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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?

#131

Post by lurker14 »

xyzzyz wrote: post #127
dxbydt wrote: post #124

Earlier quoted context omitted.

The notion of a^n definitely belongs to group theory, whether you call that n an exponent or an annihilator or a period is a matter of some contention. See: http://mathoverflow.net/questions/44393/notation-exponent-of... http://mathoverflow.net/questions/32116/exponent-of-a-group

Only if n is an integer. Real exponentiation is something completely different, because it involves all aspects of real numbers - addition, multiplication, order and continuity, which are all interconnected, and the language of group theory is too weak to describe it. For instance, while 2^pi makes perfect sense in the realm of real numbers, it makes none in Z_3.

Hmm.

(Warning, ascii math is confusing and ambiguous to read. Sorr.)

Exponentiation of group "multiplication" does not immediately seem amenable to the reals, sure. But real exponentation does form a group, as shown here:

Define x_g(r) = the function that raises a Real/{0} (non-zero real) number r to the exponent x (in the sense of of some reasonable definition of exponentiaton of continuous functions). Define X = the set x_g() functions corresponding to all reals (including 0)

Define x_g y_g as composition: y_g(x_g(r)) = (r^x)^y = r^ (xy). Then we have 0_g x_g = (r^0)^y = r ^ (0 y) = 1 = r ^ (y * 0) = (r^y)^0) = y_g 0_g -> identity

y_g (1/y)_g = r^y ^ (1/y) = r^0 = 0_g -> inverse

(x_g y_g) (z_g) = ((r^x)^y)^z = (r^(xy))^z = r^((xy)z) = r^(x(yz) = (r^x)^(yz) = x_g (r_g r_g) -> Associativity

That makes a group.

Now, I explicitly left out the 0^x case. Can we fit it back in?

Not particularly cleanly, as thoroughly discussed in this thread.

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

#132

Post by lurker14 »

dxbydt wrote: post #60
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…

You write: "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."

(Prologue: I encountered group theory first by drawing pictures of pegboards and strings to illustrate permutations (before I knew the word!), not by reading the rules up front.)

You are describing a schism between pure/formal and applied mathematics (and between formalism and intuition, to some extent). It's completely cool for you to have an interest in pure math completely separated from applied math, real-world physics, programming etc. It is also cool for someone to pursue applied math,physics, etc without any pure math, but that is sad because a lot of beautiful symmetry and cross-disciplinary value would be last. (Goodbye, encryption!)

I personally strive to connect pure and applied mathematics. After getting burned (in an emotional/psychological way) by chasing pure math study beyond my ability to intuit and apply it, I now commit myself to learning theory and application in tandem. (I'll certainly appreciate the fact that pure theorists such as your ideal have gone several steps ahead and I can study their results without trying to discover them from scratch.)

In fact, my most recent flight of fancy / big dream is to write math/CS tutorials that provide such a tight integration of theory and application, abstract and concrete, general and specific. And I want to use modern web tools (hyperlink, animation, multi-dimensional page layout) to do so.

I'd love to talk to anyone interested in working with me on that :-)

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

#133

Post by amalcon »

jkent wrote: post #129
amalcon wrote: post #92

Earlier quoted context omitted.

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.

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!

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.

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

#134

Post by kalid »

lurker14 wrote: post #132
dxbydt wrote: post #60

Earlier quoted context omitted.

>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…

You write: "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." (Prologue: I encountered group theory first by drawing pictures of pegboards and strings to illustrate permutations (before I knew the word!), not by reading the rules up front.) You are describing a schism between pure/formal and applied…

Hi, I'm really interested in developing tutorials / explanations that help merge intuition and rigor, theory and application. You can reach me at kalid.azad@gmail.com.

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

#135

Post by xyzzyz »

lurker14 wrote: post #131
xyzzyz wrote: post #127

Earlier quoted context omitted.

Only if n is an integer. Real exponentiation is something completely different, because it involves all aspects of real numbers - addition, multiplication, order and continuity, which are all interconnected, and the language of group theory is too weak to describe it. For instance, while 2^pi makes perfect sense in the realm of real numbers, it makes none in Z_3.

Hmm. (Warning, ascii math is confusing and ambiguous to read. Sorr.) Exponentiation of group "multiplication" does not immediately seem amenable to the reals, sure. But real exponentation does form a group, as shown here: Define x_g(r) = the function that raises a Real/{0} (non-zero real) number r to the exponent x (in the sense of of some reasonable definition of exponentiaton of continuous functions). Define X = th…

The group you described does not inherit any interesting structure from exponentiation -- indeed, one can easily see that it is isomorphic to the multiplicative group of reals. You could similarly construct a group isomorphic to an additive group of reals. This is an example of the fact that real exponentiation connects different aspects of real numbers, as well as the fact that just abstract algebra language is not enough to express properties of real numbers. You need to somehow relate the algebraic structure of reals to a topologic one, which stems from order imposed on reals and its continuity.

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

#136

Post by rfurmani »

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.…

> 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.

Technically, 0^y is 0 only if you approach it from the right: y>0. To the left of 0 it is indeterminate or infinity depending on how you look at it. x^0, however, makes sense for all x and is always 1

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