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.
Yee,
Dividing by zero is an operation. Operations are never wrong.
0^0=0/0 is not a definition. It is a fact.
The Cool Dude,
You are wrong.
0^0=0/0 is a definition, a bad definition.
Operations need proving depending on definitions.
You say operations are never wrong.
(a+b)^2=a^2+2*a*b+b^2
Is this operation right?
Yee,
That is a replacement, not an operation.
(a+b)^2
Is an operation, where in you muliply a+b by its self.
Regardless of that, yes, your replacement will always be at least partially correct.
The Cool Dude,
If (a+b)^2=a^2+2*a*b+b^2 is replacement,
what are operations?
(a+b) to (a+b)^2 is an operation.
Operation implies a potential change in the actual value of that being operated on.
Replacement is switching an operation or set of operations with an equivalent operation or set of operations, as to simplify, expand, or otherwise translate to an alternate but equivalent form.
The Cool Dude,
You think x^2 is an operation, that is never wrong.
If multiplication is undefined in some cases, x^2 is wrong.
Yee,
Math is never wrong,
Only the mathematician can be wrong.
It is not math that lacks definitions, it is the mathematician.
If the mathematician lacks definition, they aren’t trying hard enough.
The Cool Dude,
And then?
Are operations never wrong?
Is defining 0^0=0/0 good?
Is defining 0^0=0/0 good?
CSP: It’s not a definition. That’s The Cool Dude’s point. It is a result from doing an operation. As I have said before, just because the result doesn’t exist as a point on the complex plane of numbers, does not automatically invalidate the operation.
Yee has tried to define a result in which operations simply do not support. That definition is therefore incorrect. The Cool Dude correctly points out that when valid operations produces results that are difficult to comprehend, that an expansion of meaning needs to be done, not an invalidation of the operation that produced the result.
We are never going to get anywhere until we can agree on that. Once we do agree on that, then there is no need to invalidate an operation. There is only the task of trying to understand the result.
Finally, since definitions are not required, the concept of “good” or “bad” definitions are unneeded.
CS Prof,
It’s a matter of definition.
Operations depends on defintions and can be wrong.
0^0=0/0 is not a good definition.
Cool dude of math is never wrong and the mathematician is wrong where do i go wrong?
1+1=x
1=x-1
1^2=(x-1)^2
1=x^2-2x+1
0=x^2 -2x
x=2+/- Sqrt of 2^2 -4(1)(0) all divided by 2
x= 2+/- Sqrt of 4 all divided by 2
x= 2+/- 2 all divided by 2
x= 2 OR 0
So….. 1+1 = 2 or 0
It’s a matter of definition.
Operations depends on defintions and can be wrong.
0^0=0/0 is not a good definition.
CSP: Let’s try a different tack. Tell us why Operations depends on definitions. Maybe I can agree with you if you make a case for it.
Yee,
Is 2*3=6 a definition, or an answer?
CS Prof
(n+1)! = (n+1)*n! is valid for n >= 1
Define 0! = 1, then 1! = 1*0!
(n+1)! = (n+1)*n! is valid for n >= 0
If 0! is not defined as 1, 1! = 1*0! is wrong.
It is not definition that is wrong.
It is operation that is wrong.
The Cool Dude,
2*3=6 is the result of definition of multiplication.
Denote combination number as C(m,n).
C(m,n)=m!/n!/(m-n)!
It depends on the definition 0!=1.
If 0! is not defined as 1.
C(m,n) =
m!/n!/(m-n)! when 0<n<m
1 when n=0 or n=m
This is not wrong, but inconvenient.
Therefore 1 is the best definition of 0!.
The Wonderer,
Your entire point could be compressed down to
x=1
x^2=1
x=±1
x=1 OR -1
Without losing any meaning. Putting it through a wider variety of operations doesn’t actually change the main idea, it just makes the fallacy harder to identify.
This is not actually a fallacy, however, because “or” does not mean the same thing as “and”.
