I'm relatively new to Graph Theory, so please be gentle.

Here's what I have. Any feedback on how I could improve and finalise my little proof would be amazing. Otherwise, if it is a sound proof, let me know! Thanks for your time, all.

Every graph $G$ has a path with $\delta(G)$ edges.

$\mathbf{Proof.}$ Let $L$ be the longest possible path in $G$ and assume that $|L|<\delta(G)$. Consider vertex $v$, an endpoint of $L$. The vertex $v$ will have at least $\delta(G)$ vertices in its neighbourhood. As $L$ is the longest path in $G$, all vertices in $N_G(v)$ will be elements of path $L$'s vertex set. But this is a contradiction, as $L$ has at least $\delta(G)+1$ vertices, and so also contains at least $[\delta(G)+1]-1=\delta(G)$ edges. So $|L|\geq\delta(G)$, as was to be shown.

  • 1
    $\begingroup$ What is $\delta(G)$? Is it the minimum degree $\min_{v\in G} \deg(v)$? $\endgroup$
    – AlexR
    Oct 6, 2014 at 7:34
  • $\begingroup$ $\delta(G)$ is the minimum degree of graph $G$. Thanks, I'll tidy that up so clarity is achieved. $\endgroup$
    – Old mate
    Oct 6, 2014 at 7:35
  • 3
    $\begingroup$ Your proof looks good. $\endgroup$
    – Paul
    Oct 6, 2014 at 7:36
  • $\begingroup$ That's good to hear. I was worried there'd be a hole in my logic or something I overlooked. Thank you both. $\endgroup$
    – Old mate
    Oct 6, 2014 at 7:37
  • 1
    $\begingroup$ I'm used to seeing $|L|$ used to mean the number of vertices of $L$ (this is how it's used in Diestel for example) rather than the number of edges. If you've encountered it, notation like $|E(L)|$ or $e(L)$ might be less ambiguous. (Though obviously if you're doing some sort of course then stick to the suggested notation.) $\endgroup$
    – Raoul
    Oct 6, 2014 at 8:31

1 Answer 1


Remove the 'Proof by contradiction' part and you have a direct proof.

Take any vertex $v_1$. It has at least degree $\delta$. Take one of the $\delta$ edges to $v_2$. Remove $v_1$ from $G$. Remaining graph has degree at least $\delta - 1$. Repeat.

  • $\begingroup$ Very nice. Thanks heaps, all! $\endgroup$
    – Old mate
    Oct 6, 2014 at 9:52

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