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.
So rather than carve out exceptions to some of our favorite theorems and definitions, let us consider the following formulation in algebra textbooks (at least going forward):
x^0 ≡ 1 (for any real number x)
0^x ≡ 0 (for any positive number x)
I leave it as a challenge to any remaining quibblers to show how the first definition above leads to any contradiction by rigorous logical reasoning.
1) The indeterminate form applies to categories of limits, not values.
Consider the limit
lim(h->oo) (1+x/h)^h
This is a limit of the form 1^oo. The answer is e^x. This does not mean that 1^oo is e^x. Similarly, lack of a unique limit for 0^0 does not imply that 0^0 is undefined.
(Hmmm, I thought I’ve read a number of times that limits are always unique!)
The “indeterminate form” 1^oo is simply being used as a sensible symbol or label for the category of limits where the base approaches 1 and the exponent approaches oo as the limit is taken.
2) Even if you still insist on using limits, you can only do that if the function is continuous. Why does 0^x need to be continuous? It’s already a problem for x<0, and probably for imaginary numbers, too.
3) These are definitions. As I've asked before, how do you derive y^x = e^(x ln y)?
Well, you'd like things to obey the exponent laws. So you show that the log definition does so. But consider y^m y^n = y^(m+n) for even roots.
For 4^(1/2) you can say it's 2 or -2. Indeterminate!!! You sensibly omit the negative root and get on with things. Physics is quite happy with it too.
(Furthermore, my analysis book (_Introduction to Analysis_ by Rosenlicht) says it is defined, not derived.)
For 0^0: To satisfy the exponent rules, 0^0 has to be 0 or 1. We sensibly throw out 0, because 1 is immensely useful and 0 isn't. Power series, mappings, derivative of x^n, and more. It's 1. 0^0 = 1 works with all of these, and 0 doesn't. It's 1.
Why is one any less intrinsic or derivable than the other? They're really just definitions.
4) As Howard pointed out, what is the sense of saying, "Whenever it comes up, substitute 1, but it really is undefined"?
5) Please read the posts by Howard Ludwig, betaneptune, and me and let us know where we went wrong.
Regarding calls to close this thread:
I welcome arguments showing where Howard Ludwig, betaneptune, I, and now STEPHAN H. have gone wrong.
I also welcome new arguments. Please, please, please — no more division by zero arguments. I’ve answered that too many times. Thank you.
Lacking either, I’m pretty much done. (I reserve the right to reverse this!)
@Alan Feldman
When I suggested that it’s getting to be time to wrap up the discussion, I didn’t mean officially closing it to any further postings. I just meant that we all seem to be making the same points and stating the same arguments over and over again, and I don’t see much point in doing that. But there’s always the possibility someone will bring up a previously unconsidered aspect of this, or a new contributor may have something to say.
=========================
@STEPHAN H. has written:
“Let us agree, for the sake of convenience and epistemological honesty, that it is axiomatic (a useful definition and not a theorem) that 0º ≡ 1”
*If I am interpreting you correctly*, then we may have something we can actually agree on. I agree that, for most of the situations where 0^0 comes up, it is convenient to consider it to be 1. Calling it “a useful definition and not a theorem” to me implies that we’re defining it to be 1 as a matter of practicality rather than claiming that it can be rigorously proven to be 1 just as surely as a^(1/2) can be rigorously proven to be the square root of (a). If that’s more or less what you mean, then we’re basically in agreement, at least concerning that statement.
This is getting a bit circular for those involved, and difficult to follow for everyone else.
For the sake of casual readers, I’m going to close this to comments that rehash the same ideas. If anyone would like to make their point again, I’ll consider short, clear comments with explicit, directly supporting, linked references.
The definition of x^n is quite clear when n is a positive integer. Saying x^n = exp(n log x) for x > 0 and 0^n = 0 for n > 0 is a matter of “extending” the definition of x^n. So these are not proofs, derivations, or theorems. Reference:
108.243.42.146:8080/get/pdf/1039
pp. 130-131
That puts them on an equal standing with 0^0 = 1, except for the controversy about it.
so what about
0^1/0^1
This is technically not possible because zero exponent one is zero; you cannot divide by zero.
