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,
Saying 0/0=X is the same as 0X=0.
Name a single real, or imaginary number, that when multiplied by zero, is never zero.
“Saying 0/0=X is the same as 0X=0.”
I don’t disagree with you Cooldude, which is why proofs that rely on those concepts and ultimately show that 1=2, are invalid.
Ron,
The issue is comparing multiple results, not dividing and multiplying by zero. 1=2 is just as invalid as 2=-2, but both come up in totally different situations. What’s the difference between these two statements? Why is 1=2 more incorrect than 2=-2?
Ron,
lim (x->0+) f(x)^g(x)=0^0
This is wrong because it is a discontinuous point.
There is no reason that 0^0 may equal another value.
Therefore 0^0 equals only 1.
This site is full of fun for knowledge.
“lim (x->0+) f(x)^g(x)=0^0
This is wrong because it is a discontinuous point.”
This is a right side limit… Discontinuity is not an issue. If you have references that contradict this point, cite them.
“The issue is comparing multiple results….”
Which means indeterminate.
Ron,
No matter right side or left side or whatever side,
discontinuous points do not have this property.
If 0^0=0,0^0=1,0^0=2,…
then 0=1=2=…
It is obviously incorrect.
If 0^0=1,
lim (x->0+) f(x)^g(x) does not necessarily equal 1,
lim (x->0+) f(x)^g(x) = 0^0 is incorrect.
If 0^0 is indeterminate,
it can not equal anything.
lim (x->0+) f(x)^g(x) = 0^0 is incorrect.
Therefore
lim (x->0+) f(x)^g(x) = 0^0 is incorrect in all cases.
While the result may not be very elegant, strictly speaking, I think you can still prove the equivalent of the binomial theorem assuming 0^0 is undefined.
For a=0 and b=/= 0 (your example), consider it as a special case:
(a+b)^n=(0+b)^n = b^n
Then consider as a special case, any situation that makes both the base and exponent equal to zero anywhere in the final result or in any intermediate result of the proof.
Dan Christensen,
If not define 0^0,
binomial theorem can be written in a complicated form
or add some constraints to be correct,
but it is not convenient.
Defining 0^0 = 1 is good.
Dan, you are exactly right. It is just plain absurd to apply the binomial theorem to (x+0), especially in instances like (x+0)^14439984. Even from a computer science point of view this is absurd and inefficient.
“binomial theorem can be written in a complicated form
or add some constraints to be correct,
but it is not convenient.”
Yee, This argument simply boils down to “It looks prettier on paper.”
Why is “being prettier” more important than being correct? Besides, defining a binomial as: (x+y), where the domain of x,y≠{0} still lets you have your prettier equation.
If you are going to argue definition, simply for the sake of prettiness, then you must be willing to accept other “definition solutions” that will solve the same problem. Why go with one that is wrong and inconsistent with the rest of math, when there are other “definition solutions” that are correct and are consistent with the rest of math?
Mathematicians, the world over, do not define it as 1 anyway. They also don’t define it as indeterminate either. It’s VALUE simply IS indeterminate, as demonstrated time and time again throughout this thread. Since this thread is about it’s VALUE, arguing a non-existent definition is meaningless as a means of assigning value.
Ron,
It is no good to deny binomial theorem.
Take complex multiplication for example.
z1=a+b*i,z2=c+d*i
z1*z2=(a*c-b*d)+(a*d+b*c)i
Is it valid for real numbers that is b=d=0?
If you think it valid,
Tell me who will do so many multiplications with 0?
If you think it invalid,
you have to write it as
z1*z2=
a*c if b=d=0
(a*c-b*d)+(a*d+b*c)i otherwise.
Which do you think is better?
Ron,
Who will calculate (1+1)^100 with binomial theorem?
Well, is (1+1)^100 valid for binomial theorem?
Ron, just saying “intermediate” is just as misleading as saying 0^0=1.
Intermediate is correct because it has an infinite number of solutions.
And 0^0=1 is convenient because calculating infinite results is impractical.
But
0^0=1 is misleading because it implies there is only one solution.
And intermediate is misleading because it implies there is no actual solution.
Both of these statements are technically correct. Both of them are clearly misleading.
Frankly, I don’t see the problem. If we’re all correct, why don’t we cut our losses and just let the other guy think whatever he wants to.
The Cool Dude,
You say it has an infinite number of solutions.
But, solution of what?
Solution of something that has to be proven depending on the definition of 0^0.
Pleas tell me solution of what.
