route inspection

Let's say there's an odd number of vertices of odd order. Then the sum of their orders is the sum of an odd number of odd numbers, which will be odd (as odd *odd = odd). Then we add on the orders of the even vertices, so we get odd + a bunch of evens, giving odd. But the total of the vertices' orders can't be odd. Thus there can't be an odd number of vertices of odd order.

I understand the last question: how about this?

for part d.

we know the sum of degrees of a graph in equal to n x 2, which is equal to m.

if we use the above graph as an example: for this graph, n = the total number of node (9), and m = (9 x 2 = 18).

to total number of arcs is 17.

for part d,i) how can to total number of arcs be equal to: m/2?

The sum of the degrees (orders) of the nodes equals two times the number of arcs. This is called the Handshaking Lemma. I'm not sure what you're asking exactly, as your post is not really coherent.

oh, its equal to the sum of the arcs x 2, not the sum of the nodes x 2.

"Sum of the arcs" and "sum of the nodes" are both meaningless. To put it as clearly as I can, let's say there are n nodes, and they each have a degree - we'll use d_k to refer to the degree of node k, or in other words, the number of arcs that go through from node k.

Then we have d_1 + d_2 + d_3 + ... + d_n = 2*the number of edges.

the sum of the valencies/degrees of a graph is: 2 x number of arcs

This is just a restatement of what I just said. Remember that 'arc' and 'edge' are synonymous. "The sum of the degrees" is d_1 + d_2 + d_3 + ..., where d_i is the degree of vertex i. d_1 is the degree of the first vertex, d_2 is the degree of the second vertex, and so on, and we go on adding them up.

The proof you wrote after this is correct.