Clever student:
I know!
= 
= 
= 
= 
.
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:
= 
= 
= 
= 
which is true since anything times 0 is 0. That means that
= .
Cleverest student :
That doesn’t work either, because if 
then
is 
so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function 
and see what happens as x>0 gets small. We have:
= 
= 
= 
= 
= 
= 
= 
= 
=
So, since = 1, that means that = 1.
High School Teacher:
Showing that approaches 1 as the positive value x gets arbitrarily close to zero does not prove that 
. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that is undefined. does not have a value.
Calculus Teacher:
For all 
, we have
.
Hence,
That is, as x gets arbitrarily close to (but remains positive), stays at .
On the other hand, for real numbers y such that 
, we have that
.
Hence,
That is, as y gets arbitrarily close to , 
stays at .
Therefore, we see that the function 
has a discontinuity at the point 
. In particular, when we approach (0,0) along the line with x=0 we get
but when we approach (0,0) along the line segment with y=0 and x>0 we get
.
Therefore, the value of 
is going to depend on the direction that we take the limit. This means that there is no way to define that will make the function 
continuous at the point .
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 
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:
:= 
where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get
= 
.
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
=
which holds for any y. Hence, when y is zero, we have
.
Look, we’ve just proved that ! But this is only for one possible definition of . What if we used another definition? For example, suppose that we decide to define as
:= 
.
In words, that means that the value of is whatever 
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 in our definition of , which seems to be recursive. The reason it is okay is because we are working here only with 
, and everyone agrees about what 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
= 
= 
=
Hence, we would find that 
rather than . Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what 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 really? Well, for x>0 and y>0 we know what we mean by . But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what 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 
? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use 
or if we say that is undefined. For example, consider the binomial theorem, which says that:
= 
where 
means the binomial coefficients.
Now, setting a=0 on both sides and assuming 
we get
= 
= 
= 
= 
= 
where, I’ve used that 
for k>0, and that 
. Now, it so happens that the right hand side has the magical factor . Hence, if we do not use then the binomial theorem (as written) does not hold when a=0 because then does not equal .
If mathematicians were to use , or to say that 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 .
There are some further reasons why using 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.
In middle you say we can define y^x as
\lim_{z \to x^{+}} y^{z}.
This is basically saying “let’s define y^x so that it’s always continuous.” Is it not?
As a function of two variables (x and y), y^x is still not continuous even with this definition. The problem is that along the line x=0 we have that \lim_{z \to 0^{+}} y^{z} = 1 for y>0, and \lim_{z \to 0^{+}} y^{z} = 0 for y = 0.
Could there be a practical calculation whose result is 0^0 ? In that case, you couldn’t claim the answer is just a matter of convenience…
I think there is some problem with the second definition in the mathematician’s answer.
It says:
“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.”
I might be misunderstanding it, but isn’t that a bit like begging the question ?
I mean, it tries to define the function of y^x in terms of the function itself, as in the right hand side, it uses the term “y^z”, which is an instance of the function that is being defined!. And even if we consider it to be a “Recursive Definition”, the definition provides no base case.
This might be more philosophical than mathematical (It’s like Define “Define”), but I wish you can clarify it.
Sorry of the late comment.
I’m going on a sidetrack here.
We all know anything divided by zero is impossible. We usually call this particular answer “undefined”. So, 2/0 = undefined.
However, there’s an even MORE screwed up problem. What is 0/0? Well, we know n/n= 1, 0/n = 0, and furthermore, n/0=und.. There’s many different answers to this problem, including an und. one. We call these kinds of problems “indeterminate”. Think of it as an an extra step up in the mindf**k scale.
(For you calculus geeks out there, we can define these as: an undefined limit can still approach a “nice” value, [i.e. a hole at the limit or an asymptote], whereas an indeterminate limit does not approach a “nice” value, [i.e. you can’t do anything with it].)
tl;dr 0/0 is indeterminate. There is no answer, although in some cases, we just give it an arbitrary 0 or 1 or und. or something
Actually no one talks about the imaginary number (i). Then division by zero would be possible.
Hi, Mahmood, thanks for your comment. I should have explained this better (in fact, maybe I’ll add a clarification to the post). The reason that definition works is because we all know what we mean by y^z for y>0 and z>0. The ambiguity crops up when z=0 and y=0. Hence, it’s okay to use y^z to define y^x because the y^z in the definition only ever uses values of z>0, whereas the y^x that it is defining allows for x=0. The idea is that we are extending y^z to handle the case of z=0 by using limits.
Complex numbers give you to answers some questions about the roots of polynomials that would otherwise be unattainable. But it doesn’t help out much here.
Google says 0^0 = 1
DONE.
You don’t question google.
Doesn’t defining 0^0 as anything other than 1 create problems elsewhere? If you look at it from the perspective of exponentiation as antilogarithm, x^y=z is the same as log_x(z)=y, which is the same as ln(z)/ln(x) = y. We say y=0 and x=0, so ln(z)/ln(0)=0. We know that ln(0) = 1 because we can calculate it as a integral using limits. Therefore ln(z)/1 = 0, for which to hold ln(z) equal zero. If we solve ln (z) = 0 for z we get z=1.