Definition is irrelevant ot right and wrong.
Definition is irrelevant ot true and false.
It is also correct to leave 0 undefined.
Give an arbitrary equation and find many solutions and say it is indeterminate,
it is ridiculous.
0*1=0,1 has many solutions.
0*2=0,2 has many solutions.
.
.
.
As a result, no number exists.
“Definition is irrelevant ot right and wrong.
Definition is irrelevant ot true and false.”
True – that would be the case if it were defined, however there is no definition that states 0^0=1, so your ENTIRE point is irrelevant.
“Well, is (1+1)^100 valid for binomial theorem?”
According to the definition I offered for a binomial : x+y, where the domain of x,y≠{0}, That would be OK as a binomial. It wouldn’t create any 0^0 issues. I had also considered another additional requirement where the domain of (x+y) cannot contain only one element, which would make that invalid too, but it wasn’t necessary to solve the 0^0 problem, so I didn’t bother.
Of course, I personally wouldn’t solve it that way. I would stick with 2^100
“0*1=0,1 has many solutions.”
Actually, 0 times anything has only one solution. that solution is zero. Zero is a number. It is exactly the same distance on the number line from 1 as 1 is from 2.
“Take complex multiplication for example.
z1=a+b*i,z2=c+d*i
z1*z2=(a*c-b*d)+(a*d+b*c)i”
Yee, this has nothing to do with 0^0. You are arguing nonsense.
Ron,
0^0=1 is good definiton.
It is useful in some fields.
You should not turn deaf ear to it.
If 0^0 is undefined,
you have to add many constraints to binomial theorem.
If 0^0 is defined as 1,
you don’t need theses constraints.
Which do you think is better?
It is user’s business to spread (2+0)^100 with binomial theorem.
Why do you deny it?
Do you deny multiplication of complex numbers when imaginary parts are 0?
Ron,
You add a constraint x,y≠{0} to binomial theorem.
Defining 0^0=1 is enough to solve it.
When did we suddenly stop talking about math. It clearly speaks for its self.
For all A≠1
X^0=A
X^0*X=XA
X^1=XA
X-X=XA-X
0=X(A-1)
X=0
If I plug zero back in:
For all A≠1
0^0=A
0^0*0=0A
0^1=0A
0-0=0A-0
0=0(A-1)
0=0
The equation is fine.
I don’t need the binomial theorem, although useful.
I don’t need to use limits, although useful.
I don’t need to define it, although useful.
I don’t need to express all infinite solutions at once. Nobody wants this. Nobody.
If we are ignoring simple algebraic mathematics, we are doing it wrong.
Ron,
You don’t need to explain how to construct binomial theorem without defining 0^0=1.
The question is : which definition is better.
“The question is : which definition is better.”
The one that is consistent with all of math, of which 0^0=1 is not.
“If we are ignoring simple algebraic mathematics, we are doing it wrong.”
exactly… which is why 0^0=0/0 can not be ignored.
Yee said, “If 0^0 is undefined, you have to add many constraints to binomial theorem.”
As far as I can tell, in the natural numbers (including 0), you only need to consider 4 cases:
(a+b)^n = {the usual binomial formula if a=/=0, b=/=0 (no zeroes)
{a^n if a=/=0, b=0 (one zero)
{b^n if a=0, b=/=0 (one zero)
{0 if a=b=0, n=/=0 (two zeroes)
The last 3 cases are trivial. The case a=b=n=0 is left undefined.
Is this really all that onerous? Onerous enough to justify inventing 0^0=1 simply because it is convenient?
Ron,
You don’t understand definition.
Math is constructed according to definition.
Definition is first, then there are theorems.
If we define 0^0=1, we won’t construct a theorem that is inconsistent with it.
0^0=0/0
According to what?
According to abuse of index laws.
0=0^1=0^(2-1)=0^2/0^1=0/0
0 is inconsistent with simple algebraic mathematics.
Is it funny?
Dan Christensen,
Yes.
Defining 0^0=1 is better.
Take complex multiplication for example.
z1=a+b*i,z2=c+d*i
z1*z2=(a*c-b*d)+(a*d+b*c)i
Do you prefer writing it so or considering 4 cases:
a≠0,b≠0
a=0,b≠0
a≠0,b=0
a=0,b=0
Of course, writing it in a simplest form is better.
Considering 4 cases is onerous.
The mathematical beauty we pursue is to write math in a simplest form.
