
0^0 - hansy
http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/
======
rickhanlonii
It's very important to note here that 0^0=1 is a _shorthand_ and not a
_truth_.

Mathematicians are absolutely not stating that they have proven, or that it is
true, that 0^0=1. It is a definition, not a claim of equality. They're not
saying "0^0 _is_ 1" in the sense that they say "1+1 _is_ 2" or "0.999... _is_
1". They're saying "we define 0^0 to be 1". The difference is more than just
pedantry, it strikes at the core of why mathematics is the most powerful tool
for determining truth that humanity has either discovered or invented (hat tip
to anyone familiar with that debate).

One of the cold, austere, beauties of mathematics* is that if you do not
accept a definition, you can reject it as false and reason with the result. To
give the traditional example, if you accept as true that two parallel lines
never intersect, then along with the other 4 of Euclid's Axioms you can prove
all of Euclidian Geometry (what you learn in high school). If you do not
accept it as true (an explicitly accept it as false), then you can prove all
of Hyperbolic Geometry (one form of non-Euclidian Geometry).

In the case here, you're free to reject the convention that defines 0^0 as 1
and reason with the result; __you will not break any mathematics __. But you
should know that mathematicians have never run into any issues with this
convention--or can handle it trivially when they arise--so you 're only adding
a lot of work for yourself.

When you really think about it, it fills you with awe. It's awesome.

*“Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.” -Bertrand Russell

~~~
mathgrad
This is not true. A very natural way to define a^b is as the number of
functions for a set with "b" elements to a set with "a" elements. In this case
there is exactly one function from the empty set to the empty set, and we have
proven 0^0 = 1. This is no different than defining addition and then proving
1+1 = 2.

~~~
Twisol
I prefer thinking about 0^0 = 1 as an empty product
([https://en.wikipedia.org/wiki/Empty_product](https://en.wikipedia.org/wiki/Empty_product)),
since it generalizes nicely to any operation with an identity element. That
is, if you apply any operation zero times, the result is that operation's
identity. It's interesting that the analogous empty sum, 0*1 = 0, is a
complete non-issue.

~~~
einhverfr
But here's your problem:

infinity * n = infinity, right?

0 * n = 0, right?

0 * infinity = ?

Ok, this is relevant here particularly because:

Lim 1/x as x -> 0 from the positive side is infinity, right?

So 0 * that is.....

lim 1/x as x -> from the negative side is negative infinity, right?

So 0 * that is.....

That's why 0/0 doesn't work as such. You don't know how 0 is derived or what
it means. If we have x^2/x, and take the limit as x -> 0, we get 0. If we take
x/x^2 and take the limit as x -> 0 we get +/\- infinity.

But that doesn't mean that x/x or x^2/x are not continuous functions, any more
than 1 or x are not continuous functions.

~~~
Twisol
You are conflating the limiting behavior of a function with the value of a
function. In (standard) analysis, there is no actual value called infinity -
it's just used to describe how a function behaves arbitrarily close to a given
value.

When you have a 0^0 limiting form (or 0/0, or 0 * infinity), the function's
behavior is indeterminate. The "0" and "infinity" you're looking at aren't
precisely 0 or infinity, but only arbitrarily close. The actual behavior of
the function depends on the expressions that approach 0 and infinity, hence
`x/(x+1)` and `x/e^x` having different limiting behaviors as x increases
without bound.

But if you are literally considering the function at a specific point which
produces a so-called indeterminate form, the answer is simpler. In some cases,
the function is undefined (e.g. 1/x where x=0). In the case of 0^0, there is a
precise value we can assign: 1. And this doesn't conflict with the 0^0
limiting form: the limit of a function at a point can be different from the
actual value of the function at said point.