So to answer your question, “where do i go wrong?”,
You didn’t. x equals 0 or 2 is 100% accurate. x is in fact one of the two.
Yee,
Is 0^0=1 the result of the definition of exponentiation, or just a definition?
Yee,
Convenience in mathematics is incorrect.
Incorrect is inconvenient.
“Close enough” is convenient.
0^0=1 is not close enough.
The Cool Dude
It’s a definition and a part of definition of exponentiation.
Yee,
If it is only partially based on the definition of exponentiation, then what is the remainder based on?
The Cool Dude,
http://en.wikipedia.org/wiki/Exponentiation
Yee,
It is much more convenient to express correctness in your definition, rather than to define to fit a select few scenarios, which you just so happen to use more often. When you use those scenarios, you can treat 0^0 as 1, but for the vast majority of mathematics, it is incorrect.
Convenience applies to situations. Defining 0^0 as 1 for all situations simply because it fits most of what you personally deal with is extremely inappropriate.
If it is still a question if 0^0 is actually indeterminate, consider that the log base zero of all non-zero finite numbers is only zero, and n to the power of the log base n of x has at least one possible solution of x, and so 0^0 must be all solutions x to ln(x)/ln(0)=0, which is all x.
The Cool Dude,
Definiton is irrlevant to corect or incorrect.
Definiton is irrlevant to right or wrong.
Definiton is irrlevant to true or false.
log(0) is undefined.
Calculating with log(0) is wrong.
Yee,
Definition is relevant to convenience
Convenience is relevant to proximity to correct
0^0=1 is nowhere near correct when dealing with such things as logarithms in exponents.
log(0) is defined.
It is defined as 1/0
Which is further defined as the reciprocal of zero.
0^0 is undefined.
Calculating with 0^0 is wrong.
To both of you,
You are both arguing about the definition on undefined value. Often in math it is necessary to define and undefined object for convenience of bookkeeping in theorems. Don’t over think it on either side. As pointed out above, the same thing happens with undefined limits. Ie, a 2 dimensional function may have existing limits from both sides but if these are not equal the total limit is undefined. Your arguments between whether it is the definition or operation at fault is analogous. You must choose which is appropriate based on case.
Voice of Reason,
You are only overemphasizing limit.
If the limit does not exist,
function value can still be defined.
Limit is not a good reason not to define.
Except limit, 0^0=1 is valid for all cases.
The Cool Dude,
1/0 is undefined.
Nor is ln(0).
It is generally accepted.
Yee,
1/0 is the reciprocal of zero.
It has a definition because there is a method both to get, and to return this value to and from the real plane.
1/0 is no different, definition wise, as √(-1).
The only practical difference, the only reason it tends not to be defined, is due to the confusing and seemingly infinite nature of it, so it is a difficult concept to grasp by those who are not exceptionally mathematically disciplined, of which the both of us are.
General acceptance is a pointless concept because it relies on the opinion of the masses.
The topic is “What does 0^0 equal”
Not “What to define 0^0 as”
For certain situations, it is perfectly reasonable to define 0^0 as 1, and in fact would be extremely inappropriate to define it differently, but otherwise the value of 0^0 is a complete set of all real numbers, and further definition of 0^0 relies on further qualifiers based on how you obtained it.
For instance, taking the area from -∞ to 0 under the curve (|x|^x)*(log(|x|)+1), the integral of which is |x|^x. The actual value of 0^0, when x=0, can be treated as its limit, 1, because the limit of the area is in essence the exact value of the area anyway.
The Cool Dude,
Division is the inverse operation of multiplication.
So, 1/0 is the only solution of 0*x=1.
But this equation has no solution.
What 0^0 equals depends on definition.
Limit is a part of mathematics, but not all.
It is not a good reason not to define.
Mathematics has many other areas.
All other reasons support to define 0^0=1.
If you want to discuss 0^0 only in limit,
please prove discontinuous functions don’t exist.
