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Math Help - Counting numbers proof

  1. #1
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    Counting numbers proof

    Prove: If each of m and n is a counting number, then m+n is a counting number.

    A counting number is defined as a number in the minimal induction set.
    The minimal induction set is the set of all counting numbers and is given by C.
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  2. #2
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    Quote Originally Posted by noles2188 View Post
    Prove: If each of m and n is a counting number, then m+n is a counting number.

    A counting number is defined as a number in the minimal induction set.
    The minimal induction set is the set of all counting numbers and is given by C.
    Below is the sketch of the proof I've tried,

    Lemma 1. For a transitive set t, \bigcup(t^+) = t .
    Proof.
    \bigcup t^+ = \bigcup (t \cup \{t\})<br />
=\bigcup t \cup \bigcup \{t\}<br />
=\bigcup t \cup t<br />
=t <br />
    The last step is followed from \bigcup t \subseteq t for a transitive set t.

    1. Show that the minimal induction set (a.k.a "a set of natural numbers") is transitive using the above lemma.

    For a set theoretic construction of natural numbers, the followings hold (since it is transitive)

    0 \in 1 \in 2 \in 3 ....
    0 \subseteq 1 \subseteq 2 \subseteq 3 \subseteq....

    2. Using a recursion theorem, show that for each m \in \omega, there exists a unique function
    A_m:\omega \rightarrow \omega (guarantees that it is a function from a set of natural numbers to a set of natural numbers) such that

    A_m(0) = m, A_m(n) = m + n .

    Now, combine 1 & 2 show that if each of m and n is a counting number, then m+n is a counting number.
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  3. #3
    MHF Contributor kalagota's Avatar
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    you may also check this:

    see this page.. in particular, look on the second post (post by jhevon) and click the hyperlink there...
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