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,
lim x^y does not exist.
x->0,y->0
This is the very reason not to define 0^0.
This is why I say continuity is overemphasized.
“lim x^y does not exist.
x->0,y->0”
Nobody denies that, nor is that the reason that it is indeterminate. However, right side limits do exist for f(x)^g(x) for various functions that equal zero that demonstrate that 0^0 has a range of [0,1].
You still need to explain why you think continuity with the rest of math doesn’t matter since mathematicians the world over have rejected your position for more than 200 years.
Ron,
It is to overemphasize continuity.
You who overemphasize continuity have to explain it is better to do so.
I don’t need to explain anything.
Ron,
Do you know the difference between limit and function value?
If you do, you won’t say 0^0 is indeterminate.
Ron,
What does discontinuity matter?
Continuity has been overemphasized more than 200 years.
But, is it reasonable?
“Do you know the difference between limit and function value?
If you do, you won’t say 0^0 is indeterminate.”
If the limit of a function =0^0, (a=b)
and the limit of that function also equals 1/n, n>1 (a=c)
then it is clear that 0^0=1/n, n>1 (a=b, a=c, therefore a=b)
The limit of x^x when x approaches zero is not what is at question. That wouldn’t prove anything. The bottom line is that 0^0=0/0 because x^0=x/x.
“Continuity has been overemphasized more than 200 years.”
“Overemphasized” is an opinion, not an argument. You still haven’t justified the opinion when it is clear that the rest of the math world doesn’t agree.
Mathematics, like all rationalism systems, are built entirely on logic. The must be true internally.
Ron,
If the limit exists,
function value need not equal the limit.
If the limit does not exist,
function value can also be defined.
Not defining the function value just because the limit does not exist
is not a good reason.
x^0=x/x is invalid for x=0.
What does discontinuity disagree with the rest of math world?
This is what you have to show me.
“x^0=x/x is invalid for x=0”
Actually x^0=1 is invalid for x=0, however x^0 always equals x/x, even if it gives you an answer you don’t like.
“function value can also be defined.”
Functions are explicitly defined, but values for functions are never explicitly defined.
For example… sin θ is defined as the ratio of the side opposite a given angle (in a right triangle) to the hypotenuse. Each value for sin θ is a consequence of the FUNCTION’S definition. The values themselves are never arbitrarily defined.
The same thing applies to all other functions and operators to include Cos, ln, +, -, *, /, exponents, factorial, etc… 0^0 = indeterminate is a consequence of the definition of exponents, not a consequence of arbitrary value assignment.
Ron,
Is x^1 equal to x^2/x?
Do you like it?
x^0=1 is very useful in many functions.
I have shown some examples above.
Ron,
0^0 = 1 is not arbitrarily defined.
It is a consequence of math operation.
Overemphasizing continuity and not defining 0^0 = 1 is ridiculous.
Raising a number to a power is simply multiplying it by itself.
Anything multiplied by zero is zero.
Zero multiplied by itself zero times never happened.
Ruzho,
Does a nonzero number multiplied by itself zero times happen.
.
.
.
a^2=1*a*a
a^1=1*a
a^0=1
a^(-1)=1/a
a^(-2)=1/a/a
.
.
.
It is most reasonable to define
0^n=
0 if n > 0
1 if n = 0
undefined if n < 0
“It is most reasonable to define
0^n=
0 if n > 0
1 if n = 0
undefined if n < 0"
Therefore you define 0^0 to be greater than, for example, 0^100.
That may be mathematically functional, but it is certainly not "reasonable".
“not defining 0^0 = 1 is ridiculous.”
0^0 doesn’t always equal 1, therefore it is ridiculous to define it as 1.
You may think it is ‘reasonable’ to say 0^0=1, but it is absolutely not reasonable to say 1=0^0.
The definition of equality DEMANDS that when a=b, b=a. How reasonable could it ever be to make 1=indeterminate??? for 1=0/0???? THAT is what is ridiculous, unreasonable, and absolutely absurd about defining 0^0=1
Glenn,
Why isn’t it reasonable?
0.5^0 > 0.5^100
Is it unreasonable?
Ron,
The value of 0^0 depends on definition.
It is
lim x^y
x->0,y->0
that doesn’t always equal 1, but not 0^0.
0^0 doesn’t equal 0/0.
“Is x^1 equal to x^2/x?
Do you like it?”
Yup. Even for x=0….after all, indeterminate means it still equals something, we just don’t know what it is…something *0 is still 0.
“The value of 0^0 depends on definition.”
