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.
You just said, as I agree with, that there are times in which it is perfectly logical to define 0^0 = 1. But I say to only define it that way in such areas in which it makes sense and to not generalize this for it is not always logical to define 0^0 = 1. Also, I would think (so my own opinion) that it would be more convenient for 0^0 = 1 so you can deal with it instead of leaving it as undefined. I do not do it because I have yet to be convinced that it always holds true and should be done.
S K,
Don’t overemphasize continuity,
distinguish lim (x->c)f(x) and f(c),
then 0^0=1 will always holds true.
Nothing should be done.
0! can also be undefined.
Nor is 0 itself.
But it is good to define.
If we use that logic, then we might as well say that since x/x = 1 for all x not equal to 0, and lim (x->0) of x/x =1, why not define 0/0 =1 for convenience? And the reason why I stress the continuity is because the few times I did not, it turned out that it was super important and I came up with stuff that was always not true. Thus, learning form that lesson, I will adhere strictly to it, unless there is a specific case that says “here W.L.O.G. we can assume this” I will always adhere to it.
And 0! is a specific case, for when we talk about factorial, it is normally with combinatorics and counting. Also I have derived arguments, and seen some arguments derived, that adhered strictly to all rules that show 0! should = 1. And assuming so does not mess up any other concept in math, thus the definition can hold, in general, the rest of algebra and the real number ring.
S K,
Defining x^0 as x/x is not a good way.
And this is not the reason not to define 0^0=1,
but the result.
Directly defining x^0 as 1 is good.
If you think 0^0=0/0 is undefined,
I can also say 0! = 0*(-1)!
Since (-1)! is undefined, 0! is undefined.
Conclusively, you are just describing the mathematical structure which does not define 0^0.
But you did not explain this is good.
I think the real mystery is how popular this post is. The question is nearly two years old and yet the comments don’t stop from coming, luckily email still exists so I can read everything.
First, there should be a very definite difference between undefined and intermediate. 1/0 is undefined, and is not strictly the same as 0/0, or 0^0
Second, 0/0 is a mathematical entity that could be defined as the solution to 0*x=0.
Finally, to appropriately define 0^0, we have to decide what that we know of math is a law, and what is more of a guideline.
Multiplication, for instance, is defined as repeated addition. X*7=x+x+x+x+x+x+x, in all cases, no matter what, because that is what multiplication is defined as.
Exponents are defined as repeated multiplication. X^7=x*x*x*x*x*x*x, in all cases, no matter what, because that is what exponents are defined as.
Subtraction is defined as reverse addition, so x-x=0*x
Division is reverse multiplication, so x/x=x^0
These are rules of math, they are always true.
Guidelines, however, present themselves in many forms. Examples:
X*0=0
X^0=1
1^x=1
-1≤Cos(x)≤1
0≤X^2
Guidelines always have solutions that make them either wrong, or misleading, and usually it involves putting negatives, zeroes, or the reciprocal of two in strange places. I am of course talking about imaginary and undefined terms. Laws, however, are always true, in at least some sense, because they are laws, they are the baseline of math, and they are what everything else is derived from. Laws are never not true.
My point is that x/x=x^0 is a law, and not a guideline.
x-x=0*x and x/x=x^0 are not always true.
They are not laws.
Defining 0*x as identity element of addtion
and defining x^0 as identity element of multiplication
are more reasonable.
Because to do no operation is equivalent to operate with identity element.
why not yet reached a general consensus on this matter?
If continuity is still overemphasized,
no consensus will be reached.
Mr. Clever Student’s equation doesn’t make sense.
I would say :
x^0 = x^1-1= x^1-x^1=0^1-0^1=0
Even calculators would get it wrong because:
There are thousands upon thousands of little microscopic gremlins inside… they each have one of every possible math question memorized. I agree that this all l0oks kinda stupid but I think it’s right. x/x = 1 only if x doesn’t equal 0.
Ex: 4/4 = 1 because 1 x the denominator equals the numerator. 0/0 = 0 because nothing / nothing does not equal 1 even though 1 x the denominator equals the numerator ( in this case any number x the denominator equals the numerator).
b^x=e^(x*ln(b))
0^0=e^(0*ln(0))
limit ln(x) x->0 =-inf
(0=1/infinity)
0*ln(0)=(1/inf)*-inf=-inf/inf
0^0=e^(inf/inf)=indeterminate
MR. Walrus,
b^x=e^(x*ln(b))
is invalid for b=0.