So by extension, 0^0 should be undefined because it is equivalent to 0^1/0^1
or 0^1-1.
Zero is non-scalar and an empty place-holder. Not an empty set though, because it holds only one value zero. Thus, when you exponentize it to powers greater than one, it still evaluates to zero.
Below one; it seems like things get undefined. I heard once that 0 is like a point singularity. Please explain as simply as you can.
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I have proved that 0^0=2
0^0=0^m-m=0^m/0^m
Since anything to the power 0 is 0
Hence,0^m/0^m=0/0=100-100/100-100
=10^2-10^2/10(10-10)
=(10+10)(10-10)/10(10-10)
=10+10/10
=20/10=2 Hence 0^0=2 (proved)
A clever bit with that proof of 0^0 = 2 is that along with 0^0 = 1, and 0^0 = 0, it actually lets you define it as any value you like. If you can do it for 2, you can do it for 3 and every other positive integer, using a bit of induction.
See also, this method for all real numbers greater than 0:
log_b(0)=-inf
log_b(0^0)=-inf/-inf=1
b^1 = b so for every value of b, you get 0^0 = b
Note that log_b(x) = log_a(x)/log_a(b) and if b<0 then you get complex results.
Added these just for a bit of fun.
For a, b, c real numbers:
By definition a quotient a/b =c to be correct, two conditions must be satisfied.
1) a = c*b (c times b)
2) c must be unique
0^0 is the same as 0/0, so
0/0 = c, it satisfies the first condition because 0= 0*c for any real number c.
Obviously , the second condition ( the uniqueness) is not satisfied because c could be any real number to satisfy the first condition.
Therefore, we should not say that 0^0 is undefined, but we must say that 0^0 is “INDETERMINATE “. we say undefined when the first condition is violated, like for example A/0 ( A divided by zero) for A different from zero is undefined.
A similar situation when you try to solve a linear system of two equations of two unknown such as:
2x + 3y =5 and 4x + 6y = 10.
the solution (x,y) of that system is defined but not unique.
However, the system 2x+3y =5 and 4x+6y = 12 will be undefined, that will be equivalent to violating the first condition above but in two dimensions.
In the context of fields, it is more common to define division of a by b as the product of a by the multiplicative inverse of b. In a field there is a multiplicative inverse of b if and only if b is not the additive identity. For the ordinary arithmetic on real numbers, the additive identity is 0. This is where the idea that division by 0 is undefined, because the multiplicative inverse is not defined (does not exist) for 0. It does not matter whether the dividend is 0 or nonzero–division by 0 is undefined.
Your use of a/b being c if and only if c = ab is an alternative way of looking at things, but it is not so clean and easy to apply in the context of general fields, not just the ordinary arithmetic on real numbers. In any case, the problem of a multiplicity of possible answers for 0/0 is referred to as not being “well-defined” in this context; indeterminance is the term used in the case of ambiguity only in the context of limits.
You have made a totally unjustified assertion that 0^0 = 0/0–where you want to go with that is that 0/0 is undefined (or at least not well-defined, depending on your approach to division), so 0^0 must have the same issue. However, you cannot validly reach that conclusion because you cannot even get past the starting line. The only rationale for regarding 0^0 = 0/0 is based on an erroneous application of a division law of powers. The use of b^x / b^y = b^(x – y) can be done only if you completely avoid any reciprocal of 0 or division by 0 anywhere in that equation. Therefore, you cannot validly use that formula for b = 0 and either y > 0 or y > x. (Because in actuality 0^0 = 1, you can use it for b = x = y = 0 because 1/1 =1.
Here is another example that might more clearly show you the false logic:
0¹ = 0;
0² = 0 × 0 =0;
0^4 = (0²)² = 0² = 0;
0³ = 0^(4-1) = 0^4 / 0^1 = 0/0, which is undefined.
Since 0³ is the same as 0/0, 0³ must be likewise undefined.
However, 0^0 is no more identical to 0/0 than 0³ is.
