
What does 0^0 equal? Why do mathematicians and high school teachers disagree? - ColinWright
http://www.askamathematician.com/?p=4524
======
kalid
Technically, 0^0 is an indeterminate form and has no specific solution.
Accurate but unhelpful.

Practically, 0^0 highlights the issue that most of us don't have a good
conceptual model for what exponents really do. How would you explain to a
10-year old why 3^0 = 1 beyond "it's necessary to make the algebra of powers
work out".

I use an "expand-o-tron" analogy

[http://betterexplained.com/articles/understanding-
exponents-...](http://betterexplained.com/articles/understanding-exponents-
why-does-00-1/)

to wrap my head around what exponents are really doing: some amount of growth
(base) for some amount of time (power). This gives you a "multiplier effect".
So, 3^0 means "3x growth for 0 seconds" which, being 0 seconds, changes
nothing -- the multiplier is 1. "0x growth for 0 seconds" is also 1, since it
was never applied. "0x growth for .00001 seconds" is 0, since a miniscule
amount of obliteration still obliterates you.

This can even be extended to understand, intuitively, why i^i is a real number
([http://betterexplained.com/articles/intuitive-
understanding-...](http://betterexplained.com/articles/intuitive-
understanding-of-eulers-formula/)).

~~~
jkent
I successfully managed to explain 3^0 to 10 year olds (as a recovering high
school math teacher) as:

    
    
      3^2 = 9
      3^1 = 3  (divide 9 by 3)
      3^0 = 1  (divide 3 by 3)
      3^-1 = 1/3  (divide 1 by 3)
      etc
    

This can logically be explained as n^0=1 for all real numbers.

Unfortunately this doesn't really handle 0^0 but fortunately 10 year olds are
rarely that difficult.

~~~
amalcon
This explanation works pretty well on adults: Just include the multiplicative
identity (1) in the expansion.

    
    
      3^3 = 3*3*3*1 = 27
      3^2 = 3*3*1   = 9
      3^1 = 3*1     = 3
      3^0 = 1       = 1
    

and likewise:

    
    
      0^3 = 0*0*0*1 = 0
      0^2 = 0*0*1   = 0
      0^1 = 0*1     = 0
      0^0 = 1       = 1
    

I haven't yet tried this on an actual 10 year old, though.

~~~
jkent
Also good, but the last line doesn't work (in my opinion): to get from 0x1 to
1 you need to undo the x0, i.e. dividing by 0. Which is undefined ... Zero,
confusing 10 year olds for centuries!

~~~
amalcon
True. If you approach it from a standpoint of extrapolation from known values,
you have a division by zero one way or the other.

What I'd intended was that the value of 1 was reached by applying the same
algorithm that was applied to arrive at the other values: start with 1,
multiply by the base once per instance of the exponent. No division involved.

It's still incorrect if we want to be strict, of course. That algorithm is not
quite the definition of exponentiation, because that algorithm can't really be
extended to work outside rational exponents. Exponentiation is defined across
complex numbers (ignoring 0^0 for the moment). I think this is acceptable
because I'm only shooting for an explanation, which doesn't need to be strict.

------
yaks_hairbrush
The high school teacher in the link is a B.S. in math education. They're
usually reflexive Platonists, believing that math is out there, and we merely
discover it. This is a result of the teaching of undergrad math as essentially
a series of completed works, with little history attached to it. This teacher
probably hasn't thought critically about why, say, 1/x^2 = x^(-2).

By contrast, a mathematician has a Ph.D. in math, and has had to do original
research and look at some history of topics. Thus, the mathematician knows
that math is not a finished product, but is under constant refinement.

Now, let's talk about negative exponents. Students are taught that 1/x^2 =
x^(-2). High school teachers often don't understand why. It is so ingrained
that even asking "Why?" seems almost grammatically incorrect.

The reason why is that we know the following things about exponents: 1) x^n =
x* x* ...* x for n a positive integer 2) As a consequence of 1, x^m* x^n =
x^(m+n) for m,n positive integers

Now, the question is not "what is x^n if n is negative?" (which is what a high
school teacher might ask). Rather, the question is "Can we define (!) x^n for
negative n in a way consistent with the item two above?" (mathematician's
framing). And, of course, we can. If x^n = 1/x^(-n) for n negative, then item
two works.

