My Math Forum ordinal arithmetic

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 November 19th, 2008, 08:23 AM #1 Newbie   Joined: Nov 2008 Posts: 2 Thanks: 0 ordinal arithmetic Hello. Definition: $a,b$ are ordinals and $a\leq b$. Then there exists a unique $c$ such that $a+c=b$. We write $c=b-a$. Show that this subtraction definition coincides with the following operation defined by recursion: (1) $a-a=0$ (2) $(b+1)-a=(b-a)+1$ for a <= b (3) $c-a= \text{sup}\{b-a:b there exists a $c$ s.t a+c=b and so c=b-a. (from the first defn). LHS of (2): (b+1)-a = ((a+c)+1)-a = (a+(c+1))-a RHS of (2): (b-a)+1 = c+1 I can't seem to show they're equal. Am I even supposed to do that? Thanks for any help.
 November 27th, 2008, 03:00 PM #2 Senior Member   Joined: Nov 2008 Posts: 199 Thanks: 0 Re: ordinal arithmetic Hello. This question was posted a week ago so maybe an answer isn't needed anymore. Nevertheless, in case anyone's interested here is my attempt at a solution: Let the first definition given be labeled (1). Let the second be labeled (2). It is sufficient to show that a value c for b - a obtained from (2) satisfies a + c = b and thus is the unique c obtained by definition (1). The way to do this is (transfinite) induction on b. Base case: b = 0. trivial. Successor step: Suppose true for b. Then (b+1) – a = (b-a) +1 = c + 1 (by definition) and by inductive hypothesis a + c = b and c is unique with this property. So (a + c) + 1 = b + 1, And so a + (c+1) = b+1, i.e., the value obtained by (1) satisfies condition in (2) and thus is unique with this property. Limit Step: Let k be a limit ordinal. k – a = sup{b – a: b < k} (using (1)) and a + sup{b – a: n < k} = sup{a + (b – a):b < k} = sup{b: b < k} (by inductive hypothesis) = k.

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