If we can’t decide on what it is, then we just have to decide what it isn’t. We should note the consequences of the zero in the exponent. For any solution we can take from 0^0, it must also be that number to any power, since 0^(0*n)=0^0, for any n that solves n*0=0. Unfortunately, no single number in existence satisfies only this condition, and I only need two exponents to show it, starting with zero.
x^0=x
So long as x isn’t also 0, then
1=x
The second number is one half, the square root. Since I can raise 0^0 to any power, so long as that power times zero is undoubtedly still 0, then that resulting number is still a solution to 0^0, and as we all know, square roots have two solutions, both positive and negative.
If 0^0=1, then
0^(0/2)=1^(1/2)
0^0=1,-1
Now, consider Euler’s identity, e^(iπ)=-1.
0^0=-1
0^0=e^(iπ)
0^(0/(iπ))=e
0^0=e
Since the function e^n intersects all numbers, we have an infinite number of solutions. There is no single number that is the same number when raised to any power, which means 0^0 must be every number.
But the actual value of 0^0 is somewhat impossible to prove if you refuse to define the rules of math. Math does have laws, actual definable understandable laws. If you don’t accept our laws, then tell us yours.
0^0 = 0
if you think about it from a philosophic point of view, you would see that zero means nothing. and 1 is something. recognizing that having 2 “nothings” no matter what mathematical application you do to them, you will eventually get nothing. this is why 0+0=0; 0-0=0; 0*0=0 and of the same reason 0/0=0 although u can’t divide by zero. and this is also a proof that God exists. if there was nothing, no way we can get something out of nothing. therefore we need a 1 in order to build all the numbers.
this is what i think lol.
This response section is getting notoriously off topic.
0^0=1 is the best definition.
I can respect that.
A seventh grade student asked: If x^0 = 1, what if x is a negative number? Will x (a negative number) ^0 = -1? They were just taught that the exponent applies to the figure directly to it’s left. She argues that if x is negative that means -1 * x^0, so the answer is -1 and not 1 (provided that everyone agrees that x^0 = 1.
Hmmm….
Jean
For any x
x^0=x/x
If x were negative, the negatives would cancel out.
i think it is indeterminate…
The Cool Dude,
x^0=x/x is invalid when x=0.
Why not say for any x,x^1=x^2/x?
Yee,
I was referring to negative numbers, thus, not zero.
The Cool Dude,
You said “For any x”
Of course including 0.
Yee,
I was implying negative numbers, as I was replying to a comment about negative numbers. If I was misleading, I apologize.
On a somewhat related topic: although when working in calculus or above level mathematics, it is improper to reduce (x^2)/x to x, since there is a discontinuity of the prior at zero, however, would it not be proper to reduce (x^3)/(x^2) to (x^2)/x, since they both intersect x at all values besides zero, and at zero, both resolve to the same discontinuity? I think it is important to note this property, being simplified as (x^n)/(x^(n-1))=x*(x^0).
i dont get this but maybe there is something between 0 and 0. You may think this is imposible but note what did tinkerbell say to peter pan. the anwer was a combination of a choke,snort,and a fart.
1) Bothering to write out some law Cool Dude said.
R : x_1 = x_2 = x_3 … = x_n
A)
x_1 + x_2 + x_3 … + x_n = x*n
and
x_1 – x_2 – x_3 … – x_n = x*(2-n)
therefore
x_1 – x_2 = x*(2-2) = x*0
and
0 – 0 = 0*(2-2) = 0*0
B)
x_1 * x_2 * x_3 … * x_n = x^n
and
x_1 ÷ x_2 ÷ x_3 … ÷ x_n = x^(2-n)
therefore
x_1 ÷ x_2 = x^(2-2) = x^0
and
0 ÷ 0 = 0^(2-2) = 0^0
2)
A simple re-arrangement of xn=x where x = 0 and n = 1 shows us that if 0/0 = 1 then 1/0 = 1/0
Conclusion:
Among the solutions of 0/0, the difference is wholly arbitrary. The choice of making the solution not equal to 1 or equal to 1 was without cause. It was merely convenient.
We can clearly see one argument will not debunk any other or provide proof.
