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November 19th, 2008, 08:23 AM   #1
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ordinal arithmetic


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:

(2) for a <= b
(3) when c is a limit ordinal.

I am aware of the addition definition of two ordinals:
(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=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.
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November 27th, 2008, 03:00 PM   #2
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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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