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alexb
Charter Member
2270 posts 
May2605, 00:34 AM (EST) 

1. "RE: Numbers raised to the power of 0"
In response to message #0

>Please can anyone explain to me why, when any real number is >raised to the power of zero, the result is 1. By definition, or you can say by common agreement. It's a convenience actually since then a^{b}·a^{b} = a^{b  b} = 1. However, most of the time 0^{0} is left undefined.


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DWCantrell
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May0406, 09:25 AM (EST) 

5. "RE: Numbers raised to the power of 0"
In response to message #1

>>Please can anyone explain to me why, when any real number is >>raised to the power of zero, the result is 1. > >By definition, or you can say by common agreement.When someone asks "Why?", I've never been very fond of answers of that sort, although such answers are indeed often given. I'm not fond of them because, to me at least, they seem to evade the basic issue: _Why_ is it that we commonly agree to define something in a certain way? >It's a convenience actually since then > >a^{b}·a^{b} = a^{b  b} = 1. Yes, that's a reason for why we define it that way. But in my opinion, the "real" reason is much more compelling than mere "convenience". Below, I give my favorite answer to the question "Why is x^{0} = 1?" I will also try to make my answer appropriate to a middle school audience. Unfortunately, trying to do that will make the answer lengthy, so please bear with me. >However, most of the time 0^{0} is left undefined. I'm not at all sure that's done most of the time _by mathematicians_. But I regret to say that I agree that it is done most of the time in elementary (pregraduate school level) texts. Therefore, I also briefly address the issue of 0^0 below.  As a preliminary, please consider the simple pattern 20 = 5*4 = 5 + 5 + 5 + 5 15 = 5*3 = 5 + 5 + 5 10 = 5*2 = 5 + 5 5 = 5*1 = 5 What does the next line in the pattern look like? How about 0 = 5*0 = perhaps? I grant that it looks strange. But my point is that, since we were removing a term of 5 on the right side each time, there are now _no_ terms of 5 remaining there. It's an "empty sum", a sum of no terms. (Of course, you shouldn't really write it as a blank space. In doing that, I was exercising my pedagogical license!) Why should an empty sum be 0? Well, 0 is the additive identity (that is, x + 0 = x for all x). In a sense then, 0 is the additive "background", what's always there, whether there are any other terms or not. So after all the term of 5 have been removed, we're left with just that background, the additive identity. Of course, having used terms of 5 was not important to the example; any empty sum must be 0 by the same reasoning, and so we often say "the empty sum", rather than "an empty sum". Now let's think about products instead of sums. Consider the pattern 81 = 3^{4} = 3*3*3*3 27 = 3^{3} = 3*3*3 9 = 3^{2} = 3*3 3 = 3^{1} = 3 1 = 3^{0} = 1 I went ahead and wrote out the line involving 3^{0}. (You already knew what it would be. Right?) We were removing a factor of 3 on the right side each time, and so, by the last line, there were no factors of 3 remaining. We got an empty product, a product of no factors. Now, since x*1 = x for all x, 1 is the multiplicative identity, and so 1 is our multiplicative "background", what we can always think of as being present in a product, whether there