Intuition behind a counterexample to $|A+A|\leq |A-A|$, where $A$ is a finite set

Question

Define

$$A+A=\{a+b:a,b \in A\}, A-A = \{a-b:a,b \in A\}$$

Then prove or disprove the following

$$|A+A|\leq |A-A|$$

Intuitively, it should be true, as

$$a+b=b+a$$

$$a-b \neq b-a$$

However, I accidentally discovered the following counterexample by Conway proposed in 1969

$$A=\{1,2,3,5,8,9,13,15,16\}$$

where

$$A+A=\{2,3,...,32\}-\{27\}, |A+A|=30$$ $$A-A=\{-15,-14,...,15\}-\{\pm 9\}, |A-A|=29$$

Incredible! How did he come up with this? (Of course, I can do it using a computer program...)

Answer 1

Let me mention a general result: Theorem 4 in "Many Sets Have More Sums Than Differences" by Greg Martin and Kevin O'Bryant:

For every integer $x$, there is a set $A\subseteq\{0, 1,\dots , 17|x|\}$ with $|A+A|−|A−A|=x$.

Its proof gives you an idea of how these sets can be obtained.

Note that for $x=1$, we have $A=\{0, 2, 3, 4, 7, 11, 12, 14\}$. Then $$A+A=\{0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 28\}$$ and $$A-A= \{-14, -12, -11, -10, -9, -8, -7, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14\}.$$ Hence $|A+A|=26>25=|A-A|$.

For a generalization to abelian groups see "Sets with more sums than differences by Melvyn B.Nathanson.