Is G will form a group? [closed]
Given $G = {f : [0, 1] rightarrow mathbb{R} ; f text{continuous}}$
with respect to the operation defined by $(f.g)(x) = f(x)g(x)$ for all $x in [0, 1].$
My question is that Is G will form a group ?
i know that $G$ will form a ring but im not sure about group ?
abstract-algebra group-theory
asked Dec 27 '18 at 13:15
jasmine
1,536416
closed as off-topic by Henrik, Paul Frost, Cesareo, callculus, Dietrich Burde Dec 27 '18 at 19:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, callculus, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
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Given $G = {f : [0, 1] rightarrow mathbb{R} ; f text{continuous}}$
with respect to the operation defined by $(f.g)(x) = f(x)g(x)$ for all $x in [0, 1].$
My question is that Is G will form a group ?
i know that $G$ will form a ring but im not sure about group ?
abstract-algebra group-theory
asked Dec 27 '18 at 13:15
jasmine
1,536416
closed as off-topic by Henrik, Paul Frost, Cesareo, callculus, Dietrich Burde Dec 27 '18 at 19:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, callculus, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
All rings are groups, it can't be a ring if it's not even a group.
– ÍgjøgnumMeg
Dec 27 '18 at 13:20
Why do you think it's a ring?
– Henrik
Dec 27 '18 at 13:22
1
Rings (typically) have two binary operations. (I can't think of any that don't; in fact, I think a consequence of requirement that $0neq 1$ is that the two binary operations of a ring must be distinct.)
– Shaun
Dec 27 '18 at 13:24
1
@Jasmine which is also a group, under addition of real numbers. I'd suggest you go back over the definition of a ring.
– ÍgjøgnumMeg
Dec 27 '18 at 13:26
1
@jasmine: That's no argument, that it's a ring. And it seems you're mixing up your operations. A ring has two binary operations, and is a commutative group with respect to the first, so if your question should make any sense, the first one should be (pointwise) multiplication, and the second should be something that distributes over that, not (pointwise) addition.
– Henrik
Dec 27 '18 at 13:29
|
show 2 more comments
Given $G = {f : [0, 1] rightarrow mathbb{R} ; f text{continuous}}$
with respect to the operation defined by $(f.g)(x) = f(x)g(x)$ for all $x in [0, 1].$
My question is that Is G will form a group ?
i know that $G$ will form a ring but im not sure about group ?
abstract-algebra group-theory
asked Dec 27 '18 at 13:15
jasmine
1,536416
Given $G = {f : [0, 1] rightarrow mathbb{R} ; f text{continuous}}$
with respect to the operation defined by $(f.g)(x) = f(x)g(x)$ for all $x in [0, 1].$
My question is that Is G will form a group ?
i know that $G$ will form a ring but im not sure about group ?
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 27 '18 at 13:15
jasmine
1,536416
asked Dec 27 '18 at 13:15
jasmine
1,536416
asked Dec 27 '18 at 13:15
jasmine
1,536416
asked Dec 27 '18 at 13:15
jasmine
1,536416
asked Dec 27 '18 at 13:15
jasmine
1,536416
1,536416
closed as off-topic by Henrik, Paul Frost, Cesareo, callculus, Dietrich Burde Dec 27 '18 at 19:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, callculus, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, Paul Frost, Cesareo, callculus, Dietrich Burde Dec 27 '18 at 19:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, callculus, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
All rings are groups, it can't be a ring if it's not even a group.
– ÍgjøgnumMeg
Dec 27 '18 at 13:20
Why do you think it's a ring?
– Henrik
Dec 27 '18 at 13:22
1
Rings (typically) have two binary operations. (I can't think of any that don't; in fact, I think a consequence of requirement that $0neq 1$ is that the two binary operations of a ring must be distinct.)
– Shaun
Dec 27 '18 at 13:24
1
@Jasmine which is also a group, under addition of real numbers. I'd suggest you go back over the definition of a ring.
– ÍgjøgnumMeg
Dec 27 '18 at 13:26
1
@jasmine: That's no argument, that it's a ring. And it seems you're mixing up your operations. A ring has two binary operations, and is a commutative group with respect to the first, so if your question should make any sense, the first one should be (pointwise) multiplication, and the second should be something that distributes over that, not (pointwise) addition.
– Henrik
Dec 27 '18 at 13:29
|
show 2 more comments
All rings are groups, it can't be a ring if it's not even a group.
– ÍgjøgnumMeg
Dec 27 '18 at 13:20
Why do you think it's a ring?
– Henrik
Dec 27 '18 at 13:22
1
Rings (typically) have two binary operations. (I can't think of any that don't; in fact, I think a consequence of requirement that $0neq 1$ is that the two binary operations of a ring must be distinct.)
– Shaun
Dec 27 '18 at 13:24
1
@Jasmine which is also a group, under addition of real numbers. I'd suggest you go back over the definition of a ring.
– ÍgjøgnumMeg
Dec 27 '18 at 13:26
1
@jasmine: That's no argument, that it's a ring. And it seems you're mixing up your operations. A ring has two binary operations, and is a commutative group with respect to the first, so if your question should make any sense, the first one should be (pointwise) multiplication, and the second should be something that distributes over that, not (pointwise) addition.
– Henrik
Dec 27 '18 at 13:29
All rings are groups, it can't be a ring if it's not even a group.
