Below I have a few proofs that I'd like to get checked. Thanks.
a. Prove every tree with $v$ vertices has exactly $v - 1$ edges
We want to show that $v - e = 1.$ We use the process of tree pruning(removing a vertex of degree $1$ from the tree along with the edge that occurs at that vertex). This process is well-defined because removing a vertex and an edge still leaves us with a tree since trees contain no cycles and removing edge/vertex pair doesn't add any cycle to the tree.
Every tree with more than one vertex can be pruned because every tree with two or more vertices has at least two vertices of degree $1.$ (Consider a longest simple path $P$ in the tree whose length is $x.$ The degree of its endpoints is at most $x.$ Otherwise $P$ is no longer the longest simple path).
The difference $v - e$ stays constant throughout the pruning process. At some point the tree will be reduced down to a single vertex so that $v - e$ is $1.$ Since the the difference $v - e$ stays constant all through the pruning, we can say that $v - e = 1$ for a tree.
b. In a forest with $v$ vertices and $k$ components, the number of edges is $v - k.$
By (a) above, $v - e = 1$ for each component tree meaning $v - e$ is constant in every component. Since there are $k$ components $v - e = k$ and so the number of edges is $v - k.$
c. Suppose that a graph has $20$ vertices, $15$ edges, and contains no cycles. Then how many components does the graph contain?
This graph has no cycles so it must be a tree, but a tree must contain exactly $v - 1$ edges if $v$ is the number of vertices. Thus we are dealing with a forest. By (b) above, the number of components is $20 - 15 = 5.$
d. Suppose that a graph contains an equal number of vertices and edges. Show that the graph must contain at least one cycle
A graph with equal number of vertices and edges is not a tree by (a) above, but perhaps a forest. Then this forest has $0$ components meaning this graph is not a forest either. Thus this graph must contain at least one cycle.