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







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














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






share|cite|improve this question















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

votes


















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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012335%2fprove-that-r-is-an-equivalence-relation-on-f%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012335%2fprove-that-r-is-an-equivalence-relation-on-f%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more