Let G be a plane graph with n vertices, m edges and f faces. Using Euler’s formula, prove
that n-m+f=k)G)+1
where k(G) is the number of connected components of G.

To prove the equation \( n - m + f = k(G) + 1 \) for a plane graph \( G \) with \( n \) vertices, \( m \) edges, \( f \) faces, and \( k(G) \) connected components, we will use Euler's formula and some basic properties of graph theory.

### Step-by-Step Proof

1. **Euler's Formula for a Connected Plane Graph**:
   
   For any connected plane graph, Euler's formula states:
   
   \[
   n - m + f = 2
   \]
   
   where \( n \) is the number of vertices, \( m \) is the number of edges, and \( f \) is the number of faces.

2. **Graph with \( k(G) \) Connected Components**:
   
   Let \( G \) be a plane graph with \( k(G) \) connected components. Each connected component \( G_i \) of \( G \) satisfies Euler's formula individually:
   
   \[
   n_i - m_i + f_i = 2
   \]
   
   where \( n_i \), \( m_i \), and \( f_i \) are the number of vertices, edges, and faces in the \( i \)-th connected component, respectively.

3. **Summing Euler's Formula for Each Component**:
   
   Since \( G \) has \( k(G) \) connected components, we can sum the Euler's formula for each component:
   
   \[
   \sum_{i=1}^{k(G)} (n_i - m_i + f_i) = \sum_{i=1}^{k(G)} 2
   \]
   
   Simplifying, we get:
   
   \[
   \left( \sum_{i=1}^{k(G)} n_i \right) - \left( \sum_{i=1}^{k(G)} m_i \right) + \left( \sum_{i=1}^{k(G)} f_i \right) = 2k(G)
   \]

4. **Total Vertices, Edges, and Faces**:
   
   Since the total number of vertices \( n \), edges \( m \), and faces \( f \) in \( G \) are the sums over all components, we can write:
   
   \[
   n = \sum_{i=1}^{k(G)} n_i
   \]
   
   \[
   m = \sum_{i=1}^{k(G)} m_i
   \]
   
   \[
   f = \sum_{i=1}^{k(G)} f_i - (k(G) - 1)
   \]
   
   The adjustment \(- (k(G) - 1)\) in the faces count arises because when considering the whole graph, some faces will be shared between different components.

5. **Substitute and Simplify**:
   
   Now substitute these total values back into our summed Euler's formula:
   
   \[
   n - m + f = 2k(G)
   \]
   
   However, note the adjustment in the faces count:
   
   \[
   f = \left( \sum_{i=1}^{k(G)} f_i \right) - (k(G) - 1)
   \]
   
   So we get:
   
   \[
   n - m + \left( \left( \sum_{i=1}^{k(G)} f_i \right) - (k(G) - 1) \right) = 2k(G)
   \]
   
   Simplifying:
   
   \[
   n - m + f - k(G) + 1 = 2k(G)
   \]
   
   Thus:
   
   \[
   n - m + f = 2k(G) - k(G) + 1
   \]
   
   \[
   n - m + f = k(G) + 1
   \]

Therefore, we've proved the desired formula:

\[
n - m + f = k(G) + 1
\]

This shows that the Euler characteristic of a plane graph with \( k(G) \) connected components is given by \( n - m + f = k(G) + 1 \).

Would you like more details or have any questions?

Here are 5 questions you might want to ask next:
1. How does the concept of dual graphs relate to Euler's formula?
2. Can Euler's formula be extended to higher dimensions?
3. What is the significance of the Euler characteristic in topology?
4. How can we apply Euler's formula to non-planar graphs?
5. Are there specific graph algorithms that rely on Euler's formula?

Tip: Understanding how Euler's formula adapts to different graph structures can provide deeper insights into graph theory and its applications in various fields.

Explore how Euler's formula n - m + f = k(G) + 1 applies to plane graphs with multiple connected components. This problem delves into graph theory, specifically Euler's formula, and provides a detailed proof step-by-step. Suitable for advanced high school students and anyone interested in graph theory and its applications.

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