are any other factors or not. Once all the factors of 3 were gone, only the multiplicative background, 1, remained. But using 3 in our example was unimportant of course. Any empty product must be the multiplicative identity, 1, just as any empty sum must be the additive identity, 0. Thus we normally speak of "the empty product", rather than "an empty product". So let's think about x^{0} in general. How many factors of x does it have? Well, none, of course; the exponent tells us that it has 0 factors of x. Since there are _no_ factors of x involved in the product, our result should be _independent_ of x. And of course that's exactly what we find from the reasoning in the previous paragraph: x^{0} is the empty product, and must therefore be 1, regardless of x.  Digression about 0!  Have you ever heard of factorials? (I don't know if factorials are encountered in middle school or not.) For example, 6!, read as "six factorial", means 1*2*3*4*5*6, and so its value is 720. Generally, for a positive integer N, we could write N! = 1*2*3*4*...*N. So let's define N! as being the product of the first N positive integers. Now at first glance, most people would be inclined to think that the definition in the previous sentence would be applicable only when N is a _positive_ integer. But that's not quite true. Our definition still works when N = 0, and that's good because it turns out that, in using factorials, we often need to deal with 0! (Before reading further, can you give the value of 0! and explain why it must have that value?) If we use our definition of factorial, we find that 0! should be the product of the first 0 positive integers. Ah! It's the product of no factors, and thus it's the empty product, and so it must be 1. I included this digression because students often ask "Why is 0! = 1?" and teachers often reply simply "By definition." But of course such an answer is ultimately unsatisfying because it doesn't tell the students _why_ 0! is defined to be 1. My favorite explanation is that in the paragraph above. So 0! and x^{0} are seen to be 1 for exactly the same reason: They're both the empty product.  End digression  Finally, I need to address the matter, raised by Alex, of 0^{0}. To deal with this quickly, I could merely repeat what I had said right before the digression: x^{0} is the empty product, and must therefore be 1, _regardless_ of x. Then, as a special case, we have 0^{0} = 1. I wish I could just stop with that and say "Case closed." But then you would wonder why Alex had said "However, most of the time 0^{0} is left undefined." Since I've already presented an argument presumably showing that 0^{0} must be 1, perhaps I should also now, in fairness, present reasons for leaving 0^{0} undefined. But I can't! I am aware of not a single compelling mathematical argument for leaving 0^{0} undefined. So then you're wondering: Why is 0^{0} often left undefined, at least in lower level texts? Unfortuantely, to explain that gaffe would take us far beyond what's appropriate for a middle school audience. (If anyone's curious, I'd be willing to give my opinion about this matter in either the College Math or This and That part of CTK.) I'll close with a quotation from _Concrete Mathematics_ by Ronald Graham, Donald Knuth, and Oren Patashnik (AddisonWesley, 2nd ed., 1994) p. 162: "Some textbooks leave the quantity 0^{0} undefined... But this is a mistake. We must define x^{0} = 1, for all x..." Regards, David W. Cantrell 

