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please help me make this into a contradiction or a direct proof please.

i put the question, my answer, and the textbook i used.

thank you

also please write neatly
proof 2.5 Prove har a Simple sraph and 13 cdges cannot be bipartite CHint ercattne gr apn in to ertex Sets and Court tne忤of e
Claim Splitting the graph into two vertex, Sets ves you a 8 Ver ices So if we Change tne书 apn and an A bipartite graph Cant
be bipartite
3.5 Bipartite Graphs Definition 3.8 A graph is bipartite if its vertex set V can be partitioned into two sets B, W in such a
3. Introduction to Graphs W B W Figure 3.9 Bipartite graphs If we interpret B and W as black and white, we see that a graph i
Discrete Mathematics 54 from w to z and the part of P(v) from u to y must be of equal length, and, since they have only w in
proof 2.5 Prove har a Simple sraph and 13 cdges cannot be bipartite CHint ercattne gr apn in to ertex Sets and Court tne忤of edges
Claim Splitting the graph into two vertex, Sets ves you a 8 Ver ices So if we Change tne书 apn and an A bipartite graph Can't nave 7 Oh verbces Even verces 8 en Creating trio vertex Setsyou must breat 5 2 13 + 4 3 14 15ヤ 2 tnar tney add up to equal while doing tnis, every llo cle and one 1午ャ even Odd Cucte So the graph Can not
be bipartite
3.5 Bipartite Graphs Definition 3.8 A graph is bipartite if its vertex set V can be partitioned into two sets B, W in such a way that every edge of the graph joins a vertex in B to a vertex in W. The partition V- BUW is called a bipartition of the vertex set. Example 3.9 Labellings show that the graphs in Figure 3.9 are bipartite. In both graphs, each edge joins a B to a W.
3. Introduction to Graphs W B W Figure 3.9 Bipartite graphs If we interpret B and W as black and white, we see that a graph is bipartite precisely when the vertices can be coloured using two colours so that no edge joins two vertices of the same colour. For this reason, bipartite graphs are sometimes called bichromatie partite must be even Example 3.10 The cycle Cn is bipartite if and only if n is even Theorem 3.5 A connected graph is bipartite if and only if it contains no cycle of odd length P Proof PA If a graph G contains an odd cycle (Le a cycle of odd length) then it cannot possibly be bipartite. So suppose now that G contains no odd cyclewe shal show how to colour its vertices B and W Choose any vertex y of G, and partition V as BuW where B- (uEV: shortest path from e to u has even length) w (uEV: shortest path from u to u has odd length) We have u B since 0 is even; we have to check that no edge of G has both ends in B or both ends in W Suppose there is an edge ry with E B and y B. Then, denoting the length of the shortest path from vertex vi to vertex v by di,),we have d(v,z) 2m and d(v,y)-2n for some integers m, n. But there is a walk from e to y via z of length 2m +1, so 2n S 2m+1. Similarly 2m S2n +1, so m-n Denote the shortest paths from e to z and y by Pz) and Pu) respectively. Then, since m n, both P(z) and P(u) have equal lengths. Let u be the last vertex on P(z) which is also on P(v) (possiblywv). Then the part of P(z)
Discrete Mathematics 54 from w to z and the part of P(v) from u to y must be of equal length, and, since they have only w in common, they must, with edge zy, form an odd cycle. But G has no odd cycles, so the assumption of the existence of the edge ry must be false. So there is no edge with both edges in B: similarly there is no edge with both edges in W Corollary 36 All trees are bipartite. Definition 3.9 (Complete bipartite graphs) A simple bipartite graph with vertex set VBUW is complete if every vertex in B is joined to every vertex in W. If IBl # m and Iwi n, the graph is denoted by Km., or by K.m. For example, the utilities graph of Figure 3.2 is K3,3, and the methane graph of Figure 3.6 is Ki. Clearly, Km, has m+n vertices and mn edges; m of the vertices have degree n, and n of the vertices have degree m. The complete graphs Kn and the complete bipartite graphs Kmn play im- portant roles in graph theory, particularly in the study of planarity to which we now turn
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4X16=16 =6+11 を+10 6x11 66 and honeLa.ohntle?冲ん5n-. Irxnemtie g is Re CAM

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please help me make this into a contradiction or a direct proof please. i put the question, my answer, and the textbook i used. thank you also please write neatly proof 2.5 Prove har a Sim...
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