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".
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.
What is "indeterminate form"? What does it mean for expression to "have a specific solution"?
You see, 0^0 = 1, and it's obvious to a mathematician. The only problem is that the function f: [0, \infty) x R -> R, f(x, y) = x^y is discontinuous in (0, 0) and that's what causes problems -- for instance, this is the source of the whole "indeterminate form" notion. If a function f is continuous in (a, b), then for every two sequences a_n, b_n, such that lim a_n = a, lim b_n = b, we have lim f(a_n, b_n) = f(a, b). That's why lim (a_n)^(b_n) = a^b if (a, b) != (0, 0), and this is "determinate form". But if (a, b) = (0, 0), then no matter how we define 0^0, it does not follow that lim (a^n)^(b^n) = a^b = 0^0, because in this case, lim (a_n)^(b_n) can be every positive value, and so mathematicians used to call it "indeterminate form" (it's not common today, though). So, since this problem is unsolvable in a consistent (continuous) way, we define 0^0 = 1, to be consistent with exponentiation rules, at least.
I've never seen a need for an "intuitive" explanation of exponentiation -- the usual definition is as intuitive as one can get. The thing is, most people do not know, _why_ expressions like pi^e are supposed to make sense -- they just take exponentiation as given. Only then they need to make up some explanation why "exponentiation rules" are like this, and what exponentiation is about. Hell, people don't even know what real numbers are! How are they supposed to make sense of exponentiation with exponent other than natural number?
>most of us don't have a good conceptual model for what exponents really do
Instead of matching math to real world objects (1= one banana, 2 = two bananas, 1+2 = 3 bananas etc. ) and building up to exponentiation, multiplication etc. thereby introducing all sorts of paradoxes, group theory dodges all that and treats the whole thing as a very consistent rule-based system. Things fall into place quickly once the rules are laid out explicitly.
Consider:
finite abelian group with only 3 elements a,b,c.
Given a+b=c, a+c=a, what's b+b ?
Hmmm...okay, if a plus c is a, then c is acting like zero. So b+c must be b.
since addition is commutative (abelian gp), b+a must be a+b which you said was c.
So now we know b+a=c, b+c=b, so b+b better be a !
Students are easily convinced because you've laid out the rules very explicitly. In fact, they'll try to convince you that b plus b better be a because that's the only way to make the cayley table work out!
(http://en.wikipedia.org/wiki/Cayley_table)
There are several books that argue that the teaching of Abstract Algebra must precede Calculus for this very reason. With Calculus, the mapping of math to real-world objects leads to all sorts of messy realities. With group theory, you dodge that mess by simply stating rules upfront.
> Technically, 0^0 is an indeterminate form and has no specific solution.
Precisely, as this is the true mathematician answer: "it depends where 0^0 comes from".
As a f(x,y): RxR->R function, come from the top of the R² plane and 0^0 is 0 but come from the right side and it's 1. Limits and extension by continuity give us this easily enough for fh:x->x^0 and fv:y->0^y.
Writing this I asked myself, what if we came from some funky other path, like the diagonal, or a curve?
h: R->RxR, x->(x, 0) defines "coming from the top", and foh = fh
v: R->RxR, x->(0, y) defines "coming from the top", and fov = fv
d: R->RxR, x->(x, x) defines coming along the diagonal, where things could get interesting.
s: R->RxR, t->(e^(at)sin(t), e^(at)cos(t)) defines coming along a log spiral whose tangent at t=0 is vertical, so fos looks like fun around t=0.
Now what happens if we build a path function p: RxR->RxR, (t, z)->? that endlessly approaches v when z->0? the log spiral with z=1/a as a parameter is a possible one. With such a p function, what does lim fop(x) when x->0 (which is a function of z) look like when subsequently z->0?
> 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".
Actually, that's exactly the reason 3^0=1: it was the definition that preserved the most identities. Agreed that this explanation doesn't really help intuition.
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.
> 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.
It's not entirely clear what you mean by this, but in the most obvious interpretation this idea is correct.
Here's what I mean by that: math deals with pure logic. All logical deductions (derivations from axioms to conclusions in formal systems) are in a sense "out there", waiting to be discovered. Certainly they have always been true, before people knew about them, and there are deductions that are true even though no one knows about them yet.
The other interpretation, which you probably meant, was the question of mathematical style and interest: which axioms do we study, and why do we care about them, and why are some theorems important and others not important? Certainly the definition that 1/x^2 = x^(-2) was made for consistency, which is an aspect of style. You could make some other definition and do formal reasoning just as well (although you'd have to change some other things - maybe the definition of exponentiation, maybe your interpretation of symbols on paper).
But either way, I think you are wrong that high school teachers don't think critically about why 1/x^2 = x^(-2). Some may be bad teachers, but I have known a lot of incredibly good high school math teachers, and I suspect they have thought about this. In fact, when I talk to mathematicians about high school education, they usually agree that people doing research know more parts of mathematics than high school teachers, but good high school teachers have a much deeper understanding of elementary math than people doing research, because they have had to approach it from many different angles in order to teach different students.
