How to prove $A=R-{-1}$ and $a*b = a+b+ab $ is a binary operation?












0












$begingroup$



$A=R-{-1}$ and $a*b = a+b+ab $





  1. Show that * is a binary operation on A

  2. Show that * is associative

  3. Show that there is an identity element in A for *

  4. Show that every element in A has an inverse with respect to *


I'm new to this lesson. All I know to prove is associativity,



let $a,b,cin A$



Consider:$$(a*b)*c=(a+b+ab)*(c)$$
$$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
$$(a*b)*c=a+b+c+ab+ac+bc+abc$$



Now consider:
$$a*(b*c)=a*(b+c+bc)$$
$$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
$$a*(b*c)=a+b+c+ab+ab+ac+abc$$
Since we get the same result it is associative.



How to prove part 1? and How do I prove that identity exist? Thanks in advance!



P.S: also I'm curious to know why -1 is omitted in the set?






binary-operations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    $A=R-{-1}$ and $a*b = a+b+ab $





    1. Show that * is a binary operation on A

    2. Show that * is associative

    3. Show that there is an identity element in A for *

    4. Show that every element in A has an inverse with respect to *


    I'm new to this lesson. All I know to prove is associativity,



    let $a,b,cin A$



    Consider:$$(a*b)*c=(a+b+ab)*(c)$$
    $$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
    $$(a*b)*c=a+b+c+ab+ac+bc+abc$$



    Now consider:
    $$a*(b*c)=a*(b+c+bc)$$
    $$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
    $$a*(b*c)=a+b+c+ab+ab+ac+abc$$
    Since we get the same result it is associative.



    How to prove part 1? and How do I prove that identity exist? Thanks in advance!



    P.S: also I'm curious to know why -1 is omitted in the set?






    binary-operations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      $A=R-{-1}$ and $a*b = a+b+ab $





      1. Show that * is a binary operation on A

      2. Show that * is associative

      3. Show that there is an identity element in A for *

      4. Show that every element in A has an inverse with respect to *


      I'm new to this lesson. All I know to prove is associativity,



      let $a,b,cin A$



      Consider:$$(a*b)*c=(a+b+ab)*(c)$$
      $$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
      $$(a*b)*c=a+b+c+ab+ac+bc+abc$$



      Now consider:
      $$a*(b*c)=a*(b+c+bc)$$
      $$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
      $$a*(b*c)=a+b+c+ab+ab+ac+abc$$
      Since we get the same result it is associative.



      How to prove part 1? and How do I prove that identity exist? Thanks in advance!



      P.S: also I'm curious to know why -1 is omitted in the set?






      binary-operations







      share|cite|improve this question









      $endgroup$





      $A=R-{-1}$ and $a*b = a+b+ab $





      1. Show that * is a binary operation on A

      2. Show that * is associative

      3. Show that there is an identity element in A for *

      4. Show that every element in A has an inverse with respect to *


      I'm new to this lesson. All I know to prove is associativity,



      let $a,b,cin A$



      Consider:$$(a*b)*c=(a+b+ab)*(c)$$
      $$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
      $$(a*b)*c=a+b+c+ab+ac+bc+abc$$



      Now consider:
      $$a*(b*c)=a*(b+c+bc)$$
      $$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
      $$a*(b*c)=a+b+c+ab+ab+ac+abc$$
      Since we get the same result it is associative.



      How to prove part 1? and How do I prove that identity exist? Thanks in advance!



      P.S: also I'm curious to know why -1 is omitted in the set?






      binary-operations




      binary-operations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 15 '18 at 5:57









      emilemil

      431410




      431410






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
          $$a*b+1=a+b+ab+1ne0.$$



          For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
          needs is that $a*e=a$, that is
          $$a+e+ae=a.$$
          Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?



          Once you have done that, do inverses. You then need to solve $a*b=e$ for
          $b$ in terms of $a$, that is
          $$a+b+ab=e$$
          where you now know $e$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
            $endgroup$
            – emil
            Dec 15 '18 at 6:11












          • $begingroup$
            For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
            $endgroup$
            – emil
            Dec 15 '18 at 6:44










          • $begingroup$
            @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
            $endgroup$
            – bof
            Dec 15 '18 at 6:55






          • 1




            $begingroup$
            @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:14








          • 1




            $begingroup$
            @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:18













          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%2f3040213%2fhow-to-prove-a-r-1-and-ab-abab-is-a-binary-operation%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
          $$a*b+1=a+b+ab+1ne0.$$



          For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
          needs is that $a*e=a$, that is
          $$a+e+ae=a.$$
          Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?



