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Prove: If G is a tree than vertices of G can be numbered v1,v2....,vn in such a way that $(\forall i>1)|{vk:vkvi \in E(G)\land}k

So my idea is to do it by PHP (Pigeonhole principle):

v is number of vertices

e- number of edges

e=v-1 by definition.

so |e|<|v| than by PHP it's true

Is it a good idea?

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    What is "PHP" in this context? Is your tree oriented or non-oriented? – 2017-02-04
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    @HenningMakholm I mean Pigeonhole principle – 2017-02-04
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    Your idea doesn't sound convincing. If all you use is that $e=v-1$, then it would seem that your argument would work just as well for a graph consisting of a triangle and an isolated vertex -- but that graph doesn't have the property you're trying to prove! – 2017-02-04
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    so is there any way to do it by PHP? all I need to use some other proof? – 2017-02-04
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    Pick up the tree by any vertex and let it dangle. Number the vertices starting at the top, making arbitrary choices for vertices the same distance from the top. – 2017-02-04
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    can you write it in proof form? I don't know how to write this as math proof. – 2017-02-04

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