So, a high school teacher most likely thinks that the negative exponent rule
is simply a rule, handed down from the Gods of math. A mathematician
recognizes that it is a convention, and such a smooth convention that there is
simply no better choice.

Now, about 0^0: The HS teacher asks "What is 0^0?" and is therefore under the
impression that 0^0 is undefined because according to certain reasonings it
could be 0 or it could be 1. Textbooks (not written by mathematicians)
wouldn't correct this. The TI-86 gives a domain error when 0^0 is input. The
mathematician asks "What value of 0^0 makes my preferred formulae continue
working?" and thereby defines 0^0=1.

~~~
jpadkins
wait.. you mean math is simply made up? It's not the language of the universe?

well that's just depressing.

~~~
sliverstorm
Although my mathematician friends would probably yell at me if they heard
this, as far as I see things it's hardly any different from physics. Just a
model we create to describe observed phenomenon.

~~~
xyzzyz
Oh, math only works this way in intuitive fields, like elementary calculus,
elementary probability and staticstis, graph theory or Euclidean geometry.
However, there are fields in math that are different -- for instance, there
are topological spaces that exhibit phenomenons unseen anywhere else and that
are very hard to grasp intuitively (I had very hard time trying to imagine
what Cech-Stone compactification construction based on ultrafilters look like,
and eventually I gave up after I understood that this space is only supposed
to satisfy an universal property, and not have any intuitive shape). Deep
results in algebraic methods in topology say interesting things about the
models, while being completely indescribable outside of them, apart from some
trivial examples (for instance, what _exactly_ are cohomology classes?).

The bottom line is, if the only things you've seen are geometry and calculus,
with intuitive concepts like speed of change, area, length, then yeah, it's
only making our intuitions more formal. Otherwise, it's something completely
different.

~~~
sliverstorm
Again, I would argue this is no different than physics. The strength and
maturity of a model is not only based on its ability to describe what we can
verify, but predict what we have never (and may never) seen, and sometimes
what we have never even conceived of- and I think both physics and mathematics
are doing that.

~~~
xyzzyz
My point is, many concepts in math do not exist outside of the models
describing them, while if we throw away physical models, the reality does not
go away.

------
ihodes
The site isn't loading, but I find discussions about mathematical curiosities
(though they're not that curious) come about because people are looking for a
deeper meaning in mathematics. Math is playing with ontological objects in a
system of definitions. There doesn't need to be an answer that "makes sense"
for 0^0.

Depending on the context (are you working in set theory? are you making a new
definitions for exponentiation?) you might have a different definition. But
such operations are often defined recursively (e.g. in set theory, roughly,
where S(x) = x+1 (or successor of x) Exp(x, 0) = 1, and Exp(x, S(y)) = x *
Exp(x, y). Here you'll have 0^0 = 1, clearly.

For high school, 0^0 should be 1. It's necessary for problems high school
students might encounter in calculous, and is the way it is defined in almost
any field you'd be working in before graduate school.

High school teachers who insist that 0^0 != 1 likely don't understanding that
it's a definition.

------
dxbydt
Try these for size ( actual interview questions )

PITA interviewer > What's bigger, e^pi or pi^e ?

If you get past that one,

PITA interviewer > What's i^1 ?

Clever student> Its just 1 unit on the imaginary axis.

PITA interviewer >Good! So then, whats i^i ?

Clever student > Probably a few more units on the imaginary axis!

PITA interviewer >Then why does google say 0.207 (
<http://www.google.com/search?q=i^i> )

Clever student> hmmm...ohhh...aaahhh....WTF...I hate math I don't want this
stupid stupid job lemme go back to coding monads in Haskell for my ubercool
startup.

PITA interviewer >Don't let the door hit you on your way out.

~~~
jackpirate
What were you interviewing for?

------
kbutler
I'll stick with the grade-school math approach, at least until I need to
approach it differently.

