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Prove that a simple graph is connected if and only if it has a spanning tree.
Ans: First assume that a simple graph G has a spanning tree T. T consists of every node of G. By the definition of a tree, there is a path among any two nodes of T. As T is a subgraph of G, there is a path among each pair of nodes in G. Hence G is connected.
Here now let G is connected. If G is a tree then nothing to prove. If G is not a tree, it must consist of a simple circuit. Let G has n nodes. We can choose (n - 1) arcs from G in such type of a way that they not form a circuit. It results into a subgraph comprising all nodes and only (n - 1) arcs. So by definition this subgraph is a spanning tree.
Limit Comparison Test Assume that we have two series ∑a n and ∑b n with a n , b n ≥ 0 for all n. Determine, If c is positive (i.e. c > 0 ) and is finite (i.e. c
altitude 35000 @ 9:30 9;42 alt 17500 increase speed by factor of 3 level out at 2500= how much time will it take
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Assume that (X, d) is a metric space and let (x1, : : : , x n ) be a nite set of pointsof X. Elustrate , using only the denition of open, that the set X\(x1, : : : , x n ) obtain
If the squared difference of the zeros of the quadratic polynomial x 2 + p x + 45 is equal to 144 , find the value of p.
Area Problem Now It is time to start second kind of integral: Definite Integrals. The area problem is to definite integrals what tangent & rate of change problems are to d
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sin10+sin20+sin30+....+sin360=0 sin10+sin20+sin30+sin40+...sin180+sin(360-170)+......+sin(360-40)+sin(360-30)+sin(360-20)+sin360-10)+sin360 sin360-x=-sinx hence all terms cancel
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