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













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          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




















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          .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 ...
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