Your example of dependent linear equations is related to the question of what values of x satisfy 0 = 0, and perhaps the value of 0/0. It has nothing to do with the question at hand pertaining to 0^0.
0^0+0^0+0^0=3 that,s not true and. refuse
I,m don,t thinking that 0^0=1
@Shuvam dey:
“0^0=0^m-m=0^m/0^m”
This is where your proof falls apart. 0^m=0, thus 0^m/0^m=0/0 which is indeterminate. I could just as easily say 0^m/0^m=0^2m/0^m=0^2m-m=0^m=0.
Here’s my argument why 0^0=1. Consider the following probability problem:
You have an opaque bag with n balls, each a different colour. You reach into the bag and extract a random ball, then place it back in the bag. Then, once again, extract a random ball from the bag and place it back in. You repeat this until you have made n extractions. What is the probability that you never extract the same ball more than once?
On your first extraction, there are n balls to choose from and n of them have not yet been extracted, so the probability of extracting a new ball is n/n=1. On your second extraction, there are still n balls to choose from, but only n-1 of them have not yet been extracted, so the probability of extracting a new ball is (n-1)/n. On your nth extraction, there is only one new ball left because all the rest were previously extracted in some order, so the probability of extracting a new ball is 1/n. If everything went according to plan, you extracted a new ball every time; the probability thereof is the product of the probabilities on each individual pick, which is n/n * (n-1)/n * … * 1/n = n!/n^n.
So, for example:
5!/5^5=120/3125=24/625=0.0384
4!/4^4=24/256=3/32=0.09375
3!/3^3=6/27=2/9=0.22222…
2!/2^2=2/4=1/2=0.5
1!/1^1=1/1=1
What about when n=0? If you have an empty bag, continuing this pattern would mean the probability of never drawing the same ball more than once is 0!/0^0. But since you have no balls to extract in the first place, the probability of never drawing any ball more than once is clearly 100%! Which means 0!/0^0=1 and since 0!=1, we have 1/0^0=1 and thus 0^0=1.
One thing that no one has brought up is the actual value of zero. Zero is a term that is used to signify in between 1 and -1. The easiest way to explain it would be for someone to find the decimal value above and below 0, but to do that it continues on infinite, so you could say that 0 actually is infinite and if you take infinity and take it to the power of infinity then you can say that 0^0=infinity, not 1 or 0.
The correct value of $0^0$ is 1.
Consider the set of natural numbers: 0, 1, 2, 3, …
They are cardinals of finite sets.
Let us consider two non-empty finite sets $A$ and $B$ whose respectiva cardinals are $a$ and $b$, and compute the cardinal of $A^B$, that is, the set of functions having domain $B$ and codomain $A$.
As is well known, this cardinal number is equal to $a^b$.
Now, it is enough to consider what happen when the cardinal of $A$ or $B$ is 0.
If the cardinal $a$ of $A$ is 0, and the cardinal $b$ of $B$ is not 0,
it means that $A$ is empty and $B$ is not empty.
Then, the set $A^B=\emptyset^B=\emptyset$ has cardinal 0,
which is consistent with the fact that $a^b=0^b=0$.
(*) Proof: The set $A^B$ has to be empty, otherwise, it would have an element $f$ having domain $B$ and codomain $A$; and then, there would be elements $\alpha\in A$, $\beta\in B$, such that $\alpha=f(\beta)$, which is not possible since $A$ is empty.
If $a>0$ and $b=0$, it means that $A$ is not empty and $B$ is empty,
so that $A^B=\{\emptyset\}$, since the empty set is the empty function having domain $\emptyset$ and codomain $A$.
In this way, the set $A^B$ has exactly 1 element, which is consequent with the fact that $a^0=1$ for $a>0$.
Finally, let us consider the case $a=0$ and $b=0$.
In this case, $A$ and $B$ are empty sets.
Again, we have that $A^B=\{\emptyset\}$, since the domain $B$ is empty,
which allows the existence of the “empty” function.
We do not obtain a contradiction as above, since the argument used in (*) used that $A$ is not empty there. But here, $A$ is, indeed, empty.
Thus, the cardinal of $\emptyset^\emptyset$ is 1,
and we can assert that $0^0=1$.