    
    
      4^2 = 2 fours multiplied = 4 * 4 = 16
    

divide by 4 - so you take away one of the 4s by division(canceling like terms
like we do in grade school fraction math):

    
    
      4^1 = 4*4/4 = 4
    

divide by 4 again

    
    
      4^0 = 4*4/(4*4) = 1
    

divide by 4 again!

    
    
      4^-1 = 4*4/(4*4*4) = 1/4
    

Now try it with 0:

    
    
      0^2 = two zeros multiplied = 0*0 = 0
    

Divide by 0. Uh-oh. Well, let's keep following grade school fraction math and
cancel like terms:

    
    
      0^1 = 0*0 / 0 = 0
    

Divide by 0. Hmm - keep canceling like terms.

    
    
      0^0 = 0*0 / (0*0) = 1
    

But what's 0^-1? _grin_

~~~
rorrr
0^-1 is definitely infinity

x^(-1) doesn't have an upper bound as x approaches zero.

~~~
kbutler
Try approaching 0 from both the positive and negative sides.

~~~
rorrr
So you get a +∞ or -∞.

~~~
kbutler
But what's 0^-1? grin

------
3am
Fun digression late on a Wednesday :)

The indeterminate form seems the most correct based on the analysis of the
limit of f(x,y) = x^y as x approached zero from different paths.

I had always thought of it more of an algebraic identity thing; x^n * x^m =
x^(n+m). Obviously x^(n) = x^(n+0) = x^n * x^0 which can only be satisfied if
x^0 = 1. But this article (and really, the wikipedia treatment that beej71
linked to) made me think more about it.

~~~
Someone
"Obviously x^(n) = x^(n+0) = x^n * x^0 which can only be satisfied if x^0 =
1."

...or if x^n = 0, which is true if x=0 and n != 0

------
beej71
Since the site seems to be dead, here is a bit of wikipedia discussion on it:

<http://en.wikipedia.org/wiki/Exponent#Zero_to_the_zero_power>

~~~
yahelc
Time for my handy "Open this page in Google cache" bookmarklet:

    
    
        javascript:window.open("http://webcache.googleusercontent.com/search?q=cache:"+encodeURIComponent(location.href))

~~~
dminor
I think you're missing a quote after cache: in there.

Otherwise, very handy, thanks.

~~~
yahelc
Weird. It got stripped out somehow. Thanks for pointing it out; fixed.

------
nova
I like my high school math consistent with set theory.

ø : empty set, 1 : {ø}, A : nonempty set, ~= : isomorph to.

A^ø ~= 1, because there is only one function ø->A, the empty function.

ø^A ~= ø, because there is no function with empty codomain and nonempty
domain.

ø^ø ~= 1, because there is again one function ø->ø, the empty one.

So yes, 0^0 = 1.

~~~
psykotic
This! The empty function is the function whose graph is the empty set. When
you stop thinking of functions as symbolic expressions and make friends with
the empty set, it's all clear as crystal. The empty set has perplexed a lot of
people over the ages, so it's unsurprising that 0^0 prompts puzzlement. But
with the benefit of modern hindsight, there's really no reason to stay
confused. Set theory in the large is still a great mystery but we got the
empty set well figured out by now.

------
rikthevik
The engineer in me says, "Why don't you guys pick something, and I'll go ahead
and use it in my calculations (where it's appropriate)? I don't care much
about the theory behind it."

------
TrevorBurnham
Wolfram Alpha says "indeterminate," which is a good example of why Wolfram
Alpha needs an "Oh yeah? Prove it" button:
<http://www.wolframalpha.com/input/?i=0^0>

------
alephNaught
Doesn't a simple application of l'hospital solve this?

