I want to prove this claim using contradiction:

Let $G$ be a simple graph with $n$ vertices. Prove by contradiction that $G$ has two vertices having the same degree (Consider the vertex of smallest degree and vertex of largest degree).

I think I know where to start. The vertex of the largest degree is $n-1$ in this graph and the smallest is 1. How can I use contradiction to prove the statement above? I studied the proof of this using pigeonhole but I wanna use contradiction this time.


Suppose such a graph existed. Each vertex in the graph can have a degree from 0 to $n-1$ (simple graphs do not forbid a degree-0 vertex, connected graphs do). Since this range spans $n$ values in total and each vertex degree is different, the degrees are distributed one-per-vertex. In particular, there must exist a vertex with degree $n-1$ and one with degree 0.

Now note that the degree-$n-1$ vertex is connected to all other vertices because the graph is simple, including the degree-0 vertex. But the latter is not connected to any vertex, and this induces the contradiction.

Therefore, two vertices in the graph must have the same degree.

  • $\begingroup$ Beat me to it! +1 $\endgroup$ – Draconis Mar 14 '18 at 0:38

It would really be the same argument turned around a bit. Assume $G$ is a simple graph on $n$ vertices with all vertices having a different degree. The maximum degree of a vertex is $n-1$, so there must be a vertex of degree $0$. If there is a vertex of degree $0$, it is not a simple graph.


Take a graph $G$ where all vertices have distinct degrees. So, $$d(1)<d(2)<\cdots<d(n),$$ and $$d(1)>0, d(2)>d(1)>0, \ldots, d(n)>d(n-1)>\ldots>0.$$

Hence we have just one vertex of degree $\geq n-1$.

However, in a simple graph with $n$ vertices a vertex can have a maximum degree of $n-1$. So, $d(n)=d(n-1)$. So, our assumption of having all degrees different was wrong and we have a contradiction.

So, in a simple graph $G$, we must have at least $2$ vertices of the same degree.

  • $\begingroup$ Welcome to Math.SE! Please format you post using MathJax. This page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. $\endgroup$ – Brian Mar 16 at 16:46
  • $\begingroup$ I'm confused about why you have $d(1)>0$. Could you explain further? $\endgroup$ – Mauve Mar 16 at 17:04
  • $\begingroup$ We are showing that we want two vertices of the same degree. So, why would we bother to have d(1) = 0 if we are in a simple graph? I think we can just say that we are considering graphs with at least one edge because in a graph of just one vertex this will never apply. $\endgroup$ – user654759 Mar 17 at 10:38
  • $\begingroup$ There are graphs with at least one edge that also have isolated vertices: for example, consider the graph $•\phantom{==} •—•—• \phantom{==} • \phantom{==} •—•$. So you cannot just omit the 0-degree case. $\endgroup$ – Mauve Mar 17 at 17:09

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