Let $F$ be a field. (a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$. (b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$Some more questions on field theoryShow an Ideal is the principal ideal for some polynomial.In an infinite cyclic field of non zero units, characteristic $neq 2$, can an element $-u neq u$ be expressed as $u^t$ for some finite integer $t$?Field Extension Question for PolynomialsShow that $x^n-1 =1$ for all non-zero element in a fieldShow that if $gcd(q,q')neq1$, then fields are isomorphic to sub-fields of some finite fieldLet $F$ be a field and $K$ a splitting field for some nonconstant polynomial over $F$. Show that $K$ is a finite extension of $F$.Show that $Q[sqrt3]$ is a fieldShow that $a + bsqrt5 mid a,b in mathbfQ $ is a field.Show that for any element α of some extension of F, E(α) is a splitting field of f over F(α)
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Let $F$ be a field. (a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$. (b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
Some more questions on field theoryShow an Ideal is the principal ideal for some polynomial.In an infinite cyclic field of non zero units, characteristic $neq 2$, can an element $-u neq u$ be expressed as $u^t$ for some finite integer $t$?Field Extension Question for PolynomialsShow that $x^n-1 =1$ for all non-zero element in a fieldShow that if $gcd(q,q')neq1$, then fields are isomorphic to sub-fields of some finite fieldLet $F$ be a field and $K$ a splitting field for some nonconstant polynomial over $F$. Show that $K$ is a finite extension of $F$.Show that $Q[sqrt3]$ is a fieldShow that $a + bsqrt5 mid a,b in mathbfQ $ is a field.Show that for any element α of some extension of F, E(α) is a splitting field of f over F(α)
$begingroup$
Let $F$ be a field.
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
I have found proof's for $1+1=0$ but I am not sure if it is the right proof for this question.
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
anyone who can show me the answer and how to do it would be greatly appreciated.
abstract-algebra ring-theory field-theory
edited Mar 13 at 18:37
6005
37k751127
asked Mar 13 at 9:58
john smithjohn smith
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
Let $F$ be a field.
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
I have found proof's for $1+1=0$ but I am not sure if it is the right proof for this question.
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
anyone who can show me the answer and how to do it would be greatly appreciated.
abstract-algebra ring-theory field-theory
edited Mar 13 at 18:37
6005
37k751127
asked Mar 13 at 9:58
john smithjohn smith
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Can you describe your attempt?
$endgroup$
– Lucas Corrêa
Mar 13 at 10:03
$begingroup$
Should "a 6= O" be $a neq 0$ in first line?
$endgroup$
– coffeemath
Mar 13 at 10:06
$begingroup$
The problem is I am very new to these sort of proofs and way of thinking and don't have anyone to help me.
$endgroup$
– john smith
Mar 13 at 10:07
1
$begingroup$
yes it should be a isnt equal to 0
$endgroup$
– john smith
Mar 13 at 10:08
1
$begingroup$
@Lucas Corrêa. Why do you say $ 0,1 $ is not a field? I presume you mean it's not a field under normal addition and multiplication of integers. But it's not obvious (at least, not to me) that the OP was assuming those definitions, and with the definitions of addition and multiplication implied by the conditions he states, it becomes the field $ GFleft(2right) $. He cannot, of course, assume that $ F $ is that field, since it might be some other field of characteristic 2.
$endgroup$
– lonza leggiera
Mar 13 at 10:45
|
show 3 more comments
$begingroup$
Let $F$ be a field.
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
I have found proof's for $1+1=0$ but I am not sure if it is the right proof for this question.
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
anyone who can show me the answer and how to do it would be greatly appreciated.
abstract-algebra ring-theory field-theory
edited Mar 13 at 18:37
6005
37k751127
asked Mar 13 at 9:58
john smithjohn smith
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $F$ be a field.
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
I have found proof's for $1+1=0$ but I am not sure if it is the right proof for this question.
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
anyone who can show me the answer and how to do it would be greatly appreciated.
