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.
Zero actually is only a representation of no value. This problem is not solvable. No value to the no value power can not equal value. Energy cannot be created or destroyed only transfered. Proving creative intelligence,”GOD”; the great mathmatician exist.
If we look at all value equals energy. Tangible measurable energy. And 0’s value as the absence of energy, 0 to the 0 power can NOT EQUAL ANYTHING OTHER THAN 0. Energy can not be created or destroyed only transferred. No energy to the no energy power can’t equal energy.
chad cole,
0 is a value.
0^0=1 is reasonable.
Actually, Chad…x/x is the same thing as x^0(division is the subtraction of exponents), which obviously equals 1, but only when x≠0. When x=0, we get 0/0, which is undefined.
Ok, even though this is a mathematical conversation, and has nothing to do with physics, philosophy, or otherwise related topics, I will still respond to that statement with something that equals its value.
We are all balls.
Technically, that is not nothing, it just doesn’t contribute to anything useful in the entire universe. Zero is this way. It doesn’t do NOTHING, it just doesn’t do SOMETHING. Sure, MAYBE I was referring to the fact that we are all made of ball-shaped cells, or ball shaped atoms, or ball shaped protons neutrons and electrons, or ball shaped quarks, but when you boil it down, all I really said was “We are all balls”, and there is nothing you can do about it.
So, in short, 0^0 is the universe’s/god’s/(Insert all powerful entity here)’s way of trolling us.
Ron,
You are wrong.
x^0 is different from x/x.
Actually Yee,
Not one single mathematically valid demonstration of 0^0=1 has been made by anyone. In my limit analysis approaching infinity demonstration one page ago, it was demonstrated using valid math that it could equal anything between zero and one. However, at no point was it ever able to equal zero or one within the confines of that demonstration…
Even your binomial example didn’t quite hit the nail on the head. The binomial formula you talked about isn’t a formula that evaluates numerically. It is a formula that evaluates to another formula where the exponents are used to determine how many terms are within a particular portion of the equation. ie x^0 isn’t meant to mean 1, it is meant to mean that there are zero x’s in that particular portion of the equation.
0^0=1 is unreasonable. It gives wrong answers.
Cool dude, when you say “Well, it depends on how you define U. If you define U as equaling 1/0…” We are still left with 0 * 1/0. zero times undefined does not equal one. As much as I found your “definitions” list of ‘U’ amusing, the rest of the math world doesn’t share your enthusiasm for defining math rules for working with undefined. This is the same kind of logic that has been used to “prove” that 1=2. You were correct to say that kind of math is invalid in your original post on this “proof” of yours.
1. Multiplying both sides of an equation by zero leads to invalid proofs.
2. Using 0/0 expressions lead to invalid proofs.
3. Using 1/0 expressions lead to invalid proofs.
4. Using invalid functions that can return infinite values lead to invalid proofs.
5. Using made-up axioms leads to ridiculous proofs.
Your proof used ALL of these!!
Ron,
I have to emphasize again and again.
It is a matter of definition.
It is irrelevant to right and wrong.
It is wrong to prove it.
0^0=1 is reasonable, convenient, useful in many fields.
And limit is not a necessary consideration.
Ron, if we are going to explain 0^0, but we can’t move it around at all without the equation being instantly wrong, what else is one to do.
I will just give a simple representation of how easy it is to say why 0^0=1.
0=-0
0^0=0^(-0)
0^0=1/(0^0)
(0^0)^2=1
0^0=1
With this, I kept the expression to itself. I didn’t take it apart, or divide by zero, or any of that. If this is wrong too, then I don’t know what else to tell you. All I did was point out that negative zero is zero, and that x=1/x at one.
cool dude, Your error is here:
0^(-0) does NOT equal 1/0^0… (0^0)^(-1) would equal 1/0^0.
As for 0=-0…That is about as meaningful and useful as 0=25(0).
Yee,
You are right… It is a matter of definition…0^0 is defined as indeterminate.
Ron,
0^(-0)=1/0^0 is reasonable.
Defining 0^0 as indeterminate is unreasonable.
Defining 0^0 as 1 is reasonable.
Ron,
What about defining 0! as indeterminate?