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alexb
Charter Member
2270 posts 
May0406, 09:58 AM (EST) 

6. "RE: Numbers raised to the power of 0"
In response to message #5

>>>Please can anyone explain to me why, when any real number is >>>raised to the power of zero, the result is 1. >> >>By definition, or you can say by common agreement. > >When someone asks "Why?", I've never been very fond of >answers of that sort, although such answers are indeed >often given. I'm not fond of them because, to me at least, >they seem to evade the basic issue: >_Why_ is it that we commonly agree to define something in a >certain way? Well, no one disagrees with you. As you have noticed, the claim about the fact of the definition has been followed by an explanation, however short. I think it is a good practice to start with 12 sentences underlying the jist of the coming response, and then proceed with additional verbiage. >>It's a convenience actually since then >> >>a^{b}·a^{b} = a^{b  b} = 1. > >Yes, that's a reason for why we define it that way. But in >my opinion, the "real" reason is much more compelling than You mean an "additional" reason? >I'm not at all sure that's done most of the time _by >mathematicians_. But I regret to say that I agree that it is >done most of the time in elementary (pregraduate school level) >texts. ... and any level Calculus book, of course. >Therefore, I also briefly address the issue of 0^0 below. > >we can always think of as being present in a product, >whether there are any other factors or not. This is a delicate point. We are not exactly talking of the integers. Would you call x a factor in x×1/x = 1. (I understand what you wanted to say and accept that. The above is to underscore a possible ambiguity in your argument.) >x^{0} is the empty product, and must >therefore be 1, regardless of x. The argument is nice, but by the same token, 0^{x} = 0 regardless of x. >Since I've already presented an argument presumably showing >that 0^{0} must be 1, perhaps I should also now, in >fairness, present reasons for leaving 0^{0} >undefined. But I can't! I am aware of not a single >compelling mathematical argument for leaving 0^{0} >undefined. I do not believe that. To me, the ambiguity created by "0^{x} = 0 for all x" is sufficiently compelling to have second thoughts. >So then you're wondering: Why is 0^{0} often left >undefined, at least in lower level texts? With a view to Calculus and for the same reason that 0/0 is left undefined: there are ways and ways to approach 0. >Unfortuantely, to >explain that gaffe would take us far beyond what's >appropriate for a middle school audience. This is why ... Is it not? I can't be 100% sure, but perhaps the sources chose to avoid giving an explanation that "would take the audience far beyond what's appropriate for a middle school audience." >I'll close with a quotation from _Concrete Mathematics_ by >Ronald Graham, Donald Knuth, and Oren Patashnik >(AddisonWesley, 2nd ed., 1994) p. 162: > >"Some textbooks leave the quantity 0^{0} >undefined... But this is a mistake. We must define >x^{0} = 1, for all x..." With all due respect, I understand their argument and agree that in Discrete Mathematics it's rather a great convenience to have 0^{0} = 0. But to say it's a mistake to have Calculus in mind, is a misteak. (See B. Cipra's or E. J. Barbeau's books.) 