A quote of Knuth, taken from Wikipedia
([http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_powe...](http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_power_of_zero)):

> Knuth (1992) contends strongly that 0^0 "has to be 1", drawing a distinction
> between the _value_ 0^0, which should equal 1 as advocated by Libri, and the
> _limiting form_ 0^0 (an abbreviation for a limit of f(x)^g(x) where f(x),
> g(x) -> 0), which is necessarily an indeterminate form as listed by Cauchy:
> "Both Cauchy and Libri were right, but Libri and his defenders did not
> understand why truth was on their side."

~~~
einhverfr
Regarding Knuth's point, that's a good one, since f(x) and g(x) are not
necessarily the same.

------
didgeoridoo
Students: Let's come up with some crazy proofs based on our individual levels
of understanding.

Teachers: Let's do it by the book and come up (somehow) with conflicting
answers.

Mathematicians: Yeah, sorry guys. We made it all up.

Pretty much captures most mathematicians I know.

~~~
Bahamut
Definitions in mathematics are used to make the language describing abstract
concepts elegant, as explained in the entry.

However, definitions aren't chosen all willy-nilly - there are good arguments
why definitions are adopted, as should have been seen in the article.

~~~
kazagistar
Sort of like how we use terse symbols for everything because ink and papyrus
is expensive.

~~~
marcosdumay
Not really. We have some very good reasons for using the definitions we use
every day. Those reasons are more related to us than our technology, although
our tech is getting so good that we already need to think about it. And by the
way, people changed almost all the usual definitions at the XX century.

Now, mathematicians do throw the usual definitions away all the time. It's
important to know when to reuse other people's coding, or roll your own.

------
stephencanon
The real problem here is that x^y is a single shorthand which refers to a few
fundamentally different mathematical concepts (which happen to have
significant overlap with each other).

First, it refers to a function f:C x N --> C, defined in terms of repeated
multiplication. f(x,0) is 1 for all x != 0, and so we adopt the convention
that f(0,0) is also 1.

But it also refers to a function g:C x C --> C, defined as g(x,y) = exp(y
log(x)), which has a branch cut on the negative real axis of the first
argument and an essential singularity at (0,0), and so g(0,0) is necessarily
undefined.

The value of 0^0 depends entirely on _what sort of mathematics_ one is doing
at the time, and therefore _which_ function one is referring to.

~~~
Sniffnoy
But if you define 0^0=1 in general, it doesn't cause a problem here -- that
definition never _disagrees_ with x^y=exp(ylog(x)), it just defines it at the
point 0^0, while the latter leaves it undefined. In other words, it's possible
to make a common extension of the two; they don't actually give _different_
values in any case.

Of course, doing this makes exponentiation discontinuous at (0,0), but seeing
as it already had an essential singularity there, this isn't really a loss.

~~~
stephencanon
Right, but when you're working with the function on C x C, it doesn't really
add anything either to arbitrarily pick 0^0 = 1, except for consistency with
that other function that happens to be written the same way.

I'm not really opposed to saying 0^0 = 1; it's the only reasonable choice if
we're going to insist on using the same notation for these two functions, and
I don't expect that's going to change.

------
SilasX
\- "0^0. Why? Because mathematicians said so. No really, it’s true."

\- [Detailed explanation of the tradeoffs involved in choosing different
definitions of exponentiation.]

So, it's not "because mathematicians said so", it's because of a deep review
of the tradeoffs of defining how exponentiation generalizes, the kind of thing
that mathematicians happen to study more than other identifiable groups.

~~~
dhaivatpandya
It is still a "said so" because it is a definition. Even if thought went into
the definition, it doesn't make any less of a definition.

------
yaks_hairbrush
A word from Knuth on the matter (warning: PDF):

[http://arxiv.org/pdf/math/9205211v1.pdf](http://arxiv.org/pdf/math/9205211v1.pdf)

See page 6.

~~~
davyjones
Are PDF warnings relevant anymore? Of late Chromium and Firefox display PDF
natively, sandboxed(?).

~~~
RexRollman
I expect it has more to do with filesize.

~~~
Dylan16807
Nobody says 'warning, images' and most pdfs are pretty lightweight. This one's
200KB.

------
Tyr42
I feel like in discrete mathematics (especially combinatorics), since we don't
use continuous functions, it's useful to say 0^0 is 1, along with 0! = 1, and
so on. Makes a lot of things around Binomial theorem and the like easier. I'm
not so sure if it's safe to use that when doing and calculus proofs or
anything along those lines, but there, you have more useful tools for dealing
with limits that might approach 0^0.

~~~
anonymoushn
0! really is 1 in a much more reasonable sense. The empty product is the
multiplicative identity. The continuous notion of ! also agrees:
[http://en.wikipedia.org/wiki/Gamma_function](http://en.wikipedia.org/wiki/Gamma_function)

~~~
gordaco
And it's far easier to prove: 1!=1, (n+1)!=n!·(n+1), thus 0!=1!/1=1.