I’d like to take a different approach. We know conventionally that 0 does not equal 1. I want to use a few different notations here and show that 0 and 1 are completely opposites yet are of the same entity and by claiming this, I’m not saying that 0 equals one in value but 0 and 1 are equal in the fact that they are congruent and of the same object. In order to do this we will need to use some vector notation, some set properties and a specific number system.
For vector notation we can state that 0 represents the 0 vector and only contains an arbitrary position without magnitude and the value of 1 represents the unit vector which has magnitude. The direction of the vectors in a sense are irrelevant here but in some sense will come into play later.
For properties of sets we can use the fact that 0 will represent the empty set and that 1 will present the full set. This will support the final concepts after we combine these definitions with the use of a specific number system which I will describe below.
For the number system we are all familiar with decimal where it is defined by having a base of 10 as there are 10 digits in this number system that range from [0,9]. We know that when we take a number in decimal such as 420 that each digit has a weight of its digit value within the column or digit placement of the number. In this case 420 = 4 * 10^2 + 2 + 10^1 + 0*10^0 = 4 * 100 + 2 * 10 + 0 * 1 = 400 + 20 + 0 = 420 . We can represent the decimal number system by value placement of log 10 in which I will denote it as L:10. This is not the number system that I’m going to use.
We can see from above that L:10 is the decimal number system so we can state that L:2 is the same as Log 2 which is Binary. I’m not using the Log function to evaluate any numbers or values here, I’m only using the properties of Logs to represent number systems. So L:2 would be binary L:3 would be trinary, L:4 would be quad … L:8 octal, L:16 hexadecimal and so on… Binary is the most simplest and smallest of all number systems as it only has two digits being 0 and 1. You can not have a number system with only 1 digit for if you did no operations could be defined.
What makes binary so special is that almost all computations done in computer hardware is represented by binary values. It is gives rise to Boolean Algebra and the properties of Logic such as True and False statements. So here we can represent a single digit as being either 0 or 1 but not both at the same time. Within digital circuits a single memory cell either it be a SR – Latch, D – Latch, D – Flip Flop, JK – Flip Flop etc. can only store a single bit at a given time. In the circuitry when electricity flows through the transistors that make up the logic gates that are used to build these circuits if the voltage is high say close to 5 volts and above a certain threshold value we consider it to be high, on, closed gate or true and is a bit value of 1. When the voltage is below a certain threshold or close to 0 volts we say it is low, off, opened gate or false and is a bit value of 0. Anything in between is considered a floating voltage or i/o but we are not concerned with that here. We are only concerned with the two state system.
So a binary digit (bit) can typically only ever be in one state or another at a moment in time depending on the voltage across the wires and what voltages are allowed through the gates. When the bit stores a low voltage it is in the 0 state and when it stores a high voltage it is in the 1 state. The states are opposites states. If we look at one of the logic gates an inverter it will invert the voltage from low to high or high to low. This proves that they are opposite states.
Knowing this we can say that the bit is the empty set or the 0 vector and has no value nor magnitude when the switch is in the off position, and we can say that it is the full set or unit vector and does have value and magnitude when the switch is in the on position. So congruently 0 and 1 are opposites but are mutually of the same thing. As in the case of the switch (transistor) within an electrical circuit specifically a digital circuit the switch has two states that are opposing states and both 0 and 1 are properties of that same switch or that same memory cell.
The reason I chose this approach is because I decided to use a real world application to demonstrate this duality of 0 and 1. They are complete opposites yet are of the same entity. It’s kind of like a coin with heads and tails. This duality doesn’t just exist here, but is at the foundation of all things. In Chemistry + and – charges (protons and electrons), temperature hot and cold, acids and bases, anions and cations… In Physics magnetism + and – poles and Opposites Directions North and South or East and West or Up and Down, Newtons Law: Equal and Opposite forces. In Logic True and False. In Biology X and Y chromosomes, Life and Death. In Politics and Governments Democrat – Republican, or Democratic Republic vs. Communism or Socialism. In religious faiths Good and Evil. In Nature Day and Night or Winter and Summer and so on.
