If $a^2=a$ find the inverse of $1+a$ [closed]
2
$begingroup$
If $a^2=a$ find the inverse of $1+a$.
Note that $ain R$ where $R$ is a ring and inverse of $1+a$ exists in $R$
I tried $(1-a)(1+a)=1-a^2=1-a$
But I need $1$ in the RHS.How to get it?
abstract-algebra ring-theory idempotents
edited Dec 13 '18 at 18:33
Shaun
9,339113684
$endgroup$
closed as unclear what you're asking by quid♦ Dec 15 '18 at 2:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 15 more comments
2
$begingroup$
If $a^2=a$ find the inverse of $1+a$.
Note that $ain R$ where $R$ is a ring and inverse of $1+a$ exists in $R$
I tried $(1-a)(1+a)=1-a^2=1-a$
But I need $1$ in the RHS.How to get it?
abstract-algebra ring-theory idempotents
edited Dec 13 '18 at 18:33
Shaun
9,339113684
$endgroup$
closed as unclear what you're asking by quid♦ Dec 15 '18 at 2:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
It need not have an inverse. For example, $a=1$ in $Bbb{Z}_2$. If $a=0$, then the inverse of $1-a$ is $1$ (assuming your ring has unity and 1 neq 0$).
$endgroup$
– Anurag A
Dec 13 '18 at 18:04
2
$begingroup$
@Join_PhD then please edit the OP and add all assumptions needed and why $(1+a)^{-1}$ is not a good answer.
$endgroup$
– dan_fulea
Dec 13 '18 at 18:10
2
$begingroup$
When you have a critical piece of information the changes the entire question you should put it in the question statement.
$endgroup$
– rschwieb
Dec 13 '18 at 18:10
1
$begingroup$
@freakish;how $a=ab+b$ implies that??
$endgroup$
– user596656
Dec 13 '18 at 18:13
3
$begingroup$
@Join_PhD Can you please clarify what the actual wording of the question is? Does it assume $a+1$ is a unit for all idempotents $a$? You really need to be more careful when writing, or else it's going to cause a lot of problems.
$endgroup$
– rschwieb
Dec 13 '18 at 20:17
|
show 15 more comments
2
2
2
1
$begingroup$
If $a^2=a$ find the inverse of $1+a$.
Note that $ain R$ where $R$ is a ring and inverse of $1+a$ exists in $R$
I tried $(1-a)(1+a)=1-a^2=1-a$
But I need $1$ in the RHS.How to get it?
abstract-algebra ring-theory idempotents
edited Dec 13 '18 at 18:33
Shaun
9,339113684
$endgroup$
If $a^2=a$ find the inverse of $1+a$.
Note that $ain R$ where $R$ is a ring and inverse of $1+a$ exists in $R$
I tried $(1-a)(1+a)=1-a^2=1-a$
But I need $1$ in the RHS.How to get it?
abstract-algebra ring-theory idempotents
abstract-algebra ring-theory idempotents
edited Dec 13 '18 at 18:33
Shaun
9,339113684
edited Dec 13 '18 at 18:33
Shaun
9,339113684
edited Dec 13 '18 at 18:33
Shaun
9,339113684
edited Dec 13 '18 at 18:33
Shaun
9,339113684
edited Dec 13 '18 at 18:33
Shaun
9,339113684
9,339113684
asked Dec 13 '18 at 18:02
user596656
closed as unclear what you're asking by quid♦ Dec 15 '18 at 2:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by quid♦ Dec 15 '18 at 2:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
It need not have an inverse. For example, $a=1$ in $Bbb{Z}_2$. If $a=0$, then the inverse of $1-a$ is $1$ (assuming your ring has unity and 1 neq 0$).
$endgroup$
– Anurag A
Dec 13 '18 at 18:04
2
$begingroup$
@Join_PhD then please edit the OP and add all assumptions needed and why $(1+a)^{-1}$ is not a good answer.
$endgroup$
– dan_fulea
Dec 13 '18 at 18:10
2
$begingroup$
When you have a critical piece of information the changes the entire question you should put it in the question statement.
$endgroup$
– rschwieb
Dec 13 '18 at 18:10
1
$begingroup$
@freakish;how $a=ab+b$ implies that??
$endgroup$
– user596656
Dec 13 '18 at 18:13
3
$begingroup$
@Join_PhD Can you please clarify what the actual wording of the question is? Does it assume $a+1$ is a unit for all idempotents $a$? You really need to be more careful when writing, or else it's going to cause a lot of problems.
$endgroup$
– rschwieb
Dec 13 '18 at 20:17
|
show 15 more comments
5
$begingroup$
It need not have an inverse. For example, $a=1$ in $Bbb{Z}_2$. If $a=0$, then the inverse of $1-a$ is $1$ (assuming your ring has unity and 1 neq 0$).
$endgroup$
– Anurag A
Dec 13 '18 at 18:04
2
$begingroup$
@Join_PhD then please edit the OP and add all assumptions needed and why $(1+a)^{-1}$ is not a good answer.
$endgroup$
– dan_fulea
Dec 13 '18 at 18:10
2
$begingroup$
When you have a critical piece of information the changes the entire question you should put it in the question statement.
$endgroup$
– rschwieb
Dec 13 '18 at 18:10
1
$begingroup$
@freakish;how $a=ab+b$ implies that??
$endgroup$
– user596656
Dec 13 '18 at 18:13
3
$begingroup$
@Join_PhD Can you please clarify what the actual wording of the question is? Does it assume $a+1$ is a unit for all idempotents $a$? You really need to be more careful when writing, or else it's going to cause a lot of problems.
