A simple graph G must have atleast 2 vertices, prove that G must contain two or more vertices of the same degree
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Suppose we have n vertices and they all have different degree. Then the degree sequence must be \(0,1,2,3,4,5,6,...,n-1\) (from the pigeon hole property).
If there is a vertex of degree \(n−1\) then there can't be one of degree \(0\).
sorry i kinda dont get it.
The question is asking to show that in every simple graph there must be at least 2 vertices with the same degree.
In other words, it is NOT the case that all the vertices in a simple graph have different degrees.
So we take a graph with \(n\ge 2\) vertices and assume by contradiction that no two vertices have the same degree. In other words, every vertex has a different degree than every other vertex.
This means that the degree sequence is \(0,1,2,...,n-1\).
Why? Well take any vertex, the highest degree it can have is \(n-1\) because that's how many other vertices there are, and the lowest degree it will have is \(0\).
Well suppose we have 6 vertices and we pick some vertex and it has degree 4, then if we pick another it cant have degree 4, so it has some other degree, say 2. Then pick a third and it must have degree different from 2 and 4, say 3. Then pick a 4th vertex....if you keep doing this you will arrive at the degree sequence 0,1,2,3,4,5.
Does this make sense?