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 July 18th, 2014, 02:39 AM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 429 Thanks: 18 To infinity and beyond A very common question i see on the internet is 'What is infinity?' And a very common answer i see is 'Infinity is a concept.It is not a number.' But no reasons are given why this is so. So I tried to find a reason and this is what came to mind A number is a result of completed counting process. With infinite sets you can begin counting but you cannot complete the job. So, infinity is not a number. Is this a good reason to say infinity is not a number but a concept? Thanks
 July 18th, 2014, 02:51 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 A number, in a way, is also a concept. Infinity is not a number because of the way numbers are defined. To take the whole numbers: each number can be defined as the successor of the previous number, starting with 1. So, you eventually get every whole number by this process and the process never ends but you never reach "infinity". Your argument is correct: there is an infinite number of whole numbers; so infinity as a concept in this case is the measure of the size of a set, but not a number itself. Thanks from shunya
 July 18th, 2014, 06:56 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I think that "infinity is a concept, not a number" is a terrible description. What do "infinity", "concept", and "number" mean? It's not at all obvious. Better would be: There are no infinite members of $\mathbb{R}.$ There are some kinds of objects that I would call numbers which include infinite elements. For example, the extended real $+\infty$ is infinite, and I'd certainly call $+\infty$ a number in that context. Thanks from topsquark and shunya
 July 18th, 2014, 08:09 AM #4 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 Could an infitesimal be considered to be an element of the set of real numbers? For example, could 1 - .999... be equal to an inifitesimal rather than to 0? It seems that Robertson in the 1960s established that the caculus could be defined on this basis.
 July 18th, 2014, 08:53 AM #5 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions TLDR: Nope. $\displaystyle 0.\dot{9}$ and $\displaystyle 1$ are identical. No ifs or buts. There was a painful thread a few months back debating this over and over again, amongst other limit-based issues. Just so we don't repeat that, I'll give the following excerpt from "Introductory Real Analysis" A.N. Kolmogorov & S.V. Fomin, p15, which is a note attached to the proof showing that the set of real numbers in the closed unit interval [0,1] is uncountable: "... certain numbers, namely those of the form $\displaystyle p/10^q$, can be written as decimals in two ways, either with an infinite run of zeros or an infinite run of nines. For example, $\displaystyle \frac{1}{2} = \frac{5}{10} = 0.5000...=0.4999...,$ so that the fact that two decimals are distinct does not necessarily mean that they represent distinct real numbers. However, this difficulty disappears if in constructing [the number] we require that [the number] contain neither zeroes nor nines ... ". An explanation then follows that's related specifically to the proof and not the point we are addressing, but you get the idea. Thanks from shunya
July 18th, 2014, 09:09 AM   #6
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Quote:
 Originally Posted by Timios Could an infitesimal be considered to be an element of the set of real numbers?
Depends on your definition. I consider 0 to be an infinitesimal, the only one in the real numbers. But some definitions exclude 0 in which case there are no infinitesimals in the reals.

Quote:
 Originally Posted by Timios For example, could 1 - .999... be equal to an inifitesimal rather than to 0?
1 - 0.999... = 0 in the real numbers.

Quote:
 Originally Posted by Timios It seems that Robertson in the 1960s established that the caculus could be defined on this basis.
Yes. Nonstandard analysis uses a number system with infinitely many infinitesimals, and this can be used as a basis for calculus.

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