Given a set of nonzero real number S= $\{d_1,\cdots, d_N\}$, can I find a number $d_e \in S$ such that there does not exist $d_m,d_j\in S$ such that $d_e = d_m+d_j$ ? Assume $m$ and $j$ can be equal and $m,e,j$ denote the indices with $e\neq m,j$.
Asked
Active
Viewed 111 times
0
-
This depends a lot on $S$. What are the constraints? – Mario Carneiro Apr 23 '14 at 03:09
-
S has finite number of elements, if it helps. S can be any sets. – widapol Apr 23 '14 at 03:13
2 Answers
1
This is true for any set of real numbers with a positive minimum. If $S\subseteq\Bbb R$ and $0<x\in S$ such that for all $y\in S, y\ge x$, then for any $y,z\in S$, $x\le y<y+z.$
This is false for the set $S=A\cup(A+1)$ where $\{-2,1\}\subseteq A$. If $x\in S$, then $x\in A$ or $x-1\in A$. In the first case, $x+1\in S$, so $x=(x+1)+(-1)$ is a decomposition, otherwise $x-1\in S$, so $x=(x-1)+1$ where $1\in S$.
Mario Carneiro
- 27,438
-
-
@widapol In the first paragraph I am making an example where the property is true. The step $y<y+z$ would no longer be guaranteed if $S$ had negative numbers. In the second paragraph I give a counterexample, that is, a set such that every two numbers is decomposable into two other numbers in the set. – Mario Carneiro Apr 23 '14 at 03:25
-
Thanks for pointing out examples. However, we need to find a number that cannot be decomposed into any two other numbers in the set. Assume these two other numbers can be equal. – widapol Apr 23 '14 at 03:32
-
-
@widapol And the counterexample also shows that ${-2,-1,1,2,\dots,N}$ is also closed under decompositions, so there are sets of any size $\ge4$ that are closed under decompositions. – Mario Carneiro Apr 23 '14 at 11:19
0
Not always.
Let $S= \{1, 2, 2^2, ..., 2^N\}$ and consider the case when $d_e = 1$.
Mustafa Said
- 4,137
-
-
I think you misread the question. The OP asks whether it is possible to find a $d_e$ which is not the sum of two elements of $S$. So your example is not a counterexample. – Alex Becker Apr 23 '14 at 21:18