We define $D(\Bbb C)=\cup_n P(M_n(\Bbb C))/\sim$,where $P(M_n(\Bbb C)) $ is the set of projections in $M_n(\Bbb C)$.$\sim$ is the equivalence relation as follows:suppose $p$ is a projection in $P(M_n(\Bbb C))$,$q$ is a projection in $P(M_m(\Bbb C))$, $p\sim q$ if there is an element $v $ in $M_{m,n}(\Bbb C^n)$ with $p=v^*v,q=vv^*$.
How to show that $D(\Bbb C)\cong \Bbb Z^{+}$?
The traces on $M_n(\mathbb C)$ gives a well-defined additive homomorphism $tr:D(\mathbb C)\to\mathbb Z_{\geq0}$ by $tr([p])=tr(p)$. Well-definedness comes directly from the trace property, and to show that it is a homomorphism, given $[p],[q]\in D(\mathbb C)$, we have $$tr([p]+[q])=tr(p\oplus q)=tr(p)+tr(q)=tr([p])+tr([q]).$$ We will show that this is an isomorphism.
That this map is surjective is easy: If $n\in\mathbb Z_{\geq0}$, the class of the $n\times n$ identity matrix maps to $n$. For injectivity, if $tr([p])=tr([q])$, then $\dim(\operatorname{Range}(p))=\dim(\operatorname{Range}(q))$, so you can argue similarly to the answer in this question to show that $p\sim q$.