Yee,
Equations tend to have an infinite number of solutions, but in most cases, the vast majority of these solutions are sets. One of these solutions will always be 0^0, regardless of how many extraneous solutions the equation has.
Most equations have at least one solution which is a single number. For instance, x^2=1 has three main solutions. One solution is 1, a second solution is -1, and a third solution is a set of both 1 and -1.
In mathematics, you tend to want to isolate all the solutions which are not sets, otherwise, equations like x=1/x would be impossible to solve, because it would have an infinite number of two value set solutions, where the first value could be anything, and the second value, the reciprocal of the first.
“x^0 is only 1” is a crude generalization.
“x*0 is only 0” is also a crude generalization.
“So, 1/0 is the only solution of 0*x=1.
But this equation has no solution.”
First: 1/0 is the only non-set solution.
Second: You *just* said the solution. It has no real solution, but still a solution.
“What 0^0 equals depends on definition.”
Who’s definition? What is the definition based on? Is there any logic to this statement? You keep saying this, but it means nothing to me without explanation.
“All other reasons support to define 0^0=1.”
You mean the case by case basis that is your only argument, which is the exact last thing you should ever use to conclude anything in mathematics?
“Limit is a part of mathematics, but not all.
It is not a good reason not to define.”
“If you want to discuss 0^0 only in limit,
please prove discontinuous functions don’t exist.”
I have literally never once used limits to conclude anything ever other than things that already rely on limits. I am not using limits now. I used limits to solve ln(0), which is a function that already relies on limits. I did not use limits before that.
Stop talking about limits.
0^0=0/0 for a completely absurd number of actual mathematical laws, and I have proven it far more times than any sane person has the patience to follow through with.
Your only argument against this is “x^0=x/x is invalid for x=0” with no support or reasoning. Please stop declaring things without explanation.
0^0=0/0, and is a complete set of all values, real or otherwise.
The Cool Dude,
What equations determine the value?
It’s not difficult to list equations which has many solutions.
But what on earth does it mean?
Defining √(-1) is to confirm.
Defining 0^0 as indeterminate is to negate.
How can you negate 0^0 in the very beginning?
This attitude is irrational.
0^0=0/0 is a bad definition.
Proving or disproving a definition is ridiculous.
You did not prove anything.
Yee,
“It’s not difficult to list equations which has many solutions.
But what on earth does it mean?”
Equations tend to model real world scenarios. If I throw a ball up in the air, it can be represented as a parabola, y as height, x as time. If I solve that parabola for a certain y value “h”, I will tend to get two real results. That means it goes “h” units high, two different times. That’s what multiple solutions mean.
“Defining √(-1) is to confirm.
Defining 0^0 as indeterminate is to negate.”
Defining 0^0=1 negates all other values of 0^0. Your statement is backwards.
“0^0=0/0 is a bad definition.”
0^0=0/0 is 100% correct. It may be inconvenient, but it is correct.
Considering (a+b)^16 when a is 1000 and b is .001, a+b is very close to just 1000, but defining it as 1000 for convenience is wrong. If your function doesn’t expand to:
b^16+16*a*b^15+120*a^2*b^14+560*a^3*b^13+1820*a^4*b^12+4368*a^5*b^11+8008*a^6*b^10+11440*a^7*b^9+12870*a^8*b^8+11440*a^9*b^7+8008*a^10*b^6+4368*a^11*b^5+1820*a^12*b^4+560*a^13*b^3+120*a^14*b^2+16*a^15*b+a^16
You are wrong, even if every term, hold a^16 is very, very small. You are wrong. If you intend to approximate, then okay, but you are still wrong. And for the record, the difference between 1000^16 and 1000.001^16 is about 1.6*10^43
“Proving or disproving a definition is ridiculous.”
I define 1+1=7, because in some certain infinite sum or two, it is “convenient” to do so.
If you can not disprove the definition, and I don’t need to prove it or show any support at all, what do you do?
