Question

Let G be an undirected graph with n vertices and m edges. Use a contradiction argument to prove that if m<n−1, then G is not connected

Answer #1

Let G be a connected simple graph with n vertices and m edges.
Prove that G contains at least m−n+ 1 different subgraphs
which are polygons (=circuits). Note: Different polygons
can have edges in common. For instance, a square with a diagonal
edge has three different polygons (the square and two different
triangles) even though every pair of polygons have at least one
edge in common.

Let G = (V,E) be a graph with n vertices and e edges. Show that
the following statements are equivalent:
1. G is a tree
2. G is connected and n = e + 1
3. G has no cycles and n = e + 1
4. If u and v are vertices in G, then there exists a unique path
connecting u and v.

A simple undirected graph consists of n vertices in a single
component.
What is the maximum possible number of edges it could have?
What is the minimum possible number of edges it could have?
Prove that your answers are correct

In lecture, we proved that any tree with n vertices must have n
− 1 edges. Here, you will prove the converse of this statement.
Prove that if G = (V, E) is a connected graph such that |E| =
|V| − 1, then G is a tree.

Show that if G is connected with n ≥ 2 vertices and n − 1 edges
that G contains a vertex of degree 1.
Hint: use the fact that deg(v1) + ... + deg(vn) = 2e

Call a graph on n vertices dendroid if it has n edges and is
connected. Characterize degree sequences of dendroids.

Suppose G is a simple, nonconnected graph with n vertices that
is maximal with respect to these properties. That is, if you tried
to make a larger graph in which G is a subgraph, this larger graph
will lose at least one of the properties (a) simple, (b)
nonconnected, or (c) has n vertices.
What does being maximal with respect to these properties imply
about G?G? That is, what further properties must GG possess because
of this assumption?
In this...

Graph Theory
.
While it has been proved that any tree with n vertices must have
n − 1 edges. Here, you will prove the converse of this statement.
Prove that if G = (V, E) is a connected graph such that |E| = |V |
− 1, then G is a tree.

please solve it step by step. thanks
Prove that every connected graph with n vertices has at least
n-1 edges. (HINT: use induction on the number of vertices
n)

Make a general conjecture about the minimum number of edges in a
graph with n vertices and r components, where n, r >= 1. Then
prove this conjecture.

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