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
edited Mar 13 at 18:37
6005
37k751127
asked Mar 13 at 9:58
john smithjohn smith
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 13 at 18:37
6005
37k751127
asked Mar 13 at 9:58
john smithjohn smith
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Mar 13 at 18:37
6005
37k751127
edited Mar 13 at 18:37
6005
37k751127
edited Mar 13 at 18:37
6005
37k751127
37k751127
asked Mar 13 at 9:58
john smithjohn smith
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 9:58
john smithjohn smith
64
asked Mar 13 at 9:58
john smithjohn smith
64
64
New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
john smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
2
$begingroup$
Can you describe your attempt?
$endgroup$
– Lucas Corrêa
Mar 13 at 10:03
$begingroup$
Should "a 6= O" be $a neq 0$ in first line?
$endgroup$
– coffeemath
Mar 13 at 10:06
$begingroup$
The problem is I am very new to these sort of proofs and way of thinking and don't have anyone to help me.
$endgroup$
– john smith
Mar 13 at 10:07
1
$begingroup$
yes it should be a isnt equal to 0
$endgroup$
– john smith
Mar 13 at 10:08
1
$begingroup$
@Lucas Corrêa. Why do you say $ 0,1 $ is not a field? I presume you mean it's not a field under normal addition and multiplication of integers. But it's not obvious (at least, not to me) that the OP was assuming those definitions, and with the definitions of addition and multiplication implied by the conditions he states, it becomes the field $ GFleft(2right) $. He cannot, of course, assume that $ F $ is that field, since it might be some other field of characteristic 2.
$endgroup$
– lonza leggiera
Mar 13 at 10:45
|
show 3 more comments
2
$begingroup$
Can you describe your attempt?
$endgroup$
– Lucas Corrêa
Mar 13 at 10:03
$begingroup$
Should "a 6= O" be $a neq 0$ in first line?
$endgroup$
– coffeemath
Mar 13 at 10:06
$begingroup$
The problem is I am very new to these sort of proofs and way of thinking and don't have anyone to help me.
$endgroup$
– john smith
Mar 13 at 10:07
1
$begingroup$
yes it should be a isnt equal to 0
$endgroup$
– john smith
Mar 13 at 10:08
1
$begingroup$
@Lucas Corrêa. Why do you say $ 0,1 $ is not a field? I presume you mean it's not a field under normal addition and multiplication of integers. But it's not obvious (at least, not to me) that the OP was assuming those definitions, and with the definitions of addition and multiplication implied by the conditions he states, it becomes the field $ GFleft(2right) $. He cannot, of course, assume that $ F $ is that field, since it might be some other field of characteristic 2.
$endgroup$
– lonza leggiera
Mar 13 at 10:45
2
2
$begingroup$
Can you describe your attempt?
$endgroup$
– Lucas Corrêa
Mar 13 at 10:03
$begingroup$
Can you describe your attempt?
$endgroup$
– Lucas Corrêa
Mar 13 at 10:03
$begingroup$
Should "a 6= O" be $a neq 0$ in first line?
$endgroup$
– coffeemath
Mar 13 at 10:06
$begingroup$
Should "a 6= O" be $a neq 0$ in first line?
$endgroup$
– coffeemath
Mar 13 at 10:06
$begingroup$
The problem is I am very new to these sort of proofs and way of thinking and don't have anyone to help me.
$endgroup$
– john smith
Mar 13 at 10:07
$begingroup$
The problem is I am very new to these sort of proofs and way of thinking and don't have anyone to help me.
$endgroup$
– john smith
Mar 13 at 10:07
1
1
$begingroup$
yes it should be a isnt equal to 0
$endgroup$
– john smith
Mar 13 at 10:08
$begingroup$
yes it should be a isnt equal to 0
$endgroup$
– john smith
Mar 13 at 10:08
1
1
$begingroup$
@Lucas Corrêa. Why do you say $ 0,1 $ is not a field? I presume you mean it's not a field under normal addition and multiplication of integers. But it's not obvious (at least, not to me) that the OP was assuming those definitions, and with the definitions of addition and multiplication implied by the conditions he states, it becomes the field $ GFleft(2right) $. He cannot, of course, assume that $ F $ is that field, since it might be some other field of characteristic 2.
