Prove that $R$ is an equivalence relation on $F$.












0














Question: Let $S = {1, 2, 3, 4}$. Let $F$ be the set of all functions $f: S to S$. Let $R$ be the relation on $F$ defined by



For any $f, g in F$, $fRg$ if and only if $f (1) + f (2) = g (1) + g (2)$.



Prove that $R$ is an equivalence relation on $F$.



I understand that to do this we must prove that $R$ is reflexive, symmetric, and transitive. I'm just having trouble using the definitions of these 3 properties to make an actual proof.






discrete-mathematics







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    Please use MathJax in future :)
    – Shaun
    Nov 25 at 1:59






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    Nov 25 at 2:14
















0














Question: Let $S = {1, 2, 3, 4}$. Let $F$ be the set of all functions $f: S to S$. Let $R$ be the relation on $F$ defined by



For any $f, g in F$, $fRg$ if and only if $f (1) + f (2) = g (1) + g (2)$.



Prove that $R$ is an equivalence relation on $F$.



I understand that to do this we must prove that $R$ is reflexive, symmetric, and transitive. I'm just having trouble using the definitions of these 3 properties to make an actual proof.






discrete-mathematics







share|cite|improve this question




















  • 1




    Please use MathJax in future :)
    – Shaun
    Nov 25 at 1:59






  • 1




    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – Shaun
    Nov 25 at 2:14














0












0








0







Question: Let $S = {1, 2, 3, 4}$. Let $F$ be the set of all functions $f: S to S$. Let $R$ be the relation on $F$ defined by



For any $f, g in F$, $fRg$ if and only if $f (1) + f (2) = g (1) + g (2)$.



Prove that $R$ is an equivalence relation on $F$.



I understand that to do this we must prove that $R$ is reflexive, symmetric, and transitive. I'm just having trouble using the definitions of these 3 properties to make an actual proof.






discrete-mathematics







share|cite|improve this question















Question: Let $S = {1, 2, 3, 4}$. Let $F$ be the set of all functions $f: S to S$. Let $R$ be the relation on $F$ defined by



For any $f, g in F$, $fRg$ if and only if $f (1) + f (2) = g (1) + g (2)$.



Prove that $R$ is an equivalence relation on $F$.



I understand that to do this we must prove that $R$ is reflexive, symmetric, and transitive. I'm just having trouble using the definitions of these 3 properties to make an actual proof.






discrete-mathematics




discrete-mathematics






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 2:35









Shaun

8,686113680




8,686113680










asked Nov 25 at 1:56









Humdrum

134




134








  • 1




    Please use MathJax in future :)
    – Shaun
    Nov 25 at 1:59






  • 1




    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – Shaun
    Nov 25 at 2:14














  • 1




    Please use MathJax in future :)
    – Shaun
    Nov 25 at 1:59






  • 1




    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    – Shaun
    Nov 25 at 2:14








1




1




Please use MathJax in future :)
– Shaun
Nov 25 at 1:59




Please use MathJax in future :)
– Shaun
Nov 25 at 1:59




1




1




After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Shaun
Nov 25 at 2:14




After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Shaun
Nov 25 at 2:14










2 Answers
2






active

oldest

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1














I'll try and get you started:



Reflexivity:



Let $fin F$. Then $f(1)+f(2)=dots$.



You need $f$ in $g$'s place.



Symmetry:



Let $f,gin F$. Then we have



$$begin{align}
fRg &iff f(1)+f(2)=g(1)+g(2) \
&iff g(1)+g(2)=f(1)+f(2)quadtext{ (by symmetry of equality)} \
&iff dots
end{align}$$



You need to conclude $gRf$ (preferably using "if and only if" statements, although implication is sufficient).



