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Peano axioms for natural numbers - prove 0.5 ∉ N

  1. Mar 1, 2013 #1

    ato

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    i am studying real analysis from terence tao lecture notes for analysis I. http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/

    from what i understand , property is just like any other statment. for example P(0.5) is P(0) with the 0s replaced with 0.5 . so the notes says (assumes ?),

    ##P(0.5)\textrm{ is unprovable}\Rightarrow0.5\notin N##

    i mean its alright to assume something like but i just wanna make sure that what i understood is correct. if it is why not just assume something like this,
    ##0.5\in N\textrm{ is unprovable}\Rightarrow0.5\notin N##

    but i might be wrong, so in that case could you prove 0.5 ∉ N.

    thank you
     
  2. jcsd
  3. Mar 1, 2013 #2
    Since you did not give a page number, I cannot tell what property P you are referring to, but I would presume that P(x) is something like x =0 [itex]\vee[/itex][itex]\exists[/itex]y (y[itex]\in[/itex]N [itex]\wedge[/itex] x=y+1).

    Before I answer the question further, please tell me what P(.) is.
     
  4. Mar 1, 2013 #3

    ato

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  5. Mar 1, 2013 #4
    OK, all clear: P is simply the variable in the axiom of induction, which can be seen formally here: http://en.wikipedia.org/wiki/Mathematical_induction#Axiom_of_induction. (Note that this does not require the domain of P to be N; it can only prove P for non-negative integers.) He omits a statement (or leaves it implicit) that "this is all the natural numbers": that is, if a number does not satisfy all the axioms, then it is not a natural number. (Here I follow the convention that 0 is included in the set of natural numbers. Some places don't, and just call N the set of non-negative integers.) So the author's reasoning is basically that 0.5 does not satisfy the axioms, hence is not a natural number. Therefore, the statements you listed are equivalent if you take P(x) being "x[itex]\in[/itex]N".
     
  6. Mar 1, 2013 #5

    ato

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    i was following alright until this. do you mean this is correct,
    ##P(x)\textrm{ is unprovable for }x\in N\Rightarrow x\notin N##
    but then P(x) would never be unprovable (hence redundant) because ##P(x)## is true for ##x\in N##.
    why would x∈N assumed as condition ? would not this require N to be known.

    please clarify.
     
  7. Mar 1, 2013 #6
    Remember my parenthetical remark that x does not have to be in N. You have introduced "[itex]\in[/itex]"N where it wasn't before. That is, you had two statements
    (1)
    and
    (2)
    This latter quote is an instance of the addition that I mentioned was implicit,
    (3)
    and hence not surprising.
    I suggested
    (4)
    Applying (4) to (2), (2) morphs into (1). (And, of course, the contrary.) That's what I meant by the equivalence.
    Applying (4) to your new statement
    would give
    (x∈N is unprovable for x∈N)⇒x∉N
    which is a completely different statement, and would take us into that interesting area about true but unprovable statements..... to quote from The Never-Ending Story (a good title for mathematics), "But that is another story and shall be told another time."
     
  8. Mar 1, 2013 #7

    ato

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    got it, thanks
     
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