Combinatorial Number Theory Problem

[Mathguy15]

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

 

[mathman]

You haven't defined |A| for this situation.

 

[Mathguy15]

For a set C, |C|=the cardinality of C.

 

[DonAntonio]

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 $\{2,3,4\},\{0,2,3,4\}$ , otherwise...perhaps by induction on $|A|+|B|$ ...

DonAntonio

 

[catellus]

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.

 

[Mathguy15]

Great Scott, you may be right catellus!