~~~
webspiderus
This is completely off-topic, but did you mean to call it l'hospital as
opposed to L'Hôpital? I remember even my calculus book had similar errors, and
I always wondered if it was just because the two looked so similar (or if
there was any more reasoning behind it). didn't mean to nitpick, your comment
just triggered a repressed train of thought :)

~~~
_delirium
His original name was actually Guillaume de l'Hospital; French spelling
reforms later did away with a number of cases of silent 's' (which had been
silent for a long time already), replacing it with a circumflex over the
preceding vowel.

See:
[http://en.wikipedia.org/wiki/Use_of_the_circumflex_in_French...](http://en.wikipedia.org/wiki/Use_of_the_circumflex_in_French#Disappearance_of_.22s.22)

and:
[http://en.wikipedia.org/wiki/Guillaume_de_l%27H%C3%B4pital#c...](http://en.wikipedia.org/wiki/Guillaume_de_l%27H%C3%B4pital#cite_note-0)

~~~
webspiderus
ah, thanks - I had a feeling it was something like that! I figured a Calculus
book would probably get it right :)

------
foysavas
Only after seeing how many comments on the article attempted to genuinely
refute the article did I get a sense of how few people grasp the foundation of
mathematics (that is, the composition of arbitrary assumptions to agreeable
statements).

------
imminentdomain
Doesn't the first argument have a logic error, in that x^(1-1) = x^1 _x^(-1)
has a caveat for x not equal to 0, so it shouldn't prove anything for x = 0.
In other words, it just says x^0 = 1 for any x =/= 0.

Using Abstract Algebra, I think 0^0 = 1 is completely accurate. The power
function (y^x) could be defined to be the amount you times (x times) you apply
the _ operation between the y on the identity element. In our usual numbers
that looks like y _(y_ (y... _(y_ 1)...)). When x is negative y becomes the
multiplicative inverse of y and everything else remains the same.

------
jberryman
If you're interested in this topic and the discussion here, do yourself a
favor and pick up David Foster Wallace's "Everything and More".

I'm about 1/2 way through. It's a real gift.

------
bdr
The right convention is also "obvious" to practitioners of combinatorics. The
exponentiation x^y, for integer x and y, is the number of possible strings of
length y from a set of letters of cardinality x. (Hence 2 __8 possible bytes.)
How many ways are there to make a string of length 0, regardless of the
alphabet size? Just one... you don't do anything.

------
antihero
Nullity? :)

[http://en.wikipedia.org/wiki/Transreal_arithmetic#Transreal_...](http://en.wikipedia.org/wiki/Transreal_arithmetic#Transreal_arithmetic)

We had this guy as a lecturer. Whether nullity exists or not (though James's
argument is that it's as valid as j), and I have to admit his arithmetic
_does_ make some things simpler.

------
perfunctory

      Python: 0**0 == 1
      Javascript: Math.pow(0, 0) == 1
      Java: Math.pow(0, 0) == 1
    

Therefore 0^0 = 1

------
parallel
> How would you explain to a 10-year old why 3^0 = 1

You draw the line 3^x. It "passes through" 1 when x = 0. So don't think about
the point, think about the line. It's not rigorous but it's intuitive.

<http://fooplot.com/index.php?q0=3^x>

edit: added link and fixed typos

~~~
vessenes
This is precisely the problem, though.

The same 10 year old draws two lines: 0^x, and x^0.

They clearly do not meet.

~~~
parallel
Yeah, I was just talking about the easier concept of x^0. Going the other way,
0^x it makes a lot less sense to me. I know that for all positive x, 0^x = 0.
That it pops up to 1 at the origin means the thing is not smooth so drawing
lines isn't going to help us.

Plotting z = y^x could be interesting but we're going to see asymptotic
behaviour at x = y = 0 (when going in the x direction). So no easy wins there.

------
NHQ
As a definitive answer is not available to us via mathematics, the real answer
is that you can't give an exponent to nothing.

Sometimes you have to use another language to make sense of something.

------
S_A_P
apparently it equals:

Error establishing a database connection

that is kind of funny... sorta

------
bumbledraven
A empty product is the multiplicative identity (1), just like an empty sum is
the additive identity (0).

------
buff-a
Mathematica says "Indeterminate" =)

------
leif
Let's do it with complex numbers next!

------
biftek
if 0^0=1 than ex nihilo

------
rorrr
0.000000000001^(0.000000000001) = 1

------
user911302966
0 ^ 0 = 0, on all architectures.

~~~
splicer
but not all languages ;)