Not entirely fair. While there is a lot of choice about how to define stuff around the edges of mathematics, after the axiomatic choices have been made, there's quite a lot to discover in their structure. Particularly if the relevant axioms relate to something outside of mathematics (the examples are too many to even scratch but start out thinking of the definitions of natural and rational numbers) then you really are discovering things in the same way as a physicist - in fact, this is how many physicists go about discovering things, for better or worse.
In most situations it makes sense to define exponentiation as repeated multiplication, and a^0 as the absence of multiplication by a, hence a^0 = 1 as the multiplicative identity. I wouldn't introduce the idea of anything else to a student unless they specifically asked me about one of the problems which can arise in choosing 0^0 = 1.
Math is out there. I don't understand how anyone can say that the Mandelbrot set was created or is formed from arbitrary axioms. It was discovered full stop.
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.
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.
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.
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.
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."
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 :)
No. What would you apply it to in 0^0? The best you could do is come up with two functions, f(x) and g(x), with f(x) and g(x) going to 0 as x goes to 0, and try to use it to evaluate f(x)^g(x), but the result is going to depend on exactly what your choices are for f(x) and g(x).
A precondition of L'Hôpital's Rule is that the limit in question exists. So: Prove that the limit exists, and then you can bust out L'Hôpital's Rule to prove that it's 1.
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).
Doesn't the first argument have a logic error, in that x^(1-1) = x^1x^(-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...(y1)...)). When x is negative y becomes the multiplicative inverse of y and everything else remains the same.
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 28 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.
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.
[+] [-] kalid|14 years ago|reply
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-...
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-...).
[+] [-] jkent|14 years ago|reply
Unfortunately this doesn't really handle 0^0 but fortunately 10 year olds are rarely that difficult.
[+] [-] xyzzyz|14 years ago|reply
You see, 0^0 = 1, and it's obvious to a mathematician. The only problem is that the function f: [0, \infty) x R -> R, f(x, y) = x^y is discontinuous in (0, 0) and that's what causes problems -- for instance, this is the source of the whole "indeterminate form" notion. If a function f is continuous in (a, b), then for every two sequences a_n, b_n, such that lim a_n = a, lim b_n = b, we have lim f(a_n, b_n) = f(a, b). That's why lim (a_n)^(b_n) = a^b if (a, b) != (0, 0), and this is "determinate form". But if (a, b) = (0, 0), then no matter how we define 0^0, it does not follow that lim (a^n)^(b^n) = a^b = 0^0, because in this case, lim (a_n)^(b_n) can be every positive value, and so mathematicians used to call it "indeterminate form" (it's not common today, though). So, since this problem is unsolvable in a consistent (continuous) way, we define 0^0 = 1, to be consistent with exponentiation rules, at least.
I've never seen a need for an "intuitive" explanation of exponentiation -- the usual definition is as intuitive as one can get. The thing is, most people do not know, _why_ expressions like pi^e are supposed to make sense -- they just take exponentiation as given. Only then they need to make up some explanation why "exponentiation rules" are like this, and what exponentiation is about. Hell, people don't even know what real numbers are! How are they supposed to make sense of exponentiation with exponent other than natural number?
[+] [-] dxbydt|14 years ago|reply
Instead of matching math to real world objects (1= one banana, 2 = two bananas, 1+2 = 3 bananas etc. ) and building up to exponentiation, multiplication etc. thereby introducing all sorts of paradoxes, group theory dodges all that and treats the whole thing as a very consistent rule-based system. Things fall into place quickly once the rules are laid out explicitly.
Consider: finite abelian group with only 3 elements a,b,c. Given a+b=c, a+c=a, what's b+b ? Hmmm...okay, if a plus c is a, then c is acting like zero. So b+c must be b. since addition is commutative (abelian gp), b+a must be a+b which you said was c. So now we know b+a=c, b+c=b, so b+b better be a !
Students are easily convinced because you've laid out the rules very explicitly. In fact, they'll try to convince you that b plus b better be a because that's the only way to make the cayley table work out! (http://en.wikipedia.org/wiki/Cayley_table)
There are several books that argue that the teaching of Abstract Algebra must precede Calculus for this very reason. With Calculus, the mapping of math to real-world objects leads to all sorts of messy realities. With group theory, you dodge that mess by simply stating rules upfront.
[+] [-] lloeki|14 years ago|reply
Precisely, as this is the true mathematician answer: "it depends where 0^0 comes from".
As a f(x,y): RxR->R function, come from the top of the R² plane and 0^0 is 0 but come from the right side and it's 1. Limits and extension by continuity give us this easily enough for fh:x->x^0 and fv:y->0^y.