– ÍgjøgnumMeg
Dec 27 '18 at 13:20
All rings are groups, it can't be a ring if it's not even a group.
– ÍgjøgnumMeg
Dec 27 '18 at 13:20
Why do you think it's a ring?
– Henrik
Dec 27 '18 at 13:22
Why do you think it's a ring?
– Henrik
Dec 27 '18 at 13:22
1
1
Rings (typically) have two binary operations. (I can't think of any that don't; in fact, I think a consequence of requirement that $0neq 1$ is that the two binary operations of a ring must be distinct.)
– Shaun
Dec 27 '18 at 13:24
Rings (typically) have two binary operations. (I can't think of any that don't; in fact, I think a consequence of requirement that $0neq 1$ is that the two binary operations of a ring must be distinct.)
– Shaun
Dec 27 '18 at 13:24
1
1
@Jasmine which is also a group, under addition of real numbers. I'd suggest you go back over the definition of a ring.
– ÍgjøgnumMeg
Dec 27 '18 at 13:26
@Jasmine which is also a group, under addition of real numbers. I'd suggest you go back over the definition of a ring.
– ÍgjøgnumMeg
Dec 27 '18 at 13:26
1
1
@jasmine: That's no argument, that it's a ring. And it seems you're mixing up your operations. A ring has two binary operations, and is a commutative group with respect to the first, so if your question should make any sense, the first one should be (pointwise) multiplication, and the second should be something that distributes over that, not (pointwise) addition.
– Henrik
Dec 27 '18 at 13:29
@jasmine: That's no argument, that it's a ring. And it seems you're mixing up your operations. A ring has two binary operations, and is a commutative group with respect to the first, so if your question should make any sense, the first one should be (pointwise) multiplication, and the second should be something that distributes over that, not (pointwise) addition.
– Henrik
Dec 27 '18 at 13:29
|
show 2 more comments
1 Answer
1
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votes
Not necessarily, since it the elements may not have an inverse, consider the example
$$
f(x) = (x - 1/2)^2 ~~~mbox{for}~~~ 0 leq x leq 1
$$
it does not have an inverse but $f in G$. So the element $f^{-1}$ does not exist, and $G$ is not a group
answered Dec 27 '18 at 13:19
caverac
13.8k21030
1
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Not necessarily, since it the elements may not have an inverse, consider the example
$$
f(x) = (x - 1/2)^2 ~~~mbox{for}~~~ 0 leq x leq 1
$$
it does not have an inverse but $f in G$. So the element $f^{-1}$ does not exist, and $G$ is not a group
answered Dec 27 '18 at 13:19
caverac
13.8k21030
1
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
add a comment |
Not necessarily, since it the elements may not have an inverse, consider the example
$$
f(x) = (x - 1/2)^2 ~~~mbox{for}~~~ 0 leq x leq 1
$$
it does not have an inverse but $f in G$. So the element $f^{-1}$ does not exist, and $G$ is not a group
answered Dec 27 '18 at 13:19
caverac
13.8k21030
1
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
add a comment |
Not necessarily, since it the elements may not have an inverse, consider the example
$$
f(x) = (x - 1/2)^2 ~~~mbox{for}~~~ 0 leq x leq 1
$$
it does not have an inverse but $f in G$. So the element $f^{-1}$ does not exist, and $G$ is not a group
answered Dec 27 '18 at 13:19
caverac
13.8k21030
Not necessarily, since it the elements may not have an inverse, consider the example
$$
f(x) = (x - 1/2)^2 ~~~mbox{for}~~~ 0 leq x leq 1
$$
it does not have an inverse but $f in G$. So the element $f^{-1}$ does not exist, and $G$ is not a group
answered Dec 27 '18 at 13:19
caverac
13.8k21030
answered Dec 27 '18 at 13:19
caverac
13.8k21030
answered Dec 27 '18 at 13:19
caverac
13.8k21030
answered Dec 27 '18 at 13:19
caverac
13.8k21030
13.8k21030
1
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
add a comment |
1
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
1
1
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
The operation is pointwise multiplication (not composition). So $f(x)=0$ would do. All $f$ has to do is be zero somewhere., as $frac1{f(x)}$ would have to be the inverse.
– Chris Custer
Dec 27 '18 at 14:13
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
@ChrisCuster You're absolutely right, I overcomplicated the example. Thanks
– caverac
Dec 27 '18 at 15:40
add a comment |
All rings are groups, it can't be a ring if it's not even a group.
– ÍgjøgnumMeg
Dec 27 '18 at 13:20
Why do you think it's a ring?
– Henrik
Dec 27 '18 at 13:22
1
Rings (typically) have two binary operations. (I can't think of any that don't; in fact, I think a consequence of requirement that $0neq 1$ is that the two binary operations of a ring must be distinct.)
– Shaun
Dec 27 '18 at 13:24
1
@Jasmine which is also a group, under addition of real numbers. I'd suggest you go back over the definition of a ring.
– ÍgjøgnumMeg
Dec 27 '18 at 13:26
1
@jasmine: That's no argument, that it's a ring. And it seems you're mixing up your operations. A ring has two binary operations, and is a commutative group with respect to the first, so if your question should make any sense, the first one should be (pointwise) multiplication, and the second should be something that distributes over that, not (pointwise) addition.
– Henrik
Dec 27 '18 at 13:29