It is not defined by mathematicians as 1, so that point is moot. This thread is about it’s value, not what you would like it to be defined as.
“lim x^y….”
The limit of a function is not the same as it’s value.
The only clear cut reality of what 0^0 equals is 0/0, which is indeterminate and case specific in value. x^a/x^a=x^(a-a)=x^0= {1, for x≠0, and 0/0, for x=0}
Ron,
Why can you operate something indeterminate?
0^0 is not defined by the majority of mathematicians,
however, it does not mean it is better to do so.
On the contrary, it is ridiculous.
It is the result of overemphasis of continuity.
Index laws need to be reconsidered according to the definition of 0^0.
You shouldn’t write x^a/x^a=x^(a-a)=x^0 without meticulous consideration.
Yee –
“Why isn’t it reasonable?
0.5^0 > 0.5^100
Is it unreasonable?”
No, that example is not unreasonable, because you have introduced a fraction to the discussion.
Try this.
Open a spreadsheet, and put the Function
=POWER(B1,C1)
in cells A1 – A11, and then fill in B1 – B11 and C1 – C11 as follows –
A B C
1 0.5 0
0.5 0.5 1
0.25 0.5 2
0.125 0.5 3
0.0625 0.5 4
0.03125 0.5 5
0.015625 0.5 6
0.0078125 0.5 7
0.00390625 0.5 8
0.001953125 0.5 9
0.000976563 0.5 10
So with 0.5 as your base, and with a series of exponents from 10 to 0 inclusive, it is very clear that the result tends to 1.
Now do the same exercise with 0 as your base, and again with a series of exponents from 10 to 0 inclusive.
In this case, there is no tendency in the results, the result in ALL cases is 0, EXCEPT in the case of base = 0 and exponent = 0 where the result is undefined (shown as #NUM! in an Excel spreadsheet).
Now do the same exercise with 0 as your base, and with a series of exponents of halving fractional value starting 0.5, 0.25, 0.125 and so on down to 0.000976563 in the eleventh cell.
Once again, there is no tendency in the results, the result in ALL cases is 0.
So there is no “reasonable” justification for saying 0^0 = 1.
It may be mathematically functional to do so, in some circumstances, but there is no logical reasoning which demonstrates this.
If you think otherwise, please provide the equivalent base and exponent spreadsheet parameters which you believe make your case.
The Excel data has been re-formatted unclearly.
Perhaps it would be better the other way round, showing base, exponent and then result –
0.5 0 1
0.5 1 0.5
0.5 2 0.25
0.5 3 0.125
0.5 4 0.0625
0.5 5 0.03125
0.5 6 0.015625
0.5 7 0.0078125
0.5 8 0.00390625
0.5 9 0.001953125
0.5 10 0.000976563
P.S. Obviously, I mean put the Function
=POWER(Bn,Cn)
in cells A1 – A11, where n = the row number.
Let’s take it a stage further.
Using 0 as your base, and with a series of exponents of halving fractional value starting 0.5, 0.25, 0.125 and so on down to 0.000…0001 in the billionth or so cell.
I.e. about one billion zeroes before the 1.
Once again, there is no tendency in the results, the result in ALL cases is 0, even when the exponent is as near to zero as possible.
But we are then supposed to believe that when the exponent actually becomes zero, suddenly the result jumps to 1.
There really is no “reasonable” justification for saying 0^0 = 1.
And the only mathematically functional reason why we should consider that 0^0 = 1 is a logically unsupported extension of the simple postulate that any other value to the power of zero = 1, therefore zero to the power of zero = 1.
In my view 0^0 should simply be considered undefined.
We actually have conflicting postulates –
1. Any value to the power of zero = 1, therefore zero to the power of zero = 1
2. Zero to the power of any value = 0, therefore zero to the power of zero = 0
The only “reasonable” course is to decide that 0^0 is undefined.
Glenn,
1.
You are talking about limit.
Limit is not a necessary consideration.
2.
0 to only positive power = 0
not to power of any value.
Do you think 0 to negative power = 0?
3.
Define 0^0 = 1,
then x^0 = 1 for any x.
It is very good in mathematics.
It is reasonable.
I’m still not seeing the reason for lack of acceptance of my perfectly viable proof. I don’t need to divide by zero, it is an incredibly simple equation, and can be solved like any other.
X^0=N
X=NX
0=(N-1)X
0/(N-1)=X
Whenever N≠1, X must be 0.
The Cool Dude,
You hypothesize N≠1 in advance.
