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

          104k1160132




          104k1160132












          • $begingroup$
            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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          Dieringhausen

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          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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