Let $G$ be a graph with $n$ vertices where every vertex has a degree of at least $\frac{n}{2}$. Prove that G is connected. 
Let $G$ be a graph with $n$ vertices where every vertex has a degree of at least $\frac{n}{2}$. Prove that $G$ is connected.

First question, if the problem uses a fraction such as $\frac{n}{2}$, would we round down in case $n$ is odd?
As for the actual problem, I'm trying to do this with induction and contrapositive and was wondering if my methods are correct. I'm also wondering if anyone can give a proof by contradiction, I feel like it would be pretty similar to a proof by contrapositive but can't quite put my finger on it.
Induction

A base case of $n=2$ gives a degree of one for all vertices.
Assume $G$ with $n$ vertices where every vertex has a degree of at least $\frac{n}{2}$ is connected.
Show that $G$ with $n+1$ vertices has a degree of at least $\frac{n+1}{2}$ is connected.
If we remove a vertex $v$, then by the inductive hypothesis, the graph is connected.
Now we add the $v$ back in. We now need to ensure that $v$ and all the other vertices have at least a degree of $\frac{n+1}{2}$. This means that we need to connect $v$ to the other vertices.
It follows that $G$ is connected.

Contrapositive

Prove that if $G$ is disconnected, then $G$ cannot be a graph with $n$ vertices where every vertex has a degree of at least $\frac{n}{2}$.
Assume that $G$ is disconnected.
We can then examine the case where there are the smallest number of connected components - $2$.
Each one can have at most $\frac{n}{2}$ vertices. Thus, each vertex can have a degree of at most $\frac{n}{2}-1$ since it cannot connect to itself.
It follows that this is a contradiction and $G$ is connected.


Suppose $G$ is not connected and has $n$ vertices.  Some component of $G$ then has at most $n/2$ vertices.  A vertex in this component then has degree at most one less, i.e. $\lt n/2$.
To make this a "proof by contradiction" we need only begin with assuming $G$ has degrees at least $n/2$ for every vertex.