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DWCantrell
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May0506, 08:20 AM (EST) 

7. "RE: Numbers raised to the power of 0"
In response to message #6

>>>>Please can anyone explain to me why, when any real number is >>>>raised to the power of zero, the result is 1.>>>It's a convenience actually since then >>> >>>a^{b}·a^{b} = a^{b  b} = 1. >> >>Yes, that's a reason for why we define it that way. But in >>my opinion, the "real" reason is much more compelling than > >You mean an "additional" reason? I must confess to being somewhat of a Platonist. From that viewpoint, I meant _the real reason_. :) Convenience can be a reason for doing certain things, no doubt. Given a choice of several things, their only differences being in matters of convenience, we would certainly choose the most convenient one. But if there are differences of a _fundamental_ nature between the alternatives, then convenience should not govern our choice. Concerning leaving 0^{0} undefined: >>I'm not at all sure that's done most of the time _by >>mathematicians_. But I regret to say that I agree that it is >>done most of the time in elementary (pregraduate school level) >>texts. > >... and any level Calculus book, of course. No. For example, just look in their sections on power series: x^{0} = 1, even when x = 0. (Perhaps you can find some texts which don't do that, but that is done in all the texts I've taught from.) >>Therefore, I also briefly address the issue of 0^0 below. >> >>we can always think of as being present in a product, >>whether there are any other factors or not. > >This is a delicate point. We are not exactly talking of the >integers. I'm not certain that I understand your point. Are you perhaps wishing to distinguish between the integer 0 and the real number 0 when used as exponents? > Would you call x a factor in > >x×1/x = 1. Yes, the expression on the left does have a factor of x. Indeed, it has factors of x in both numerator and denominator. (Furthermore, I'll note that those factors of x can legitimately be "cancelled", giving 1 as the result, except when x = 0.) >(I understand what you wanted to say and accept that. The >above is to underscore a possible ambiguity in your >argument.) What is that possible ambiguity? >>x^{0} is the empty product, and must >>therefore be 1, regardless of x. > >The argument is nice, but by the same token, 0^{x} = >0 regardless of x. I don't know what you mean by "the same token". 0^{x} is not an empty product unless x happens to be 0. We should have 0^{x} = 0 if x > 0, 1 if x = 0, and (assuming we want to stay in the real number system) undefined if x < 0. >>Since I've already presented an argument presumably showing >>that 0^{0} must be 1, perhaps I should also now, in >>fairness, present reasons for leaving 0^{0} >>undefined. But I can't! I am aware of not a single >>compelling mathematical argument for leaving 0^{0} >>undefined. > >I do not believe that. Well, it's true. Note the important word "compelling". >To me, the ambiguity created by >"0^{x} = 0 for all x" is sufficiently compelling to >have second thoughts. I'd rather that you had second thoughts about 0^{x} = 0 for negative x! ;) >>So then you're wondering: Why is 0^{0} often left >>undefined, at least in lower level texts? > >With a view to Calculus and for the same reason that 0/0 is >left undefined: there are ways and ways to approach 0. Approach? There is no "approaching" going on, at least not in what I've been talking about. I have been talking about 0^{0}, which is an _arithmetic_ expression. The base and the exponent are _constant_. (I cannot stress that point too strongly.) But I now see what is perhaps causing your consternation, Alex. When we consider limits in Calculus, we have certain forms which are called indeterminate. For example, knowing merely that x and y both approach 0, we cannot determine the limit of x^{y}. Depending on _how_ x and y approach 0, many different answers are possible. For this reason, this form of limit, in which base and exponent both approach 0, is called indeterminate. Now for better or worse  and I think the latter  this indeterminate limit form is often denoted simply as 0^{0}. But this shorthand can easily cause confusion. It _looks_ like a mere arithmetic expression, but it's not. It would be somewhat better, I think, if this indeterminate limit form were abbreviated as "0^{0}", in the hope that the quotation marks would provide a clue to the reader that we are not literally talking about 0^{0}. In summary: When we refer to the limit form "0^{0}", we are talking about a situation in which the base and exponent _approach_ 0. This form is rightly called indeterminate. But in contrast, in the arithmetic expression 0^{0}, the base and exponent are _fixed_. One should not confuse the arithmetic expression and the limit form! >>Unfortuantely, to >>explain that gaffe would take us far beyond what's >>appropriate for a middle school audience. > >This is why ... Is it not? I can't be 100% sure, but perhaps >the sources chose to avoid giving an explanation that "would >take the audience far beyond what's appropriate for a middle >school audience." Yes, I suspect that confusion between the arithmetic expression and the limit form could well contribute to the problem. >>I'll close with a quotation from _Concrete Mathematics_ by >>Ronald Graham, Donald Knuth, and Oren Patashnik >>(AddisonWesley, 2nd ed., 1994) p. 162: >> >>"Some textbooks leave the quantity 0^{0} >>undefined... But this is a mistake. We must define >>x^{0} = 1, for all x..." > >With all due respect, I understand their argument and agree >that in Discrete Mathematics it's rather a great convenience >to have 0^{0} = 0. You intended to say 1. >But to say it's a mistake to have Calculus in mind, is a misteak. Of course, it's no mistake to have Calculus in mind. The mistake is to think that Calculus would keep us from defining 0^{0} as it should be! The fact that "0^{0}" is an indeterminate limit form has no bearing on whether 0^{0} should be defined or not. >(See B. Cipra's or E. J. Barbeau's books.) I'd be happy to look at any particular passages you'd care to mention, if you still think they're relevant. Regards, David W. Cantrell 

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alexb
Charter Member
2270 posts 
May0506, 10:44 AM (EST) 

