Find classes of relation equivalence












0














$S={3n+1:n∈N} = {1,4,7,10,...}$ and relation is defined as:



$(x,y) ∈ ρ text{ def }⇔ 4|(x + 3y)$



I need to prove that relation is relation of equivalence (that means that it is reflexive, symmetric and transitive.) I know how to do that, and once I prove that it is relation of equivalence, I need to find the equivalence classes.




My question is: For example: class of $1$ is defined as $$1={x∈S : xtext{ is related to }1} = {x∈S: 4|(x + 3*1)} = {3n+1:n∈Ntext{ and }4|(3n+1)+3text{ or just }4|(3n+3)}$$ ???




Please help, thank you!






equivalence-relations







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    0














    $S={3n+1:n∈N} = {1,4,7,10,...}$ and relation is defined as:



    $(x,y) ∈ ρ text{ def }⇔ 4|(x + 3y)$



    I need to prove that relation is relation of equivalence (that means that it is reflexive, symmetric and transitive.) I know how to do that, and once I prove that it is relation of equivalence, I need to find the equivalence classes.




    My question is: For example: class of $1$ is defined as $$1={x∈S : xtext{ is related to }1} = {x∈S: 4|(x + 3*1)} = {3n+1:n∈Ntext{ and }4|(3n+1)+3text{ or just }4|(3n+3)}$$ ???




    Please help, thank you!






    equivalence-relations







    share|cite|improve this question



























      0












      0








      0


      0





      $S={3n+1:n∈N} = {1,4,7,10,...}$ and relation is defined as:



      $(x,y) ∈ ρ text{ def }⇔ 4|(x + 3y)$



      I need to prove that relation is relation of equivalence (that means that it is reflexive, symmetric and transitive.) I know how to do that, and once I prove that it is relation of equivalence, I need to find the equivalence classes.




      My question is: For example: class of $1$ is defined as $$1={x∈S : xtext{ is related to }1} = {x∈S: 4|(x + 3*1)} = {3n+1:n∈Ntext{ and }4|(3n+1)+3text{ or just }4|(3n+3)}$$ ???




      Please help, thank you!






      equivalence-relations







      share|cite|improve this question















      $S={3n+1:n∈N} = {1,4,7,10,...}$ and relation is defined as:



      $(x,y) ∈ ρ text{ def }⇔ 4|(x + 3y)$



      I need to prove that relation is relation of equivalence (that means that it is reflexive, symmetric and transitive.) I know how to do that, and once I prove that it is relation of equivalence, I need to find the equivalence classes.




      My question is: For example: class of $1$ is defined as $$1={x∈S : xtext{ is related to }1} = {x∈S: 4|(x + 3*1)} = {3n+1:n∈Ntext{ and }4|(3n+1)+3text{ or just }4|(3n+3)}$$ ???




      Please help, thank you!






      equivalence-relations




      equivalence-relations






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      edited Nov 28 at 17:43









      Mason

      1,8971528




      1,8971528










      asked Nov 28 at 17:29









      Haus

      307




      307






















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          Well, let's say that the equivalence class of $1$ is
          $$bar 1={xin S colon (x,1)in rho}=$$$$={xin Scolon 4|x+3cdot 1}={xin mathbb N_0colon x=3n+1, 4|(3n+1)+3}.$$



          So $bar 1$ is the set of natural numbers such that $x=3n+1$, $(nin mathbb N_0)$ and $4|3n+4$.



          Since this is the same that $4|3n$, this implies $4|n$, so $n=4k$ with $kin mathbb N_0$.



          So the elements of $bar 1$ are those of the form $3n+1=12k+1$ with $kin mathbb N_0$, that is,
          $$bar 1={1,13,25,37,ldots}$$
          (note that, in fact, $bar 1subset S$) and you can figure out what (and how many) other equivalence classes there are.






          share|cite|improve this answer





















          • Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
            – Haus
            Nov 28 at 18:28






          • 1




            No, because ${0,4,8,16,ldots}notsubset S$.
            – Alejandro Nasif Salum
            Nov 28 at 18:43












          • So how would class of 4 look like?
            – Haus
            Nov 28 at 18:45






          • 1




            You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
            – Alejandro Nasif Salum
            Nov 29 at 21:17











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          1 Answer
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          1 Answer
          1






          active

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          active

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          1














          Well, let's say that the equivalence class of $1$ is
          $$bar 1={xin S colon (x,1)in rho}=$$$$={xin Scolon 4|x+3cdot 1}={xin mathbb N_0colon x=3n+1, 4|(3n+1)+3}.$$



          So $bar 1$ is the set of natural numbers such that $x=3n+1$, $(nin mathbb N_0)$ and $4|3n+4$.



          Since this is the same that $4|3n$, this implies $4|n$, so $n=4k$ with $kin mathbb N_0$.



