$f:\Bbb N\dashrightarrow\Bbb N$ is partial recursive $\Leftrightarrow$ its graph is a recursively enumerable?
A $k$-ary relation $R \subseteq \mathbb{N}^k$ is recursively enumerable if its indicator function $f_R$ defined by

$$f_R(n_1,\ldots,n_k) = \begin{cases} 1 &amp; \text{if } R(n_1,\ldots,n_k) \\
0 &amp; \text{otherwise} \end{cases}$$

is a partial recursive function. 

So if $f: \mathbb{N} \rightarrow \mathbb{N}$ is a partial recursive function, it is easy to see that the graph

$$R_f = \{ (m,n) \mid f(m) = n \}$$

is a recursively enumerable set. We have $f_{R_f}(m,n) = 1 - \mathrm{sg}(f(m) - n)$, where $\mathrm{sg}(n)$ is the primitive recursive function that is $0$ for $n=0$ and $1$ otherwise. Since we can express the indicator function $I_{R_f}$ of the graph as the composition of partial recursive functions, we therefore conclude that it is also a partical recursive function.

Conversely, if $$R_f = \{ (m,n) \mid f(m) = n \}$$

is a recursively enumerable set with indicator function $I_{R_f}$, then 

$$f(m) = \mu n. [ (1 - I_{R_f}(m,n)) = 0 ] $$

Here $\mu n.P(n)$ denotes unbounded mineralization over the unary predicate $P(n)$.