\paragraph{22.}
Show that in any simple graph there is a path from any vertex of odd degree to some other vertex of odd degree.\newline
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Any simple graph can be broken up into a number of connected subgraphs. According to the handshaking theorem, the sum of the degrees of all the vertices in a graph must be even, as $\sum{deg(v)} = 2e$. It follows from this fact that the connected subgraphs have an even number of vertices with odd degree, as the theorem would be violated if any of the subgraphs had an odd number of vertices with odd degree. (This would make the sum of the vertex degrees odd, which is impossible. For example, $2e = 5$ is impossible in a discrete system). As each connected subgraph has more than one vertex of odd degree, and (by the definition of connected) there is a path between any two vertices in all the connected subgraphs, it follows that there exists a path from any vertex with odd degree to a different vertex with odd degree in any graph.