          So the elements of $bar 1$ are those of the form $3n+1=12k+1$ with $kin mathbb N_0$, that is,
          $$bar 1={1,13,25,37,ldots}$$
          (note that, in fact, $bar 1subset S$) and you can figure out what (and how many) other equivalence classes there are.






          share|cite|improve this answer





















          • Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
            – Haus
            Nov 28 at 18:28






          • 1




            No, because ${0,4,8,16,ldots}notsubset S$.
            – Alejandro Nasif Salum
            Nov 28 at 18:43












          • So how would class of 4 look like?
            – Haus
            Nov 28 at 18:45






          • 1




            You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
            – Alejandro Nasif Salum
            Nov 29 at 21:17
















          1














          Well, let's say that the equivalence class of $1$ is
          $$bar 1={xin S colon (x,1)in rho}=$$$$={xin Scolon 4|x+3cdot 1}={xin mathbb N_0colon x=3n+1, 4|(3n+1)+3}.$$



          So $bar 1$ is the set of natural numbers such that $x=3n+1$, $(nin mathbb N_0)$ and $4|3n+4$.



          Since this is the same that $4|3n$, this implies $4|n$, so $n=4k$ with $kin mathbb N_0$.



          So the elements of $bar 1$ are those of the form $3n+1=12k+1$ with $kin mathbb N_0$, that is,
          $$bar 1={1,13,25,37,ldots}$$
          (note that, in fact, $bar 1subset S$) and you can figure out what (and how many) other equivalence classes there are.






          share|cite|improve this answer





















          • Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
            – Haus
            Nov 28 at 18:28






          • 1




            No, because ${0,4,8,16,ldots}notsubset S$.
            – Alejandro Nasif Salum
            Nov 28 at 18:43












          • So how would class of 4 look like?
            – Haus
            Nov 28 at 18:45






          • 1




            You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
            – Alejandro Nasif Salum
            Nov 29 at 21:17














          1












          1








          1






          Well, let's say that the equivalence class of $1$ is
          $$bar 1={xin S colon (x,1)in rho}=$$$$={xin Scolon 4|x+3cdot 1}={xin mathbb N_0colon x=3n+1, 4|(3n+1)+3}.$$



          So $bar 1$ is the set of natural numbers such that $x=3n+1$, $(nin mathbb N_0)$ and $4|3n+4$.



          Since this is the same that $4|3n$, this implies $4|n$, so $n=4k$ with $kin mathbb N_0$.



          So the elements of $bar 1$ are those of the form $3n+1=12k+1$ with $kin mathbb N_0$, that is,
          $$bar 1={1,13,25,37,ldots}$$
          (note that, in fact, $bar 1subset S$) and you can figure out what (and how many) other equivalence classes there are.






          share|cite|improve this answer












          Well, let's say that the equivalence class of $1$ is
          $$bar 1={xin S colon (x,1)in rho}=$$$$={xin Scolon 4|x+3cdot 1}={xin mathbb N_0colon x=3n+1, 4|(3n+1)+3}.$$



          So $bar 1$ is the set of natural numbers such that $x=3n+1$, $(nin mathbb N_0)$ and $4|3n+4$.



          Since this is the same that $4|3n$, this implies $4|n$, so $n=4k$ with $kin mathbb N_0$.



          So the elements of $bar 1$ are those of the form $3n+1=12k+1$ with $kin mathbb N_0$, that is,
          $$bar 1={1,13,25,37,ldots}$$
          (note that, in fact, $bar 1subset S$) and you can figure out what (and how many) other equivalence classes there are.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 at 17:57









          Alejandro Nasif Salum

          3,999117




          3,999117












          • Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
            – Haus
            Nov 28 at 18:28






          • 1




            No, because ${0,4,8,16,ldots}notsubset S$.
            – Alejandro Nasif Salum
            Nov 28 at 18:43












          • So how would class of 4 look like?
            – Haus
            Nov 28 at 18:45






          • 1




            You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
            – Alejandro Nasif Salum
            Nov 29 at 21:17


















          • Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
            – Haus
            Nov 28 at 18:28






          • 1




            No, because ${0,4,8,16,ldots}notsubset S$.
            – Alejandro Nasif Salum
            Nov 28 at 18:43












          • So how would class of 4 look like?
            – Haus
            Nov 28 at 18:45






          • 1




            You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
            – Alejandro Nasif Salum
            Nov 29 at 21:17
















          Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
          – Haus
          Nov 28 at 18:28




          Does it means that the elements of class 4 will have form 3n+1 = 4k with k ∈ N and that is {0, 4, 8, 16,...} ? Thanks!
          – Haus
          Nov 28 at 18:28




          1




          1




          No, because ${0,4,8,16,ldots}notsubset S$.
          – Alejandro Nasif Salum
          Nov 28 at 18:43






          No, because ${0,4,8,16,ldots}notsubset S$.
          – Alejandro Nasif Salum
          Nov 28 at 18:43














          So how would class of 4 look like?
          – Haus
          Nov 28 at 18:45




          So how would class of 4 look like?
          – Haus
          Nov 28 at 18:45




          1




          1




          You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
          – Alejandro Nasif Salum
          Nov 29 at 21:17




          You can do the same math for $4$ instead of $1$, and you would get$$bar 4={4,16,28,40,ldots}.$$ Also $$bar7={7,19,31,43,ldots}$$ and $$bar{10}={10,22,34,46,ldots},$$ and those are the only four equivalence classes in $S$.
          – Alejandro Nasif Salum
          Nov 29 at 21:17


















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