Proof - edges = $v \choose 2$ iff G is complete and simple I am self-learning graph theory from Bondy and Murty's Graph Theory with Applications and want some clarification on if the following proof is correct. Exercise 1.2.7 asks to show that for a simple graph G, $\epsilon = {v \choose 2}$ if and only if G is complete. Here $\epsilon$ and $v$ stand for the number of edges and vertices of G respectively.
One direction I found quite simple. If G is complete and simple, every vertex is connected to each other by a single edge and so the number of edges corresponds to the number of pairs that can be chosen using $v$ distinct objects, hence ${v \choose 2}$ edges.
It's the second direction I struggled with a bit. I'd like to know if the following is rigorous and also if there is a more elegant way of proving it:
By using induction on $v$, we will construct a complete graph for all $v$ vertices.
Base case: $v = 1 \implies \epsilon = 0$, this is the trivial complete graph.
IH: given that a simple graph G with $k$ vertices and $k \choose 2$ edges is complete, we want to construct a new graph G' with $k+1$ vertices and $k+1 \choose 2$ edges , meaning we have to add ${k+1 \choose 2} - {k \choose 2} = k$ edges to G.
To construct G', we add a new vertex $w'$ and
as the subgraph G is already complete, the only way to add these edges and keep G' simple is to create edges between the new vertex and all the $k$ vertices of G. This means G' is also complete.
I think the argument above relies on the fact the complete graphs are isomorphic which is why I'm unsure if its rigorous but it was the only way I managed to come up with a proof. Originally I tried to prove it directly using the contrapositive but I thought my argument ended up being circular.
Any help and comments are appreciated, cheers.
 A: I think the key thing here is that it is assumed that $G$ is simple, meaning there is at most one edge connecting any two vertices, and no vertice can connect to itself. Since there are $v$ vertices, that implies the graph must have at most $\frac{v(v-1)}{2}={{v}\choose{2}}$ edges. If $G$ has a total of ${{v}\choose{2}}$ edges, then that implies it has every edge a simple graph can have, meaning it must be complete.
A: In the following we assume all graphs are simple without further mention.
First, consider the vertex set $[n]=\{1,\dots,n\}$. Let $\binom{[n]}{2}=\{e\subseteq[n]:|e|=2\}$ be the set of all possible edges. In the following we restrict to graphs with vertex set $[n]$.Let $K_n=([n],\binom{[n]}{2})$ be the graph that includes all edges.

*

*Every graph $G=([n],\mathcal E)$ is a subgraph of $K_n$: For $e\in\mathcal E$ we have $e=\{i,j\}$ for some $i,j\in[n]$ with $i\neq j$, so $e\in\binom{[n]}{2}$ and thereby $G$ is a subgraph of $K_n$.

*The graph $K_n$ is the only complete graph with vertices $[n]$: Let $G=([n],\mathcal E)$ be complete, then $G$ is a subgraph of $K_n$ by the first item. On the other hand, for every $\{i,j\}\in\binom{[n]}{2}$ we have $\{i,j\}\in\mathcal E$ because $G$ is complete, so $K_n$ is a subgraph of $G$.

*The graph $K_n$ is the only graph with vertices $[n]$ and $\binom{n}{2}$ edges: Let $G=([n],\mathcal E)$ be a graph with $|\mathcal E|=\binom{n}{2}$. By the first item $G$ is a subgraph of $K_n$, so $\mathcal E\subseteq\binom{[n]}{2}$. So, with $|\mathcal E|=\binom{n}{2}=|\binom{[n]}{2}|$ we have $\mathcal E=\binom{[n]}{2}$.

This shows that $K_n$ is the only complete graph and the only graph with $\binom{n}{2}$ edges on vertices $[n]$. For an arbitrary graph $G=(\mathcal V,\mathcal E)$, let $n=v(G)$, fix a bijection $\iota:\mathcal V\rightarrow[n]$, and let $G'=([n],\mathcal E')$ denote the image of $G$ under the graph isomorphism induced by $\iota$. Clearly, we have $e(G')=e(G)$ and $G'$ is complete if and only if $G$ is complete. Thus, $G$ is complete if and only if $G'$ is complete if and only if $e(G')=\binom{v(G')}{2}$ if and only if $e(G)=\binom{v(G)}{2}$.