Transitivity:



Let $f, g,hin F$ with $fRg$ and $gRh$. Then, by definition of $R$, we have $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=dots$



You need to conclude that $fRh$.






share|cite|improve this answer































    2














    Reflexivity: For all $f in F$, we have $f(1)+f(2)=f(1)+f(2)$ so $f R f$



    symmetric: Let $f, g in F$ and $fRg$, then $f(1)+f(2)=g(1)+g(2) Rightarrow g(1)+g(2)=f(1)+f(2) Rightarrow g R f$



    Transitivity: Let $f,g,h in F$ and $f R g$ and $g R h$ then $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=h(1)+h(2)$ $Rightarrow$ $f(1)+f(2)=h(1)+h(2) Rightarrow f R h.$



    So $R$ is an equivalence relation on $F$.






    share|cite|improve this answer



















    • 1




      NB: It's "equivalence relation", not "equivalent relation".
      – Shaun
      Nov 25 at 2:40






    • 1




      Ah thank you for correcting me!
      – mathnoob
      Nov 25 at 2:41






    • 1




      You're welcome :)
      – Shaun
      Nov 25 at 2:41











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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

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    1














    I'll try and get you started:



    Reflexivity:



    Let $fin F$. Then $f(1)+f(2)=dots$.



    You need $f$ in $g$'s place.



    Symmetry:



    Let $f,gin F$. Then we have



    $$begin{align}
    fRg &iff f(1)+f(2)=g(1)+g(2) \
    &iff g(1)+g(2)=f(1)+f(2)quadtext{ (by symmetry of equality)} \
    &iff dots
    end{align}$$



    You need to conclude $gRf$ (preferably using "if and only if" statements, although implication is sufficient).



    Transitivity:



    Let $f, g,hin F$ with $fRg$ and $gRh$. Then, by definition of $R$, we have $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=dots$



    You need to conclude that $fRh$.






    share|cite|improve this answer




























      1














      I'll try and get you started:



      Reflexivity:



      Let $fin F$. Then $f(1)+f(2)=dots$.



      You need $f$ in $g$'s place.



      Symmetry:



      Let $f,gin F$. Then we have



      $$begin{align}
      fRg &iff f(1)+f(2)=g(1)+g(2) \
      &iff g(1)+g(2)=f(1)+f(2)quadtext{ (by symmetry of equality)} \
      &iff dots
      end{align}$$



      You need to conclude $gRf$ (preferably using "if and only if" statements, although implication is sufficient).



      Transitivity:



      Let $f, g,hin F$ with $fRg$ and $gRh$. Then, by definition of $R$, we have $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=dots$



      You need to conclude that $fRh$.






      share|cite|improve this answer


























        1












        1








        1






        I'll try and get you started:



        Reflexivity:



        Let $fin F$. Then $f(1)+f(2)=dots$.



        You need $f$ in $g$'s place.



        Symmetry:



        Let $f,gin F$. Then we have



        $$begin{align}
        fRg &iff f(1)+f(2)=g(1)+g(2) \
        &iff g(1)+g(2)=f(1)+f(2)quadtext{ (by symmetry of equality)} \
        &iff dots
        end{align}$$



        You need to conclude $gRf$ (preferably using "if and only if" statements, although implication is sufficient).



        Transitivity:



        Let $f, g,hin F$ with $fRg$ and $gRh$. Then, by definition of $R$, we have $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=dots$



        You need to conclude that $fRh$.






        share|cite|improve this answer














        I'll try and get you started:



        Reflexivity:



        Let $fin F$. Then $f(1)+f(2)=dots$.



        You need $f$ in $g$'s place.



        Symmetry:



        Let $f,gin F$. Then we have



        $$begin{align}
        fRg &iff f(1)+f(2)=g(1)+g(2) \
        &iff g(1)+g(2)=f(1)+f(2)quadtext{ (by symmetry of equality)} \
        &iff dots
        end{align}$$



        You need to conclude $gRf$ (preferably using "if and only if" statements, although implication is sufficient).