Ron, 0^(-0)=(0^0)^-1. The exponents multiply. They do so in the same way that:
(2^3)^2=2^6
2^3=8
8^2=2^6
64=64
The argument isn’t that 0^0 ONLY equals 1, but rather that it is the explanation that makes the most sense. For instance, if someone asked “What is the square root of four” you would probably say two and be done with it, when in fact, negative two is also a correct answer.
The simple fact is, X^0 always has at least one answer, and that answer is one, regardless of what X actually is. The only difference with zero is that 0^0 might also have other answers, but 1 is still the most useful for it to be.
We can prove that X^0=1 like so:
X^0=1
X^0*X=1*X
X^1=X
X=X
All we do is multiply by X, increasing the exponent on X from 0 to 1, and multiplying 1 by X to get X. X=X, therefore, X^0=1.
The only difference is, say for instance I say that:
X^0=-1
X^1=-X
2X=0
X=0
This shows that X^0=-1 at X=0.
I can also show this being true for five:
X^0=5
X^1=5X
4X=0
X=0
Or just about any other number in existence:
X^0=16777216
X=16777216X
0=16777215X
X=0
So 0^0 really only is as intermediate as the square root of four is.
The definition of negative is less than zero. The definition of positive is greater than zero. Zero is neither positive nor negative. -1 *0 does NOT = -0… There is no such thing as -0. It equals just zero.
In the end, your previous proof demonstrated that 0^0=0^0 and nothing more.
Multiplying both sides of an equation by zero ends all proofs. If it didn’t, we would end up with 1=2 proofs.
Dividing by zero ends all proofs… If it didn’t we would end up with 1=2 proofs.
As for your last demonstration….The flaw is clear when you view your proof in reverse order….
x=0
4x=0
x^1=5x fine so far…zero equals zero….
x^1=5x^1
x^1/x^1, which is x^(1-1), or x^0=5 Which is exactly where you started…
Unfortunately, we already know that x=0, which means that the equations is:
0/0=5
We can not ignore the rules of math in proofs. 0/0 is indeterminate, as is 0^0.
Basically all it boils down to is this:
N=X^0
N*X=X^1
(N-1)X=0
X=0
I multiply by X, then subtract X, then divide by (N-1).
Unless (N-1)=0, I don’t divide by zero, so X=0.
What if (N-1)=0, you might ask? That means that N=1. What if N=1 you might ask?
1=X^0
1X=X^1
X=X
The only way we get incorrect answers is if we divide by zero. Multiplying by zero does nothing wrong. Only division. Since the only instance of division by zero is when N=1, and when N=1, we get an identity, there is nothing wrong with this.
The real issue at hand is this:
(-2)^2=4
2^2=4
2^2=(-2)^2
2=-2
The point here is, working an equation with multiple answers backwards is what REALLY gets incorrect answers, not dividing by zero.
The moment you say that 0/0 or 0^0 is intermediate because you can prove that 1=2, I can just as easy respond by saying that 4^.5 is intermediate, because I can prove that 2=-2.
Nobody ever said that the square root of 4 is indeterminate other than to say that it equals 2 OR -2. However, any proofs involving the square root of 4 require considering both possible values… Math is perfectly fine with it having 2 values, just as it is fine with 0^0 being indeterminate because it has an infinite number of values.
Saying that 0^0=1 ignores the fact that it doesn’t always equal 1. In fact, nobody here has even demonstrated one single instance where it does….
0^0 is indeterminate. If you want to insist that it isn’t, prove that it is ALWAYS 1.
Ron,
You have made a mistake,
that is trying to prove it.
We must focus on one thing:
which definition is better.
If 0^0 is defined as 1,
it will not equal any other values.
It doesn’t particularly make any sense to say that 0^0 is intermediate just because it has MORE values. Though I do understand what you are saying. But it’s quite easy, you see. It isn’t that it needs to always ONLY equal one, it just needs to always equal at least one, in the same way that the square root of four is always both 2 and -2, but never only one of them. That would be silly. 0^0=1, because:
0*-1=0
0^0=0^(0*-1)=(0^0)^-1=1/(0^0)
1/(0^0)=0^0
1/1=(0^0)^2
1=0^0
Or because:
N=X^0
N*X=X^(0+1)
N*X=X
N*X-X=0
(N-1)X=0
X=0/(N-1)
For all N≠1, X=0 in N=X^0, or, in other words, 0^0=N.
