Find the smallest set containing 1/2 and closed under addition

Problem: Find the smallest set $S\subset \mathbf{R}$ containing $\frac{1}{2}$ and closed under addition.

Answer: $S=\{\frac{n}{2}:n\in \mathbf{N}\}$.

Proof. We need to prove 3 things: (a) $\frac{1}{2}\in S,$ (b) $S$ is closed under addition, (c) $S$ is the smallest of all sets containing $ \frac{1}{2}$ and closed under addition.

Proof of (a). Obvious.

Proof of (b). Take $a,b\in S.$ Then $a=\frac{n}{2},b=\frac{m}{2}$ for some $n,m\in \mathbf{N}.$ Consider $a+b=\frac{n}{2}+\frac{m}{2}=\frac{n+m}{2}.$This element belongs to $S$ because $n+m\in \mathbf{N}.$ Thus $S$ is closed under addition.

Proof of (c). Suppose $T$ is another set containing $\frac{1}{2}$ and closed under addition. We need to prove that $S$ is "smaller" than $T,$ i.e. $S\subset T.$ Consider an element $a$ of $S,$ then $a=\frac{n}{2}$ for some $n\in \mathbf{N}.$ But $a=\frac{n}{2}=\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}$ ($\frac{1}{2}$ taken $n$ times). Since $\frac{1}{2}$ belongs to $T$ then so does $a=\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}$ because $T$ is closed under addition. Thus $a\in T,$ and since $a\in S$ was chosen arbitrarily, we have $S\subset T.\blacksquare $