$endgroup$
– lonza leggiera
Mar 13 at 10:45
$begingroup$
@Lucas Corrêa. Why do you say $ 0,1 $ is not a field? I presume you mean it's not a field under normal addition and multiplication of integers. But it's not obvious (at least, not to me) that the OP was assuming those definitions, and with the definitions of addition and multiplication implied by the conditions he states, it becomes the field $ GFleft(2right) $. He cannot, of course, assume that $ F $ is that field, since it might be some other field of characteristic 2.
$endgroup$
– lonza leggiera
Mar 13 at 10:45
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Hint for (a): Given $1+1=0,$ multiply by $a$ and apply distributive law. Also need that $acdot 0=0$ for any $a$ but that should be a lemma early on [also can be proved fairly easily from field axioms].
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
$endgroup$
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
1
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
add a comment |
$begingroup$
It is easily provided by the field axioms; specifically, if $a,b,cin F$ where $(F,+,cdot)$ is a field, then $acdot(b+c)= acdot b+acdot c$. Using this we have
$$1+1=0$$
$$acdot(1+1)=acdot 0$$
$$acdot 1+acdot 1=0$$
$$a+a=0$$
Note that all these steps are reversible, because in a field all non-zero elements are units, i.e., have an inverse, and for it to be a field it must be a domain so cancellation law must hold. Since all steps are reversible, this also answers $(b).$
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
$endgroup$
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
add a comment |
$begingroup$
It's very easy to show the given relations . And we are gonna work with some basic properties of a field. Let( F ,+, .) Is our field .
Now it's given, 1+1=0.
Now if , a belongs to F then by using the property,
a.(b+c)=a.b+a.c , for all , a,b,c belongs to F , we can write,
a.(1+1)=a.1+a.1 = a+a , as '1' is multiplicative identity.
So, a+a=0 ,as a.0=0 ....(p)
[ a.0= 0 because,
0+0=0 , as 0 is additive identity.
Now if a belongs to F then,
a.0 +a.0=a.0
Now as a belongs to F , -a belongs to F , as it's the additive inverse.
So , -a.0+a.0+a.0=-a.0+a.0
Or, a.0=0 ]
Now for some a≠0 , if a belongs to F, the (1/a) belongs to F as (1/a) is multiplicative inverse of a .
Now , as a+a=0 , by using a.(b+c)=a.b+a.c , we can write (1/a).(a+a)= a.1/a+a.1/a .
As, a.1/a=1 , then
1+1=0.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
Let's read through the requirements of the problem carefully. All that is given to us is: "Let $F$ be a field." So we know $F$ is a field -- we don't know how many elements it has, but all fields at least have a zero element (which is written $0$) and a one element (which is written $1$).
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
Now for this part, it does not assume that $F = 0,1$ like you said; but it's just stating a fact about $0$ and $1$, which are elements of the field. So it is stating the fact that $1 + 1 = 0$. Then we have to prove that $a + a = 0$, for any $a in F$. To do this, try multiplying by $a$ and using the field axioms.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
For this part, try "dividing" by $a$ instead of multiplying. What axiom of the fields would let us divide by $a$? Remember that "dividing" really means multiplying by the multiplicative inverse.
answered Mar 13 at 18:40
60056005
37k751127
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint for (a): Given $1+1=0,$ multiply by $a$ and apply distributive law. Also need that $acdot 0=0$ for any $a$ but that should be a lemma early on [also can be proved fairly easily from field axioms].
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
$endgroup$
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
1
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
add a comment |
$begingroup$
Hint for (a): Given $1+1=0,$ multiply by $a$ and apply distributive law. Also need that $acdot 0=0$ for any $a$ but that should be a lemma early on [also can be proved fairly easily from field axioms].
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
$endgroup$
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
1
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
add a comment |
$begingroup$
Hint for (a): Given $1+1=0,$ multiply by $a$ and apply distributive law. Also need that $acdot 0=0$ for any $a$ but that should be a lemma early on [also can be proved fairly easily from field axioms].
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
$endgroup$
Hint for (a): Given $1+1=0,$ multiply by $a$ and apply distributive law. Also need that $acdot 0=0$ for any $a$ but that should be a lemma early on [also can be proved fairly easily from field axioms].