For N=1, however:
1=X^0
1X=X(0+1)
X=X
0=0
X=0
For a solution on N=1, we get X=X, so any number that equals its self, then raised to the power of zero, can get 1. Since 0=0, it is possible for X=0, so it is possible that 0^0=1.
It’s not as if I’m saying 0^0=1 so much that it is the only solution, but it is that I am saying that it can, in the same way that I can say 4^.5=-2. If this is what you want, however, I will express 0^0 as a domain:
0^0=(-∞,∞)
Here it is expressed as a domain. It spans all real and hypothetical planes of existence. But when you multiply all of those solutions, you get a roundabout answer of (-1, 1), because all of the reciprocals cancel out. It just so happens that 1 is the multiplicative identity, and fits the X^0 theme, so it makes sense to say that 0^0=1 as the dominant answer.
All of the sense. All of it.
Cool dude,
X=X
0=0
X=0
fails because you multiplied each side of the equation by zero, destroying the identity of x. It doesn’t work to say that you subtracted either because:
x=x
1x=1x
1x-1x=1x-1x
0x=0x
0=0
further more….
0=0 doesn’t follow to x=0 in your last step because x is no longer apart of the proof.
cool dude,
N=X^0
N*X=X^(0+1)
N*X=X
For all N≠1
fails because it is obvious that
N*X=X leads to
N=X/X
N=1 (For all N≠1)
Which is clearly false and breaks your proof.
Ron,
1) Saying X=X is the same as saying all solutions. Since 0=0, 0 is allowed to be X. Same goes for 1, 2, 3, 4, 5, and any other number that equals its self.
2) We already know that X=0, and you can only break my proof if 0/0=1, and then you would be agreeing with me in the first place, and that breaks your entire argument.
3) N≠1 is a conditional. The conditional is merely a tool to prove a fact WITHOUT exception. I later came back and also proved the exception with an identity.
Yee,
I am perfectly fine with your comment: “We must focus on one thing: which definition is better.” So the real question becomes, What criteria are we using to determine which definition is better?
Do we wish to use the appeal-to-authority logical fallacy? That is, so-and-so said its true, therefore that should be good enough for all of us?? If that is the case, I definitely want to know who that person is, and even then, I want to know what their argument actually is. Did they REALLY say that 0^0=1 or did they say it is only useful to think of it that way for very specific applications? Also, how do we deal with the possibility that multiple ‘experts’ may say different things? How do we compare the disagreements??
Do we go with the idea that it is useful in some cases?? i.e. binomial theorem? If that is the case, It must be pointed out that in that case, the binomial theorem doesn’t evaluate numerically, it evaluates to an equation which can then evaluated numerically. Also, is this a problem with how we define x^0 or how we have worded the binomial theorem? A simple fix is to remind people that y^n means that there are n y factors in that portion of the equation as apposed to having a numerical value… in other words, x^3(0)^0 in a polynomial generated from (x+0)^3 by the binomial theorem would mean 3 x’s and zero zeros, or x^3. This isn’t a stretch either. X^n actually does mean n x factors.
Is the criteria that it makes our life easier the best route?? To that end, why not just define pi=1?? Now we don’t even need a calculator to solve circumference equations. Does it matter that it is correct??? Not by this way of thinking, but it leads to the last idea….
And finally, should our definition be actually true and correct?? Is that the best criteria for determining what the best definition is?? To that end, 0^0 simply does not equal 1 no matter how much we would like it to and pi does not equal 1 no matter how much I would love for it to when I don’t have a calculator handy. Naturally, this is the criteria that I think is most important. After all, what is the value of a definition if it leads to demonstrably incorrect solutions??
