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Imagine I have a graph G = (V,E), where exactly 2*n vertices have odd degree. (Obviously, there cannot be odd number of vertices with odd degree in a graph). How can I prove that set of edges E of the graph G can be partitioned into n open trails? (ie all the edges can be divided into n chains).

Probably, this can be proved using properties of euler graphs. For example, we can add n edges to the graph, so that all the vertices get even degree. And then we can build the euler circuit. Then we remove added edges and get n open trials. This proof seems good, but it doesn't work in some cases, for example, when we have a complete graph, where all the vertices have odd degrees and we cannot add any more edges.

I'll be really grateful for any clues how to prove this statement or how to improve my own proof!

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    Related: http://math.stackexchange.com/questions/103536/prove-a-graph-containing-2k-odd-vertices-contains-k-distinct-trails?rq=1 – 2017-01-07
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    Isn't this false when $n=0$? – 2017-01-07
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    Isn't this clearly false for disconnected graphs? like for example a graph consisting of one $K_2$ and five $K_3$'s. The edges clearly cannot be partitioned into one path. – 2017-01-07
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    The linked question includes in it the additional hypothesis that the graph is connected and the linked proof using adding of edges seems to take out outside of just simple graphs and requires the potential use of multigraphs. I expect that the OP simply forgot to write the question accurately @JorgeFernándezHidalgo. Still the OP really should be the one to fix that himself. – 2017-01-07

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