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
answered Mar 13 at 10:11
coffeemathcoffeemath
2,9071415
2,9071415
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
1
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
add a comment |
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
1
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
$begingroup$
so I just say that a(1 + 1) = a . 0 distributivity
$endgroup$
– john smith
Mar 13 at 10:16
1
1
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
then I jsut use the field axiom for anything times 1 is itself?
$endgroup$
– john smith
Mar 13 at 10:17
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith That seems like a very reasonable way to do it.
$endgroup$
– Arthur
Mar 13 at 10:29
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
$begingroup$
@johnsmith Yes, use also $acdot 1=a.$
$endgroup$
– coffeemath
Mar 13 at 10:47
add a comment |
$begingroup$
It is easily provided by the field axioms; specifically, if $a,b,cin F$ where $(F,+,cdot)$ is a field, then $acdot(b+c)= acdot b+acdot c$. Using this we have
$$1+1=0$$
$$acdot(1+1)=acdot 0$$
$$acdot 1+acdot 1=0$$
$$a+a=0$$
Note that all these steps are reversible, because in a field all non-zero elements are units, i.e., have an inverse, and for it to be a field it must be a domain so cancellation law must hold. Since all steps are reversible, this also answers $(b).$
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
$endgroup$
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
add a comment |
$begingroup$
It is easily provided by the field axioms; specifically, if $a,b,cin F$ where $(F,+,cdot)$ is a field, then $acdot(b+c)= acdot b+acdot c$. Using this we have
$$1+1=0$$
$$acdot(1+1)=acdot 0$$
$$acdot 1+acdot 1=0$$
$$a+a=0$$
Note that all these steps are reversible, because in a field all non-zero elements are units, i.e., have an inverse, and for it to be a field it must be a domain so cancellation law must hold. Since all steps are reversible, this also answers $(b).$
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
$endgroup$
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
add a comment |
$begingroup$
It is easily provided by the field axioms; specifically, if $a,b,cin F$ where $(F,+,cdot)$ is a field, then $acdot(b+c)= acdot b+acdot c$. Using this we have
$$1+1=0$$
$$acdot(1+1)=acdot 0$$
$$acdot 1+acdot 1=0$$
$$a+a=0$$
Note that all these steps are reversible, because in a field all non-zero elements are units, i.e., have an inverse, and for it to be a field it must be a domain so cancellation law must hold. Since all steps are reversible, this also answers $(b).$
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
$endgroup$
It is easily provided by the field axioms; specifically, if $a,b,cin F$ where $(F,+,cdot)$ is a field, then $acdot(b+c)= acdot b+acdot c$. Using this we have
$$1+1=0$$
$$acdot(1+1)=acdot 0$$
$$acdot 1+acdot 1=0$$
$$a+a=0$$
Note that all these steps are reversible, because in a field all non-zero elements are units, i.e., have an inverse, and for it to be a field it must be a domain so cancellation law must hold. Since all steps are reversible, this also answers $(b).$
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
edited Mar 13 at 18:19
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
answered Mar 13 at 10:30
Bruno AndradesBruno Andrades
15911
15911
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
add a comment |
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
I'd say $acdot 0 = 0$ is more elementary then $a(-a) = - a^2$. How would you prove $(-1)cdot a = - a$ without $acdot 0 = 0$?
$endgroup$
– Ennar
Mar 13 at 18:04
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
$begingroup$
@Ennar You're right, I'll just assume $acdot 0=0$
$endgroup$
– Bruno Andrades
Mar 13 at 18:20
add a comment |
$begingroup$
It's very easy to show the given relations . And we are gonna work with some basic properties of a field. Let( F ,+, .) Is our field .
Now it's given, 1+1=0.
Now if , a belongs to F then by using the property,
a.(b+c)=a.b+a.c , for all , a,b,c belongs to F , we can write,
a.(1+1)=a.1+a.1 = a+a , as '1' is multiplicative identity.
So, a+a=0 ,as a.0=0 ....(p)
[ a.0= 0 because,
0+0=0 , as 0 is additive identity.
Now if a belongs to F then,
a.0 +a.0=a.0
Now as a belongs to F , -a belongs to F , as it's the additive inverse.
So , -a.0+a.0+a.0=-a.0+a.0
Or, a.0=0 ]
Now for some a≠0 , if a belongs to F, the (1/a) belongs to F as (1/a) is multiplicative inverse of a .
