Let F be a Simple Graph with n $\displaystyle >=$2 vertices.
Proove that F contains at least two vertices of the same degree.
well i know that A graph is called simple if there is at most one edge between any two points.
let $\displaystyle d_i$ be the degree of the vertex $\displaystyle v_i.$ then either $\displaystyle d_i \in \{1, 2, \cdots , n-1 \}, \ \forall i,$ or $\displaystyle d_i \in \{0,1, \cdots, n-2 \}, \ \forall i.$ you should be able to easily finish the proof now.
let $\displaystyle d_i$ be the degree of the vertex $\displaystyle v_i.$ then either $\displaystyle d_i \in \{1, 2, \cdots , n-1 \}, \ \forall i,$ or $\displaystyle d_i \in \{0,1, \cdots, n-2 \}, \ \forall i.$ you should be able to easily finish the proof now.
thank you for this...just little bit more
well i see that you are making two sets and one less then other, but how are you supposed to put that in to proof language...
thank you for this...just little bit more
well i see that you are making two sets and one less then other, but how are you supposed to put that in to proof language...
the number of elements of both sets is $\displaystyle n-1.$ so we have $\displaystyle n$ vertices and $\displaystyle n-1$ possible values for the degrees of the vertices. thus at least two vertices must have the same degree.
the number of elements of both sets is $\displaystyle n-1.$ so we have $\displaystyle n$ vertices and $\displaystyle n-1$ possible values for the degrees of the vertices. thus at least two vertices must have the same degree.