------
gweinberg
Missing Q and A: But if mathematicians insist it is 1, why do high school
teachers act like they know more than the mathematicians do?

A: They don't. The statement that mathematicians uniformly say it is 1 is
simply false. My high school teacher had a PhD in math, I think it's fair to
say she was a mathematician. And yes, she said it was undefined.

~~~
joliv
I think it might be better to say it's indeterminant.

~~~
Sniffnoy
No, because there's no such thing. Either we can define 0^0 to have a value,
or we can not do so. Either way, there's no separate option "indeterminate"
that is different from "undefined".

Now, it's common to teach in calculus classes that 0^0 is one of the
"indeterminate forms", along with 0/0 and so forth, which, so the story goes,
is a different thing from being undefined, like 1/0\. Since, after all, if
f(x) approaches 1 and g(x) approaches 0, then the limit of f(x)/g(x) is
undefined, whereas if f(x) and g(x) both approach 0, then the limit of
f(x)/g(x) cannot be predicted in advance. So 1/0 is undefined, but 0/0 is
indeterminate.

Now this certainly is getting at a real distinction! But it's not a
distinction between the _value_ of 1/0 and that of 0/0; both are undefined.
"Indeterminate" is not some separate actual value. Rather, they are getting at
the distinction of the behavior of the division function _near_ the point
(1,0) vs. how it behaves _near_ the point (0,0). Not _at_ the points! _At_
both those points, the function is not defined.

And similarly with 0^0. Of course, it's pretty common to define that 0^0=1,
and it's a definition I'd agree with -- but this is not inconsistent with the
calculus teacher's statement that 0^0 is "indeterminate", because the latter
(once made sense of) is not really a statement about the value of 0^0 at all;
it's a statement about how the exponentiation function behaves _near_ the
point (0,0) (not at it; at it, it's equal to 1, or at least by my definition
it is, at any rate).

In short, there's no such value as "indeterminate"; the calculus teacher's
"indeterminate forms" (as opposed to "undefined"), while getting at a real
destinction, is not actually about the value of the function _at_ the point at
all.