Now , as a+a=0 , by using a.(b+c)=a.b+a.c , we can write (1/a).(a+a)= a.1/a+a.1/a .
As, a.1/a=1 , then
1+1=0.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
It's very easy to show the given relations . And we are gonna work with some basic properties of a field. Let( F ,+, .) Is our field .
Now it's given, 1+1=0.
Now if , a belongs to F then by using the property,
a.(b+c)=a.b+a.c , for all , a,b,c belongs to F , we can write,
a.(1+1)=a.1+a.1 = a+a , as '1' is multiplicative identity.
So, a+a=0 ,as a.0=0 ....(p)
[ a.0= 0 because,
0+0=0 , as 0 is additive identity.
Now if a belongs to F then,
a.0 +a.0=a.0
Now as a belongs to F , -a belongs to F , as it's the additive inverse.
So , -a.0+a.0+a.0=-a.0+a.0
Or, a.0=0 ]
Now for some a≠0 , if a belongs to F, the (1/a) belongs to F as (1/a) is multiplicative inverse of a .
Now , as a+a=0 , by using a.(b+c)=a.b+a.c , we can write (1/a).(a+a)= a.1/a+a.1/a .
As, a.1/a=1 , then
1+1=0.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
It's very easy to show the given relations . And we are gonna work with some basic properties of a field. Let( F ,+, .) Is our field .
Now it's given, 1+1=0.
Now if , a belongs to F then by using the property,
a.(b+c)=a.b+a.c , for all , a,b,c belongs to F , we can write,
a.(1+1)=a.1+a.1 = a+a , as '1' is multiplicative identity.
So, a+a=0 ,as a.0=0 ....(p)
[ a.0= 0 because,
0+0=0 , as 0 is additive identity.
Now if a belongs to F then,
a.0 +a.0=a.0
Now as a belongs to F , -a belongs to F , as it's the additive inverse.
So , -a.0+a.0+a.0=-a.0+a.0
Or, a.0=0 ]
Now for some a≠0 , if a belongs to F, the (1/a) belongs to F as (1/a) is multiplicative inverse of a .
Now , as a+a=0 , by using a.(b+c)=a.b+a.c , we can write (1/a).(a+a)= a.1/a+a.1/a .
As, a.1/a=1 , then
1+1=0.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It's very easy to show the given relations . And we are gonna work with some basic properties of a field. Let( F ,+, .) Is our field .
Now it's given, 1+1=0.
Now if , a belongs to F then by using the property,
a.(b+c)=a.b+a.c , for all , a,b,c belongs to F , we can write,
a.(1+1)=a.1+a.1 = a+a , as '1' is multiplicative identity.
So, a+a=0 ,as a.0=0 ....(p)
[ a.0= 0 because,
0+0=0 , as 0 is additive identity.
Now if a belongs to F then,
a.0 +a.0=a.0
Now as a belongs to F , -a belongs to F , as it's the additive inverse.
So , -a.0+a.0+a.0=-a.0+a.0
Or, a.0=0 ]
Now for some a≠0 , if a belongs to F, the (1/a) belongs to F as (1/a) is multiplicative inverse of a .
Now , as a+a=0 , by using a.(b+c)=a.b+a.c , we can write (1/a).(a+a)= a.1/a+a.1/a .
As, a.1/a=1 , then
1+1=0.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
answered Mar 13 at 17:50
Gogole pi MukherjeeGogole pi Mukherjee
12
12
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gogole pi Mukherjee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
Let's read through the requirements of the problem carefully. All that is given to us is: "Let $F$ be a field." So we know $F$ is a field -- we don't know how many elements it has, but all fields at least have a zero element (which is written $0$) and a one element (which is written $1$).
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
Now for this part, it does not assume that $F = 0,1$ like you said; but it's just stating a fact about $0$ and $1$, which are elements of the field. So it is stating the fact that $1 + 1 = 0$. Then we have to prove that $a + a = 0$, for any $a in F$. To do this, try multiplying by $a$ and using the field axioms.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
For this part, try "dividing" by $a$ instead of multiplying. What axiom of the fields would let us divide by $a$? Remember that "dividing" really means multiplying by the multiplicative inverse.
answered Mar 13 at 18:40
60056005
37k751127
$endgroup$
add a comment |
$begingroup$
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
Let's read through the requirements of the problem carefully. All that is given to us is: "Let $F$ be a field." So we know $F$ is a field -- we don't know how many elements it has, but all fields at least have a zero element (which is written $0$) and a one element (which is written $1$).