Doesn’t saying 0^0 is undefined lead to logical inconsistencies or problems because of this? If 0^0 ≠ 1 then something in the above proof has to also change, otherwise you arrive at a contradiction.
Hello Old Fashioned Whisky, thanks for your comment. If your analysis does lead to some insight, it is not the insight that you imagine. The problem is that if ln(0) is to be assigned a value at all, then it is -infinity, not 1. Hence, when you write ln(z)/ln(0)=0 you actually have ln(z)/(-infinity) = 0. It is generally considered to be the case that a / (-infinity) = 0 for any real number a. Hence, the formula ln(z)/ln(0)=0 tells nothing about the value of z.
You guys view this all wrong, book smarts can only get you so smart…
0 to the 0 power=0
0/0 =0
However you look at it, it equals 0, BECAUSE!….
Numbers do not actualy mean anything specific or equal anything physical, numbers are a figment of our imagination that everyone agrees to and uses to understand “how much” of “something”(something physically real) there is or was or may be…THEREFORE!….
0 is “nothing”
0 represents “literally nothing” (Aka the absence of something)
So when you look at it for what it really is (not what a book or a mathematician sais it is) you see that you are trying to play with nothing…
0/0= nothing, not 1
(it doesn’t equal a number because you have to use logic or reality also I guess you could say).
When you use just numbers and forget what they are representing you sort of stop using reality or logic, because numbers don’t exist, you always have to keep in mind what youre really “working with”…. And when it comes to the number 0… 0 represents “nothing”!!! It always represents nothing
So now, because this is a math problem and in math we use numbers to represent values so everything has to be represented by a number…. When you look at 0/0 or 0 to the 0 power, you have to use 0 as the answer because nothing x nothing = nothing…
And “0” represets “nothing”!!!!!!!
Whoow!!! I’ve been preaching the good news for years, please pass on the truth brothers and sisters, let the world know that our schooling systems that focus on just regurgitating repetitive general things written on paper and never tapping into real reasoning and understanding is messing up how we (our brains) process and view information (life)
Wake up people! There’s more to life than what you learned in a school study book!!
Numbers are used because of real things that exist, so we must keep this in mind when we use numbers, don’t just get in the number auto. Pilot mode, numbers have no intelligence to correct these kind of mistakes, this is were our brains are supposed to do that work
0^0=Indeterminate
Do not question Wolfram|Alpha
0 x anything = 0. 0x0=0. 0x0x0=0.
Why is it that the higher in ranking person you ask the question to longer the explanation gets?
What was so wrong with the clever students answer…
Wow I feel so retarded after reading this…
i thought it was 3
And while all you folk are debating this, the chances of you getting a girlfriend or even a shag, is 0
Dear god, tell me Chris was joking…
I think there are two issues being confused here, ‘sign’ and ‘value’, they are different.
All numbers have a sign and a value, 0 has neither, it is neither positive nor negative.
Therefore using limits to assess its value is wrong because, for example, a small positive value will ALWAYS be positive as it approaches zero. There may be a continuous change in ‘value’ as it gets smaller, but it will have discontinuous change in ‘sign’ if it reaches zero.
Its value is 0. Zero cannot be made larger or smaller, it is zero.
Hello James. The problem with the “clever student” answer is that it is… well… wrong. You cannot take x/x and plug in 0, as 0/0 is really undefined.
enigma, don’t worry, Chris had nothing to say.
2+2=5
I can’t believe all these people arguing over nothing, I need a Pan Galatic Gargleblaster and a patio.
Seriously, life is far too short for this stuff. Please go out and just have a really good laugh. I’m begging you guys!
wouldn’t you have to be able to measure 0^0 somehow to prove it?? I don’t believe convenient answers will help very much in the way of scientific advances, unless they lead to something productive. I’m no math expert, and I certainly don’t want to end up sounding like Chris, but I don’t see how just saying the answer is 1 or 0 because its mathematically convenient. I mean, actually measuring an answer to this problem could lead to some impressive scientific discoveries (how the universe began maybe, for example.) right?? Am I missing something???
I love the “get a life” and “you guys will never get laid” comments. These people do not understand that this thread is intellectual mental jousting, just plain goddam fun. Some people play call of duty others fool around with abstractions like numbers and enjoy seeing what kind of cloth one can weave with them. Cheers!
Yo, Clever Student. I’m really happy for you. I’m gonna let you finish, but Einstein had the best equations of all times. One of the best equations of all time!
None of this makes any sense to me. Doesn’t
0=nothing? So how can we raise nothing to the nothing power and get anything other than nothing? I obviously skipped a lot of math and cheated my way through the rest because this is not even remotely “clicking” in my brain.
I will reluctantly accept that 0^0=1, however this raises an important question:
If dividing by zero causes space-time to collapse in on itself and create a black hole, does raising zero to the zeroeth power create a universe? Food for thought…
I started off intrigued, but somehow along the way I lost interest.
