The statement you provided refers to a simple graph with the condition that the number of edges (denoted $E$) is less than or equal to the number of vertices (denoted $V$) minus 2, i.e.,

$E \leq V - 2$

This condition implies that the graph is not connected. Here's why:

Explanation:

• A simple graph is a graph with no loops or multiple edges.
• A connected graph is one where there is a path between every pair of vertices.

For a graph to be connected, it must have at least $V - 1$ edges, where $V$ is the number of vertices. This is because a tree, which is the simplest connected graph, has $V - 1$ edges.

If $E \leq V - 2$:

• The graph cannot have enough edges to guarantee a path between all vertices, meaning there will be at least one pair of vertices that cannot be reached from each other.
• In other words, if there are fewer than $V - 1$ edges, the graph will necessarily have disconnected components.

Thus, if $E \leq V - 2$, the graph cannot be connected, and it must have at least two disconnected components.

Example:

If you have a graph with 4 vertices, the graph would need at least $4 - 1 = 3$ edges to be connected. However, if the graph only has 2 edges (which satisfies $E \leq V - 2$ for $V = 4$), it will not be connected and will have two disconnected components.

Let me know if you'd like a more detailed explanation or example!

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Here are some related questions for you to explore:

1. How does the concept of a tree relate to connected graphs?
2. What happens when a graph has exactly $V - 1$ edges?
3. Can a graph with fewer edges than vertices be a spanning tree?
4. What is the minimum number of edges needed for a graph to be connected if the graph is a complete graph?
5. How does the addition of edges affect the connectivity of a graph?

Tip: When studying graph theory, always remember that a graph with $E < V - 1$ is always disconnected, while one with exactly $V - 1$ edges must be connected (and is a tree).