9. "RE: Numbers raised to the power of 0"
In response to message #7

>>You mean an "additional" reason? > >I must confess to being somewhat of a Platonist. From that >viewpoint, I meant _the real reason_. :) No doubt. I just would add an IMHO. >>... and any level Calculus book, of course. > >No. For example, just look in their sections on power >series: x^{0} = 1, even when x = 0. (Perhaps you can >find some texts which don't do that, but that is done in all >the texts I've taught from.) It was a while that I last taught Calculus (or, for that matter, anything else). If the memory serves, they did universally use the shorthand Sa_{k}x^{k} for the infinite sum a_{0} + a_{1}x + ..., but I have no recollection of any instance where the proper point has been made or the matter of x^{0} = 1 at all taken up. > >>>Therefore, I also briefly address the issue of 0^0 below. >>> >>>we can always think of as being present in a product, >>>whether there are any other factors or not. >> >>This is a delicate point. We are not exactly talking of the >>integers. > >I'm not certain that I understand your point. Are you >perhaps wishing to distinguish between the integer 0 and the >real number 0 when used as exponents? No, I just meant to point out that the terminology here became somewhat lax. With integers, there is no question about a factor being present in or absent from a product. It's not the case with real x's. You admit this in the next paragraph. Consider what you said previously: "So let's think about x^{0} in general. How many factors of x does it have? Well, none, of course; the exponent tells us that it has 0 factors of x."
In the context of real x, asking "how many factors" is different from having this question for integer x. Suggesting that along with x comes also 1/x is not altogether compelling because:
 there could be more factors, like x·y^{2}·1/(xy)·...
 this brings up the simple conventional argument that x^{a  a} = 1.
>> Would you call x a factor in >> >>x×1/x = 1. > >Yes, the expression on the left does have a factor of x. >Indeed, it has factors of x in both numerator and >denominator. (Furthermore, I'll note that those factors of x >can legitimately be "cancelled", giving 1 as the result, >except when x = 0.) > >>(I understand what you wanted to say and accept that. The >>above is to underscore a possible ambiguity in your >>argument.) > >What is that possible ambiguity? As above: too many factors may conceal a mutual cancellation of x and 1/x or just to bring up the basic x^{a  a} = 1. Ambiguity may not be the right word. But "circular reasoning" may. >>>x^{0} is the empty product, and must >>>therefore be 1, regardless of x. >> >>The argument is nice, but by the same token, 0^{x} = >>0 regardless of x. > >I don't know what you mean by "the same token". >0^{x} is not an empty product unless x happens to be >0. We should have "Going to the extreme": having an empty product, admittedly a nice concept, is analogous to taking 0 to a "no power". For any (positive) power it's 0. Hence, it is also 0 for a nopower. >0^{x} = 0 if x > 0, 1 if x = 0, and (assuming we >want to stay in the real number system) undefined if x < 0. > >>>Since I've already presented an argument presumably showing >>>that 0^{0} must be 1, perhaps I should also now, in >>>fairness, present reasons for leaving 0^{0} >>>undefined. But I can't! I am aware of not a single >>>compelling mathematical argument for leaving 0^{0} >>>undefined. >> >>I do not believe that. > >Well, it's true. Note the important word "compelling". > >>To me, the ambiguity created by >>"0^{x} = 0 for all x" is sufficiently compelling to >>have second thoughts. > >I'd rather that you had second thoughts about 0^{x} >= 0 for negative x! ;) Always have them. I even dream twice a night. Monogamy is especially hard on me. >>>So then you're wondering: Why is 0^{0} often left >>>undefined, at least in lower level texts? >> >>With a view to Calculus and for the same reason that 0/0 is >>left undefined: there are ways and ways to approach 0. > >Approach? There is no "approaching" going on, at least not >in what I've been talking about. I have been talking about >0^{0}, which is an _arithmetic_ expression. The base >and the exponent are _constant_. (I cannot stress that point >too strongly.) But I now see what is perhaps causing your >consternation, Alex. No, no. It's much rather an amusement. I simply do not believe that there is always one best way or vene one _real_ explanation. As a platonist, I believe in a beauty of an argument on its own merits. I absolutely enjoyed your "empty sum" and "empty product" concepts. It's just I can envisage that for somebody this arumentation will be outlandish, let alone un _real_. >When we consider limits in Calculus, we have certain forms >which are called indeterminate ... we are not literally >talking about 0^{0}. We are not; somebody is. >One should not confuse the >arithmetic expression and the limit form! They will, they will ... You'll yet be surprised. >>>Unfortuantely, to >>>explain that gaffe would take us far beyond what's >>>appropriate for a middle school audience. >> >>This is why ... Is it not? I can't be 100% sure, but perhaps >>the sources chose to avoid giving an explanation that "would >>take the audience far beyond what's appropriate for a middle >>school audience." > >Yes, I suspect that confusion between the arithmetic >expression and the limit form could well contribute to the >problem. >>But to say it's a mistake to have Calculus in mind, is a misteak. > >Of course, it's no mistake to have Calculus in mind. The >mistake is to think that Calculus would keep us from >defining 0^{0} as it should be! Again, I see your point and enjoy your argument but I can't accept the value judgement concealed behind "... as it should be." Look how far we (the humanity, I mean) got (developing mathematics) without this point having been settled. >>(See B. Cipra's or E. J. Barbeau's books.) > >I'd be happy to look at any particular passages you'd care >to mention, if you still think they're relevant. What's relevant here is an inordinate amount of ways the kids get confused in. The "empty sum" and "empty product", however natural they are to your (or mine) mind's organization are bound to confuse some. I can't substantiate my opinion in any way but I think that they may be enjoyable more than helpful. In other words, they provide an additional angle to what one has already accepted or they bring more confusion to one who is struggling anyway.


