2
$\begingroup$

Suppose that $A$ is a set of size $n$

Define $A+A$ as sum of two distinct members of $A$

Prove that $|A+A| \ge 2n-3$ and equality holds for arithmetic progression.

The question is an exercise of probabilistic methods in combinatorics and it's assumed to solve this way

  • 1
    Hint: Let $a_1$A$. – 2017-02-05
  • 1
    Just for clarification, do you mean for $A=\{1,4\}$, $A+A=\{2,5,8\}$? – 2017-02-05
  • 0
    @YuxiaoXie Yeah. But why this was discussed it probabilistic methods? – 2017-02-05
  • 0
    @StubbornAtom no. Just 5 – 2017-02-05
  • 1
    I don't know about probabilistic methods in combinatorics, but this problem is easily solved with the constructive approach. Maybe you could provide more context, like where did you see this problem? – 2017-02-05
  • 1
    There can only be a "sum" of distinct members of $A$ if addition on $A$ is properly defined. Information about that is needed. What do the elements of $A$ look like? – 2017-02-05
  • 1
    On [this](https://www.cmi.ac.in/admissions/sample-qp/ugmath2016-solutions.pdf) file containing the solutions of the CMI entrance exam in 2016, there is a closely related problem and its corresponding solution on pages 4 and 5. – 2017-02-05
  • 0
    @drhab sum of two member – 2017-02-05
  • 0
    @StubbornAtom Yeah the solution is right but I'm looking for a probabilistic solution – 2017-02-05

0 Answers 0