False exponentials
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From: William Elliot on 24 Mar 2010 07:05 On Wed, 24 Mar 2010, Jussi Piitulainen wrote: > William Elliot writes: >> On Tue, 23 Mar 2010, Marc Olschok wrote: >>> William Elliot <marsh(a)rdrop.remove.com> wrote: >>>> It doesn't and the article is wrong. >>>> 0^0 isn't defined, doesn't exist. >>> >>> This depends on where your 0 lives. >>> If you only deal with natural numbers, it makes sense to >>> view them as cardinalities of finite sets. >> >> Are negative numbers somehow unnatural? >> Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * oo. > > Would you say that 0^3 = 0^(7 - 4) = 0^7 * 0^-4 = 0 * oo? Or would you > rather admit that this particular argument for leaving 0^0 undefined > is bogus? > Read what I said at the beginning. Your equation says that for 0 to be a base for exponentation and for the laws of exponents for integers to hold, is bogus. That one or the other has to fail. It's much stronger than my statement and equation which backs my first statement, assuming exponentation over all integers of any integer. There are two different real valued exponentation function: g:(0,oo)xR -> (0,oo), (x,y) -> x^y and h:R\0 x Z -> R\0, (x,n) -> x^n. Real variable calculus uses g; basic algebra uses h.
From: Jussi Piitulainen on 24 Mar 2010 07:53 William Elliot writes: > On Wed, 24 Mar 2010, Jussi Piitulainen wrote: > > William Elliot writes: > >> On Tue, 23 Mar 2010, Marc Olschok wrote: > >>> William Elliot <marsh(a)rdrop.remove.com> wrote: > > >>>> It doesn't and the article is wrong. > >>>> 0^0 isn't defined, doesn't exist. > >>> > >>> This depends on where your 0 lives. > >>> If you only deal with natural numbers, it makes sense to > >>> view them as cardinalities of finite sets. > >> > >> Are negative numbers somehow unnatural? > >> Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * > >> oo. > > > > Would you say that 0^3 = 0^(7 - 4) = 0^7 * 0^-4 = 0 * oo? Or would > > you rather admit that this particular argument for leaving 0^0 > > undefined is bogus? > > Read what I said at the beginning. Sorry, I don't know where that beginning is. > Your equation says that for 0 to be a base for exponentation and for > the laws of exponents for integers to hold, is bogus. That one or > the other has to fail. It's much stronger than my statement and > equation which backs my first statement, assuming exponentation over > all integers of any integer. To me, 0^(7 - 4) = 0^7 * 0^-4 seems to use the same alleged law of exponents as your 0^(1 - 1) = 0^1 * 0^-1. What difference do you see? Do you think 0^n is defined for some n? > There are two different real valued exponentation function: > g:(0,oo)xR -> (0,oo), (x,y) -> x^y > and > h:R\0 x Z -> R\0, (x,n) -> x^n. > > Real variable calculus uses g; basic algebra uses h. I notice that your g(0,y) is undefined for all y, and h(0,n) is undefined for all n, but here you don't say anything about 0^y or 0^n for any y or n. Do you think 0^x is defined for some x?
From: William Elliot on 25 Mar 2010 02:15 On Wed, 24 Mar 2010, Jussi Piitulainen wrote: The beginning. >>>>>> It doesn't and the article is wrong. >>>>>> 0^0 isn't defined, doesn't exist. >>>>> >>>>> This depends on where your 0 lives. >>>>> If you only deal with natural numbers, it makes sense to >>>>> view them as cardinalities of finite sets. >>>> >>>> Are negative numbers somehow unnatural? >>>> Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * >>>> oo. >>> >>> Would you say that 0^3 = 0^(7 - 4) = 0^7 * 0^-4 = 0 * oo? Or would >>> you rather admit that this particular argument for leaving 0^0 >>> undefined is bogus? >> >> Read what I said at the beginning. > > Sorry, I don't know where that beginning is. > It's marked above. >> Your equation says that for 0 to be a base for exponentation and for >> the laws of exponents for integers to hold, is bogus. That one or >> the other has to fail. It's much stronger than my statement and >> equation which backs my first statement, assuming exponentation over >> all integers of any integer. > > To me, 0^(7 - 4) = 0^7 * 0^-4 seems to use the same alleged law of > exponents as your 0^(1 - 1) = 0^1 * 0^-1. What difference do you see? > > Do you think 0^n is defined for some n? > It depends upon the context or upon what you wish to toss to define it. >> There are two different real valued exponentation function: >> g:(0,oo)xR -> (0,oo), (x,y) -> x^y >> and >> h:R\0 x Z -> R\0, (x,n) -> x^n. >> >> Real variable calculus uses g; basic algebra uses h. > > I notice that your g(0,y) is undefined for all y, and h(0,n) is > undefined for all n, but here you don't say anything about 0^y or 0^n > for any y or n. > Of course, (0,y) is not in the domain. > Do you think 0^x is defined for some x? > That depends upon the context. For example, it's defined for all y in the context k:[0,oo)^2 -> [0,oo), (x,y) -> x^y.