------
olalonde
Are there any examples where the x^0=1 definition turns out to make other
definitions _more_ complicated to write down? For example, you'd have a
general definition and need a special definition for when you get an exponent
that equals 0.

~~~
xamuel
Suppose you want a compact, clever formula for the function: f(x)=1 if x!=0,
f(x)=0 if x=0

If 0^0 were defined as 0, you could write the above function as f(x)=x^0.

With 0^0 defined as 1, you're forced to write something like f(x)=1-0^|x|
(absolute values to avoid division by zero), a bit more complicated.

This is silly though, and of no importance anywhere.

------
vog
I find it a bit disappointing that the article considers only limits that
would justify "0^0 = 0" and "0^0 = 1". In fact, the termin "0^0" as a limit
can reach any value - the same way as "0/0" can reach any value.

Im stressing this because back in school, I really thought that although 0^0
is undetermined, it can only reach exactly 0 or 1, but nothing else. This is
of course wrong, but no teacher was able to tell me why. For some time I even
thought I found a new theorem and tried to prove it. Later, I told some math
guru about this, he thought about a minute, and told me two functions f(x) and
g(x) whose limit is each 0, but for which the limit of f(x)/g(x) is 2 (or any
other value, if you adjust f(x) and g(x) accordingly).

Having said that, in most cases "0^0 = 1" is a useful convention, especially
in a purely algebraic context when polynomials are involved.

------
tel
This question doesn't have an answer because, as this post suggests, the
meanings of the first 0, the second 0, and the (^) are overloaded. The reason
mathematicians don't mind the overloading is that they all have "kind of"
similar meanings.

So we can talk about a recursively defined function on naturals where it's
convenient to "define" 0^0 = 1. We can talk about various kinds of limit
techniques in the reals where 0^0 is not actually a value but instead a
shorthand for a particular kind of limit.

This probably forms the most interesting notion of what 0^0 means where we
define it as the limit of a particular kind of path in the complex plane and
then notice that 0^0 can take any value we choose---depending on exactly what
path was taken to get there. [0]

[0]
[http://en.wikipedia.org/wiki/Complex_logarithm](http://en.wikipedia.org/wiki/Complex_logarithm)

------
myhf
For the continuous use cases, it helps to look at a graph:

[http://www.wolframalpha.com/input/?i=y%3Dx%5Ex](http://www.wolframalpha.com/input/?i=y%3Dx%5Ex)

------
vlasev
This is exactly why you have things like 0! = 1, 0 choose 0 = 1, 0^0 = 1, the
empty sum is 0 and the empty product is 1, and so on. These are DEFINITIONS
and they make notation easier. They basically help the flow of mathematics.
It's often difficult to watch someone try to explain "intuitively" why some of
these things are the way they are and completely miss the point that they are
like this because they help make other things easier.

------
gameshot911
I have a problem with this part of the explanation:

>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 y^{0} = 1, which holds for any y. Hence, when y is zero, we have
0^0 = 1.

When y is zero, don't we have 0^0 = 1 x [y zero times]? Maybe I'm
conceptualizing it incorrectly, but I'm envisioning an empty space where y
would be, akin to an empty set. 1 times an 'empty set' is _not_ 1, it's one
empty set, which rubs me as another way of saying nothing/zero, not 1.

I know my language is imprecise and I'm probably describing empty sets and the
definition of zero incorrectly. The point is that the last step of his
explanation doesn't sit right with me. Just because Y exists zero times does
not mean you can just throw it out of the multiplication.

~~~
j2kun
Another convention is that the empty product is 1, and this is precisely what
is being said by putting the 1 out front.

------
textminer
Had a set theory professor who taught us that for the non-negative integers,
m^n was just the number of unique mappings from a set of cardinality n to one
of cardinality m. Ergo, for all sets A such that |A| = k, k^0 is just all
mappings from Ø, which is necessarily the one with empty image and pre-image.
So 0^0 = 1.

------
skrowyek
While high schoolers try to prove their own intuitions about their
understanding of exponents (intuition drilled into them through rote
learning), mathematicians just say "we defined it that way".

It speaks to the tragedy that is the high school math curriculum.

------
anonymoushn
I did not encounter this convention while working on my math degree. I am
surprised that none of the characters in the article said "0^0 is nothing, but
limits of the form 0^0 can be any nonnegative real number or infinite".

~~~
yaks_hairbrush
Yes, you did. It may not have been stated explicitly, but it was implied in a
couple places:

1\. Binomial theorem. Its statement does not conventionally include any
caveats about the x^0 case.

2\. D(x^n) = n x^(n-1). When n = 1 and x = 0, you get a 0^0 on the RHS. If 0^0
is taken to be 1, the derivative rule holds. Else, the statement gets messy:
D(x^n) = n x^(n-1) if (x,n) != (0,1)

------
quchen
That's a rather long text to say "it's an arbitrary -- and conveniently chosen
-- definition of a special case of the power function similar to how 1 is not
prime". I also think the presentation was chosen poorly: lots of wrong
information before the correct approach is presented.

Wikipedia is probably a better source here:
[https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer...](https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero)

~~~
forrestthewoods
I love Wikipedia. It's amazing. It makes the world a better place. I'm a
pretty decent programmer. I do video games so I do lots of 3d math. I'd say
I'm decent at that as well.

I _hate_ Wikipedia for math. Absolutely hate it. Unless you are a
mathematician by trade Wikipedia is damn near useless for learning new math
concepts. I don't even bother checking it anymore.

~~~
vlasev
I had an idea for a couple of years now of creating a wikipedia-style
mathematics textbook that will be crowd-sourced, standardized and cover all of
math in a way that's accessible to learn from on your own. It would have a
kind of a zoom function where you can expand details on explanations and
calculations to a depth that you prefer. Ideally this kind of thing would
start off with basic math and get progressively further into mathematics like
the roots of a tree.