Writing this I asked myself, what if we came from some funky other path, like the diagonal, or a curve?
h: R->RxR, x->(x, 0) defines "coming from the top", and foh = fh
v: R->RxR, x->(0, y) defines "coming from the top", and fov = fv
d: R->RxR, x->(x, x) defines coming along the diagonal, where things could get interesting.
s: R->RxR, t->(e^(at)sin(t), e^(at)cos(t)) defines coming along a log spiral whose tangent at t=0 is vertical, so fos looks like fun around t=0.
Now what happens if we build a path function p: RxR->RxR, (t, z)->? that endlessly approaches v when z->0? the log spiral with z=1/a as a parameter is a possible one. With such a p function, what does lim fop(x) when x->0 (which is a function of z) look like when subsequently z->0?
Damn. It was supposed to be a two-line comment.
[+] [-] yaks_hairbrush|14 years ago|reply
Actually, that's exactly the reason 3^0=1: it was the definition that preserved the most identities. Agreed that this explanation doesn't really help intuition.
[+] [-] MtrL|14 years ago|reply
X to the 0.5 is square root, X to the 0.3 is cube root, X to the 0.25 is fourth root, etc
Therefore X^0 is what you get if you're taking the 'infiniteth' root
= 1
[+] [-] hbt|14 years ago|reply
7^2=7 * 7=49
7^2/7^2=7 * 7/ 7 * 7=49/49=7^(2-2)=7^0=1
But 0^0 was never intuitive to me.
[+] [-] yaks_hairbrush|14 years ago|reply
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.
[+] [-] noahl|14 years ago|reply
It's not entirely clear what you mean by this, but in the most obvious interpretation this idea is correct.
Here's what I mean by that: math deals with pure logic. All logical deductions (derivations from axioms to conclusions in formal systems) are in a sense "out there", waiting to be discovered. Certainly they have always been true, before people knew about them, and there are deductions that are true even though no one knows about them yet.
The other interpretation, which you probably meant, was the question of mathematical style and interest: which axioms do we study, and why do we care about them, and why are some theorems important and others not important? Certainly the definition that 1/x^2 = x^(-2) was made for consistency, which is an aspect of style. You could make some other definition and do formal reasoning just as well (although you'd have to change some other things - maybe the definition of exponentiation, maybe your interpretation of symbols on paper).
But either way, I think you are wrong that high school teachers don't think critically about why 1/x^2 = x^(-2). Some may be bad teachers, but I have known a lot of incredibly good high school math teachers, and I suspect they have thought about this. In fact, when I talk to mathematicians about high school education, they usually agree that people doing research know more parts of mathematics than high school teachers, but good high school teachers have a much deeper understanding of elementary math than people doing research, because they have had to approach it from many different angles in order to teach different students.
[+] [-] dataduck|14 years ago|reply
In most situations it makes sense to define exponentiation as repeated multiplication, and a^0 as the absence of multiplication by a, hence a^0 = 1 as the multiplicative identity. I wouldn't introduce the idea of anything else to a student unless they specifically asked me about one of the problems which can arise in choosing 0^0 = 1.
[+] [-] Almaviva|14 years ago|reply
[+] [-] jpadkins|14 years ago|reply
well that's just depressing.
[+] [-] ihodes|14 years ago|reply
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|14 years ago|reply
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|14 years ago|reply
[+] [-] kbutler|14 years ago|reply
[+] [-] MetaMan|14 years ago|reply
[+] [-] rorrr|14 years ago|reply
x^(-1) doesn't have an upper bound as x approaches zero.
[+] [-] 3am|14 years ago|reply
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.
[+] [-] beej71|14 years ago|reply
http://en.wikipedia.org/wiki/Exponent#Zero_to_the_zero_power
[+] [-] yahelc|14 years ago|reply
[+] [-] ColinWright|14 years ago|reply
[+] [-] zedpm|14 years ago|reply
[+] [-] nova|14 years ago|reply
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|14 years ago|reply
[+] [-] rikthevik|14 years ago|reply
[+] [-] TrevorBurnham|14 years ago|reply
[+] [-] alephNaught|14 years ago|reply
[+] [-] webspiderus|14 years ago|reply
[+] [-] tzs|14 years ago|reply
[+] [-] unknown|14 years ago|reply
[deleted]
[+] [-] TrevorBurnham|14 years ago|reply
[+] [-] Rusky|14 years ago|reply
[+] [-] foysavas|14 years ago|reply
[+] [-] imminentdomain|14 years ago|reply
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...(y1)...)). When x is negative y becomes the multiplicative inverse of y and everything else remains the same.
[+] [-] jberryman|14 years ago|reply
I'm about 1/2 way through. It's a real gift.
[+] [-] bdr|14 years ago|reply
[+] [-] antihero|14 years ago|reply
http://en.wikipedia.org/wiki/Transreal_arithmetic#Transreal_...
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|14 years ago|reply
[+] [-] parallel|14 years ago|reply
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|14 years ago|reply
The same 10 year old draws two lines: 0^x, and x^0.
They clearly do not meet.
[+] [-] unknown|14 years ago|reply
[deleted]
[+] [-] NHQ|14 years ago|reply
Sometimes you have to use another language to make sense of something.