0/0 can be any n since when you multiply any n by 0 you get back to 0. But this means that you also have all other numbers divisible by 0: 1/0, 2/0, 3/0, -e/0, etc.
You can’t arbitrarily pick any real number to represent them. Each quotient must have a unique result so that when multiplied by 0 you get back to either 1, 2, 3, -e, etc.
So 0/0 = n is just a placeholder, and misleading, for what really should just stay as 0/0.
And 0/0 = 1 is just not valid.
The correct solution is:
n/n * n = n
0/n * n = 0
n/0 * 0 = n
0/0 * 0 = 0
0/0 is its own entity = 0/0.
Phil,
0/0 is different from 0^0.
If 0^0=0^(1-1)=0/0
then 0=0^1=0^2/0^1=0/0
“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.”— Truth is never simple. So the truth is that 0^0 is undefined
Jim,
Not defining 0^0 is also a choice, not truth.
This is the result of overemphasis of continuity.
It’s not a good choice.
Hello to every one:
I some how agree with “cool dude”
As you know:
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
And also we can conclude 0^0 = all of the Tensors in every Rank
So at last but not the least (even more important):
0^0 is not a number it is a collection of all of numbers and all of arrays of numbers means like Matrix or Tensors
I’d love to tell me opinions of yours here or personally
My email: immortal_ah71@yahoo.com
Reply to Yee:
As you said:
If 0^0=0^(1-1)=0/0
then 0=0^1=0^2/0^1=0/0
so 0^0 means that 0 is on of the answers of 0*x=0
so we can’t obtain that 0/0 is not equal 0^0
in deed 0/0 is the other way of showing 0^0
mostafa,
Since you can’t obtain anything,
you can’t say 0^0=0/0.
Why are we so concerned with defining 0^0 with such absoluteness? Doing so only limits the scope of our interpretation.
The following analogies may not be perfect but two come to mind:
1. The dual nature of light. In some situations, light behaves like a particle; in others, like a wave. Our interpretation of light’s nature thus depends on the broader context of the situation and neither interpretation is any “more correct” than the other.
2. Our choice of axiomatic structure. Consider for example euclidean geometry vs spherical geometry. In the former we define a line as a 1-dimensional straight object extending infinitely in two directions (positive and negative?). In the latter, lines are great circles that occupy 2 dimensions.
In either situation, “which one is correct?” is a misleading question. A more informative question might be, “which one is more convenient given the problem at hand?”. Why not think of 0^0 in this light?
D -A young, humble, aspiring mathematician
D,
Is defining 0!=1 too absolute?
Is defining 0^2=0 too absolute?
Is defining 2^0=1 too absolute?
…
Why is defining 0^0=1 too absolute?
(0^0)^2=0^(0*2)=0^0,x^2=x
0^0=0^(-0)=1/0^0,x=1/x
1 is the only solution.
Therefore 1 is the only reasonable definition of 0^0.
Continuity needs not overemphasizing.
Yee,
We can choose to define 0^0 = 1 if we find it to be, as you say,”reasonable”. There are arguments that challenge this reasonableness (see continuity) so we should be careful to make the distinction that we are choosing to define 0^0 as 1, it is not absolutely true.
If we define 0^0 to be 0 (or any other number for that matter) there will be consequences of this choice. A main reason that we choose to define it as 1 is not because it is the correct answer but because it is the most convenient answer (see author’s combinatorical argument).
Defining 0^0=1 is for convenience.
So is defining 0!=1.
Definitions are irrelevant to correct and incorrect,
irrelevant to right and wrong,
irrelevant to true and false.
Overemphasizing continuity to an extent
where a function is undefined at a point where limit does not exist
is not a good choice.
The question in the first place was “what does 0^0 equal”, rather than “what to define 0^0 as”, so the original conversation topic implies factuality, rather than convenience, or what is best.
This being said, it’s an indisputable fact that 0^0 maintains the same value, regardless of what value you multiply the exponent by, so long as that number times zero is still zero. Since 0/(i2π)=0, and 1=e^(i2π), then if 0^0=1, 0^0=e. Raising e to the power of any number that solves for n*0=0 gives an infinite range of values, since all finite numbers, imaginary or otherwise, satisfy this condition.
It’s an inescapable property of this particular mathematical entity. It either evaluates to being all numbers at once, or no number at all, depending on how you interpret math, as a function of multiple solutions, or of limited solutions. I prefer the later, especially since most, if not all sufficiently advanced subjects of mathematics have multiple solutions, including logarithms, trigonometric functions, roots, integrals, and so on.
