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Show that if a self-complementary graph contains a pendant vertex, then it must have at least another pendant vertex.

$\exists v\in V (G):d_G(v)=1 (\because \text{self-complementary graph contains a pendant vertex})$

$\implies \exists w \in V (G):d_G(v)=odd.(\because {\sum_{v\in V (G) }d_G(v)=2.\text{No.of edges}})$ I am not able to prove it has another pendant vertex.

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    $\begingroup$ Have your tried to construct a counterexample? You ought not to succeed, of course, but the trying may give you some ideas why it has to be true (if it is). $\endgroup$ – hmakholm left over Monica Aug 22 '18 at 17:24
  • $\begingroup$ no. how do i proceed with the counterexample $\endgroup$ – Math geek Aug 22 '18 at 17:27
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    $\begingroup$ x @MathGeek, if you haven't tried to find a counterexample yet, try that for at least half an hour before you declare yourself to be stuck. $\endgroup$ – hmakholm left over Monica Aug 22 '18 at 17:29
  • $\begingroup$ okay. thank you $\endgroup$ – Math geek Aug 22 '18 at 17:29

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