Cool dude,
0*-1=0
0^0=0^(0*-1)
(0^0)^-1=1/(0^0)
fails because anything multiplied by zero looses it identity and becomes unavailable for the next step in your proof. Step two instantly becomes 0^0=0^0, which is trivial, and given that you are starting from the premise that 0^0 is unknown, there is no logical reason to jump to the conclusion that any unknown value equals it’s inverse.
but, it fails for an even better reason…
x^a / x^a = x/x = x^(a-a) = x^0, which no matter how you slice it means that when x=0, x^0=0/0 (no matter how many times someone may wish to pretend that it doesn’t) and as such, is not valid for use in any proof. Expressions that evaluate to undefined can not be used in proofs. That is how we end up with 1=2 “proofs”.
This, of course, is why all “proofs” that it equals one will be found to have errors, and that reason is simple: 0^0=1 is not true.
Actually, if Pi were to equal anything but Pi, it would equal four.
https://www.askamathematician.com/2011/01/q-%CF%80-4/
But the fact of the matter is, I’ve already proven without fault that 0^0 is at LEAST all numbers that AREN’T one, so why should we dismiss my explanation for 0^0=1 as well? It has no problems, and multiplying by zero never gets incorrect results unless you’ve divided by zero already.
Cool dude,
When you say”Saying X=X is the same as saying all solutions. Since 0=0, 0 is allowed to be X. Same goes for 1, 2, 3, 4, 5, and any other number that equals its self.”
that means:
x=4
x=x
0=0 using the rationale that x=x means every solution possible leads to:
x=0
0=4
Clearly false.
When you say” N≠1 is a conditional. The conditional is merely a tool to prove a fact WITHOUT exception. I later came back and also proved the exception with an identity.”
this fails because the last step in your “proof” was:
X=0/(N-1)
if N ultimately equals one, then you are saying x=0/0
yet you said in the beginning of that proof:
N=X^0
N*X=X^(0+1)
N*X=X
from which it was clear that N=1 from n=x/x;
Of course, if x=0, you just proved exactly what I have said all along: n=0/0 and 0^0=0/0 or 0^0 is indeterminate.
Ron, you are making the mistake of comparing multiple solutions. 0=4 just as much as 2=-2. Dividing by zero is reckless. Comparing multiple solutions is incorrect.
Furthermore, N does NOT ultimately equals one. N is the independent variable, and X is the responding variable. N is whatever you want it to be, X responds to what N is.
For all N≠1, X=0
For N=1, X=X
All solutions.
X=X means all solutions. If it did not, then that means that X≠X. I do not want to live in a world where X≠X.
Cool dude, it looks like we are in agreement when you say: “For all N≠1, X=0” with respect to N=X^0.
0^0≠1
Ron,
Polynomial and binomial theorem are good reasons to define 0^0=1.
What are the reasons that make mathematicans not to define it?
The reason is limit.
But is it a good reason?
What matters to define a value at a point where the limit does not exist?
Defing 0^0=1 is much better.
And whether 0^0 is defined or not,
nothing incorrect will happen.
Ron,
(0^0)^-1=1/(0^0) is a reasonable expectation.
An expectation does not need a proof.
Whether it is correct depends on how we define 0^0.
If we define 0^0=1, it is correct.
If not, it is incorrect.
x^a/x^a=x^(a-a) is invalid when x=0.
It is not a good reason not to define 0^0.
It is irrelevant to right and wrong.
No proof is needed.
What is your reason?
You say “0^0=1 is not true.”
Is 0! = 1 true?
Ron,
The definition of π is
the ratio of a circle’s circumference to its diameter.
Therefore you cannot define it as 1.
Nor can you define it as 3.14159 or any other value.
Yee,
You asked the question: The reason is limit. But is it a good reason?
This is a great question, because it forces us to ask another question. When does 0^0 ever come up in real math?
The short answer is, it really only comes up when we are dealing with limits.
Does it happen with the binomial theorem? A better question is why should it? After all (x+0) may look like a binomial because it is in the form of one, but is it really a binomial (or should we consider it to be a binomial) given that 0 is the additive identity? (x+0)^3 simplifies to x^3 before we even consider applying the binomial theorem. This naturally means that we wouldn’t apply the theorem at all.
Of course, when we apply the binomial theorem to the more abstract (x+y)^2 to get the new formula x^2+2xy+y^2, we still never have to deal with 0^0, even if y is latter determined to equal 0 because y^0 in the first term has fallen out of the equation already.