The Cool Dude,
“Since 0/(i2π)=0, and 1=e^(i2π), then if 0^0=1, 0^0=e.”
I don’t understand what it means.
Yee,
Complex numbers are sometimes notated as A+Bi, with a real and imaginary part, but you can also express them as R*e^(iΘ), where Θ is the angle in the imaginary plane, and R is the absolute value. √(A^2+B^2)=R, arccos(A/R)=Θ=arcsin(B/R).
Since 1 has an imaginary part of zero, the angle in the complex plane is 0+n2π, for whole number n’s, meaning that 1=e^(i2π). This also stems directly from Euler’s identity, e^(iπ)=-1, since (-1)^2=1, e^(i2π)=1. Adding an imaginary number in the exponent changes the value to a number that does not have the absolute value of 1, meaning its value can be expanded further when using exponents.
The Cool Dude,
What does it have to do with 0^0?
If 0^0 is defined as 1, it must also be defined as e^n, for any number n*0=0. It can’t be only one value.
Why must it also be defined as e^n, for any number n*0=0?
Yee,
1=e^(i2π). If 0^0=1, then 0^0=e^(i2π), and since 0/(i2π)=0, 0^0=e.
It’s true that for all real n, only 1 solves for 1^n=1, but for imaginary or complex n, it does not. There is no single number that can correctly solve to fit the definition of 0^0. It’s either a Universal set, or a Void set.
I still don’t understand.
How does 0/(i2π)=0 imply 0^0=e?
Euler’s formula from complex analysis states that
e^(ix)=cos(x)+i*sin(x)
To put 1 into this form of notation, we start by putting it in polar form, in the complex plane. 1 has an absolute value of 1, so we don’t need to worry about the radius in the complex plane, and instead solve for x in 1=cos(x)+i*sin(x). We know that 1 is real, and so, i*sin(x)=0, so it follows that x=0+n2π, for any whole number n.
Plugging x=0+n2π, for any whole number n, back into cos(x)+i*sin(x) will always get 1, so we know our solution is correct. Using Euler’s formula, we can obtain the statement 1=e^(0+in2π), for any whole number n. Plugging this in to any sufficiently advanced calculator will validate this result as well. What follows is a conditional. If 0^0=1, then 0^0=e^(0+in2π), for any whole number n. Working the second half of the conditional, assuming n=1,
0^0=e^(i2π)
0^(0/i2π)=e^(i2π/i2π)
0^0=e^1
0^0=e
If 0^0=1, then 0^0=e.
The Cool Dude,
You are not denying 0^0=1,
you are denying 1 is defined.
e^0=e^(i2π)
e^(0/i2π)=e^(i2π/i2π)
e^0=e^1
1=e
According to your inference,
1 is indeterminate.
Yee,
The conclusion I’ve pulled may seem incorrect, but the reason I’m able to conclude that if 0^0=1, 0^0=e, is because 1 can be represented in an infinite variety of forms.
For all non-zero, finite n: n^0=1
For all real, finite n: 1^n=1
For all whole number n: e^(in2π)=1
I’m able to derive my conclusion because 1 can be expressed in entirely different forms, while 0^0 can not. Adding an imaginary exponent to any other number would yield multiple solutions. 0^0 is a mathematical entity that does not permit this. If we define 0^0 as 1 for convenience, it is mathematically incorrect to assume it is not also e, or any number in the range of e^n.
e^0=e^(i2π)
e^(0/i2π)=e^(i2π/i2π)
e^0=e^1
1=e
What does it have to do with 0^0?
Yee,
That math is incorrect because, again, one has multiple representations, and additionally, imaginary numbers in exponents have multiple solutions. When taking the imaginary root, you have to consider all possible angles in the complex plane of that number, but 0^0 does not have an angle in the complex plane, and thus, taking the imaginary root of it does nothing to how many solutions it has.
Doing e^(i2π/i2π) results in multiple solutions. 1=e^(n*i2π), for every single whole number n, which means that e^(i2π/i2π)=e^n, for every single whole number n.
The problem is that performing certain operations on 1 results in multiple solutions, but performing these same operations on 0^0 does nothing, and so they can not be the same. It is incorrect to define 0^0 as 1 without including an additional infinite solutions.