~~~
marcosdumay
Your "basic" in basic Math means "foundational", or "what we teach to
children"?

~~~
vlasev
That's a good question. In the spirit of making learning easier it would
perhaps "start" with what we teach children. However, that might be ultimately
misguided because that would just mimic the curriculum and force the whole
thing into a box. But then again the math knowledge itself forces a structure
on any such textbook because to know A you need to know B, C, D, etc.

This is further complicated by the fact that most subjects in math can be
approached from multiple equivalent angles. For example, complex analysis can
be taught from the complex derivative perspective or from the power series
perspective equally well. Ideally the format of the whole thing will allow the
student to naturally learn things from the angle they understand better.

------
mxms
Interesting article. I ran across a problem yesterday that was similar to
P(x)^Q(x) = 1, which then asked me to find the sum of the solutions. I noticed
that both P(x) and Q(x) share a root at some a. But I realized, 0^0 is most
often defined as 1, and carried on. Checking the answer key later on showed
that they chose to neglect that a, and call 0^0 undefined. I'm not sure how I
really felt about it.

Note, I also forgot to check when P(x) = -1, assuming that Q(x) is even there
;P

------
haddr
I think that the most intuitive way to get the idea of why 0^0 = 1 is to take
the example from combinatorics. n^k is the number of distinct sequences for
the sampling with replacement and ordering (for example ball picking from
repository of n different balls and counting the number of distinct ways that
k balls can be picked and ordered - with replacement). I think that there is
only one way of ordering results of drawing zero balls from a set of 0
different balls :)

------
socrates1998
I love math, but I have never really enjoyed these types of debates.

I guess I have always been drawn to the application of the concepts in the
real world rather than the abstract beauty of it.

------
ryannevius
I understand the "math"...the numbers...the work on paper. But how does that
translate to something useful in the real world? That, after all, is what
useful math helps us do...solve problems for the real, tangible world. Saying
that 0^0 = 1 is a cool math game; but translate 0 into something in the real
world (i.e. nothing, none, etc.)...and trying to make something out of it
other than 0 or "indeterminate" starts to make less sense.

~~~
jomtung
From the article's example the simplicity of the binomial formula is extremely
useful compared to a formula that would have to account for the case where
k=0. Another commenter pointed out the useful elegance of 0log0=0 for
physicists. These are the real world applications for mathematicians choosing
definitions directly. Saying that an idea may be defined in many ways is
correct, but choosing a working definition for the system helps to apply the
definition to appropriate concepts. This is what mathematicians are doing when
choosing a specific definition instead of saying that any definition will
work.

One could argue the entire field of Real Analysis was formed because Calculus
showed the world that we didn't really have those definitions, but they were
needed. There are cases where the integral of the derivative does not equal
the derivative of the integral (violation of the fundamental theorem of
calculus) without having a specific epsilon-delta definition of limit.

Also, zero is not always the same as nothing or none. Zero is an abstract
number that some have decided is useful to represent nothing or an empty set,
but really comes from an abstract idea that you can count nothing and have a
number. This goes back to the fact that numbers are pretty useful ideas
regardless if you may consider one of them a function or not.

~~~
djur
Supporting your final paragraph: the concepts of nothingness, emptiness,
absence, etc. are all much older and more universal than mathematical zero.

------
analog31
Perhaps a related question: How should it be defined in a math library for a
programming language? Should it return 1, or throw an exception?

~~~
biot
Why is that even a question? It should return the correct answer, of course.

Edit: At the very least, that behaviour should be a configurable option for
those who desire something other than what most mathematicians accept as being
the correct answer.

~~~
my_username_is_
The whole point of this is that there isn't necessarily a single right answer.
As another user pointed out, mathematicians define 0^0 to be 1, but you don't
necessarily have to accept that as truth in the way that 1+1=2.