        Transitivity:



        Let $f, g,hin F$ with $fRg$ and $gRh$. Then, by definition of $R$, we have $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=dots$



        You need to conclude that $fRh$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 at 2:38

























        answered Nov 25 at 2:09









        Shaun

        8,686113680




        8,686113680























            2














            Reflexivity: For all $f in F$, we have $f(1)+f(2)=f(1)+f(2)$ so $f R f$



            symmetric: Let $f, g in F$ and $fRg$, then $f(1)+f(2)=g(1)+g(2) Rightarrow g(1)+g(2)=f(1)+f(2) Rightarrow g R f$



            Transitivity: Let $f,g,h in F$ and $f R g$ and $g R h$ then $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=h(1)+h(2)$ $Rightarrow$ $f(1)+f(2)=h(1)+h(2) Rightarrow f R h.$



            So $R$ is an equivalence relation on $F$.






            share|cite|improve this answer



















            • 1




              NB: It's "equivalence relation", not "equivalent relation".
              – Shaun
              Nov 25 at 2:40






            • 1




              Ah thank you for correcting me!
              – mathnoob
              Nov 25 at 2:41






            • 1




              You're welcome :)
              – Shaun
              Nov 25 at 2:41
















            2














            Reflexivity: For all $f in F$, we have $f(1)+f(2)=f(1)+f(2)$ so $f R f$



            symmetric: Let $f, g in F$ and $fRg$, then $f(1)+f(2)=g(1)+g(2) Rightarrow g(1)+g(2)=f(1)+f(2) Rightarrow g R f$



            Transitivity: Let $f,g,h in F$ and $f R g$ and $g R h$ then $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=h(1)+h(2)$ $Rightarrow$ $f(1)+f(2)=h(1)+h(2) Rightarrow f R h.$



            So $R$ is an equivalence relation on $F$.






            share|cite|improve this answer



















            • 1




              NB: It's "equivalence relation", not "equivalent relation".
              – Shaun
              Nov 25 at 2:40






            • 1




              Ah thank you for correcting me!
              – mathnoob
              Nov 25 at 2:41






            • 1




              You're welcome :)
              – Shaun
              Nov 25 at 2:41














            2












            2








            2






            Reflexivity: For all $f in F$, we have $f(1)+f(2)=f(1)+f(2)$ so $f R f$



            symmetric: Let $f, g in F$ and $fRg$, then $f(1)+f(2)=g(1)+g(2) Rightarrow g(1)+g(2)=f(1)+f(2) Rightarrow g R f$



            Transitivity: Let $f,g,h in F$ and $f R g$ and $g R h$ then $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=h(1)+h(2)$ $Rightarrow$ $f(1)+f(2)=h(1)+h(2) Rightarrow f R h.$



            So $R$ is an equivalence relation on $F$.






            share|cite|improve this answer














            Reflexivity: For all $f in F$, we have $f(1)+f(2)=f(1)+f(2)$ so $f R f$



            symmetric: Let $f, g in F$ and $fRg$, then $f(1)+f(2)=g(1)+g(2) Rightarrow g(1)+g(2)=f(1)+f(2) Rightarrow g R f$



            Transitivity: Let $f,g,h in F$ and $f R g$ and $g R h$ then $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=h(1)+h(2)$ $Rightarrow$ $f(1)+f(2)=h(1)+h(2) Rightarrow f R h.$



            So $R$ is an equivalence relation on $F$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 at 2:37









            Shaun

            8,686113680




            8,686113680










            answered Nov 25 at 2:05









            mathnoob

            1,794422




            1,794422








            • 1




              NB: It's "equivalence relation", not "equivalent relation".
              – Shaun
              Nov 25 at 2:40






            • 1




              Ah thank you for correcting me!
              – mathnoob
              Nov 25 at 2:41






            • 1




              You're welcome :)
              – Shaun
              Nov 25 at 2:41














            • 1




              NB: It's "equivalence relation", not "equivalent relation".
              – Shaun
              Nov 25 at 2:40






            • 1




              Ah thank you for correcting me!
              – mathnoob
              Nov 25 at 2:41






            • 1




              You're welcome :)
              – Shaun
              Nov 25 at 2:41








            1




            1




            NB: It's "equivalence relation", not "equivalent relation".
            – Shaun
            Nov 25 at 2:40




            NB: It's "equivalence relation", not "equivalent relation".
            – Shaun
            Nov 25 at 2:40




            1




            1




            Ah thank you for correcting me!
            – mathnoob
            Nov 25 at 2:41




            Ah thank you for correcting me!
            – mathnoob
            Nov 25 at 2:41




            1




            1




            You're welcome :)
            – Shaun
            Nov 25 at 2:41




            You're welcome :)
            – Shaun
            Nov 25 at 2:41


















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