[math-fun] Much ado about nothing (was Re: Proof that 0^0 = 1)
As a computer guy, I consider real numbers and integers to be two very different things. The English language shares this distinction, in that real numbers go with "less" and integers go with "fewer." I'll use 0 to mean the integer zero and 0.0 to mean the real zero. I would say that 0^0 = 1, and 0.0^0.0 is undefined. If anyone disagrees, I'd like to know why. I'd also like to know if anyone has an opinion on 0^0.0 or 0.0^0. And if sticking a minus sign in anywhere would change anything. Thanks.
If I were explaining this to a skeptical student or a curious non-scientific friend, I would say that x^0 represents an expression in which x appears zero times, i.e. not at all, and thus must be the same for all x. Which would suggest 0.0^0=1 with 0^0.0 undefined. On Sat, Jun 13, 2020 at 2:57 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
As a computer guy, I consider real numbers and integers to be two very different things. The English language shares this distinction, in that real numbers go with "less" and integers go with "fewer."
I'll use 0 to mean the integer zero and 0.0 to mean the real zero.
I would say that 0^0 = 1, and 0.0^0.0 is undefined. If anyone disagrees, I'd like to know why. I'd also like to know if anyone has an opinion on 0^0.0 or 0.0^0. And if sticking a minus sign in anywhere would change anything. Thanks.
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If we have a finite set of discrete real numbers, {x_i}, then we can ask: What values of a,b,c minimize, respectively \sum_i | xi - a |^2 \sum_i | xi - b |^1 \sum_i | xi - c |^0 a = mean b = median c = mode, if we define 0^0 = 0 (or 0.0^0 = 0) I find this useful when discussing mean, median, and mode with middle-school students, as a way of looking at the values in a more abstract way. I note that there does not seem to be agreement on how to define median. -----Original Message----- From: math-fun <math-fun-bounces@mailman.xmission.com> On Behalf Of Michael Collins Sent: Saturday, June 13, 2020 3:09 PM To: math-fun <math-fun@mailman.xmission.com> Subject: [EXTERNAL] Re: [math-fun] Much ado about nothing (was Re: Proof that 0^0 = 1) If I were explaining this to a skeptical student or a curious non-scientific friend, I would say that x^0 represents an expression in which x appears zero times, i.e. not at all, and thus must be the same for all x. Which would suggest 0.0^0=1 with 0^0.0 undefined. On Sat, Jun 13, 2020 at 2:57 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
As a computer guy, I consider real numbers and integers to be two very different things. The English language shares this distinction, in that real numbers go with "less" and integers go with "fewer."
I'll use 0 to mean the integer zero and 0.0 to mean the real zero.
I would say that 0^0 = 1, and 0.0^0.0 is undefined. If anyone disagrees, I'd like to know why. I'd also like to know if anyone has an opinion on 0^0.0 or 0.0^0. And if sticking a minus sign in anywhere would change anything. Thanks.
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In Common LISP (expt 0 0)=1 (expt 0.0 0)=1 (expt 0 0.0) and (expt 0.0 0.0) return a "division by zero" error. Brent On 6/13/2020 1:57 PM, Keith F. Lynch wrote:
As a computer guy, I consider real numbers and integers to be two very different things. The English language shares this distinction, in that real numbers go with "less" and integers go with "fewer."
I'll use 0 to mean the integer zero and 0.0 to mean the real zero.
I would say that 0^0 = 1, and 0.0^0.0 is undefined. If anyone disagrees, I'd like to know why. I'd also like to know if anyone has an opinion on 0^0.0 or 0.0^0. And if sticking a minus sign in anywhere would change anything. Thanks.
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=Keith F. Lynch As a computer guy, I consider real numbers and integers to be two very different things...
Yes! Likewise, real numbers and floating-point numbers are two very different things. Alas floating-point numbers are deceptively advertised as "real". Flonums only implement a limited subset of the dyadic rationals. All else is marketing hype. Anyway, Wikipedia: https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero#Treatment_on_compute...
participants (5)
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Brent Meeker -
Cordwell, William R -
Keith F. Lynch -
Marc LeBrun -
Michael Collins