Confusing Definition of field in Sheldon Axler's Linear Algebra done right. What does “1+1 is defined to...












-1












$begingroup$


In the 2015th edition of the book, linear algebra done right. Sheldon axler, defines a field to be 



A field is a set containing at least two distinct elements called 0 and 1, along

with operations of addition and multiplication satisfying all the properties

listed in 1.3. Thus R and C are fields, as is the set of rational numbers along

with the usual operations of addition and multiplication. Another example of

a field is the set {0, 1} with the usual operations of addition and multiplication

except that 1 + 1 is defined to equal 0.


In the last sentence he says, except that 1 + 1 is defined to equal 0.



I don't understand what this means, in the context of a field.






linear-algebra







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    In the 2015th edition of the book, linear algebra done right. Sheldon axler, defines a field to be 



    A field is a set containing at least two distinct elements called 0 and 1, along
    
with operations of addition and multiplication satisfying all the properties
    
listed in 1.3. Thus R and C are fields, as is the set of rational numbers along
    
with the usual operations of addition and multiplication. Another example of
    
a field is the set {0, 1} with the usual operations of addition and multiplication
    
except that 1 + 1 is defined to equal 0.


    In the last sentence he says, except that 1 + 1 is defined to equal 0.



    I don't understand what this means, in the context of a field.






    linear-algebra







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      In the 2015th edition of the book, linear algebra done right. Sheldon axler, defines a field to be 



      A field is a set containing at least two distinct elements called 0 and 1, along
      
with operations of addition and multiplication satisfying all the properties
      
listed in 1.3. Thus R and C are fields, as is the set of rational numbers along
      
with the usual operations of addition and multiplication. Another example of
      
a field is the set {0, 1} with the usual operations of addition and multiplication
      
except that 1 + 1 is defined to equal 0.


      In the last sentence he says, except that 1 + 1 is defined to equal 0.



      I don't understand what this means, in the context of a field.






      linear-algebra







      share|cite|improve this question









      $endgroup$




      In the 2015th edition of the book, linear algebra done right. Sheldon axler, defines a field to be 



      A field is a set containing at least two distinct elements called 0 and 1, along
      
with operations of addition and multiplication satisfying all the properties
      
listed in 1.3. Thus R and C are fields, as is the set of rational numbers along
      
with the usual operations of addition and multiplication. Another example of
      
a field is the set {0, 1} with the usual operations of addition and multiplication
      
except that 1 + 1 is defined to equal 0.


      In the last sentence he says, except that 1 + 1 is defined to equal 0.



      I don't understand what this means, in the context of a field.






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 11:09









      Daksh MiglaniDaksh Miglani

      1247




      1247






















          2 Answers
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          0












          $begingroup$

          The zero element $0$ and unit element $1$ of a field $F$ form itself a field $P={0,1}$ if the addition is defined such that $1+1=0$. This is the prime field with two elements usually denoted by ${Bbb Z}_2$ (as ring) or $GF(2)$ (as Galois field).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
            $endgroup$
            – Giuseppe Negro
            Jan 14 at 11:16












          • $begingroup$
            @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:18



















          1












          $begingroup$

          Sheldon is talking about the field $mathbb{Z}_2={0,1}$, for which we have$$0+0=0, 0+1=1,, 1+0=1text{, and }1+1=0$$and$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:17










          • $begingroup$
            No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:19












          • $begingroup$
            can you simply this explanation a bit further, I'm still confused.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:28






          • 1




            $begingroup$
            In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:30












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          0












          $begingroup$

          The zero element $0$ and unit element $1$ of a field $F$ form itself a field $P={0,1}$ if the addition is defined such that $1+1=0$. This is the prime field with two elements usually denoted by ${Bbb Z}_2$ (as ring) or $GF(2)$ (as Galois field).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
            $endgroup$
            – Giuseppe Negro
            Jan 14 at 11:16












          • $begingroup$
            @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:18
















          0












          $begingroup$

          The zero element $0$ and unit element $1$ of a field $F$ form itself a field $P={0,1}$ if the addition is defined such that $1+1=0$. This is the prime field with two elements usually denoted by ${Bbb Z}_2$ (as ring) or $GF(2)$ (as Galois field).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
            $endgroup$
            – Giuseppe Negro
            Jan 14 at 11:16












          • $begingroup$
            @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:18














          0












          0








          0





          $begingroup$

          The zero element $0$ and unit element $1$ of a field $F$ form itself a field $P={0,1}$ if the addition is defined such that $1+1=0$. This is the prime field with two elements usually denoted by ${Bbb Z}_2$ (as ring) or $GF(2)$ (as Galois field).






