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Thread: Addition operator

  1. #1 Addition operator 
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    The addition operator is usually regarded as the "+" sign. So adding 1 to 2 is written as 2 + 1. However, one could denote an addition operator by some symbol such as and define its effect by,

    . So would be the result of .

    Expanding it a bit further, or .
    In general, .

    My question is, other than giving definitions such as , does anything else have to be done to show that the operator exists? Or would writing additions as I have above simply be a valid alternative to the conventional way of writing them?


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  3. #2  
    Bassaricyon neblina Olinguito's Avatar
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    I presume you are defining addition on the natural numbers as constructed from the Peano axioms. It can be defined recursively as follows: Given , define a function by:






    We write . This is the same as your defintion, but stately recursively. The existence of the operator is due to the successor axiom, while that of is guaranteed by the recursion theorem in set theory.


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  4. #3  
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    Quote Originally Posted by Olinguito View Post
    I presume you are defining addition on the natural numbers as constructed from the Peano axioms. It can be defined recursively as follows: Given , define a function by:





    We write . This is the same as your defintion, but stately recursively. The existence of the operator is due to the successor axiom, while that of is guaranteed by the recursion theorem in set theory.

    Thank you. I wasn't thinking of the Peano axioms, although I was thinking about natural numbers. The the successor axiom is clearly relevant to my question.
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  5. #4  
    Forum Junior anticorncob28's Avatar
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    This is based off the fact that n + 1 is defined for all n. This has to be proven, or accepted as an axiom, or defined to be the case.
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  6. #5  
    Bassaricyon neblina Olinguito's Avatar
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    In the Peano axioms, the successor property is that to each natural number there corresponds a unique natural number called the successor of , denoted . At this stage the has nothing to do with addition. It’s only later when addition is defined on the natural numbers that the successor of becomes the result of adding to .

    Alternatively, if you wish to construct the natural numbers from the set-theoretic axiom of infinity, then is defined as .
    Last edited by Olinguito; September 2nd, 2013 at 07:03 AM.
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  7. #6  
    Forum Radioactive Isotope MagiMaster's Avatar
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    You're free to define any operator any way you want. Just giving something a definition makes it exist as an operator. Now whether that new operator is anything interesting is a completely separate question, and you do have to keep track of which meaning each symbol has. (You can't just redefine, say, = and then use your new definition with something else that was written using an old definition.)

    In this case, the successor operator, usually written S, is actually one of the axioms (something accepted as a self-evident truth, at least for the area under discussion) of Peano arithmetic and is used to define the + operator (and all the numbers except 0) instead of the other way around. (Though other posters already mentioned most of this.)
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