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DWCantrell
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May1606, 10:38 PM (EST) 

10. "RE: Numbers raised to the power of 0"
In response to message #9

>>>... and any level Calculus book, of course. >> >>No. For example, just look in their sections on power >>series: x^{0} = 1, even when x = 0. (Perhaps you can >>find some texts which don't do that, but that is done in all >>the texts I've taught from.) > >It was a while that I last taught Calculus (or, for that >matter, anything else). If the memory serves, they did >universally use the shorthand Sa_{k}x^{k} for the >infinite sum a_{0} + a_{1}x + ..., but I >have no recollection of any instance where the proper point >has been made or the matter of x^{0} = 1 at all >taken up.Sorry for my delay in responding, Alex. When talking about empty products, it's particularly nice to consider Taylor's series. The k^{th} term involves f^{(k)}(a), k!, and (x  a)^{k}. When k happens to be 0, _all three_ of these quantities are empty "products". As mentioned earlier in this thread, the latter two (which are products, literally) are 1, the identity element for multiplication. But what about f^{(k)}(a) when k happens to be 0? In other words, what must the 0^{th} derivative of our function be? Well, instead of writing the k^{th} derivative of f as f^{(k)}, it's nicer here to write it using the alternative notation D^{k}f. In this form, it's clear that the operator D must be applied k times. (For example, D^{2}, the "product" of D and D, is an operator causing us to differentiate twice.) Then D^{0} is the empty "product" of the operator D. Therefore, it must be the _identity_ for operators, and that identity operator applied to our function f gives just f itself: D^{0}f = f. Of course, I suppose that most people's intuition tells them that, differentiating 0 times, the function f is unchanged. But I think it's also nice to see, via the empty product principle, that D^{0} must be the identity operator. >>Of course, it's no mistake to have Calculus in mind. The >>mistake is to think that Calculus would keep us from >>defining 0^{0} as it should be! > >Again, I see your point and enjoy your argument but I can't >accept the value judgement concealed behind "... as it >should be." If it would make you any happier, feel free to replace "should" by "must", above and below. Consider the statement "2^{3} = 8 is as it should be." Do you really think that there is some unacceptable value judgment concealed therein? I don't. Similarly, I don't hesitate to say that 0^{0} = 1 is as it should be. >Look how far we (the humanity, I mean) got >(developing mathematics) without this point having been >settled. Yes, I see your point. But... This morning, while pulling up a vine in my garden, I got a thorn in my finger. The thorn didn't hurt much at all. I could have continued with my gardening, without any noticeable detriment, without ever having removed the thorn. But why not remove it at the first opportunity? To me, having 0^{0} undefined is just such a thorn. Sure, we could live with it and do well enough. But why not remove the thorn? >>>(See B. Cipra's or E. J. Barbeau's books.) >> >>I'd be happy to look at any particular passages you'd care >>to mention, if you still think they're relevant. > >What's relevant here is an inordinate amount of ways the >kids get confused in. The "empty sum" and "empty product", >however natural they are to your (or mine) mind's >organization are bound to confuse some. I can't substantiate >my opinion in any way but I think that they may be enjoyable >more than helpful. In other words, they provide an >additional angle to what one has already accepted or they >bring more confusion to one who is struggling anyway. You raise pedagogical concerns. Clearly, such concerns are valid. But of course we know that what is _pedagogically_ nice cannot dictate what is _mathematically_ correct. What is mathematically correct is determined first; only subsequently may we consider how that can best be taught. But in fact, I consider having x^{0} = 1 for _all_ x to be not only a mathematical necessity, but also pedagogically easier to handle than leaving 0^{0} undefined. After all, if we explain well why _any_ 0^{th} power must be 1, then we are done. Period. (And then, much later, when the student is studying limits, he will be cautioned that the fact that the limit form called "0^{0}" is indeterminate has nothing to do with 0^{0} = 1.) But if we were to have x^{0} = 1 for all nonzero x, leaving 0^{0} undefined, we would have to explain to students not only why x^{0} should be 1 for all nonzero x, but also explain why 0^{0} should be left undefined. I could not, in honesty, do the latter. As I said before, I know of no compelling mathematical reason for leaving 0^{0} undefined. But if you know of any such reason, Alex (or anyone else), please give it! Regards, David Cantrell 

