if 0^0=1 than ex nihilo
What does 0^0 equal? Why do mathematicians and high school teachers disagree?
81–90 of 136 posts
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#82Earlier quoted context omitted.
Group theory is formed by abstracting out the observed properties of number systems. If you want to show that some collection is a group, you will have to do the computations to show that they follow the rules, in which case, it helps to have an understanding of the mechanics of the computation.
> it helps to have an understanding of the mechanics My claim is the exact opposite. I claim you don't need to understand the mechanics ( just blindly abide by the rules of the group or abelian group or finite simple group or whatever), which is why the approach is better. If you show a monkey red means stop and green means go and reinforce these rules by rewarding with a banana, eventually the monkey will stop when…
You are probably right that such an approach is better at teaching students to be able to crank out solutions to problems, but I think it also reinforces the attitude a lot of students have that math is just a bunch of arbitrary rules that don't mean anything, and is therefore a waste of time. For students that aren't destined to be math majors, the most important aspect of math is being able to map real-world concepts to abstract rules. Being able to mechanically manipulate those rules is far secondary, especially with easy access to computers.
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#830.000000000001^(0.000000000001) = 1
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#84Earlier quoted context omitted.
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…
I haven't seen any reason why -e^(i pi)=1 could be considered incorrect. It's consistent with the Taylor Series expansion of e^x. It's consistent with the view of complex exponentiation as rotation. I don't know of any particular problems that arise from taking -e^(i pi)=1. This is not the case with 0^0=1, which is inconsistent with many limits. That's why 0^0=1 is an agreed-upon convention sometimes . http://en.wiki…
This is not the case with 0^0=1, which is inconsistent with many limits.
So what? It's only a problem if you want the exponentation function to be continuous, so you escape the problem by leaving it undefined. You could place similar unbased requirements on complex exponentiation to make it seem incorrect. For instance, real exponentiation always gives a positive value for positive base, while complex does not, so e^(i pi) = -1 is wrong. I agree that this is ridiculous requirement, but leaving 0^0 undefined because the math is not as we want it to be (e.g. exponentation is not continuous) looks just as ridiculous and silly to me.
On the other hand, putting 0^0 = 1 makes it consistent many combinatoric formulas, and is also consistent with cardinal exponentation, where nobody objects to 0^0 = 1, when you look at 0 as the cardinal number.
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#85I'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^-1 is definitely infinity
x^(-1) doesn't have an upper bound as x approaches zero.
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#86Earlier quoted context omitted.
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…
> Also, mathematicians agree in this , seriously. Go and ask one. Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-p... . More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-p... , if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is sim…
But my point is, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not. 0^0 = 1 does not make calculus theorems more complicated to state or prove with modern language. It was a problem in XIX century, when mathematicians did not have a solid foundation with concepts like limit or continuity, but it's over now.
Mathematician here; we do not.
It seems I was a little too bold with my claim. All the mathematicians I know (and I'm a mathematician as well) agree with 0^0 = 1. It's a folklore specific thing, I guess.
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#87Earlier quoted context omitted.
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.
joe's reply clarified things somewhat - at least for me. Are replies always supposed to be rebuttals?
Are replies always supposed to be rebuttals?
It turns out to be so. If a reply is not a rebuttal, it is usually preceded with something like "To clarify, ..." or "I wanted to add, that ...". I just got confused without it.
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#88Wolfram 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
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#89Earlier quoted context omitted.
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…
To your main point: are irrational numbers, say, "out there"? If so, where? On the sheet of paper in front of me, as the hypotenuse of the isosceles right-angled triangle with unit length I drew a moment ago. Irrational numbers are probably a bad example of what you're talking about.
> Irrational numbers are probably a bad example of what you're talking about.
No, they're the perfect example and you're observation about the unit square is glib.
Trying to reason about the diagonal of the unit square basically destroyed the Pythagorean world view of integers as the fundamental building blocks of reality.
If mathematical entities are simply discovered and that-is-that, then why did it take nearly two millenia after this observation for western math to accept irrationals as numbers?
From M. Stifel, 1544:
"Now, that cannot be called a true number which is of such a nature that it lacks precision. Therefore, just as an infinite number is not a number, so an irrational number is not a true number, but lies hidden in a kind of cloud of infinity"
Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?
#90Earlier quoted context omitted.
> Also, mathematicians agree in this , seriously. Go and ask one. Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-p... . More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-p... , if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is sim…
But my point is, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not. 0^0 = 1 does not make calculus theorems more complicated to state or prove with modern language. It was a problem in XIX century, when mathematicians did not have a solid foundation w…
I'm a mathematician as well
As am I, by training if not by profession. As is lotharbot. You're in a thread full of mathematicians. :)
Which is what I would expect on this site, actually. I'm always timid making technical claims here unless I'm sure I'm correct; it seems to be a place frequented by arbitrarily large fish.
To answer your question, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not.
I'm not persuaded it is always the answer. I think the fact that it is an indeterminate form in limits is a forceful enough demonstration of that. It all depends on context. If I came across a 0^0 in, say, an engineering context, my first instinct would be to check whether the formula was defined in that case, not to just assume that 1 would work.
I mean, it's like 1/0. If you're working in R, that's simply illegal. If you're working in R*, it's the infinite point. If you're taking a limit, it means "unbounded". If you're working in my favorite field, the hyperreals, it could be any number of flavors of infinity depending on the flavor of zero it was.
It would be foolhardy to try to define the symbol; without a context to supply some sort of sense, it is nonsense. And that is how I feel about 0^0 as well.