$endgroup$
– rschwieb
Dec 13 '18 at 20:17
5
5
$begingroup$
It need not have an inverse. For example, $a=1$ in $Bbb{Z}_2$. If $a=0$, then the inverse of $1-a$ is $1$ (assuming your ring has unity and 1 neq 0$).
$endgroup$
– Anurag A
Dec 13 '18 at 18:04
$begingroup$
It need not have an inverse. For example, $a=1$ in $Bbb{Z}_2$. If $a=0$, then the inverse of $1-a$ is $1$ (assuming your ring has unity and 1 neq 0$).
$endgroup$
– Anurag A
Dec 13 '18 at 18:04
2
2
$begingroup$
@Join_PhD then please edit the OP and add all assumptions needed and why $(1+a)^{-1}$ is not a good answer.
$endgroup$
– dan_fulea
Dec 13 '18 at 18:10
$begingroup$
@Join_PhD then please edit the OP and add all assumptions needed and why $(1+a)^{-1}$ is not a good answer.
$endgroup$
– dan_fulea
Dec 13 '18 at 18:10
2
2
$begingroup$
When you have a critical piece of information the changes the entire question you should put it in the question statement.
$endgroup$
– rschwieb
Dec 13 '18 at 18:10
$begingroup$
When you have a critical piece of information the changes the entire question you should put it in the question statement.
$endgroup$
– rschwieb
Dec 13 '18 at 18:10
1
1
$begingroup$
@freakish;how $a=ab+b$ implies that??
$endgroup$
– user596656
Dec 13 '18 at 18:13
$begingroup$
@freakish;how $a=ab+b$ implies that??
$endgroup$
– user596656
Dec 13 '18 at 18:13
3
3
$begingroup$
@Join_PhD Can you please clarify what the actual wording of the question is? Does it assume $a+1$ is a unit for all idempotents $a$? You really need to be more careful when writing, or else it's going to cause a lot of problems.
$endgroup$
– rschwieb
Dec 13 '18 at 20:17
$begingroup$
@Join_PhD Can you please clarify what the actual wording of the question is? Does it assume $a+1$ is a unit for all idempotents $a$? You really need to be more careful when writing, or else it's going to cause a lot of problems.
$endgroup$
– rschwieb
Dec 13 '18 at 20:17
|
show 15 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
Hint: I assume that the $2$ is invertible in the ring, $(1+a)(2-a)=2-a+2a-a^2=2$.
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
1
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
3
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
1
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
2
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
1
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Hint: I assume that the $2$ is invertible in the ring, $(1+a)(2-a)=2-a+2a-a^2=2$.
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
1
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
3
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
1
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
2
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
1
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
|
show 6 more comments
0
$begingroup$
Hint: I assume that the $2$ is invertible in the ring, $(1+a)(2-a)=2-a+2a-a^2=2$.
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
1
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
3
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
1
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
2
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
1
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
|
show 6 more comments
0
0
0
$begingroup$
Hint: I assume that the $2$ is invertible in the ring, $(1+a)(2-a)=2-a+2a-a^2=2$.
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
Hint: I assume that the $2$ is invertible in the ring, $(1+a)(2-a)=2-a+2a-a^2=2$.
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
answered Dec 13 '18 at 18:26
Tsemo AristideTsemo Aristide
58.9k11445
58.9k11445
1
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
3
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
1
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
2
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
1
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
|
show 6 more comments
1
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
3
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
1
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
2
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
1
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
1
1
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
$begingroup$
Please be more precise. What do you believe is the inverse?
$endgroup$
– Bill Dubuque
Dec 13 '18 at 19:11
3
3
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
$begingroup$
@BillDubuque Given the assumptions that $1+a$ and $2$ are units, this seems pretty clear cut already... "Precision" doesn't seem to be needed here for this as a hint. At least, spelling out the obvious would destroy it as a hint, and as far as hints go it's not that bad.
$endgroup$
– rschwieb
Dec 13 '18 at 19:26
1
1
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
$begingroup$
I may be that the question source actually assumes that the ring has the property that whenever $a$ is an idempotent, then $1+a$ is invertible. In this case $2$ is invertible.
$endgroup$
– Improve
Dec 13 '18 at 19:49
2
2
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
$begingroup$
@Improve A (hint) answer should not require guessing some stronger hypothesis in order to make sense of the answer. Too many "hints" on this site lead nowhere. And certainly if one is going to assume some stronger hypothesis then that should be made explicit in the answer.
$endgroup$
– Bill Dubuque
Dec 13 '18 at 20:52
1
1
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
$begingroup$
@BillDubuque, as a teacher in particular an elementary teacher, I allow my students to make mistakes even when they are asking questions, I believe they are learning better.
$endgroup$
– Tsemo Aristide
Dec 13 '18 at 23:13
|
show 6 more comments
5
$begingroup$
It need not have an inverse. For example, $a=1$ in $Bbb{Z}_2$. If $a=0$, then the inverse of $1-a$ is $1$ (assuming your ring has unity and 1 neq 0$).
$endgroup$
– Anurag A
Dec 13 '18 at 18:04
2
$begingroup$
@Join_PhD then please edit the OP and add all assumptions needed and why $(1+a)^{-1}$ is not a good answer.
$endgroup$
– dan_fulea
Dec 13 '18 at 18:10
2
$begingroup$
When you have a critical piece of information the changes the entire question you should put it in the question statement.
$endgroup$
– rschwieb
Dec 13 '18 at 18:10
1
$begingroup$
@freakish;how $a=ab+b$ implies that??
$endgroup$
– user596656
Dec 13 '18 at 18:13
3
$begingroup$
@Join_PhD Can you please clarify what the actual wording of the question is? Does it assume $a+1$ is a unit for all idempotents $a$? You really need to be more careful when writing, or else it's going to cause a lot of problems.
$endgroup$
– rschwieb
Dec 13 '18 at 20:17