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alexb
Charter Member
2270 posts 
May1706, 07:54 AM (EST) 

11. "RE: Numbers raised to the power of 0"
In response to message #10

>When talking about empty products, it's particularly nice to >consider Taylor's series. The k^{th} term involves > >f^{(k)}(a), k!, and (x  a)^{k}. > >When k happens to be 0, _all three_ of these quantities are >empty "products". You are going to answer a question I did not ask. All I said was that I do not know of any Calculus book that took up the matter of 0^{0} = 1. >>>Of course, it's no mistake to have Calculus in mind. The >>>mistake is to think that Calculus would keep us from >>>defining 0^{0} as it should be! Who does? >>Again, I see your point and enjoy your argument but I can't >>accept the value judgement concealed behind "... as it >>should be." > >If it would make you any happier, This is besides the point. I am quite happy as it is. And my happiness or its degree is irrelevant to the discussion. >Consider the statement "2^{3} = 8 is as it should >be." Do you really think that there is some unacceptable >value judgment concealed therein? I don't. I do not. >Similarly, ??? >I don't hesitate to say that 0^{0} = 1 is >as it should be. I've been reading with my 6 year old a small book "Why did not I say that?" on a matter of a suitable repartee under various circumstances. The thing that just leaps out of my mouth is "Do you ever?"  Sorry for that. But seriously, if the above is not a value judgement, I do not know what is. >To me, having 0^{0} undefined is just such a thorn. >Sure, we could live with it and do well enough. But why not >remove the thorn? By all means. Please go ahead. >You raise pedagogical concerns. Clearly, such concerns are >valid. But of course we know that what is _pedagogically_ >nice cannot dictate what is _mathematically_ correct. We are not going to start a discussion on mathematical correctness, are we? >What >is mathematically correct is determined first; only >subsequently may we consider how that can best be taught. > >But in fact, I consider having x^{0} = 1 for _all_ x >to be not only a mathematical necessity, but also >pedagogically easier to handle than leaving 0^{0} >undefined. After all, if we explain well why _any_ >0^{th} power must be 1 This you did explain really very well. >, then we are done. But if only life were that easy >Period. Ellipses. >undefined. But if you know of any such reason, Alex (or >anyone else), please give it! There is none. Now what? As I said, I liked your argument and the notion of an empty product. All you need now is a magic wand. I wish you well, Alex 

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Janine
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Jun2607, 08:34 PM (EST) 

17. "RE: Numbers raised to the power of 0"
In response to message #5

The issue of 0^0 still intrigues me. 0^n = 0 for all n>0 0^n is undefined for n<0 But from your argument you state that 0^0 = 1. So while f(x) = x^n is usually a continuous function, the results above imply that this is only true if x does not equal 0. Is this a simple explanation as to why 0^0 is left undefined since a base of 0 breaks the rules of continuity anyway? I am very hazy on my limits but it'seems strange to contemplate that lim n approaches 0 of 0^n = 1 when the nth root of 0 is 0. 