This duality of 0 and 1 is very important to completely understand so now that we can see 0 and 1 in this manner, we can go back to the idea of any number system and look at the ones digits of any and all number systems from the base being any integer that is greater than or equal to 2. b >= 2 for any log b number system.
How does this help us with 0^0? Well it kind of doesn’t because if we have a number system with 0 digits in it where the base is 0. There is no number system at all however what is any value of any number of x that is taken with log of base 0… Well we know that through logarithms and exponential forms that log base 0 is undefined due to the fact that b^x = 0; x does not exist. So this should include the indeterminate form of 0^0 because we can not have a base of log 0.
There is ambiguity here because 0^0 can be interpreted as being either 0 or 1 depending on the context that it is being used, but in general it is indeterminate and has ambiguity. This goes back to concept of the digital circuit with voltage going through a transistor that has a voltage that is too high to be 0 and too low to be 1 that we consider in Electrical Engineering to be a floating voltage of input or output as it hasn’t settled into a defined state. So my conclusion is that 0^0 is both 0 and 1 simultaneously and within general mathematics this is something that most people can not wrap their heads around nor accept because they are always limited by the boundaries of their own limits and are not willing to look outside of the box to the see the entire box! 0 and 1 are mutually the same thing but in different states! Consider a set when it is empty it evaluates to 0 then consider that same set when it is full and it evaluates to 1. At the end of the day it is still the same set. However if we didn’t know what state that set was in and we didn’t observe it then we can say that the state of that set is floating or ambiguous and it can be in either or both states. This phenomenon has been seen with the state of electrons through quantum mechanics!
An edit or addendum to my original posting, I did neglect to state that 0^0 could also be either -infinity or +infinity but this would require some advanced calculus involving complex numbers, infinite sets and series, and extreme limits, linear algebra and vector calculus of multi variables and more… which is beyond the context of 0^0. However, since 0^0 is indeterminate we can also conclude that 0/0 is also indeterminate because I like to think that in some way 0^0 == 0/0. Just something to think about. So in truth both 0^0 and 0/0 can equal 1, 0, +/- infinity thus they are all ambiguous and therefor they are called indeterminate! Kind of like Super Position within Quantum Mechanics – Quantum Physics!
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Pablo1977 makes a good argument using the cardinality of sets.
0⁰ = any number!!!!!!!!!!!!!!!!!!!!!!!
Consider an equation
2x=x⋯⋯⋯⋯⋯⋯⋯⋯①
If we divide the whole equation by x, we get 1=2!!!!! (contradiction!!!!!!!!!!!!!!)
But if we subtract x on both sides, we get x=0.
Now, Substituting in ① , we get the true value. But if we divide the whole equation:
2 x 0 = 0 by 0, we get
2 x 0/0 = 0/0⋯⋯⋯⋯⋯⋯⋯⋯②
2 x 0 = 0. So, ② changes to
0/0 = 0/0!!!!!!!!!!!!!
Considering 0/0 as an unknown value, 0/0 has any value. Here is why:
1 x = x. Then value of x is any number. Similarly, 0/0 is any number.
I have yet another example, which I have not posted before. This is _new_ content.
In quantum mechanics, the wave function for a single-electron atom contains a factor
(2 \kappa r)^L
where r is the radius (distance from the nucleus) and L is the angular momentum.
r can have any non-negative value. L can have any non-negative integral value. (\kappa is a positive real number)
When r = 0 and L = 0, which is 100% legitimate, and if you define 0^0 as 1, everything works out nicely! No need to specify a formula for this special case! Just like in the series form of e^x. For e^0 to work, you need 0^0 = 1.
NOTE: The angular momentum number is normally represented by a lowercase l, but for clarity in reading this post, I changed it to an uppercase L.
Unfair!! What makes this right? Is it was put here by someone else. I don’t want to take a survey anymore. Going on 7 years and the survey is the same???? Make it stop!!!! I’m not a fuckin victim, but I do not engage. That’s all I have. Please leave me alone. Just because you’re bored, you can’t fuck with. It’s wrong !!!!!!!!