How to prove that at least two vertices have the same degree in any graph?












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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?






combinatorics discrete-mathematics graph-theory







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  • $begingroup$
    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.
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    – Shaun
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  • 1




    $begingroup$
    Symmetric, simple graphs?
    $endgroup$
    – Mindlack
    9 hours ago






  • 2




    $begingroup$
    Possible duplicate of "In a party people shake hands with one another"
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    – JMoravitz
    8 hours ago










  • $begingroup$
    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?
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    – Acccumulation
    4 hours ago


















5












$begingroup$


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?






combinatorics discrete-mathematics graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
    9 hours ago






  • 1




    $begingroup$
    Symmetric, simple graphs?
    $endgroup$
    – Mindlack
    9 hours ago






  • 2




    $begingroup$
    Possible duplicate of "In a party people shake hands with one another"
    $endgroup$
    – JMoravitz
    8 hours ago










  • $begingroup$
    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?
    $endgroup$
    – Acccumulation
    4 hours ago
















5












5








5


1



$begingroup$


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?






combinatorics discrete-mathematics graph-theory







share|cite|improve this question











$endgroup$




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?






combinatorics discrete-mathematics graph-theory




combinatorics discrete-mathematics graph-theory






share|cite|improve this question















share|cite|improve this question













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edited 2 hours ago









James Rettie

31




31










asked 9 hours ago









desiignerdesiigner

575




575












  • $begingroup$
    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
    9 hours ago






  • 1




    $begingroup$
    Symmetric, simple graphs?
    $endgroup$
    – Mindlack
    9 hours ago






  • 2




    $begingroup$
    Possible duplicate of "In a party people shake hands with one another"
    $endgroup$
    – JMoravitz
    8 hours ago










  • $begingroup$
    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?
    $endgroup$
    – Acccumulation
    4 hours ago




















  • $begingroup$
    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
    9 hours ago






  • 1




    $begingroup$
    Symmetric, simple graphs?
    $endgroup$
    – Mindlack
    9 hours ago






  • 2




    $begingroup$
    Possible duplicate of "In a party people shake hands with one another"
    $endgroup$
    – JMoravitz
    8 hours ago










  • $begingroup$
    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?
    $endgroup$
    – Acccumulation
    4 hours ago


















$begingroup$
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
9 hours ago




$begingroup$
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
9 hours ago




1




1




$begingroup$
Symmetric, simple graphs?
$endgroup$
– Mindlack
9 hours ago




$begingroup$
Symmetric, simple graphs?
$endgroup$
– Mindlack
9 hours ago




2




2




$begingroup$
Possible duplicate of "In a party people shake hands with one another"
$endgroup$
– JMoravitz
8 hours ago




$begingroup$
Possible duplicate of "In a party people shake hands with one another"
$endgroup$
– JMoravitz
8 hours ago












$begingroup$
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?
$endgroup$
– Acccumulation
4 hours ago






$begingroup$
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?
$endgroup$
– Acccumulation
4 hours ago












3 Answers
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$begingroup$

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?
    $endgroup$
    – Shubham Johri
    8 hours ago










  • $begingroup$
    Is it necessary to specify "no loops"?
    $endgroup$
    – Shufflepants
    4 hours ago










  • $begingroup$
    Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
    $endgroup$
    – greedoid
    4 hours ago








  • 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.
    $endgroup$
    – Shufflepants
    4 hours ago










  • $begingroup$
    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 ${0, 1, 2..., n-1}$.
    $endgroup$
    – immibis
    2 hours ago





















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$begingroup$

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.






share|cite|improve this answer








New contributor




Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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  • $begingroup$
    You are assuming the graph is simple?
    $endgroup$
    – Servaes
    4 hours ago










  • $begingroup$
    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."
    $endgroup$
    – Robert Shore
    3 hours ago



















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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. $square$






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    3 Answers
    3






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    3 Answers
    3






    active

    oldest

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    active

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    10












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Perhaps it would be better to include that the graph is simple?
      $endgroup$
      – Shubham Johri
      8 hours ago










    • $begingroup$
      Is it necessary to specify "no loops"?
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
      $endgroup$
      – greedoid
      4 hours ago








    • 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.
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      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 ${0, 1, 2..., n-1}$.
      $endgroup$
      – immibis
      2 hours ago


















    10












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Perhaps it would be better to include that the graph is simple?
      $endgroup$
      – Shubham Johri
      8 hours ago