What is more reasonable? defining 0^0=1 when it is clear that the only times 0^0 is ever really encountered in an unavoidable way is in circumstances where that definition is the least useful and often wrong, or having a more strict definition of what a binomial is that forces simplification, would never be wrong to do so, and entirely eliminates the need to define 0^0=1 when it is clearly not correct to do so?
I am perfectly fine with definitions, just as long as we focus on the definitions that are at the real root of the problem. Perhaps we should focus on the definition of a monomial? Rather than define it as: “A monomial is a variable, a real number, or a multiplication of one or more variables and a real number with whole-number exponents” We could define it more rigidly as:
A monomial is a variable, a real number, or a multiplication of one or more variables and a real number with whole-number exponents, where the domain of it’s factors does not include zero.
Note that I said domain, not the range. That means ‘X’ can be a monomial, even if it might be zero as long as it isn’t defined as such. It’s domain is still (-∞,∞). Zero would not be since the domain of a constant is the constant itself. Also x(a-a) would not be referred to as a monomial either since the factors are 0 and x.
This isn’t to say that (a-a) or adding 0 aren’t a valid steps in a proof, they are just not valid elements of a monomial, and as such, not valid elements of a polynomial when it comes to applying definitions to those constructs.
Would this more rigid definition cause other problems??? I don’t know off hand, but given that the over all effect is to demand valid simplification of expressions before moving on, I doubt it. Even if it did, it would probably only prompt trivial changes in other definitions that would go without controversy.
Why should we do this?? It would put to rest the 0^0 question. It would no longer be a problem to acknowledge that it is indeterminate.
Actually, I worded that definition wrong…
It should be “where the domain if it’s factors do not explicitly equal zero.” or something like that….
this may just be a waste but whatev
consider the matrix: [x y] (^) [x y] = [x z]
[ x y] [x y] [z x]
(cross multiply)
(2xy)^(2xy)= x^2+ z^2
z=(((2*x*y)^(2*x*y))-x^2)^(1/2)
graph equation on google search and set ranges x(-1,1) y(-1,1) z(-1,1)
make z= 0 and so that
x^2=((2xy)^(2xy))
now make y=o
x^2=(0^0), x is indeterminate therefore 0^0 is to
Thoughts?? or i am just dumb for trying
[x y] (^) [x y] = [x z]
[ x y] [x y] [z x]
ignore the matrix above in last comment, the spacing did not work out
Patty,
Thanks for the input, but I think we’re roughly agreeing on a value. Or rather, values. 0^0 is all real numbers because of the preposterous amount of math that I don’t want to type again, but you can read it. It’s all up there.
The issue isn’t particularly what 0^0 is, at this point, but rather, what it is not (Which is nothing), and, of all the values, what is the most practical to use. 0^0=1 is probably the most practical, because it follows the x^0 pattern, and it looks nice in most equations.
I personally like 0^0=3, because it ruins graphs and makes me laugh.
Ron,
It’s a matter of convenience.
In binomial theorem,
(x+y)^n=
n
ΣC(n,k)*x^(n-k)*y^k
k=0
If 0^0=1,
x,y can be any complex number.
n can be any nonnegative real number.
If not,
x,y can only be nonzero real number,
n can only be positive real number.
Obviously, defining it is much more convenient.
I have to say it again:
It’s irrelevant to right and wrong.
By the way,
0! can also be indeterminate.
Yee,
Definitions are fine. We already agree that some people have assigned it that value strictly for the reason of convenience. A better scenario, however, would be to update our definitions across math in such a way, that the “reason of convenience” becomes no longer necessary.
The issue that still must be dealt with is why not be more precise with our definition of a monomial and avoid the obviously false statement of 0^0=1 in the first place. We still end up with a useful binomial theorem for all the reasons you just stated, we just avoid the absurd idea that (x+0) is a binomial since it so readily breaks down to a monomial.
Like I said before, I am fine with definitions. But if we are going to discuss definitions as absolute, then we must be willing to deal with all definitions in a way that is serious about dealing with the inconsistencies that have crept into mathematics by poorly written definitions.
As for 0!=1…. As un-intuitive as it may be…and it definitely seems absurd!!
I am perfectly fine with it equaling one because (n-1)!=n!/n, where n=1 proves that it is, and furthermore, it is consistent with the definition of the factorial operator and invokes valid math in it’s proof.
