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.
@Ron,
“A lawyer appears before a judge. The judge says, “Where is your client?” The lawyer says, “I have no client, but I’m ready to argue a case.””
The lawyer says, “if you don’t have a client, then you are not the lawyer for the client who is not here.”
answers are very nice. but that is inderminate value
An indeterminate form is something that you get when you are taking limits. The question was not what happens when you get 0^0 during the process of taking a limit (in that case, it is indeed an indeterminate form) but rather, what is a^b when you assign a=0 and b=0.
Mathematics is a “logical” body of knowledge. And, as Euclid showed in the development of the first logical system called Geometry, the need for particular beginnings: “undefined terms and unproofed statements called axioms.” Occassionally, situations arise that call into the logical development of mathematical developments the use of Euclid’s approach. In developing “Taylor Series” the need arises to define Zero Factorial to equal One (1 ) so that the closed form of the idea can be compactly expressed. Also, in mathematics there are frequently particular conditions that must be met or else answers evolved that lead to contradictions. In the case of “Zero to the Zero Power” the attempt to “divide by zero” as well as “applying the derivative when it does not exist” are two examples of conclusions that lead to contradictions in previously developed logical ideas. So, within the set of Real Number Valued Functions, Zero to the Zero Power is “Undefined” meaning that it has no real number meaning. Harry, I hopes this helps you understand the mathematician’s view. Loyd
I now see why really clever people can bring the whole world crashing down around our heads on such a regular basis.
Since mathematics is a modeling tool , there will be times when the arbitrary rules of mathematics can not yield an accurate depiction of the real world, especially on the margins.
If you put nothing into a container repeatedly, you still have nothing in the container.
“If you put nothing into a container repeatedly, you still have nothing in the container.”
no. 0^0 means you never put nothing into the container.
it’s called a singularity
I would define this creative loafing #0-sometimes doing nothing is doing something.
does all of this mean that 0=space?
If a chicken in a half laid an egg in a half in a day and a half, how many pancakes does it take to shingle an outhouse?
It makes sense: If 0^0 means you never put nothing in a container than the double negative makes it a positive, meaning you put something in a container. Therefore 0^0 is all real numbers greater than 0
“Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory” (Kleene 1967, p. 250), aka Godel’s first incompleteness theorem. http://en.wikipedia.org/wiki/Godels_theorem. It is a riveting read.
Does this mean by corollary that the zeroth root of one equals zero?
0^0 is an emotion-con just like 0.O
done.
Unbelievable, 0.0=0, x/0 is undefined. Period end of story Math is truth, or closer than anything Diogenes every found. I took my last Calculus course 45 years ago and if we’d pulled any of this crap My Prof would have thrown me out the window. We learned something back in the sixties. I doubt if many of the above can say the same.
@ rob.
it takes 10.
Why?…. Because blue is 9 inches long.
I’d say 0^0 is both 0, 1 and infinity, just like Schrödinger’s cat is both alive and dead.
Is
0^0 = (1-1)^0 = 1^0 – 1^0 = 0
somehow valid?
Not quite. (1-1)^0 isn’t equal to 1^0 – 1^0. The only time you can pull apart stuff inside of parentheses like that is when the exponent is 1.
0 is not a value, but a symbol. There is a difference. Asking invalid question can’t bring you a valid answer. And the definition of 0 has limitations.
The symbolical interpretation of 0^0 can be treated safely as value of 1 as part of any other expression without “side effects” (like binomial coefficients). That’s assumed to be acceptable for simplicity and it is intended to be used that way as stated above.
By itself 0^0 is an indeterminate expression thus it lacks dimension/scope in which to have a valid value in which case the only valid value for practical purposes for this expression is again to be assumed 1.
As a rare exception I tend to agree with the explanation “Because mathematicians said so” for all considering this a valid question.
A long time ago a no one took nothing and multiplied it by nothing and BOING! the BIG BANG occured. So dont’t do that again!
This is simple. As I have always said: Nothing does not Exist
0 is empty, therefore it cannot exist, nor can it not exist. When you take infinite ^ infinite you get the opposite of 0^0. Infinite is everything that exists and everything that has ever existed, and everything that will ever exist. The exact inverse is simply not feasible since it would result in absolutely nothing. When you ask what 0^0 = You are really asking if the possibility of absolute nothing is true or false. Since it is impossible for absolute nothing to exist since we exist, then the real answer to 0^0 = False.
Ask a Philosopher.
“@Rob says:
March 12, 2011 at 10:26 pm
If a chicken in a half laid an egg in a half in a day and a half, how many pancakes does it take to shingle an outhouse?”
Without searching google the answer is: Ice cream has no bones.
nothing is still something. by giving it a name and talking about IT we have agreed that there is an IT. IT exists, as a concept, an idea, nothing else. when i say im doing nothing, i really mean i am existing. i am still doing something. if there is a pocket of space that has nothing in it, then there is nothing in between point A and point B, therefore stating that there is nothing separating point A and point B. they are right next to each other! also!
http://asset.soup.io/asset/1177/1797_9350_450.jpeg
0^0 =8–>
0^0 is evenly divisible by 5.
Ask a Discordian.
2 + 2 = Bushel of Potatoes
In English, zero has a quantity of one. Thus, you’ll hear the correct form, “None was injured in the accident.” By default then, we English peeps must go with the mathematicians on this one: 0^0 = 1.
I’m fairly sure that should be “None were injured . . .” or “No-one was injured. . .”
“There were no injuries . . .”
I’ve always believed any quantity other than one (including zero) to be treated as a plural when determining which verb (is/are, was/were, has/have) to use.
But I could be wrong.
