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Combinatorial Number Theory Problem

 
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Apr25-12, 01:26 PM   #1
 

Combinatorial Number Theory Problem


Hello,

I would like to see a solution to the following problem:

Let A be a finite collection of natural numbers. Consider the set of the pairwise sums of each of the numbers in A, which I will denote by S(A). For example, if A={2,3,4}, then S(A)={5,6,7}. Prove that if S(A)=S(B) for two different finite sets of natural numbers A and B, then |A|=|B|, and |A|=|B| is a power of 2.

I find this problem interesting, but I am working on other problems. Anyone have ideas?

Thanks,
Mathguy
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Apr25-12, 03:53 PM   #2
 
Recognitions:
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You haven't defined |A| for this situation.
Apr25-12, 04:28 PM   #3
 
For a set C, |C|=the cardinality of C.
Apr25-12, 04:28 PM   #4
 

Combinatorial Number Theory Problem


Quote by Mathguy15 View Post
Hello,

I would like to see a solution to the following problem:

Let A be a finite collection of natural numbers. Consider the set of the pairwise sums of each of the numbers in A, which I will denote by S(A). For example, if A={2,3,4}, then S(A)={5,6,7}. Prove that if S(A)=S(B) for two different finite sets of natural numbers A and B, then |A|=|B|, and |A|=|B| is a power of 2.

I find this problem interesting, but I am working on other problems. Anyone have ideas?

Thanks,
Mathguy


Well, if one considers zero a natural number (many do) then both claims above are false, as [itex]\{2,3,4\},\{0,2,3,4\}[/itex] , otherwise...perhaps by induction on [itex]|A|+|B|[/itex] ...

DonAntonio
Apr26-12, 09:00 PM   #5
 
Maybe I'm missing something, but it looks like for A = {1,2,3,4,5,6,7} and B = {1,2,3,5,6,7}

S(A) = {3,4, ... 12,13} = S(B), but |A|=7 and |B|=6

Maybe there should be some other restrictions on A and B?

(guessing) If neither A nor B may be a subset of the other, then consider {1,2,3,4,6,7,8,9}, {1,2,3,5,7,8,9} . Cardinalities 8 and 7.
Apr27-12, 11:23 AM   #6
 
Great Scott, you may be right catellus!
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