November 19th, 2008, 08:23 AM  #1 
Newbie Joined: Nov 2008 Posts: 2 Thanks: 0  ordinal arithmetic
Hello. Definition: are ordinals and . Then there exists a unique such that . We write . Show that this subtraction definition coincides with the following operation defined by recursion: (1) (2) for a <= b (3) when c is a limit ordinal. I am aware of the addition definition of two ordinals: (i)a+0=a (ii)a+(b+1) = (a+b)+1 (iii)a+c = sup{a+d: d<c} OK, so (1) is trivial. But (2) is bugging me to no end. a <= b ==> there exists a s.t a+c=b and so c=ba. (from the first defn). LHS of (2): (b+1)a = ((a+c)+1)a = (a+(c+1))a RHS of (2): (ba)+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 = (ba) +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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arithmetic, ordinal 
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