It’s the quantity of the subject that determines the conjugation of the verb. When you break the parts of speech down to the simplest elements, you have, “Injuries were.” With “injuries” (plural) being the subject, modified by the adjective “no,” the correct plural verb conjugation is “were,” modified by the adverb “there.” So back to my example again, the singular subject “none” requires the singular verb conjugation “was.” Cheers!
I still don’t buy that zero (or “none”) is singular.
The only time I’ve heard a singular verb conjugation used with “none” is when referring to a fluid (or similar) noun: “I went to get some meat, but none was left.” When used with a quantized noun, I’ve always heard it as “I went to get a steak, but none were left.” You may very well be technically correct. I’ve just never heard it used that way.
Of course, I’m an American. And we all know Americans speak about as much English as the Mexicans speak Spanish.
More specifically on topic though:
Does 0^0 involve multiplying zero (albeit zero times), which always ends up as zero? Or does it imply NOT multiplying zero? If you are NOT multiplying zero, and there are no other functions listed, then zero remains unchanged, if it even existed to begin with. No?
I would think that performing any function (or any number of functions), where all values = 0 simply means you can pretend the whole equation (or that part of the equation) doesn’t exist, as all the various nothings fail to affect each other.
Pingback: What is 0^0? And is math true, or just useful? « Measure of Doubt
2+2=5
It is what it is said it is.
My TI-89 Titanium says 0^0 = 1, problem solved xD
Furthermore;
solve(0^x=1 , x)
x=0
That is the newbie way to figure it out!
i was always lead to believe that when dealing with the 0^0 for simplicity it is reasonable to use the definition 0^0=1, unless dealing with continuity problems, as it is a singularity of the rule.
wen talking about a rule of math, its the same as wen we talk about the english language, what we consider to be correct english, is simply what the people decide is right, and by people i mean whoever is in charge of it all, and about that whole none is, or none are debacle, i believe it is technically none is, because it means not one is, but it has become acceptable to say none are, because everyone says it like that
The temptation is strong for some to “define” 0^0=1, just as I suppose “defining” 0/0=1 might be tempting to some. But the fact still is that these indeterminate forms can approach whatever value you’d like.
As an exercise, try to find continuous f(x), g(x) such that
lim x->0 f(x)=0
lim x->0 g(x)=0
but lim x->0 f(x)^g(x) is something OTHER than 0 or 1.
A simple (but almost cheating) example is f(x) = e^-1/x^2, g(x)=x^2. In this case we have:
lim x->0 f(x)^g(x) = 1/e
(Hint: You need to choose f(x) to be non-analytic. I think it’s a theorem that if f(x) is analytic, and lim x->0 f(x)^g(x) exists, then lim x->0 f(x)^g(x)=1.)
@ lance you assume that absolutely nothing doesn’t exist but why would you assume that we are something? A nihlist would say that we are nothing and therefore 0^0 is nothing because nothing else exists except in something you might possibly conceive as conscience. which i also think is nothing
If we found nothing, could we then prove we had it? How wold we know when nothing was there? We would have to make sure nothing was within it, which all agreed cannot be done at our current level of technology.
Im confused as to why the first students definition is invalid. If we take the function separated from x^x and break it using x^1-1 to get x/x, the take the limit as x approaches zero to get the indeterminant 0/0. we can then use L’hopitals rule to simplify the fraction to 1/1=1
Hi Tom. The logic doesn’t work because proving that
lim x-> 0 x^x = 1
does not prove that 0^0 = 1. Similarly, showing that
lim x->0 0^x = 0
does not prove that 0^0 = 0.
I think it might interesting to extend “ask the mathematician” to “ask the complex analyst”. What happens if we consider the function w=abs(y^x) where x, y /in R, that way we have a function that maps R^2 -> R and we can look at it’s behavior around the origin. Haven’t had time to check this out in more detail, but it looks like w -> 1 as (x,y) -> (0,0): http://goo.gl/2jMak
I was told ANYTHING to the power of 0 equals 1… 10^0=1, 5^0=1 100, 000, 000^0=1
so why not 0^0? DO not over complicat simple things or else you will miss the easy answer.
All of this makes perfect sense, except for the train problem. See, if lightning strikes both ends of the train, it wouldn’t matter when things are happening in Bob’s point of view because he’d be dead.
Here’s the thing. Although the math is accurate, and the limit as x–> o of x^(x) is 1, that is not valid. That means that that the closer and closer that x gets to 0, the closer and closer that function gets to 1, but it is never that exact number. For any other set of numbers, that would be true, but zero is an exception. So theoretically, 0^0 MAY be 1 due to several proven calculus theories, but I think that 0 is an exception in this case, just as you can’t divide anything by 0, and just as 0 = nothing. My guess is that 0^0 = 0
why does it even matter?
In order to really understand how something works it’s important to understand how it doesn’t work. For example, studying “poles” (singularities) makes doing complex integration easy, and yields an amazing array of useful theorems (specifically in engineering).
This particular example sheds light on what can go wrong in vector calculus, which is the backbone of damn near everything in physics.
As far as “what does it matter to the man on the street?”: nothing. But it means a lot to the people who make and design everything that the man-on-the-street uses.
OH GOD THIS CONFUSES ME SO.
Hi there.
Turtles rock.
0^0 = 1 ; 1^1 = 1
and therefor 0 = 1
u jelly ? :]
well think of it this way: if you put a number to a power any where from 1 up you are plugging that many of that number into a repeating multiplication problem
(ex. 7^2 = 7*7 and so on so forth)(In a problem with the power as 1 it appears as this 7^1 = 7 = 7). There for just as in fractions if you put it to any power less than 1 you are dividing it by itself (ex. 7^0 = 7/7 = 1) there for any number to the power of 0 is 1.
Even though you cannot divide any number by 0 because the out come is infinity, we treat it as 1 because it applies to all other numbers this way.