From: Jay Belanger on 25 Mar 2010 09:08 William Elliot <marsh(a)rdrop.remove.com> writes: > On Wed, 24 Mar 2010, Jussi Piitulainen wrote: > > The beginning. > >>>>>>> It doesn't and the article is wrong. >>>>>>> 0^0 isn't defined, doesn't exist. >>>>>> >>>>>> This depends on where your 0 lives. >>>>>> If you only deal with natural numbers, it makes sense to >>>>>> view them as cardinalities of finite sets. >>>>> >>>>> Are negative numbers somehow unnatural? >>>>> Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * >>>>> oo. >>>> >>>> Would you say that 0^3 = 0^(7 - 4) = 0^7 * 0^-4 = 0 * oo? Or would >>>> you rather admit that this particular argument for leaving 0^0 >>>> undefined is bogus? >>> >>> Read what I said at the beginning. >> >> Sorry, I don't know where that beginning is. >> > It's marked above. The beginning is >>>>>>> It doesn't and the article is wrong. >>>>>>> 0^0 isn't defined, doesn't exist. ? That isn't helpful, and not everybody agrees with that. Also, I don't see why 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * oo is a criticism of defining 0^0 while 0^3 = 0^(7 - 4) = 0^7 * 0^-4 = 0 * oo isn't a criticism of defining 0^3. >>> Your equation says that for 0 to be a base for exponentation and for >>> the laws of exponents for integers to hold, is bogus. Both equations say that. I've never heard anybody suggest that 0^n is defined for all integers n. Your equation uses 0^n for integral n to show a problem with 0^0; the next equation shows that the same argument can be used to show that there is a problem with 0^n for any integer n. The conclusion should be that 0^n shouldn't be defined for all integers n, but it doesn't say anything about 0^0.
From: Marc Olschok on 25 Mar 2010 14:59
William Elliot <marsh(a)rdrop.remove.com> wrote: > On Tue, 23 Mar 2010, Marc Olschok wrote: > > William Elliot <marsh(a)rdrop.remove.com> wrote: > >> On Mon, 22 Mar 2010, elmerturnipseed wrote: > >> > >>> Please forgive my ignorance. I failed math in school, but I just read > >>> an article about how 0^0=1, and I'm wondering if maybe there's > >>> something wrong the way that exponential notation is traditionally > >>> expressed. > >>> > >> It doesn't and the article is wrong. > >> 0^0 isn't defined, doesn't exist. > > > > This depends on where your 0 lives. > > If you only deal with natural numbers, it makes sense to > > view them as cardinalities of finite sets. > > Are negative numbers somehow unnatural? You really did not now the expression "natural number" as synonym for "nonnegative integer" ? > Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * oo. One has to be more precise. In fact exponentiation is an operation of two variables, so for general discussions like this one should make clear where both the base b and the exponent e in an expression like "b^e" are supposed to live, and also specify whether "^" should be total or not. For the quite common case of ^ : N x N --> N (i.e. base and exponent nonnegative integers) I have already outlined why 0^0 = 1 makes sense. For the related case of ^ : Z x N --> Z (base and exponent integer, exponent nonnegative) or ^ : A x N --> A where A is some ring, it also makes sense to have 0^0 = 1 in order for the the exponent laws to work properly. In all the above situations 0^0 = 1 is well established so your "0^0 isn't defined, doesn't exist." misses large parts of Mathematics. Now let's turn to your above reasoning with 0^0 = 0^(1 - 1) = 0^1 * 0^-1. If exponentiation is of type ^ : Z x Z --> Z then one could as well use 2^0 = 2^(1 - 1) = 2^1 * 2^-1 as an argument against 2^0 = 1, which is hopefully not your intention. But let's extend the type to ^ : Z x Z --> Q (with Q the rationals). Then your example still applies and illustrates that the laws of exponents do not work properly if one insists on ^ being everywhere defined. But your example also illustrates that the _negative_ exponents pose the problem, not the exponent 0. So you need to allow ^ to be a partial operation from Z x Z to Q which is not defined at the subset {0} x { z in Z | z < 0 }. The law of exponents then reads: | the equation x^(i+j) = x^i * x^j holds whenever both sides are defined. and your examples is no more. In particular 0^0 = 1 still makes sense. Of course other types of exponentiation may require different restrictions. -- Marc |