I think contradiction is fine. Assume the set is finite, so it is bounded above. That means that your set, since now S, have a max say then show that you can construct another number that belongs to the set and is greater than m
A set of numbers is defined as a1=1, a2=2, a3=3, and a number n belongs to this set if it can be written as a unique sum of three distinct members of this set e.g. a4=6, a5=9, a6=10, a7=11, a8=12 but 13, 14, 15, 16, 17 do not belong to this set. Is this set of numbers finite or infinite? Prove your claim.
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I've been trying to solve this question by contradiction. So I assume that the set is finite. I have no idea how to proceed. can anyone give me a hint?
I've tried working that out. It turns out that it isn't necessary that the sum of the three largest numbers will belong to set. Counter Example is 18.
Working:
Suppose S = {1,2,3,6,9} so far.
now 3+6+9 = 18, which should be in this set S.
BUT it is not because when we work out more elements of S, we get S = {1,2,3,6,9,10,11,12} and from here, we clearly see that 18 has another representation i.e. 11+6+1.
So 18 has two representations!
oh I think I've figured out the mistake I was making. Once you suppose that you're set is finite, you take 3 largest numbers and add them to get a number x. This x can either be in the set or must have another triple-number representation. But if it has a triple number representation then there must be a number bigger than the 3 largest numbers we took, leading to a contradiction!
thank you all for the help... I'm most grateful