    • $begingroup$
      Is it necessary to specify "no loops"?
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
      $endgroup$
      – greedoid
      4 hours ago








    • 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.
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      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 ${0, 1, 2..., n-1}$.
      $endgroup$
      – immibis
      2 hours ago
















    10












    10








    10





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 8 hours ago

























    answered 9 hours ago









    greedoidgreedoid

    41k1149101




    41k1149101












    • $begingroup$
      Perhaps it would be better to include that the graph is simple?
      $endgroup$
      – Shubham Johri
      8 hours ago










    • $begingroup$
      Is it necessary to specify "no loops"?
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
      $endgroup$
      – greedoid
      4 hours ago








    • 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.
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      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 ${0, 1, 2..., n-1}$.
      $endgroup$
      – immibis
      2 hours ago




















    • $begingroup$
      Perhaps it would be better to include that the graph is simple?
      $endgroup$
      – Shubham Johri
      8 hours ago










    • $begingroup$
      Is it necessary to specify "no loops"?
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
      $endgroup$
      – greedoid
      4 hours ago








    • 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.
      $endgroup$
      – Shufflepants
      4 hours ago










    • $begingroup$
      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 ${0, 1, 2..., n-1}$.
      $endgroup$
      – immibis
      2 hours ago


















    $begingroup$
    Perhaps it would be better to include that the graph is simple?
    $endgroup$
    – Shubham Johri
    8 hours ago




    $begingroup$
    Perhaps it would be better to include that the graph is simple?
    $endgroup$
    – Shubham Johri
    8 hours ago












    $begingroup$
    Is it necessary to specify "no loops"?
    $endgroup$
    – Shufflepants
    4 hours ago




    $begingroup$
    Is it necessary to specify "no loops"?
    $endgroup$
    – Shufflepants
    4 hours ago












    $begingroup$
    Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
    $endgroup$
    – greedoid
    4 hours ago






    $begingroup$
    Of course. Take a graph with two vertices and one connected to it self. @Shufflepants
    $endgroup$
    – greedoid
    4 hours ago






    1




    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.
    $endgroup$
    – Shufflepants
    4 hours ago




    $begingroup$
    @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.
    $endgroup$
    – Shufflepants
    4 hours ago












    $begingroup$
    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 ${0, 1, 2..., n-1}$.
    $endgroup$
    – immibis
    2 hours ago






    $begingroup$
    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 ${0, 1, 2..., n-1}$.
    $endgroup$
    – immibis
    2 hours ago













    8












    $begingroup$

    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.






    share|cite|improve this answer








    New contributor




    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      You are assuming the graph is simple?
      $endgroup$
      – Servaes
      4 hours ago










    • $begingroup$
      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."
      $endgroup$
      – Robert Shore
      3 hours ago
















    8












    $begingroup$

    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.






    share|cite|improve this answer








    New contributor




    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      You are assuming the graph is simple?
      $endgroup$
      – Servaes
      4 hours ago










    • $begingroup$
      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."
      $endgroup$
      – Robert Shore
      3 hours ago














    8












    8








    8





    $begingroup$

    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.






    share|cite|improve this answer








    New contributor




    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$



    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.







    share|cite|improve this answer








    New contributor




    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered 8 hours ago









    Robert ShoreRobert Shore

    1915




    1915




    New contributor




    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • $begingroup$
      You are assuming the graph is simple?
      $endgroup$
      – Servaes
      4 hours ago










    • $begingroup$
      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."
      $endgroup$
      – Robert Shore
      3 hours ago


















    • $begingroup$
      You are assuming the graph is simple?
      $endgroup$
      – Servaes
      4 hours ago










    • $begingroup$
      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."
      $endgroup$
      – Robert Shore
      3 hours ago
















    $begingroup$
    You are assuming the graph is simple?
    $endgroup$
    – Servaes
    4 hours ago




    $begingroup$
    You are assuming the graph is simple?
    $endgroup$
    – Servaes
    4 hours ago












    $begingroup$
    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."
    $endgroup$
    – Robert Shore
    3 hours ago




    $begingroup$
    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."
    $endgroup$
    – Robert Shore
    3 hours ago











    0












    $begingroup$

    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. $square$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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. $square$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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. $square$






        share|cite|improve this answer









        $endgroup$



        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. $square$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        PyRulezPyRulez

        4,68622469




        4,68622469






























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