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Is it possible to represent a "number" which has no beginning or end (denoted by the ellipsis ...), then apply maths operations such as addition, subtraction, etc.,...?

Some examples

plus

minus

Found related questions

As noted in the comments section, a real number has always a beginning but not necessarily an end (which usually means that the decimal expansion does not terminate). For example $\large\pi$

pi

The thought experiment here is to entertain the idea of a "number" existing without a beginning or end and what mathematical operations can be applied, if any.

vengy
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    This isn't really number theory by the way. If you mean to say: can we perform operations on the space of infinite sequences, $\Bbb K^\Bbb N$ (where you take $\Bbb K$ to be any set of your choosing) then yes, we can – FShrike Mar 31 '22 at 20:34
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    You would need an unambiguous "center" to these "numbers" to avoid ambiguity. Treat it instead as a doubly infinite sequence $\Bbb K^\Bbb Z$. Otherwise, $\dots 12121212\dots + \dots 12121212\dots$, should that have been $\dots33333\dots$ or $\dots 24242424\dots$. As for "multiplication", what should make sense for that? Termwise multiplication? If so, then what should be $\dots 555555\dots \times \dots 222222\dots$? Noting that $5\times 2 = 10$ should this have been $\dots XXXXXX\dots$ where $X$ is the character representing the number ten, or $\dots 10101010\dots$ or something else? – JMoravitz Mar 31 '22 at 20:40
  • You would have to define what you mean. Are these new objects or are you trying to represent existing number systems this way? – John Douma Mar 31 '22 at 20:40
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    The punchline is that you can define anything you want to. Whether or not it is useful is another story. Popular definitions are popular precisely because they are useful for solving some known problem, and the definitions were crafted such that they solve those specific problems. – JMoravitz Mar 31 '22 at 20:41
  • One interpretation of the double ellipsis could be as a Laurent series and it more or less works like polynomials do. https://en.wikipedia.org/wiki/Laurent_series – CyclotomicField Mar 31 '22 at 20:45
  • @CyclotomicField its worth emphasizing that you don't run into the problem of "carry digits" with polynomials. You can be perfectly content to have a coefficient not be one of the digits $0,1,2,\dots,9$. With the numbers described by OP, if we were to try to have them satisfy the same properties as other number types if it has leading zeroes or ending zeroes then we run into problems trying to make it work for the doublyinfinite type they have here. – JMoravitz Mar 31 '22 at 20:47
  • An integer has always a beginning and an end. A real number has always a beginning but not necessarily an end (which usually means that the decimal expansion does not terminate). Objects without a beginning and without an end are quite exotic (numbers do not have this property , not even the ordinal numbers !) and I doubt you can establish some meaningful with them. – Peter Apr 01 '22 at 13:26
  • Your multiplication does not match normal arithmetic. $111 \times 111 \neq 111$. – badjohn Apr 03 '22 at 19:59
  • Multiplication will end up with infinite sums of digits. Sums and differences can be made to work. The real problem is, you don’t know in this notation which digit corresponds to which. So I could add $\dots 121212\dots$ to $\dots 212121\dots$ and get $\dots 333333$ or I could shift and get $\dots242424\dots.$ It would be wise to include a decimal point location to the notation to make clear which digit corresponds to which in two numbers. – Thomas Andrews Apr 03 '22 at 20:44

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