' Prove that if a graph G = (V,E) is connected, then log |E| =…

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Step 1: Understanding the Properties of a Connected Graph Minimum Number of Edges: In a connected graph with ∣ V ∣ |V| ∣ V ∣ vertices, there must be at least ∣ V ∣ − 1 |V| - 1 ∣ V ∣ − 1 edges. This is because a tree, which is a minimal connected graph, has exactly ∣ V ∣ − 1 |V| - 1 ∣ V ∣ − 1 edges. Maximum Number of Edges: In a connected simple graph with ∣ V ∣ |V| ∣ V ∣ vertices, the maximum number of edges is ( ∣ V ∣ 2 ) \binom{|V|}{2} ( 2 ∣ V ∣ ​ ) , which represents a complete graph K ∣ V ∣ K_{|V|} K ∣ V ∣ ​ . This is given by: ∣ E ∣ ≤ ∣ V ∣ ( ∣ V ∣ − 1 ) 2 |E| \leq \frac{|V|(|V|-1)}{2} ​ Explanation: In a connected graph, there are at least ∣ V ∣ − 1 |V| - 1 ∣ V ∣ − 1 edges because a tree has ∣ V ∣ − 1 |V| - 1 ∣ V ∣ − 1 edges. Step 2: Establishing the Upper Bound for ∣ E ∣ |E| ∣ E ∣ Upper Bound Formula: We use the fact that in the worst case (a complete graph), the number of edges ∣ E ∣ |E| ∣ E ∣ is at most ∣ V ∣ ( ∣ V ∣ − 1 ) 2 \frac{|V|(|V| - 1)}{2} 2 ∣ V ∣ ( ∣ V ∣ − 1 ) ​ . Explanation: A connected simple graph with ∣ V ∣ |V| ∣ V ∣ vertices can have at most ∣ V ∣ ( ∣ V ∣ − 1 ) 2 \frac{|V|(|V|-1)}{2} 2 ∣ V ∣ ( ∣ V ∣ − 1 ) ​ edges (a complete graph). Step 3: Applying the Logarithm to the Upper Bound Taking the Logarithm: Apply the logarithm to both sides of the inequality: log ⁡ ∣ E ∣ ≤ log ⁡ ( ∣ V ∣ ( ∣ V ∣ − 1 ) 2 ) \log |E| \leq \log \left( \frac{|V|(|V| - 1)}{2} \right) Using Logarithm Properties: Use the properties of logarithms to break down the expression: log ⁡ ∣ E ∣ ≤ log ⁡ ( ∣ V ∣ 2 2 ) \log |E| \leq \log \left( \frac{|V|^2}{2} \right) Simplifying Further: Split the logarithm using the property log ⁡ ( a b ) = log ⁡ a + log ⁡ b \log(ab) = \log a + \log b log ( ab ) = log a + log b : log ⁡ ∣ E ∣ ≤ log ⁡ ∣ V ∣ 2 + log ⁡ ( 1 2 ) \log |E| \leq \log |V|^2 + \log \left( \frac{1}{2} \right) Explanation: For a complete graph, ∣ E ∣ ≤ ∣ V ∣ ( ∣ V ∣ − 1 ) 2 |E| \leq \frac{|V|(|V| - 1)}{2} ∣ E ∣ ≤ 2 ∣ V ∣ ( ∣ V ∣ − 1 ) ​ . Step 4: Simplifying the Logarithmic Expression Simplifying the Components: Recognize that log ⁡ ∣ V ∣ 2 = 2 log ⁡ ∣ V ∣ \log |V|^2 = 2 \log |V| log ∣ V ∣ 2 = 2 log ∣ V ∣ and log ⁡ ( 1 2 ) = − log ⁡ 2 \log \left( \frac{1}{2} \right) = -\log 2 log ( 2 1 ​ ) = − log 2 : log ⁡ ∣ E ∣ ≤ 2 log ⁡ ∣ V ∣ + log ⁡ ( 1 2 ) \log |E| \leq 2 \log |V| + \log \left( \frac{1}{2} \right) Explanation: Apply the logarithm to ∣ E ∣ |E| ∣ E ∣ : log ⁡ ∣ E ∣ ≤ log ⁡ ( ∣ V ∣ ( ∣ V ∣ − 1 ) 2 ) \log |E| \leq \log \left( \frac{|V|(|V|-1)}{2} \right) Step 5: Converting to Big-O Notation Big-O Notation: In Big-O notation, we focus on the dominant term and ignore constant factors and lower order terms. The dominant term here is 2 log ⁡ ∣ V ∣ 2 \log |V| 2 log ∣ V ∣ , and the − log ⁡ 2 - \log 2 − log 2 term is a constant, which can be ignored in Big-O notation. log ⁡ ∣ E ∣ = O ( log ⁡ ∣ V ∣ ) Final Answer: We've shown that for a connected graph G = ( V , E ) G = (V, E) G = ( V , E ) , the logarithm of the number of edges log ⁡ ∣ E ∣ \log |E| log ∣ E ∣ is bounded above by a constant multiple of the logarithm of the number of vertices log ⁡ ∣ V ∣ \log |V| log ∣ V ∣ . Thus, we conclude that: log ⁡ ∣ E ∣ = O ( log ⁡ ∣ V ∣ ) \log |E| = \mathcal{O}(\log |V|)

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