However, 0^0=1 is not consistent with the definitions and rules of exponents. In fact, if it weren’t for x^0=x/x , where x≠0, I would also find it counter intuitive that x^0=1, where x≠0. It would be more intuitive to think it is nothing, since it literally means zero x’s. But, valid math does show that it is what it is.
This leads to another possible “definitions” solution to 0^0 nightmare…(not that I prefer this solution, but it does demonstrate that there are alternatives to insisting on 0^0=1, while still keeping the rest of math useful.)
We could simply define a new operator for use with equations similar to the binomial theorem… I will call it the “explicit exponent”… in the form of: x^ (some symbol that isn’t taken….) The way this would work is that it has no equality associated with it. it simply means N ‘x’ terms, where zero ‘x’ terms means that ‘x’ doesn’t exist and whatever the other factors of the monomial are, they remain because they do exist.
Definitions ARE important, but they don’t relieve us of the obligation to ensure that they are consistent with the entirety of math.
Proving 0!=1 with (n-1)!=n!/n is wrong.
In fact, 0! must be define as 1 to prove (n-1)!=n!/n .
x^0=x/x is invalid when x=0
x^0=1 is valid for all numbers.
These are not proofs.
If you still think about right and wrong,
if you still try to prove and disprove,
You have not yet understood it.
Ron,
Binomial theorem are good reasons to define.
You shouldn’t say 0^0=1 is a false statement.
There is nothing right or wrong.
There is nothing true or false.
No proof is needed.
Yee is right. No proof is needed. And even if it was, I already made one.
All rationalism systems (systems built on logic and axioms) require their definitions to be deduced, ultimately, from their axioms.
In other words, definitions are NEVER based on convenience or appeals to authority. A properly constructed rationalism system will always be 100% internally consistent and true within itself. Any system that is not consistent within it’s own reality ultimately owes that inconsistency to improperly deduced(invalid logic) definitions and conclusions.
As long as one instance exists where 0^0≠1, it can NEVER be defined as 1. There is no such thing as a “reasonable” definition. It is either true in ALL cases or it is false.
Yee, when you say “x^0=x/x is invalid when x=0” You are using the wrong word: “invalid.” 0^0=0/0 is a valid conclusion. However, It can only be at the end of a proof, not in the middle of a proof as the basis of proving something else. i.e. 1=2 proofs.
If 0/0 was not allowed at the END of any proof, we would never be able to prove that ANYTHING was indeterminate. 0/0 is a valid RESULT of a proof.
Ron,
In what case, 0^0≠1?
Limit is not a necessary consideration.
Defining a value at a point where limit does not exist
is not wrong.
According to what reason do you say 0^0=0/0?
0/0 is not allowed,
but 0^0 is irrelevant to 0/0 .
Yee,
After additional thought on 0!, I agree with you that it is undefined, but not for the same reason.
In order to get from:
N!= N(N-1)(N-2)…2*1
to
(N-1)!=N!/N Where both sides are divided by N
Requires that the domain of N for the entire proof be [1,∞) to include the definition. Sure, by consequence of the math, 0!=1 when N=1, but the domain of the definition is tied to [1,∞) meaning that the result is meaningless.
It is good to define 0!=1 and 0^0=1.
Yee, You asked “In what case, 0^0≠1?”
When we consider f(x)=e^(-1/x) and g(x)=x;
We know that the limits of those functions as x->0 are both zero:
lim(x→0) f(x)=0
lim(x→0) g(x)=0
lim(x→0) f(x)^g(x)= (lim(x→0) f(x))^(lim(x→0) g(x)) = 1/e
lim(x→0) f(x)^g(x)= 0^0 = 1/e
PAY VERY CLOSE ATTENTION to the fact that the right side of the equation is no longer a limit evaluation. It is a straight forward 0^0 equals something…. In this case, it equals 1/e.
Ron,
Limit is not a necessary consideration.
Defining a value at a point where limit does not exist
is not wrong.
Besides, defining a function value which is different from its limit
is not wrong.
Because f(a) does not necessarily equal
lim f(x)
x->a
Limit is not a good reason to criticize 0^0=1 .