@Awesome Astarte, I agree!
We already know that there is nothing provable, according to whatever the lastest asro-physical discovery or whatever that I read somewhere. Can we just accept life has unaccepted curves and bends, so let’s enjoy the ride?
Even if you exist in another universe, you’re *here*, not *there*, so why worry about the *there*?
ha i dont care the world is domed anyways
I love these little kiddies that act as if math isn’t important.
I’m currently an Engineering major, and let me tell you. Math makes everything in the world work.
Just because you dicked around in High School, causing you to be fairly dumb don’t say this doesn’t matter.
0^0=Indeterminate
if you take 3 apples and divide them among nobody, you have divided by zero
Allow the old statistitian to add 0! = 1 to the mix.
It seems ridiculously simple to me, and I really don’t see how nobody else thought of it before. Look at it this way:
4^3 is essentially 4x4x4. So 0^0 is essentially 0x….. My solution just fell apart. If I ever run into this on a math test, I’m just going to say it is “undefined”.
0 x 0 = 0, 0 x (any int) = 0, 0^0 = 0.
The statement that because this has no meaning whatsoever scientifically is silly. Look at the history of science. You will see that math that seems “pure” and not able to be applied is often used years down the road. Likewise, we see many examples of physics that stalled because we ran out of math, then someone invented/discovered it and away those projects went (sometimes even the physicist himself/herself was the person to invent it!)
At some point, certain things are defined the way they are defined because it is useful to do it that way, and likewise not harmful. Consider 0!=1… If it weren’t that way, many a taylor series would totally fall apart. In many instances these ambiguous definitions have proofs on all sides of the answer.
Also I’m curious about the problem that occurs when defining the limit as x approaches zero of y^x that occurs when we limit ourselves to approaching zero from the positive, as y^x is ALWAYS positive if y is positive. Graphing f(x)=y^x for any y will never go below the x axis.
Finally, does L’Hopital have anything to say about this?
Does it help to think about it this way?
Multiplication is a binary operation – that is, you must have two operands to complete the operation. In order to evaluate y^1, you must have something to multiply y by in order to arrive at the answer. We are all in agreement that y^1 = y and the multiplicative identity is 1. Therefore, y^1 represents 1 * y (1 multiplied by y one time).
Furthermore, y^x represents 1 * y * y * y * …. however many times are specified by the exponent (x). Thus, y^0 represents 1 multiplied by y zero times.
This gives us:
2^1 = 1 * 2 = 2
3^2 = 1 * 3 * 3 = 9
4^0 = 1 (multiplied by 4 zero times) = 1
0^3 = 1 * 0 * 0 * 0 = 0
0^0 = 1 (multiplied by 0 zero times) = 1
Yes, this construction is used in the article. Unfortunately, it is not the only way to define y^x, and in particular, using other definitions (which all are equivalent when y>0 and x>0) can lead to different conclusions about 0^0, as the article points out.
Great article!
@Matthew
I felt dumb when I asked myself “why does he only considers the positive limit as x approaches 0 of f(x,0)=y^x?”
0^x does not exist when x0, 0^-a :=1/(0^a) =1/0.
Graphing f(x)=y^x may never go below the x axis, but f(x)=0^x is what we’re concerned with. He does consider both left and right limits for f(y)=y^0.
This English prof says: The 0 represents societies inability to properly empathize with the most disadvantaged peoples. Done!
Hi!
Just wanted to point out that 0 does not necessarily equal 0:
Take for example these two functions:
a) x/sin x
b) x^2/sin x
What are their values at x=0?
Well – it turns out that the former evaluates to 1 whereas the latter equals 0 – although in both cases we have an expression of the kind 0/0.
(If you don’t believe me – look up L’Hôpital’s Rule)
My point is, that one has to know where the expression 0^0 came from to be able to evaluate it.
Cheers
There is no Zero.
I stumbled upon this slightly drunk. I don’t think I woulder understand it sober anyway.
actually Chris is not entirely wrong, numbers were ‘invented’, not discovered, to signify real objects and communicate with other people…arguing about nothing raised to the power nothing can go on forever, whatever majority people (or the clever ones) decide (through mathematical proof’ or otherwise) at this particular time, is what it is… all the equations will be treated accordingly…
0^0 is undefined.
EXPLANATION 1
0^3=0
0^2=0
0^1=0
0^0=undefined because you cannot multiply nothing by nothing.
simple as that.
EXPLANATION 2
0^(1-1)
=(0^1)*(0^-1)
=0* (0/0)
=undefined
Therefore, 0^0 is undefined
To Tom:
here’s where 0^0 comes from.
f'(x) x^x=(x)*(x^-1)
now plug in 0 for x
(0)*(0^-1)
0^-1= 0/0= undefined
That’s where 0^0 derives from
0 is not a number. It is the absence of a number. 0 means that the objects being counted are not there. It is used inappropriately in mathematics, physics … (all the sciences) often. Another example is 10. 10 is the 10 th digit which uses 0 in an inappropriate way.