Yee, I must have missed something. I don’t understand how your example of complex multiplication is relevant. We are talking about exponentiation, not multiplication, right?
Dan Christensen,
The connection is :
writing theorem in the simplest form is better.
Yee said, “The connection is : writing theorem in the simplest form is better.”
Not always, not even in science. While we are talking about mathematical, not scientific theory, you might find the Wikipedia entry on “Occam’s razor” to be instructive in this case.
“The [principle of Occam’s] razor asserts that one should proceed to simpler theories until simplicity can be traded for greater explanatory power. The simplest available theory need not be most accurate….
“In science, Occam’s razor is used as a heuristic (general guiding rule or an observation) to guide scientists in the development of theoretical models rather than as an arbiter between published models. In the scientific method, Occam’s razor is not considered an irrefutable principle of logic, and certainly not a scientific result.”
http://en.wikipedia.org/wiki/Occam's_razor#Controversial_aspects_of_the_razor
If you can write a theorem in a simple form,
why don’t you do it?
You must have a better reason not to do it.
Yee wrote: “If you can write a theorem in a simple form, why don’t you do it? You must have a better reason not to do it.”
If you can write a theorem (or proof) in a more correct form, why wouldn’t you do it? There can be no valid reason for doing otherwise.
Yee, I have a hard time taking anyone seriously who’s only point is that it makes it look prettier.
Why does prettier matter?
“You must have a better reason not to do it.”
Says who?
Dan Christensen,
Definition is irrelevant to right and wrong.
Definition is irrelevant to true and false.
Definition is irrelevant to correct and incorrect.
There is nothing incorrect in defining 0^0=1.
How can you write a theorem in a more correct form?
Ron,
Looking prettier is a kind of mathematical beauty that mathematicians pursue.
This is how we choose to define.
One shouldn’t make a definition that makes a theorem ugly
although it is not wrong.
Your attitude is :
Accept the status quo,
don’t make any change,
and don’t care whether it is reasonable or not.
Yee wrote: “How can you write a theorem in a more correct form?”
You can refrain from using “definitions” that can only be justified by convenience.
Yes, I understand that no inconsistencies will arise from assuming 0^0=1. Then again, no inconsistencies will arise from assuming 0^0=0 either. You just won’t get the results you desire. 0^0=0 would be “inconvenient.”
If you don’t mind playing it fast and loose, and don’t care much for formalities, you are unlikely to get into trouble assuming 0^0=1. For me, unlikely isn’t good enough in mathematics. I would rather play it safe until the matter is better understood. Especially since it seems to be only a matter of a few extra lines of proof.
“Looking prettier is a kind of mathematical beauty that mathematicians pursue.”
Funny, I would say that their real endeavor is truth. Of course, John Keats may ultimately provide our middle ground:
‘”Beauty is truth, truth beauty,” – that is all
Ye know on earth, and all ye need to know.’ -from: Ode on a Grecian Urn, 1819
Dan Christensen,
0^0=1 had been generally accepted until continuity was overemphasized.
There may be nothing more to be understood about 0^0.
If continuity is not taken into consideration,
no one will doubt 0^0=1.
Is it reasonable to overemphasize continuity?
That is the point.
Define 0! = 1,
then n*(n-1)! = n! is valid for n = 1
If 0! is not defined as 1,
n*(n-1)! = n! is invalid for n = 1
and brings some inconveniences.
However, it is not wrong.
So is 0^0.
Yee wrote: “Define 0! = 1”
You can derive that result from the definition:
1! = 1
(n+1)! = n! * n+1
For n =0, we have 1! = 0! * 1
Therefore 0! = 1! / 1 = 1
Dan Christensen,
You must define 0! = 1 first,
then prove (n+1)! = n! * (n+1) is valid for n=0.
Ron,
If you think beauty is truth, truth beauty,
0^0 = 1 is truth.
If not, there is not beauty with 0^0 = 1.
Yee wrote: “You must define 0! = 1 first, then prove (n+1)! = n! * (n+1) is valid for n=0.”
I don’t agree, but in this case, I have no strong objection to “defining” 0! = 1 since, as I have shown, this result can be derived from what I believe is a more intuitive definition. In my opinion then, 0! = 1 is entirely justified. The same cannot be said for 0^0 = 1, which, as far as I know, is a choice based entirely on convenience.
“0^0=1 had been generally accepted until continuity was overemphasized.”
WHO says continuity is overemphasized?
Dan Christensen，
You don’t agree,
nor did I.
But I have to agree now.
That’s what definition means.