Let $G=(V,E)$ be a Graph and $k$ a positive integer, the radio-k-number of $G$ is defined to be the smallest integer $s$ for which $\exists f:V(G)\rightarrow\{0,\ldots,s\}$ such that $$|f(u)-f(v)|\geq k+1-d(u,v) \forall u,v \in V(G) |u \neq v$$ where $d(u,v)$ denotes the distance between $u$ and $v$.
We have to prove that the Radio-1-number of $P_n$ (the path graph on n vertices) is 1 for any $n\geq2$.
Here's my proof:
Consider the mapping $f:\{1,\ldots,n\}\rightarrow\{0,1\}$ given by $f(u)=0$ if u is odd and $f(u)=1$ if u is even.
CASE 1: When the vertices u and v are either both even or both odd.
$d(u,v)\geq2$ $\implies$ $0 \geq 2-d(u,v)$ $\implies$ $|f(u)-f(v)| \geq 2-d(u,v)$
CASE 2: When one of the vertices is odd and the other is even.
$d(u,v) \geq 1$ $\implies$ $0 \geq 1-d(u,v)$ $\implies$ $1 \geq 2-d(u,v)$ $\implies$ $|f(u)-f(v)| \geq 2-d(u,v)$.
Have I missed anything here? (I haven't proved here that the radio-1-number of $P_n$ cannot be $0$, but it is a true result). And is there a different way to prove this?