False exponentials
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From: elmerturnipseed on 23 Mar 2010 01:04 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. We think of 5*5 as 5^2, two fives multiplied by each other, and 5^3 as three fives multiplied. What if the act of multiplying a number by itself was considered as one occurrence of this function? So 5^1 would be 5 multiplied by itself once, 25. And 5^2 would be 5*5*5. 5^0 would be 5, with the function performed 0 times. Would this new system of notation change the result of 0^0? (I seem to recall that negative exponents are dividers. Please correct me if I'm mistaken.) Using conventional notation: 0^2 = 0*0 = 0 0^1 = 0 0^0 = 1? 0^-1 = ? 0^-2 = 0/0 = impossible Using my notation: 0^2 = 0*0*0 = 0 0^1 = 0*0 = 0 0^0 = 0 0^-1 = 0/0 = impossible It doesn't seem possible that by changing the system of notation we actually change the laws of math. I think that the error in conventional exponential notation creates false exponentials. Ones that appear to exist, but don't. I can illustrate the flaw in conventional notation by amplifying it. Suppose whoever invented conventional notation had made an even larger error, for example that 5^5=5*5*5. 5^5 = 5*5*5 5^4 = 5*5 5^3 = 5 5^2 = ? 5^1 = ? 5^0 = ? 5^-1 = ? 5^-2 = ? 5^-3 = 5? 5^-4 = 5/5 As I hope you can see, the greater the notation error, the more false exponentials. So I must making an error somewhere? Or has someone thought of this before? If not, may I boldly propose that powers of 0 and -1 be considered false exponentials. Please criticize ruthlessly.
From: Virgil on 23 Mar 2010 02:16 In article <9cd6482b-f077-4e1b-9647-7a6f31f7a85d(a)u9g2000yqb.googlegroups.com>, elmerturnipseed <sterlingramone(a)gmail.com> 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. > > We think of 5*5 as 5^2, two fives multiplied by each other, and 5^3 as > three fives multiplied. > > What if the act of multiplying a number by itself was considered as > one occurrence of this function? > > So 5^1 would be 5 multiplied by itself once, 25. And 5^2 would be > 5*5*5. 5^0 would be 5, with the function performed 0 times. > > Would this new system of notation change the result of 0^0? (I seem to > recall that negative exponents are dividers. Please correct me if I'm > mistaken.) > > Using conventional notation: > > 0^2 = 0*0 = 0 > > 0^1 = 0 > > 0^0 = 1? > > 0^-1 = ? > > 0^-2 = 0/0 = impossible > > Using my notation: > > 0^2 = 0*0*0 = 0 > > 0^1 = 0*0 = 0 > > 0^0 = 0 > > 0^-1 = 0/0 = impossible > > > It doesn't seem possible that by changing the system of notation we > actually change the laws of math. I think that the error in > conventional exponential notation creates false exponentials. Ones > that appear to exist, but don't. > > I can illustrate the flaw in conventional notation by amplifying it. > Suppose whoever invented conventional notation had made an even larger > error, for example that 5^5=5*5*5. > > 5^5 = 5*5*5 > > 5^4 = 5*5 > > 5^3 = 5 > > 5^2 = ? > > 5^1 = ? > > 5^0 = ? > > 5^-1 = ? > > 5^-2 = ? > > 5^-3 = 5? > > 5^-4 = 5/5 > > As I hope you can see, the greater the notation error, the more false > exponentials. > > So I must making an error somewhere? Or has someone thought of this > before? > If not, may I boldly propose that powers of 0 and -1 be considered > false exponentials. > > Please criticize ruthlessly. 0^0 is a bit ambiguous in that it can be imagined as the limit of a function either as lim_{x -> 0+} 0^x = 0 or lim_{x -> 0} x^0 = 1. The function f(x,y) = x^y cannot be made continuous at (0,0), no matter what value is assigned to f(0,0), so it is a matter of convenience which value it will be assumed to take, if any. It is generally agree that it is more useful to consider 0^0 to be defined by the functions of the form g(y) = x^y, for fixed x rather than by h(x) = x^y, for fixed y. In which case, 0^0 = 1 makes g(y) = x^y continuous for x = 0 . There is a discusion of this and related points in "Concrete Mathematics, by Graham, Knith and Patashink.
From: William Elliot on 23 Mar 2010 03:42 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. What's the source of the article? ----
From: Norbert_Paul on 23 Mar 2010 03:54 This is no formal argument but should help your imagination: elmerturnipseed wrote: > Using conventional notation: .... where you made errors with the negative exponentials. > > 0^2 = 0*0 = 0 OK > > 0^1 = 0 OK > > 0^0 = 1? I leave this open. Virgil already answered that. > > 0^-1 = ? Wrong. 0^-1 would be 1/0 (either disallowed or mane it "infinity") > 0^-2 = 0/0 = impossible 0^-2 would also be 1/0 "=" 1/(0*0) So 0^0 could be imaginated some "threshold" between 0 and "infinity" (in IR +{"infinity"}). Norbert
From: Virgil on 23 Mar 2010 03:57
In article <20100323003446.V94049(a)agora.rdrop.com>, 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. > What's the source of the article? > > ---- As a matter of convenience, to avoid having to deal with exceptions, it quite often is defined, either as equalling 1, or, a bit less commonly, as equalling 0. The function f(x) = x^0 becomes continuous for all real x provided one has defined 0^0 to have the value 1, and that is quite useful when dealing with functions of the form g(x,y) = x^y. |