A connected acyclic graph has $n-1$ edges
Proving 2,3 implies 1:
We have an acyclic graph $G=(V, E)$ with $n-1$ edges. We want to prove that $G$ is a connected graph.
Assume for the sake of contradiction that $G$ is not connected. This means we have $d&gt;1$ connected components, $G=\{\bigcup_{i=1}^{d}G_i\}$.
Since $G$ is acyclic, each connected component is a tree by definition. Let $V_i$ be the set of vertices of graph $G_i$. Then, the number of edges in $G_i$ is $\mid E_i \mid = \mid V_i \mid - 1$. Thus, the total number of edges in $G$ is
$\mid\ E \mid\ =\sum_{i=1}^{d}\mid E_i \mid =\sum_{i=1}^{d} (\mid V_i \mid-1)=\sum_{i=1}^{d}(\mid V_i \mid)- d = n - d &lt; n -1$.
The last inequality follows because $d &gt; 1$. This is a contradiction that $\mid\ E \mid\ = n-1$.
Proving that 1,3 implies 2: We have a connected graph $G=(V, E)$ with with $n-1$ edges. We want to prove that $G$ is acyclic. I will not write the proof, but I will hint that (one of many possible proofs) you can use induction on the number of vertices. And let your assumption be that given $G$ with $n-1$ vertices and at most $n-2$ edges, then $G$ is acyclic.