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What does 0^0 equal? Why do mathematicians and high school teachers disagree?

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Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#51

Post by jberryman »

Almaviva wrote: post #41
yaks_hairbrush wrote: post #17

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 m…

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.

Your parent never says anything about arbitrary axioms.

To your main point: are irrational numbers, say, "out there"? If so, where?

Until a few centuries ago it was mathematical standard practice to fudge 1/2 as 25/49 when taking it's square root. But then mathematicians invented (some would argue) the notion of an irrational, because it was, well... useful.

There are real metaphysical questions here; there have been since the greeks started reaching the limits of a purely geometric (read physical) understanding of math.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#52

Post by sliverstorm »

xyzzyz wrote: post #44
sliverstorm wrote: post #30

Earlier quoted context omitted.

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.

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 constru…

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.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#53

Post by 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.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#54

Post by joe_the_user »

xyzzyz wrote: post #36
kalid wrote: post #11

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-exponen…

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 t…

Mathematicians don't argue about what an expression "really is" (or at least, real mathematics doesn't involve this). They define functions and use axioms to prove theories about them.

"No really". Mathematics just isn't concerned with this stuff. Sometimes infinity it defined as single point making the real number compact, sometimes a "positive infinity" and a "negative infinity" are defined. Sometimes you add points to a given function to make it more tractable and sometimes you don't. But none of this "means" anything. The real number line can be embedded in a number of topological spaces. At least two division rings and various things (the complex numbers are most common). The way you extend a given function (say e^x) is going to vary depending on what space you're looking at as well as what topic you're interested in.

Math works with definition systems and get theorems out of them. If you want to know what something "really is", consult philosophy or something.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#55

Post by xyzzyz »

Dove wrote: post #45
xyzzyz wrote: post #36

Earlier quoted context omitted.

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 t…

What is "indeterminate form"? http://en.wikipedia.org/wiki/Indeterminate_form You see, 0^0 = 1, and it's obvious to a mathematician . . . we define 0^0 = 1, to be consistent with exponentiation rules Well, you're going to be inconsistent with them no matter how you define it, since, as you point out, x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis.…

By "exponentiation rules" I mean algebraic equalities, like a^x * a^y = a^(x+y). Most of them work no matter if you define 0^0 = 1 or 0, but some of them are cleaner with 0^0 = 1. It's also consistent with cardinal and ordinal exponentiation (look it up). "Approaching along x axis" is not algebraic notion, it's analytic one.

0^0 makes no less sense than, say, -e^(i pi). They're both 1 because we define them like this. If you think that -e^(i pi) makes more sense than 0^0, please, explain me why.

Also, mathematicians agree in this, seriously. Go and ask one.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#56

Post by xyzzyz »

joe_the_user wrote: post #54
xyzzyz wrote: post #36

Earlier quoted context omitted.

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 t…

Mathematicians don't argue about what an expression "really is" (or at least, real mathematics doesn't involve this). They define functions and use axioms to prove theories about them. "No really" . Mathematics just isn't concerned with this stuff. Sometimes infinity it defined as single point making the real number compact, sometimes a "positive infinity" and a "negative infinity" are defined. Sometimes you add poin…

Are you sure you're replying to right post? I ask, because nothing I see in yours can be seen as reply to mine -- it actually repeats my point.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#57

Post by 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

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#58

Post by Almaviva »

jberryman wrote: post #51
Almaviva wrote: post #41

Earlier quoted context omitted.

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.

Your parent never says anything about arbitrary axioms. To your main point: are irrational numbers, say, "out there"? If so, where? Until a few centuries ago it was mathematical standard practice to fudge 1/2 as 25/49 when taking it's square root. But then mathematicians invented (some would argue) the notion of an irrational, because it was, well... useful. There are real metaphysical questions here; there have been…

I don't think the word "invented" is much better an analogy for real numbers than "position" is for "where" the Mandelbrot Set is.

This goes beyond what is strictly math, but I don't think it's reasonable to say that properties of the real numbers (say, roots, pi, e, and so forth) are invented. In some sense they seem like the simplest thing that fits a few properties (and not that many). Similarly with Euclidean Geometry, and the natural numbers (with primes and their structure, etc.) I really actually think something akin to Occam's Razor applies in math!

You can of course tweak your starting points and get really interesting things too, say non-standard analysis or non-Euclidean geometries. And you can derive (I would say "discover") some rich and surprising properties and patterns in them.

But I don't think any of these things are accidental. The real numbers (and pi, e, etc) would surely be "invented" in the same (modulo shifts in convention) ways by other advanced civilizations, I think. Could you imagine this not being so?

This is also extra-mathematical, and I admit this may be hooey, but I believe (as a non-mathematician) that math gives hints at large patterns and tells you when things fit well or are funny. Like, I can argue whether Pi or 2*Pi is the more "natural" constant, and it's not a discussion completely devoid of content! Also, I think math tells me something is funny about 0^0 (because of the limit 0^x) but that 1 is the more natural fit.

I realize this puts me quite firmly into the category of people being belittled here!

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#59

Post by xyzzyz »

sliverstorm wrote: post #52
xyzzyz wrote: post #44

Earlier quoted context omitted.

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 constru…

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.

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.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#60

Post by dxbydt »

kalid wrote: post #11

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-exponen…

>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.

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