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Mark Huber
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May2605, 07:38 PM (EST) 

2. "RE: Numbers raised to the power of 0"
In response to message #0

>Please can anyone explain to me why, when any real number is >raised to the power of zero, the result is 1. As Alex noted, 0^0 is usually left undefined, but you are correct, a^0 = 1 for any real number a not equal to zero. Alex gave the practical reason: it is that way so that the rule a^(b_1)*a^(b_2) = a^(b_1 + b_2) holds. But if you go deeper into the definition of exponentiation, it also holds. Defining how to raise a real number to a nonzero integer is easy, and from there it is a short step to raising any real number to any nonzero rational number. But then there is the question, what is a^b where a is any positive real and b is any real number or zero? There are two directions that you can take in defining exponents of this form, and both are commonly used. The first is for a real number b, consider any sequence of numbers b_1,b_2,... that converge to b. Then let a^b := lim_{n goes to infinity} a^(b_n). The only tricky part here is showing that this definition makes sense: that is, for *any* sequence that converges to b, that you get the same limit. In fact this is true: any sequence that converges to b will have the same limiting value of a^(b_n). For the zero case then a^0 = lim_{n goes to infinity} a^(1/n) which is just 1. The second method used to define exponentials first defines what e^x is for all real x as the sum of a convergent series: e^x := 1 + x + x^2/2 + x^3/3! + ... This function e^x can be shown to be increasing, so the inverse exists and the natural logarithm (written ln(x)) is this inverse. Now exponentials can be defined: a^b := e^(b*ln(a)) No matter what a is, when b = 0, b*ln(a) = 0 and a^0 = e^0. But from the definition of e^x: e^0 = 1 + 0 + 0^2/2 + 0^3/3! + ... = 1. Hence a^0 = 1. The advantage of this definition of exponentiation over the first is that (while less intuitive) it extends in a natural way to deal with negative numbers raised to real exponents. 

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Ryan
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Dec1606, 06:12 PM (EST) 

16. "RE: Numbers raised to the power of 0"
In response to message #0

Consider the limit as n approaches infinite of x^(1/n). Once we get out far enough towards infinite it becomes x^0=1. What this says is that succesive square rooting of any number will get us close enough to 1. It is easy to demonstrate this numerically using a calculator. The formal proof is very ugly. 

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Karen Kaul
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Jul3007, 07:38 AM (EST) 

19. "RE: Numbers raised to the power of 0"
In response to message #0

It must be that way in order for the number system to be consistent. If you look at the pattern of numbers raised to different powers, going down to the power of zero, you will see. Example (each time you divide by 10, so you will get 1 when you divided 10 by 10) 10 to 5th power = 100000 (10x10x10x10x10) 10 to 4th power = 10000 10 to 3rd power = 1000 10 to 2nd power= 100 10 to 1st power = 10 10 to zero power = 1 likewise 10 to 1 power = 0.1It works for any number 5 to 3rd power = 125 5 to 2nd power = 25 (125 divided by 5) 5 to 1st power = 5 (25 divided by 5) 5 to zero power = 1 (5 divided by 5)


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Chris Tolley
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Sep3007, 05:31 AM (EST) 

20. "RE: Numbers raised to the power of 0"
In response to message #0

Consider any numeber x raised to a ppower of n. When x raised to n is divided by x riased to the same power n it is now equivalent to x raised to the power nn, where nn = 0, so the number x is now raised to 0. But x raised to n divided by the same number xraised to n is equal to 1 and that explains why a number raised to 0 equals 1.


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