If not, how can you divide by N-1?
The Cool Dude,
You ignore the case that N=1,
then it becomes 0=0*X
“0^0 is not defined by the majority of mathematicians,
however, it does not mean it is better to do so.”
‘better’ is an opinion, not an argument or reason.
“On the contrary, it is ridiculous.
It is the result of overemphasis of continuity.”
‘overemphasis’ is an opinion, not an argument or reason.
the value of 0^0 is indeterminate whether you LIKE it or not… after all, liking something is an opinion as well, not an argument or reason.
Ron,
The value of 0^0 is not indeterminate in itself,
but defined as indeterminate.
It is wrong to take it for granted.
Defining it as indeterminate is also an opinion.
However, in all cases except limit, you have to use 1 as its value
whether you admit or not.
0^0 begins the Fibonacci sequence of creation and a natural computer made of nothing at all… with all in every part… calculates to infinite precision all values possible. thus we live in zero dimension not even time it is a mirage.
Daniel Hazelton Waters,
Sum of nothing is 0.
Product of nothing is 1.
It is good explanation of 0! = 0^0 = 1
In all cases except limit,
you have to use 1 as the value of 0^0.
How about a graphical solution?
When y = x^0, then the graph of this line is y = 1. Looking at this line, we easily accept that for every x on the x-axis, y = 1. The question then boils down to this – is this line continuous for every x, including that most special x, x = 0?
If there is a logical reason to consider y = x^0 discontinuous, then the discussion continues. But if there is no logical proof that y = x^0 (a.k.a. y=1) is discontinuous, then that special case, 0^0 = 1.
I lean towards the latter.
That’s nice, I look at this article wanting an actual answer and all I get is opinions from different students, teachers & mathematicians? What’s the actual answer – no, really, what does 0 to the power of 0 equal? Thank you!
I’ll be checking back at times.
Or just give me a direct answer at:
dontemailthisaddress@hotmail.comI’m just joking.
I’ll check back.
Hopefully.
Okay, I’ll post my actual email address.
NO SPAM PLEASE.
bubblegumjess123@hotmail.com
I expect an answer there.
I don’t think I have enough time to check back here all the time.
I have a life too.
It’s a matter of definition.
It’s irrelevant to right and wrong.
It’s irrelevant to correct and incorrect.
It’s irrelevant to true and false.
Overemphaszing continuity and defining 0^0 as indeterminate is an opinion.
Not overemphaszing continuity and defining 0^0 as 1 is also an opinion.
There is not actual answer.
Whatever value 0^0 is defined, there is no continuity.
Wattever value 0^0 is defined,
you have to use 1 as the value of 0^0 in all cases except limit.
0^0 = 1^0 = .618033…^0 = 3.14…^0 = (PI*PHI)^0 = One ultimate hyper analog qubit of all possibility interacting with itself in my opinion. If you look at all it would be one nothing/everything.
Or maybe it starts with zero an absence of anything at all because it is the most economical way to explain creation. But simultaneously it creates it’s opposite one or all but they are equal 0^0 = 1^0. This is actually the fundamental unit of a quantum computer. I believe it may act like infinite qubits because of the number of states in between 0 and 1 is indefinite.
(0^0)^2=0^(0*2)=0^0
0^0=0^(-0)=1/0^0
Therefore 0^0=1.
It is much easier to understand than these abstract concepts.
Yes zero over zero equals one. But one what?!? All of nothing is not nothing it is made of.
Daniel Hazelton Waters,
Zero over zero is indeterminate,
Zero to power zero is one.
What do you mean by asking one what?
What answer do you want?
Daniel Hazelton Waters:
Seems that you’re asking for the meaning of the base vs the meaning of the exponent, as in 2^3 (the meaning of 2 vs the meaning of 3), which is easy to grasp at the gut level.
Except, in our case, we’re working with two very special numbers, or more precisely, two absences of value: 0^0. The absence of value raised to the absence of value, equals a value of one – what does it all mean? Is this what you’re asking?
The absence of value raised to no value equals a value. That is what I am trying to get at what is this value consisting of?
Daniel Hazelton Waters,
0 is a value, not no value.
Daniel Hazelton Waters,
How do you explain 0! = 1?
Let f(x) = exp(-1/x^2), and g(x) = x^2
as x–>0 from either side we have x^2 –>0 thus g(x) –>0
and
-1/x^2 –> -oo and thus f(x) –>0
then f^g is of the form in limits of 0^0
BUT f(x)^g(x) = exp (-1) = e^-1 AND NOT 1