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
Now for this part, it does not assume that $F = 0,1$ like you said; but it's just stating a fact about $0$ and $1$, which are elements of the field. So it is stating the fact that $1 + 1 = 0$. Then we have to prove that $a + a = 0$, for any $a in F$. To do this, try multiplying by $a$ and using the field axioms.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
For this part, try "dividing" by $a$ instead of multiplying. What axiom of the fields would let us divide by $a$? Remember that "dividing" really means multiplying by the multiplicative inverse.
answered Mar 13 at 18:40
60056005
37k751127
$endgroup$
add a comment |
$begingroup$
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
Let's read through the requirements of the problem carefully. All that is given to us is: "Let $F$ be a field." So we know $F$ is a field -- we don't know how many elements it has, but all fields at least have a zero element (which is written $0$) and a one element (which is written $1$).
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
Now for this part, it does not assume that $F = 0,1$ like you said; but it's just stating a fact about $0$ and $1$, which are elements of the field. So it is stating the fact that $1 + 1 = 0$. Then we have to prove that $a + a = 0$, for any $a in F$. To do this, try multiplying by $a$ and using the field axioms.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
For this part, try "dividing" by $a$ instead of multiplying. What axiom of the fields would let us divide by $a$? Remember that "dividing" really means multiplying by the multiplicative inverse.
answered Mar 13 at 18:40
60056005
37k751127
$endgroup$
I am a bit unsure as to that the question is asking, do I assume $F$ is a field $0,1$?
Let's read through the requirements of the problem carefully. All that is given to us is: "Let $F$ be a field." So we know $F$ is a field -- we don't know how many elements it has, but all fields at least have a zero element (which is written $0$) and a one element (which is written $1$).
(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a in F$.
Now for this part, it does not assume that $F = 0,1$ like you said; but it's just stating a fact about $0$ and $1$, which are elements of the field. So it is stating the fact that $1 + 1 = 0$. Then we have to prove that $a + a = 0$, for any $a in F$. To do this, try multiplying by $a$ and using the field axioms.
(b) If $a + a = 0$ for some $a neq 0$, show that $1 + 1 = 0$
For this part, try "dividing" by $a$ instead of multiplying. What axiom of the fields would let us divide by $a$? Remember that "dividing" really means multiplying by the multiplicative inverse.
answered Mar 13 at 18:40
60056005
37k751127
answered Mar 13 at 18:40
60056005
37k751127
answered Mar 13 at 18:40
60056005
37k751127
answered Mar 13 at 18:40
60056005
37k751127
37k751127
add a comment |
add a comment |
john smith is a new contributor. Be nice, and check out our Code of Conduct.
john smith is a new contributor. Be nice, and check out our Code of Conduct.
john smith is a new contributor. Be nice, and check out our Code of Conduct.
john smith is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
Can you describe your attempt?
$endgroup$
– Lucas Corrêa
Mar 13 at 10:03
$begingroup$
Should "a 6= O" be $a neq 0$ in first line?
$endgroup$
– coffeemath
Mar 13 at 10:06
$begingroup$
The problem is I am very new to these sort of proofs and way of thinking and don't have anyone to help me.
$endgroup$
– john smith
Mar 13 at 10:07
1
$begingroup$
yes it should be a isnt equal to 0
$endgroup$
– john smith
Mar 13 at 10:08
1
$begingroup$
@Lucas Corrêa. Why do you say $ 0,1 $ is not a field? I presume you mean it's not a field under normal addition and multiplication of integers. But it's not obvious (at least, not to me) that the OP was assuming those definitions, and with the definitions of addition and multiplication implied by the conditions he states, it becomes the field $ GFleft(2right) $. He cannot, of course, assume that $ F $ is that field, since it might be some other field of characteristic 2.
$endgroup$
– lonza leggiera
Mar 13 at 10:45