          Once you have done that, do inverses. You then need to solve $a*b=e$ for
          $b$ in terms of $a$, that is
          $$a+b+ab=e$$
          where you now know $e$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
            $endgroup$
            – emil
            Dec 15 '18 at 6:11












          • $begingroup$
            For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
            $endgroup$
            – emil
            Dec 15 '18 at 6:44










          • $begingroup$
            @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
            $endgroup$
            – bof
            Dec 15 '18 at 6:55






          • 1




            $begingroup$
            @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:14








          • 1




            $begingroup$
            @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:18


















          4












          $begingroup$

          For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
          $$a*b+1=a+b+ab+1ne0.$$



          For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
          needs is that $a*e=a$, that is
          $$a+e+ae=a.$$
          Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?



          Once you have done that, do inverses. You then need to solve $a*b=e$ for
          $b$ in terms of $a$, that is
          $$a+b+ab=e$$
          where you now know $e$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
            $endgroup$
            – emil
            Dec 15 '18 at 6:11












          • $begingroup$
            For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
            $endgroup$
            – emil
            Dec 15 '18 at 6:44










          • $begingroup$
            @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
            $endgroup$
            – bof
            Dec 15 '18 at 6:55






          • 1




            $begingroup$
            @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:14








          • 1




            $begingroup$
            @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:18
















          4












          4








          4





          $begingroup$

          For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
          $$a*b+1=a+b+ab+1ne0.$$



          For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
          needs is that $a*e=a$, that is
          $$a+e+ae=a.$$
          Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?



          Once you have done that, do inverses. You then need to solve $a*b=e$ for
          $b$ in terms of $a$, that is
          $$a+b+ab=e$$
          where you now know $e$.






          share|cite|improve this answer











          $endgroup$



          For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
          $$a*b+1=a+b+ab+1ne0.$$



          For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
          needs is that $a*e=a$, that is
          $$a+e+ae=a.$$
          Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?



          Once you have done that, do inverses. You then need to solve $a*b=e$ for
          $b$ in terms of $a$, that is
          $$a+b+ab=e$$
          where you now know $e$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 6:46

























          answered Dec 15 '18 at 6:02









          Lord Shark the UnknownLord Shark the Unknown

          102k1059132




          102k1059132












          • $begingroup$
            It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
            $endgroup$
            – emil
            Dec 15 '18 at 6:11












          • $begingroup$
            For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
            $endgroup$
            – emil
            Dec 15 '18 at 6:44










          • $begingroup$
            @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
            $endgroup$
            – bof
            Dec 15 '18 at 6:55






          • 1




            $begingroup$
            @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:14








          • 1




            $begingroup$
            @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:18




















          • $begingroup$
            It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
            $endgroup$
            – emil
            Dec 15 '18 at 6:11












          • $begingroup$
            For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
            $endgroup$
            – emil
            Dec 15 '18 at 6:44










          • $begingroup$
            @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
            $endgroup$
            – bof
            Dec 15 '18 at 6:55






          • 1




            $begingroup$
            @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:14








          • 1




            $begingroup$
            @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
            $endgroup$
            – bof
            Dec 15 '18 at 7:18


















          $begingroup$
          It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
          $endgroup$
          – emil
          Dec 15 '18 at 6:11






          $begingroup$
          It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
          $endgroup$
          – emil
          Dec 15 '18 at 6:11














          $begingroup$
          For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
          $endgroup$
          – emil
          Dec 15 '18 at 6:44




          $begingroup$
          For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
          $endgroup$
          – emil
          Dec 15 '18 at 6:44












          $begingroup$
          @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
          $endgroup$
          – bof
          Dec 15 '18 at 6:55




          $begingroup$
          @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
          $endgroup$
          – bof
          Dec 15 '18 at 6:55




          1




          1




          $begingroup$
          @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
          $endgroup$
          – bof
          Dec 15 '18 at 7:14






          $begingroup$
          @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
          $endgroup$
          – bof
          Dec 15 '18 at 7:14






          1




          1




          $begingroup$
          @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
          $endgroup$
          – bof
          Dec 15 '18 at 7:18






          $begingroup$
          @emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
          $endgroup$
          – bof
          Dec 15 '18 at 7:18




















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3040213%2fhow-to-prove-a-r-1-and-ab-abab-is-a-binary-operation%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more