Is an inverse element of binary operation unique? If yes then how?
$begingroup$
I am trying to prove it but not getting any clue how to start it!
$$a*b=b*a=e,$$
$$a*c=c*a=e$$
How to show $b=c$?
abstract-algebra binary-operations nonassociative-algebras
edited Feb 14 at 9:38
Shaun
9,366113684
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
$endgroup$
add a comment |
$begingroup$
I am trying to prove it but not getting any clue how to start it!
$$a*b=b*a=e,$$
$$a*c=c*a=e$$
How to show $b=c$?
abstract-algebra binary-operations nonassociative-algebras
edited Feb 14 at 9:38
Shaun
9,366113684
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
$endgroup$
add a comment |
$begingroup$
I am trying to prove it but not getting any clue how to start it!
$$a*b=b*a=e,$$
$$a*c=c*a=e$$
How to show $b=c$?
abstract-algebra binary-operations nonassociative-algebras
edited Feb 14 at 9:38
Shaun
9,366113684
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
$endgroup$
I am trying to prove it but not getting any clue how to start it!
$$a*b=b*a=e,$$
$$a*c=c*a=e$$
How to show $b=c$?
abstract-algebra binary-operations nonassociative-algebras
abstract-algebra binary-operations nonassociative-algebras
edited Feb 14 at 9:38
Shaun
9,366113684
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
edited Feb 14 at 9:38
Shaun
9,366113684
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
edited Feb 14 at 9:38
Shaun
9,366113684
edited Feb 14 at 9:38
Shaun
9,366113684
edited Feb 14 at 9:38
Shaun
9,366113684
9,366113684
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
asked Feb 14 at 9:06
Vivek DhingraVivek Dhingra
141
141
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that
$ac=ca=e$ and $ab=ba=e$
then
$c=ce=c(ab)=(ca)b=eb =b$
so
$c=b$
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
3
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
1
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
add a comment |
$begingroup$
Well, I think you are refering to the shortening rule:
$$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
$a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
$endgroup$
add a comment |
$begingroup$
It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that
$ac=ca=e$ and $ab=ba=e$
then
$c=ce=c(ab)=(ca)b=eb =b$
so
$c=b$
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
3
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
1
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
add a comment |
$begingroup$
If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that
$ac=ca=e$ and $ab=ba=e$
then
$c=ce=c(ab)=(ca)b=eb =b$
so
$c=b$
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
3
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
1
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
add a comment |
$begingroup$
If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that
$ac=ca=e$ and $ab=ba=e$
then
$c=ce=c(ab)=(ca)b=eb =b$
so
$c=b$
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that
$ac=ca=e$ and $ab=ba=e$
then
$c=ce=c(ab)=(ca)b=eb =b$
so
$c=b$
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
edited Feb 14 at 9:53
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
answered Feb 14 at 9:08
Federico FalluccaFederico Fallucca
2,270210
2,270210
3
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
1
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
add a comment |
3
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
1
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
3
3
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
All binary operations are no Associative
$endgroup$
– Vivek Dhingra
Feb 14 at 9:19
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
$begingroup$
Sorry, you’re right, I’ve corrected my answer
$endgroup$
– Federico Fallucca
Feb 14 at 9:54
1
1
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
$begingroup$
You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
$endgroup$
– Dannie
Feb 14 at 10:00
add a comment |
$begingroup$
Well, I think you are refering to the shortening rule:
$$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
$a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
$endgroup$
add a comment |
$begingroup$
Well, I think you are refering to the shortening rule:
$$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
$a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
$endgroup$
add a comment |
$begingroup$
Well, I think you are refering to the shortening rule:
$$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
$a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
$endgroup$
Well, I think you are refering to the shortening rule:
$$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
$a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
answered Feb 14 at 13:42
WuestenfuxWuestenfux
4,9821513
4,9821513
add a comment |
add a comment |
$begingroup$
It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
add a comment |
$begingroup$
It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
add a comment |
$begingroup$
It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
answered Feb 14 at 9:40
J.-E. PinJ.-E. Pin
18.5k21754
18.5k21754
add a comment |
add a comment |
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