Actually, what was said earlier is correct, mathematicians just chose zero to have certain properties so that it would fit into the structure. Some ancient societies didn’t have zero most likely because they were not able to define it practicle terms (Sumerian maths). What zero should be in maths is an old argument and is likely to continue. It all rather depends on what we consider to exist because as humans we can understand what we cannot classify. If we don’t know which class to ‘put something’ in we get confused. But we are inventive and sometimes went invent a ‘ class’ just to be able to move on. We all know that the are zero Unicorns but having said that, what I do I say about the British Coat of Arms?
Yee, I’m not a mathematician, but I look at it a slightly different light.
Imagine you are going to construct a number system. Your first job is to make certain assumptions about the symbols you intend to use. Suppose you decide to include ‘0’ so that every other symbol multiplied by it, is changed to ‘0’. Every other symbol added to ‘0’ remains unchanged. These are your two starting definitions. It is then up to you define the other operations so that the first two are not violated.
Then what would 0^0 be? It wouldn’t be 0/0; your first definition takes care of it because an index is simply a shorthand for a single symbol multiplication so the answer has to be 0.
The answer to the next question, what is ‘symbol’^0, would depend on how you define your law of indices, which must follow your first two definitions.
The important point is that all your symbols would be abstract entities themselves used only by you to represent some reality.
If only the philosophical issues were as simple. For example if ‘0’ or ‘1’ etc doesn’t exist, what are those things on the blackboard?
In what law does 0^0=0/0. You guys said index laws earlier but when looking that up none of the laws applied to that.
y = (1/x)^(1/lnx)
limit as x tends to infinity = 0^0
however if we graph the function y = (1/x)^(1/lnx)
and evaluate it geometrically
limit as x tends to infinity y = (1/x)^(1/lnx) = 0,3679
limit(x–>infinity) (1/x)^(1/lnx) = 0^0=0,3679
Check it out, a different argument 🙂
The Cool Dude,
List the two equations to determine 0^0:
(0^0)^2=0^(0*2)=0^0,x^2=x
0^0=0^(-0)=1/0^0,x=1/x
The only solution is 1.
Therefore, 1 is the only reasonable definition of 0^0.
These two equations model real world scenarios of 0^0.
To define is to affirm(or hypothesize) it is a single value.
Defining √(-1) is also to affirm it is a single value, isn’t it?
Definition is irrelevant to correct or incorrect.
Proving or disproving a definition is ridiculous.
You say you define 1+1=7.
Addition has its definition.
If you define 1+1=7, how can you define other parts of addition?
As a result, this definition brings nothing more than trouble.
LarryD,
I prefer defining power zero as “identity element of multiplication”.
Jake D,
Limit is a part of mathematics, but not all.
If limit exists, it is natural to define it as the limit.
If not, it does not mean it is better not to define it.
And if there are other ways to define it,
don’t deny it because of limit.
Limit should not be overemphasized.
Yee,
0^0 is 1, but it is not only 1.
1/x=x^2=x has one solution that is not a set, and it is 1, but it has an infinite number of solutions when x is a set.
If x were to be a set of all numbers, it would still be correct, which means your point is invalid, because it can be either.
You keep saying to prove discontinuous functions don’t exist. If discontinuous functions are proven to not exist, then this means that the limit of a function must equal that value at that point, correct? Then if functions have multiple limits from multiple directions, in must be all of those limits, correct? If discontinuous functions are proven to not exist, then all functions equal all of their limits?
The Cool Dude,
Hypothesize 0^0 has a single value is reasonable.
It is ridiculous to say 0^0 has many values in the very beginning.
Can I say 0! has many values?
If a functions has many limits at a point from different directions,
the limit does not exist.
Yee,
Square roots have multiple values
Inverse trig functions have multiple values
All solutions to polynomials have multiple values
Logarithms have multiple values
Integrals have multiple values
I don’t see why it’s such an impossible concept for you to grasp, that things with multiple solutions and different values are everywhere in mathematics.
You can pretend the multiple solutions don’t exist
But the multiple solutions can pretend that you don’t exist too.