~~~
xamuel
If the calculator is powerful enough (as in the case of Mathematica) to
evaluate Taylor series, for example the Maclaurin series for e^x when x=0,
then in doing so it implicitly admits 0^0=1. If it simultaneously says 0^0 is
not 1, then the calculator is inconsistent. (Mathematica IS inconsistent in
this example)

------
calcsam
0 ^ any positive power = 0

0 ^ any negative power = 1/0 = undefined = +- inf

So strictly only the _right limit_ as n --> 0 of 0^n = 0. Not the _limit_.

~~~
Bahamut
0 is not defined as 0 * x = 0 - that is a derived formula. 0 is defined as the
additive identity, i.e. x + 0 = x = 0 + x. It is a unique number.

0 * x = 0 is proven by noting that 0 * x = (0 + 0) * x = 0 * x + 0 * x
(distributive property) and then subtracting the additive inverse of 0 * x
from both sides to get that 0 = 0 * x.

However, this says nothing about 0^0, and one cannot talk about 0^(-n) for
natural number n since 0 has no multiplicative inverse.

Also your argument on limits is not correct - you chose a particular path of
approach for the expression y^x fixed along y = 0. Looking at another angle,
the limit of x^x as x approaches 0 is clearly 1 as reasoned in the article, so
this causes a clear disagreement here since you can argue for different values
to make sense by tweaking the path of approach of the two dimensional function
y^x appropriately.

------
bumbledraven
Although standard mathematica notation glosses over the difference, it's
important to distinguish between the following two versions of the power
operation, which I'll call pow1 and pow2: pow1(x, y) = x^y where y is any
integer pow2(x, y) = x^y where y is any real

The value of 0^0 is 1 for (pow-a) and 0 for (pow-b).

~~~
SamReidHughes
But lim x->0 (x^x) = 1. Why should 0^0 be 0 then?

------
t__r
To me the matter looks similar to taking the conjunction of the empty set,
which is defined to be TRUE. This, however, starts to make sense when you see
that the taking the conjunction of a set X can be interpreted as "forall x in
X, x is TRUE". If X is empty then this statement is trivially true.

------
blackrabbit
Can anyone explain why this is a valid operation?
[http://wp.com/latex.php?latex=\lim_{x%20\to%200^{+}}%20x^{x}...](http://wp.com/latex.php?latex=\\lim_{x%20\\to%200^{+}}%20x^{x}%20%20=%20\\lim_{x%20\\to%200^{+}}%20\\exp\(\\log\(x^{x}\)\)&bg=ffffff&fg=000&s=0)

~~~
my_username_is_
it's using log to mean log base e, also known as the natural log or ln. you
can do that because they're inverse operations (I think that's the right term
for it, it's been a while since a formal math class) so they undo each other

~~~
clebio
Yes. It's an identity relationship. x = exp(log(x))

[http://en.wikipedia.org/wiki/List_of_logarithmic_identities#...](http://en.wikipedia.org/wiki/List_of_logarithmic_identities#Canceling_exponentials)

------
Totient
"...they boil down to that choice being more useful than the alternative
choices, leading to simpler theorems, or feeling more “natural” to
mathematicians."

Along these lines, my preferred definition of the natural numbers" is "the set
of all the positive integers and, when convenient, 0".

------
nemasu
I don't get what the big deal is. I put 0^0 in Windows Calculator and it gives
me 1 ... problem solved!

------
im3w1l
It also plays nice with the convention of 0log 0=0, used in for example
formulas for entropy.

~~~
olalonde
Why? With the x^0=1 definition, wouldn't log 0 be undefined?

~~~
j2kun
0 log 0 = log (0^0) = log (1) = 0

------
rkowalick
My favorite explanation:

y^x is the number of functions from a set with x elements to a set with y
elements.

Since there is only one function from the empty set to the empty set, namely,
the empty function, we get that 0^0 = 1.

------
deepakjc
I read upto "High School Teacher"... skimmed through "Calculus Teacher"... and
scrolled through "Mathematician"...

------
EGreg
Mathematics is about generalizing concepts and principles to ever larger
domains.

In this case, x^y is defined for all pairs (x, y) of real numbers except (0,
0). The question is what limit is "closer" to the set of outputs in the
neighborhood of (0, 0) than any other.

0^x is defined for all x except 0, and same for x^0. We can define 0^0 as the
limit of one or the other as x goes to 0. and one is constant and more
"stable" than the other, so it is typically taken to be that, i.e. 1.

