Why do mathematicians and high school teachers disagree? I love stuff like this. | bigjim.org
Yee says:

November 27, 2011 at 7:20 pm

In complex variables,
e^(i*x)=1
x=2*k*π
k can be any integer.
patty says:

November 29, 2011 at 12:40 am

x=2*k*π
I understand this, “i” is not 0, but it acts like it. So say integer k was a really high negative # (like negative infinity), would we then assume that 0^i=1?
Yee says:

November 29, 2011 at 5:45 am

e^(0*x)=1 is correct for any number.
e^(i*x)=1 is correct only for 2*k*π.
They act very differently.
Yee says:

November 29, 2011 at 6:46 pm

If 0^0 is undefined,
How to express binomial theorem correctly?
patty says:

November 29, 2011 at 8:40 pm

I might just be dumb saying this but lets say have we have (a+b)^-2=???
the answer is obvious, but my point is that does the binomial theorem, (a+b)^n = a^(n-k)b^(k), work with negative n?and how would Pascal’s triangle look with negatives’s on top ? would it look like an hour glass?

I honestly don’t have any beliefs on what 0^0 is still, i heard good arguments for it being 1, undefined(not so much), and indeterminate. I just want a solid answer with dominate reasoning
patty says:

November 29, 2011 at 10:24 pm

here is the best argument for indeterminate I think:
0^2*0^0=0^(2+0)
0 * 0^0= 0
what does 0^0 have to be? any number
Yee says:

November 29, 2011 at 11:31 pm

(a+b)^n
a,b are complex numbers,
n is a nonnegative integer.
Do not exceed this range now.

0^0 is unsolved.
There is no solid answer.

0*1=0
Can 1 be any number?
patty says:

November 30, 2011 at 1:33 am

“Do not exceed this range now.” why?does the theorem not work for negative numbers?

0^0=1
(0^0)^i=1^i
e^(-inf*pi*i)=1^i
e^(-inf*pi*i)=e^(2*i*pi)^i (1=e^(2*i*pi))
e^(-inf*pi*i)=e^(-2pi)
(-inf*pi*i)=(-2pi)
2=i*infinity ?????????

-I’m enjoying this conflict we are having if you want to call it that 🙂
Yee says:

November 30, 2011 at 1:36 am

In complex variables,
e^x=e^y
does not imply x=y.
patty says:

November 30, 2011 at 1:38 am

ops correction:

0^0=1
(0^0)^i=1^i
e^(-inf*pi*i*0)=1^i
e^(-inf*pi*i*0)=e^(2*i*pi)^i (1=e^(2*i*pi))
e^(-inf*pi*i*0)=e^(-2pi)
(-inf*pi*i*0)=(-2pi)
1=0*i*infinity ?????????
AnushaManne says:

November 30, 2011 at 8:37 am

0^0 is infinity!!!
AnushaManne says:

November 30, 2011 at 8:39 am

0^0= 0^(1-1)=0/0=infinity!!!

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

1,178 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Leave a Reply Cancel reply

Send your questions about math, physics, or anything else you can think of to:

Subscribe via Email

Search the Archives

Categories

Archives

Recent