“Can I say 0! has many values?”
Yes, actually you can. See, you define 0! as one for the exact same reason you define 0^0 as one, you just don’t realize it. 0! shares a few properties with the function of x^x, one is that both x^x and x! approach 1 as x goes to zero, and between x=0 and x=1, it curves down below y=1 at a slow rate briefly, before ascending at an astronomical rate after it passes the point (1,1). However, there is a second property that x! shares with x^x. They are both indeterminate at x=0. For ALL values of x!, there is at least one solution on either side to always support x!=x*(x-1)!, even if x<1. When considering negative factorials, it is a little known fact that:
(-x)!=πx/(sin(πx)*x!)
Given this, it is immediately apparent that the factorial of all negative integers is simply 1/0, though for negative non-integers, there is a real solution. However, relevant to 0!, this in combination with the law that the factorial function is founded on proves that 0!=0*(-1)!, which given the previous observation of negative factorials, results in 0/0. Once again, you are substituting the indeterminate form as 1 because it is convenient to you.
"If a functions has many limits at a point from different directions,
the limit does not exist."
Maybe not by your definition, but your definitions have been wrong before.
How else is a function supposed to approach a set anyway?
The Cool Dude,
You list some examples that have multiple values.
But they are not defined to have multiple values in the very beginning.
As for 0^0, it is not good to define it to have multiple values.
1 is its unique reasonable value.
All the problems of ‘0’ really have their root in the question, ‘is 0 a number?’ If 0 is not a number then it doesn’t belong in a Number System. If it IS a number then what does it represent (in the sense that 1 may represent 1 apple, 1 car, 1 person…)?
Set theory tells us that the Natural Numbers must follow a sequence that ‘counts’ naturally from the smallest onward; that each number must combine with each other number to give ‘something new’ although remaining in the universe of counting numbers. For example, 1+2 = 3, 101+104 = 205, etc. But 0+1 = 1,0+5 =5, fails here because ‘something new’ is not produced!?
However, the Peano Axioms 0 is ASSUMED as a natural number.
If one follows Aristotle then Quantities have no Contrary so clearly here there is another problem with Zero.
So like it or not there are and will remain, problems with including 0 in any system. Mathematicians decided to use 0 and that it should have certain properties so as to make the system complete. The rest of just need to accept this and move on.
Fundamental to exponential notation is the base and the exponent. A base represents “something” as opposed to “nothing”. If we select 0 as our base, then we must look at 0 as “something” even if we define it as “the absence of anything else” or as “the presence of something unrecognizable”. When we do this, we have no problem allowing 0^0=1 (just like any other base raised to zero). This is not a matter of convention or simple definition, this is just one of the demonstrations of how Philosophy underlies Mathematics. You may wish to call it “Definitions and Assumptions” rather than “Philosophy”, but when extrapolated, it all comes down to the fundamental challenge of rectifying the concepts of discrete properties and the continuum. When applied to physical reality, this is manifest in the search for the elegant equation.
Yee,
“But they are not defined to have multiple values in the very beginning.”
Who cares what they are in the beginning. What matters is what they are in the end.
“As for 0^0, it is not good to define it to have multiple values.
1 is its unique reasonable value.”
No opinion is worth anything unless backed with enough genuine information.
This is all the more true when dealing with a subject as objective as mathematics.
If you want factual truth, 0^0 is a set of all real numbers, and a function at that point can only be defined for other intersecting curves. Factual truth is some times applicably pointless.
If you want real world applicability, then you should actually treat 0^0 as its limit. It’s most common and simple limit is 1, but it has been shown numerous times to be able to have a limit that is anything. Real world applicability is some times factually wrong.
Anything other than the factual truth or real world applicability, is both pointless and wrong.
The Cool Dude,
It is irrelevant to true or false.
What is the factual truth?
You said 0! can also be a set.
Well, is 0!=1 a factual truth?
You are just overemphasizing continuity.
The most intuitive and natural understanding of power zero
is identity element of multiplication.