          share|cite|improve this answer









          $endgroup$



          The zero element $0$ and unit element $1$ of a field $F$ form itself a field $P={0,1}$ if the addition is defined such that $1+1=0$. This is the prime field with two elements usually denoted by ${Bbb Z}_2$ (as ring) or $GF(2)$ (as Galois field).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 11:13









          WuestenfuxWuestenfux

          5,5561513




          5,5561513












          • $begingroup$
            Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
            $endgroup$
            – Giuseppe Negro
            Jan 14 at 11:16












          • $begingroup$
            @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:18


















          • $begingroup$
            Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
            $endgroup$
            – Giuseppe Negro
            Jan 14 at 11:16












          • $begingroup$
            @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:18
















          $begingroup$
          Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
          $endgroup$
          – Giuseppe Negro
          Jan 14 at 11:16






          $begingroup$
          Correct, but $0$ and $1$ are not, a priori, elements of a field. They are just symbols. We could have written $a$ and $b$ instead, and said that, by definition, $a+a=a, a+b=b$ and $b+b=a$.
          $endgroup$
          – Giuseppe Negro
          Jan 14 at 11:16














          $begingroup$
          @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
          $endgroup$
          – Daksh Miglani
          Jan 14 at 11:18




          $begingroup$
          @GiuseppeNegro can you verify my comment on the answer by José Carlos Santos.
          $endgroup$
          – Daksh Miglani
          Jan 14 at 11:18











          1












          $begingroup$

          Sheldon is talking about the field $mathbb{Z}_2={0,1}$, for which we have$$0+0=0, 0+1=1,, 1+0=1text{, and }1+1=0$$and$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:17










          • $begingroup$
            No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:19












          • $begingroup$
            can you simply this explanation a bit further, I'm still confused.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:28






          • 1




            $begingroup$
            In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:30
















          1












          $begingroup$

          Sheldon is talking about the field $mathbb{Z}_2={0,1}$, for which we have$$0+0=0, 0+1=1,, 1+0=1text{, and }1+1=0$$and$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:17










          • $begingroup$
            No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:19












          • $begingroup$
            can you simply this explanation a bit further, I'm still confused.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:28






          • 1




            $begingroup$
            In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:30














          1












          1








          1





          $begingroup$

          Sheldon is talking about the field $mathbb{Z}_2={0,1}$, for which we have$$0+0=0, 0+1=1,, 1+0=1text{, and }1+1=0$$and$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$






          share|cite|improve this answer











          $endgroup$



          Sheldon is talking about the field $mathbb{Z}_2={0,1}$, for which we have$$0+0=0, 0+1=1,, 1+0=1text{, and }1+1=0$$and$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 11:20

























          answered Jan 14 at 11:15









          José Carlos SantosJosé Carlos Santos

          176k24136245




          176k24136245












          • $begingroup$
            okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:17










          • $begingroup$
            No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:19












          • $begingroup$
            can you simply this explanation a bit further, I'm still confused.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:28






          • 1




            $begingroup$
            In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:30


















          • $begingroup$
            okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:17










          • $begingroup$
            No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:19












          • $begingroup$
            can you simply this explanation a bit further, I'm still confused.
            $endgroup$
            – Daksh Miglani
            Jan 14 at 11:28






          • 1




            $begingroup$
            In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
            $endgroup$
            – José Carlos Santos
            Jan 14 at 11:30
















          $begingroup$
          okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
          $endgroup$
          – Daksh Miglani
          Jan 14 at 11:17




          $begingroup$
          okay so since there is no other element in the field ({0, 1}) except 0 and 1, 1+1 will go to the next possible element in the field but since there's none, it goes back to 0. right?
          $endgroup$
          – Daksh Miglani
          Jan 14 at 11:17












          $begingroup$
          No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
          $endgroup$
          – José Carlos Santos
          Jan 14 at 11:19






          $begingroup$
          No. It's because $1+1=1iff1+1=1+0implies1=0$ but, by definition, in a field you always have $1neq0$.
          $endgroup$
          – José Carlos Santos
          Jan 14 at 11:19














          $begingroup$
          can you simply this explanation a bit further, I'm still confused.
          $endgroup$
          – Daksh Miglani
          Jan 14 at 11:28




          $begingroup$
          can you simply this explanation a bit further, I'm still confused.
          $endgroup$
          – Daksh Miglani
          Jan 14 at 11:28




          1




          1




          $begingroup$
          In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
          $endgroup$
          – José Carlos Santos
          Jan 14 at 11:30




          $begingroup$
          In any field, you have$$a+b=a+cimplies b=c.$$In particular, if we had $1+1=1$, we would have $1+1=1+0$, but then it would follow from the rule above that $1=0$, which cannot happen, by the definition of field.
          $endgroup$
          – José Carlos Santos
          Jan 14 at 11:30


















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