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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 state

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 13y ago
I think it might be better to say it's indeterminant.
- Sniffnoy 13y ago
  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.
- j2kun 13y ago
  You may be confusing terms from calculus, where the label "indeterminate form" is given to 0^0 when encountered in a limit.
kazagistar 13y ago
I thought "most of the time" was implied in these sorts of things by now. And you will note the mathematicians don't "say it is 1"... they say they are choosing to follow the convention of defining it as 1, with reasons listed below. I could come up with whatever crazy definitions in math I wanted to, and sometimes even get useful results. However, this view is not... convenient for teaching, so they pretend that you are learning "math", as opposed to "a particular math".
j2kun 13y ago
Often teachers (esp high school teachers) don't have the time to give the same kind of in depth analysis of the tradeoffs.
xamuel 13y ago
>"The statement that mathematicians uniformly say it is 1 is simply false" Not really. No serious mathematician would dispute the fact that for every real x, e^x=sum(n=0 to infinity)x^n/n!. But the above fails at x=0 if we don't define 0^0=1. So even mathematicians who claim to not use 0^0=1, can almost always be convinced to admit that they do indeed use 0^0=1, using the Taylor series for e^x.

joliv 13y ago

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

Sniffnoy 13y ago
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.
j2kun 13y ago
You may be confusing terms from calculus, where the label "indeterminate form" is given to 0^0 when encountered in a limit.

kazagistar 13y ago

I thought "most of the time" was implied in these sorts of things by now. And you will note the mathematicians don't "say it is 1"... they say they are choosing to follow the convention of defining it as 1, with reasons listed below. I could come up with whatever crazy definitions in math I wanted to, and sometimes even get useful results. However, this view is not... convenient for teaching, so they pretend that you are learning "math", as opposed to "a particular math".

j2kun 13y ago

Often teachers (esp high school teachers) don't have the time to give the same kind of in depth analysis of the tradeoffs.

xamuel 13y ago

>"The statement that mathematicians uniformly say it is 1 is simply false" Not really. No serious mathematician would dispute the fact that for every real x, e^x=sum(n=0 to infinity)x^n/n!. But the above fails at x=0 if we don't define 0^0=1. So even mathematicians who claim to not use 0^0=1, can almost always be convinced to admit that they do indeed use 0^0=1, using the Taylor series for e^x.