Up next ... if P(X) = false, what is "P(X) for all X in Ø"?

~~~
cbright
Correction: x^y is defined for all pairs of real numbers with x!=0. 0^x is not
defined for negative x.

------
azinman2
Another good reminder on how math itself is arbitrary and made up by humans
(often for what's simplest/easiest), and not handed down to us by God. Luckily
it's an extremely useful and extendable made up system.

I see this all the time with AI/machine learning. Most algorithms are based on
assumptions that make the math work out better rather than being aligned with
some "fundamental truth." The world is not linear, but it's much easier to
approach it as if it were!

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gus_massa
This is a problem of definitions. The definitions are arbitrary and are chosen
to make the life (of the mathematicians) easier. It’s easier to write a lot of
results if we define 0^0=1.

On the other hands, he proofs express a fundamental truth and are handed down
to us by God (or whatever deity you believe in).

~~~
xamuel
The definitions only decorate the truths. By unwrapping the definitions, the
truths can be expressed using just the barest predicates and function symbols
of the background language, and logical operators. Whether you define 0^0 to
be 1 or not doesn't change the _unwrapped_ truth.

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Oculus
A pretty interesting question(the value of 0^0), in a very understandable
format for the layman (me).

Very nice blog post!

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bakhy
best thing i read in quite a while!

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Eleutheria
Whenever you try to cross-breed an apple tree with a donkey, you get nonsense.

But for convenience, whatever the offspring is, we may call it a donkapple.

That's they beauty of math.

~~~
noonespecial
The utility of math is that you can then do a bunch of calculations with
donkapples and at the end, get a meaningful real-world answer.

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LanceH
Well meaningful as soon as someone finds a donkapple.

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dagw
Not at all. Complex numbers are used to solve real world engineering problems,
despite the fact that you cannot "find" any complex numbers in nature.

~~~
JonnieCache
I thought quantum mechanics was riddled with complex numbers? What is it about
QM which means we're "using" complex numbers without having "found" them in
nature?

(I realise that this is kindof a horrible question to try and answer on an
internet forum, please try your best physicists ;)

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mcweaksauce
This is ridiculous. Doesn't this seem counter-intuitive for it to be anything
else besides 0 or undefined?

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baddox
If we don't have a specific mathematical context, then saying it's undefined
is intuitive to me.

Without context, 0 is no more intuitive to me than 1. These two statements are
equally intuitive to me, but they give different results for 0^0:

"Zero raised to any power is still just zero."

"Any number raised to the zeroth power is one."

~~~
dllthomas
For me, intuition-wise, I'd order it "undefined, 1, 0".

There are 3 cases for 1 and one case for 0 that immediately spring to my mind
when considering the problem:

    
    
        0) Limit of 0^x, as x approaches 0 (from above).
        1a) Limit of x^0 as x approaches 0 (from either direction).
        1b) Limit of x^x as x approaches 0 (from above).
        1c) "What did you multiply by 3 once, to get 3^1?  So, multiplying 1 by zero, zero times..."
    

Limits here, simplified to intuition level, being "what would you need to fill
that hole in the graph?" The fact that these disagree would be why I'd assume
undefined, but the case for 1 seems stronger (to me).

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baddox
The limit in 1b is 1 from below as well, right? I'm not sure how limits work
with complex numbers, but the imaginary part of x^x approaches zero as x
approaches zero from below, so can we say that the limit of x^x as x
approaches zero from below is also zero?

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dllthomas
Yes, but getting there steps out of the realm of "intuition" for me.

~~~
baddox
Yes, perhaps "intuition" isn't the best word. Formal limits certainly aren't
"intuitive" to me, at least by one definition of the word. I suppose I used
"intuitive" to mean "according to my mathematical understanding, ignoring the
mathematics explicitly dealing with 0^0."

~~~
dllthomas
Sure. I was limiting it to things I could do in my head in tens of seconds.

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elwell
In any case, _zero_ is a different animal than any other number. It can't
actually exist in the physical world like other numbers; it's definition is
non-existence. _Zero_ is merely conceptual. Therefore, it can't easily fit
into the picture of mathematics; at least not without requiring the definition
of special cases.

