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Consider the result from lecture that for any graph G = (V, E), 2|E| = degree(v). VEV We proved this informally in lecture, which was good enough for our purposes; here we will construct the formal proof by induction, for rigour and practice. Some of the proof is provided for you; your job is to fill in the missing pieces, namely: 1. the base case; 2. the inductive step. Proof. For any integer m≥ 0, let P(m) denote that in any graph G = (V, E) on m edges, 2m =Evev degree(v). We want to show that Vm > 0: P(m). As a base case, consider m = 0. Please fill in the proof of this base case. Consider k>0. For the induction hypothesis, assume P(k); that is, that any graph G = (V, E) on k edges has 2k=Evev degree(v). = For the inductive step, we want to show P(k+ 1); that is that any graph H edges has 2(k+ 1) = Evew degree(v). (W, F) on k + 1 Let H = (W, F) be any graph on on k+ 1 ≥1 edges. Let {u, w} be any edge of H. Consider the graph G constructed from H by removing that one edge {u, w}; that is, G = (W, F \ {{u, w}}). Note that we keep the same set of vertices; we only remove a single edge. For any vertex v W, let dega(v) denote the degree of v in graph G, and degн (v) denote the degree in graph H, as now these might be different. Please fill in the rest of the inductive step using the provided setup and notation. Thus we showed P(0) and that for all k ≥ 0 if P(k) then P(k+1). Together, these show that for all m ≥ 0, P(m).

Consider the result from lecture that for any graph G = (V, E), 2|E| = degree(v). VEV We proved this informally in lecture, which was good enough for our purposes; here we will construct the formal proof by induction, for rigour and practice. Some of the proof is provided for you; your job is to fill in the missing pieces, namely: 1. the base case; 2. the inductive step. Proof. For any integer m≥ 0, let P(m) denote that in any graph G = (V, E) on m edges, 2m =Evev degree(v). We want to show that Vm > 0: P(m). As a base case, consider m = 0. Please fill in the proof of this base case. Consider k>0. For the induction hypothesis, assume P(k); that is, that any graph G = (V, E) on k edges has 2k=Evev degree(v). = For the inductive step, we want to show P(k+ 1); that is that any graph H edges has 2(k+ 1) = Evew degree(v). (W, F) on k + 1 Let H = (W, F) be any graph on on k+ 1 ≥1 edges. Let {u, w} be any edge of H. Consider the graph G constructed from H by removing that one edge {u, w}; that is, G = (W, F \ {{u, w}}). Note that we keep the same set of vertices; we only remove a single edge. For any vertex v W, let dega(v) denote the degree of v in graph G, and degн (v) denote the degree in graph H, as now these might be different. Please fill in the rest of the inductive step using the provided setup and notation. Thus we showed P(0) and that for all k ≥ 0 if P(k) then P(k+1). Together, these show that for all m ≥ 0, P(m).

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter1: The Six Trigonometric Functions
Section: Chapter Questions
Problem 1GP: The diagram shown in Figure 1 was used by the Hindu mathematician Bhaskara to prove the theorem in...
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Problem 5 Consider the result from lecture that for any graph G = (V, E), degree(v). 2|E| = VEV We proved this informally in lecture, which was good enough for our purposes; here we will construct the formal proof by induction, for rigour and practice. Some of the proof is provided for you; your job is to fill in the missing pieces, namely: 1. the base case; 2. the inductive step. Proof. For any integer m≥ 0, let P(m) denote that in any graph G = (V, E) on m edges, 2m = Evey degree(v). We want to show that Vm > 0: P(m). As a base case, consider m = 0. Please fill in the proof of this base case. Consider k > 0. For the induction hypothesis, assume P(k); that is, that any graph G = (V, E) on k edges has 2k=Evev degree(v). = For the inductive step, we want to show P(k+ 1); that is that any graph H edges has 2(k+ 1) = Evew degree(v). (W, F) on k + 1 Let H = (W, F) be any graph on on k +1≥1 edges. Let {u, w} be any edge of H. Consider the graph G constructed from H by removing that one edge {u, w}; that is, G = (W, F\ {{u, w}}). Note that we keep the same set of vertices; we only remove a single edge. For any vertex v € W, let dega (v) denote the degree of u in graph G, and degí(v) denote the degree in graph H, as now these might be different. Please fill in the rest of the inductive step using the provided setup and notation. Thus we showed P(0) and that for all k ≥ 0 if P(k) then P(k+1). Together, these show that for all m > 0, P(m).
Transcribed Image Text:Problem 5 Consider the result from lecture that for any graph G = (V, E), degree(v). 2|E| = VEV We proved this informally in lecture, which was good enough for our purposes; here we will construct the formal proof by induction, for rigour and practice. Some of the proof is provided for you; your job is to fill in the missing pieces, namely: 1. the base case; 2. the inductive step. Proof. For any integer m≥ 0, let P(m) denote that in any graph G = (V, E) on m edges, 2m = Evey degree(v). We want to show that Vm > 0: P(m). As a base case, consider m = 0. Please fill in the proof of this base case. Consider k > 0. For the induction hypothesis, assume P(k); that is, that any graph G = (V, E) on k edges has 2k=Evev degree(v). = For the inductive step, we want to show P(k+ 1); that is that any graph H edges has 2(k+ 1) = Evew degree(v). (W, F) on k + 1 Let H = (W, F) be any graph on on k +1≥1 edges. Let {u, w} be any edge of H. Consider the graph G constructed from H by removing that one edge {u, w}; that is, G = (W, F\ {{u, w}}). Note that we keep the same set of vertices; we only remove a single edge. For any vertex v € W, let dega (v) denote the degree of u in graph G, and degí(v) denote the degree in graph H, as now these might be different. Please fill in the rest of the inductive step using the provided setup and notation. Thus we showed P(0) and that for all k ≥ 0 if P(k) then P(k+1). Together, these show that for all m > 0, P(m).
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