I want to prove that at least two vertices have the same degree in any graph (with 2 or more vertices). I do have a few graphs in mind that prove this statement correct, but how would I go about proving it (or disproving it) for ALL graphs?
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1$\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$– Shaun ♦Commented Feb 4, 2019 at 17:56
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1$\begingroup$ Symmetric, simple graphs? $\endgroup$– AphelliCommented Feb 4, 2019 at 17:56
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1$\begingroup$ Possible duplicate of If n is a natural number ≥2 how do I prove that any graph with n vertices has at least two vertices of the same degree? $\endgroup$– AcccumulationCommented Feb 4, 2019 at 22:23
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3 Answers
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Say graph is simple with no loops and that all vertices have different degree $d_1,d_2,...d_n$, then $$\{d_1,d_2,...d_n\} = \{0,1,2...,n-1\}$$
So there is a vertex with degree $n-1$ and a vertex with degree $0$. A contradiction.
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$\begingroup$ Perhaps it would be better to include that the graph is simple? $\endgroup$ Commented Feb 4, 2019 at 18:04
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$\begingroup$ Is it necessary to specify "no loops"? $\endgroup$ Commented Feb 4, 2019 at 22:24
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$\begingroup$ Of course. Take a graph with two vertices and one connected to it self. @Shufflepants $\endgroup$– nonuserCommented Feb 4, 2019 at 22:25
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1$\begingroup$ @greedoid Then by "no loops" did you mean "no self connection"? Otherwise I'd interpret "simple with no loops" as being equivalent to a tree. But in any case, the OP didn't specify "simple" or "no loops". So, it would seem the answer to the question is: it's not provable because it's not true. $\endgroup$ Commented Feb 4, 2019 at 22:32
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1$\begingroup$ Because I didn't see it at first: This follows because in a simple graph the degree of any vertex must be less than the number of vertices. Thus if the vertex degrees are all different and less than $n$ they must be $\{0, 1, 2..., n-1\}$. $\endgroup$ Commented Feb 5, 2019 at 0:02
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I assume we're talking about finite graphs. I'm pretty sure your statement is false for infinite graphs.
Assume that a finite graph $G$ has $n$ vertices. Then each vertex has a degree between $n-1$ and $0$. But if any vertex has degree $0$, then no vertex can have degree $n-1$, so it's not possible for the degrees of the graph's vertices to include both $0$ and $n-1$. Thus, the $n$ vertices of the graph can only have $n-1$ different degrees, so by the pigeonhole principle at least two vertices must have the same degree.
answered Feb 4, 2019 at 17:59
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$\begingroup$ You are assuming the graph is simple? $\endgroup$– ServaesCommented Feb 4, 2019 at 22:07
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1$\begingroup$ Yes. That is assumed in the definition of graph that I learned. I learned to call a graph with loops or multiple edges between vertices a "multigraph." $\endgroup$ Commented Feb 4, 2019 at 23:04
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EDIT: Graphs are typically defined as being finite. Infinite graphs are a generalization. I did not know this at the time of the post. I will leave this answer up though in case anyone finds it useful.
Here is a counterexample.
Let $G$ be a graph on the positive integers where there is an edge from $x$ to $y$ if $x < y \le 2x$. Note that we will ignore the direction of the edges. So $2$, for example, is neighbored by $1$, $3$, and $4$.
Let $j$ and $k$ be distinct positive integers. Without loss of generality assume that $j < k$. Note that $deg(j) = j + \lfloor j/2 \rfloor$ and $deg(k) = k + \lfloor k/2 \rfloor$. We have that
$$j < k$$ $$\lfloor j/2 \rfloor \le \lfloor k/2 \rfloor$$ $$j + \lfloor j/2 \rfloor < k + \lfloor k/2 \rfloor$$ $$deg(j) < deg(k)$$ $$deg(j) \neq deg(k)$$
Therefore, no two different vertices will have the same degree. $\square$
answered Feb 4, 2019 at 22:49
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$\begingroup$ As noted at en.wikipedia.org/wiki/Graph_theory#Definitions, "V and E are usually taken to be finite, and many of the well-known results are not true (or are rather different) for infinite graphs because many of the arguments fail in the infinite case." $\endgroup$– ruakhCommented Feb 5, 2019 at 3:11
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$\begingroup$ @ruakh Huh, didn't know that. I edited a disclaimer into my answer. $\endgroup$ Commented Feb 5, 2019 at 3:44