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-ary set operations.

  1. Jul 7, 2009 #1
    Hello all.

    Along the lines of unary and binary operations, could someone describe what 0-ary operation is and possibly give an example. I have only seen such an operation mentioned once, and that was early in the first chapter of Grillet's Abstract Algebra (2007), and so presumably it does not figure figure very highly, but is of some interest. The explanation in the above mentioned book was not very clear to me.

    Thanks. Matheinste.
  2. jcsd
  3. Jul 7, 2009 #2
    A 0-ary (or "nullary") operation is also known as a "constant symbol".

    Example. The first-order logical theory of "fields" can be formulated in many ways.One of them has...
    two binary operations:
    [tex](x,y) \mapsto x+y[/tex]
    [tex](x,y) \mapsto x\cdot y[/tex]
    two unary operations:
    [tex] x \mapsto -x[/tex]
    [tex] x \mapsto 1/x[/tex]
    and two nullary operations:
    then some axioms that these operations must satisfy. Actually, [tex]1/x[/tex] is not defined for [tex]x=0[/tex], so some adjustment would have to be made for that.
  4. Jul 7, 2009 #3
    Thanks for your reply.

    In the case you have cited are the zero and one nullary operators the same as the identity elements for the two field operations?

  5. Jul 7, 2009 #4
    Yes. Among the axioms would be
    [tex](\forall x)\; (x+0=x)[/tex]
    [tex](\forall x)\; (x\cdot 1 = x)[/tex]
  6. Jul 7, 2009 #5

    But aren't these binary operations.

  7. Jul 7, 2009 #6
    + and * are, but 0 and 1 aren't.
  8. Jul 7, 2009 #7
    Yes, so they are.

    My problem, mental block, is that a binary operation on a set is an operation needing two inputs from the set which returns a member of the set. A unary operation has a single input and returns a member of the set. And so a nullary operation has no inputs and returns a member of the set. Does this equate to just picking a member of the set.

  9. Jul 7, 2009 #8


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    A unary operation can be identified with its output, yes. That's why g_edgar wrote 0 and 1 instead of 0() and 1().
  10. Jul 8, 2009 #9
    Hello again

    ----A 0-ary or constant operation on a set S is a mapping f : {0} −→ S and simply selects one element f (0) of S. -----

    Can anyone please explain what exactly does the "element f(0) of S" mean in the above sentence, from Grillet, Abstract Algebra, .

  11. Jul 9, 2009 #10


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    f(0) denotes the value assigned to 0 by f. The value will be a member of f's range, which is S.
  12. Jul 9, 2009 #11
    Does this leave us free to assign any member of the range, which will be a member of the set S, as the image of the function.

  13. Jul 11, 2009 #12


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    When you define the function, you can choose any value from S. But the requirement that a function assign only one value to each member of the domain means that you cannot assign any more values after one has been chosen. You would have to define a new function and give it a new name.
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