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
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
$endgroup$
add a comment |
-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
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
$endgroup$
add a comment |
-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
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
$endgroup$
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
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
asked Jan 14 at 11:09
Daksh MiglaniDaksh Miglani
1247
1247
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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).
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
$endgroup$
$begingroup$
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
add a comment |
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$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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).
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
$endgroup$
$begingroup$
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
add a comment |
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).
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
$endgroup$
$begingroup$
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
add a comment |
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).
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
$endgroup$
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).
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
answered Jan 14 at 11:13
WuestenfuxWuestenfux
5,5561513
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
add a comment |
$begingroup$
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
add a comment |
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$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
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
add a comment |
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$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
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
add a comment |
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$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
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$$0times0=0, 0times1=0, 1times0=0text{, and }1times1=1.$$
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
edited Jan 14 at 11:20
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
answered Jan 14 at 11:15
José Carlos SantosJosé Carlos Santos
176k24136245
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
add a comment |
$begingroup$
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
add a comment |
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