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A graph G has tne property that every edge joins an odd vertex (a vertex of odd degree) with an even vertex: Show that G is bipartite:...

Question

A graph G has tne property that every edge joins an odd vertex (a vertex of odd degree) with an even vertex: Show that G is bipartite:

A graph G has tne property that every edge joins an odd vertex (a vertex of odd degree) with an even vertex: Show that G is bipartite:



Answers

Let $G$ be an undirected graph with a loop at every vertex. Show that the relation $R$ on the set of vertices of $G$ such that $u R v$ if and only if there is an edge associated to $\{u, v\}$ is a symmetric, reflexive relation on $G .$

So for honey, What is it you with a degree? Consider connected. The component which contains the word is you enacted. The component's up graph H contending you. And here we assume it has. What is this? You won it to the dark end. A one is just you. Then we have a degree. You last degree a two last at a double plus degree. N is equal to two times the number of badges in this. A graph which is even number and sayings degree you This is odd. So we have there exists and I such a sight agree A I It's odd. Then sees this graph is connected. So they exist. A path? Um, you two. Aye. Aye, you're a I. It's a degree ard. So we are done.

Were given a by apartheid simple graph and were asked two proven inequality between the number of edges and the number of Vergis ease of this graph. So let G. B A simple bi partite graph. Let the D the number of overseas of this craft and E v number of edges. Now, since G is by partite, it follows that the Enver Toussie's not an Verses v. Vergis ease partition into two sets. V one and V two. And so if we have that who's set V one contains and Vergis ease, then it follows that set V two will contain total number of overseas v my ass. Remember Verte Season V one, which is n so v minus m. We have every vertex in the set V one can only connect with for season two, and vice versa. So it follows that if V one lies in set the one, then it follows that the degree of V one must be less than or equal to V minus end. And they're Enver Toussie's in V one. So the max number, uh, edges and this is one be a function depends on that value n that we chose for the size of the one. This is going to be the end Vergis ease of the number of urgencies and view one and times the maximum number of edges which is going to be V minus and per vertex. So we get n times B minus and and just to simplify this All right, this is a function after then and we want to find an upper bound on the number of edges. And to do this you want to find the maximum number of edges and maximize it So it follows that the derivative of the maximum number of edges f prime respect to end this is going to be the minus end plus happens to have a plus. This really should be a minus and which is V minus two n. And we want to find the critical values. So said this first derivative equal to zero, and we have that and is equal to V over to It's the maximum we know occurs at and equals the over to we have that f of V over to is equal to V over two times the minus v over to this is the same as have you ever two squared, which is B squared over four. It's the maximum number of edges is be square before, so it follows that the number of edges in the graph e is less than or equal to B squared over four. This is what we wanted to prove.


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