Show that in every simple graph there is a path from....

[hyderman]

I need help understanding this question:

Show that in every simple graph there is a path from any vertex of odd degree to some other vertex of odd degree.
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Edited by stapel -- Reason for edit: Converting image to text.

 

[pka]

Show that in every simple graph there is a path from any vertex of odd degree to some other vertex of odd degree.

Hyderman, you have posted several statements of questions without any evidence of work on your part. We need to see some of your effort. At least tell us what you have done. Now, it is not good enough to just say “I don’t know where to even begin”! Because then we will tell you to read your textbook. Learn the definitions and theorems.

 

[hyderman]

assume graph G, assume vertex v of odd degree.

lets take subset of vertex v through some path is v; name it Gv

now Gv is a graph, with sum degree even....... now how can i connect this ot odd graph?

 

[pka]

O.K. If there is an odd vertex A we can take a component (which is a maximal connected subgraph) that contains A. Thus we can assume the graph is connected to begin with. In any simple graph the number of odd vertices is even. Thus if A is a odd vertex there must be at least one other odd vertex call it B. Because the graph is connected there is a path from A to B.

 

[hyderman]

thanx much .... explained it very clear