Abstract Algebra: Prove if it is an abelian group. Need Clarification
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I need clarification of what it means when they say Let $S = mathbbR setminus -1$. Does this mean all the real numbers except $-1$.
Problem: Let $S = mathbbR setminus -1$ and define a binary operation on $S$ by $a*b = a + b + ab$. Prove that $(S,*)$ is an abelian group.
What I know: I know I need to show closure, identity, inverse, assoc, and comm.
For closure I want to prove using contrapositive. So I know I have to start with assuming that $a*b$ is not in S and show that $a$ is not in S or $b$ is not in $S$. But I'm really confused with $S = mathbbR setminus -1$. Because I know I can say $a*b = a+b+ab$ if it was in S but I'm assuming it is not.
abstract-algebra proof-writing proof-explanation abelian-groups
edited Sep 2 at 5:09
Guido A.
4,728727
asked Sep 2 at 5:08
Rose
1348
add a comment |Â
up vote
1
down vote
favorite
I need clarification of what it means when they say Let $S = mathbbR setminus -1$. Does this mean all the real numbers except $-1$.
Problem: Let $S = mathbbR setminus -1$ and define a binary operation on $S$ by $a*b = a + b + ab$. Prove that $(S,*)$ is an abelian group.
What I know: I know I need to show closure, identity, inverse, assoc, and comm.
For closure I want to prove using contrapositive. So I know I have to start with assuming that $a*b$ is not in S and show that $a$ is not in S or $b$ is not in $S$. But I'm really confused with $S = mathbbR setminus -1$. Because I know I can say $a*b = a+b+ab$ if it was in S but I'm assuming it is not.
abstract-algebra proof-writing proof-explanation abelian-groups
edited Sep 2 at 5:09
Guido A.
4,728727
asked Sep 2 at 5:08
Rose
1348
You know that the reals are closed by sum and multiplication, and so $a * b$ will always lie on $mathbbR$. Thus, $a * b not in S$ can only occur when $a * b = -1$.
– Guido A.
Sep 2 at 5:11
2
Note that $ab+a+b=(a+1)(b+1)-1$, and consider some of the identities related to $0$ in $Bbb R$...
– abiessu
Sep 2 at 5:12
Your question is already has an answer here
– Chinnapparaj R
Sep 2 at 5:21
Have you tried calculating a few things with this operation? Like what do you get for $0*0$, $1*1$, or $-2*-2$? What about $-2*0$ or $10*1$? Just so a few random ones. Then you can try some general stuff, like $0*x$. Familiarise yourself with the operation. Only once you've done that should you try to prove general properties like associativity and commutativity. It will probably be a lot easier by then.
– Arthur
Sep 2 at 5:21
Here's a hint: if $a, b in mathbbR setminus -1$, then $(a + 1)(b + 1) neq 0$.
– Theo Bendit
Sep 2 at 5:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need clarification of what it means when they say Let $S = mathbbR setminus -1$. Does this mean all the real numbers except $-1$.
Problem: Let $S = mathbbR setminus -1$ and define a binary operation on $S$ by $a*b = a + b + ab$. Prove that $(S,*)$ is an abelian group.
What I know: I know I need to show closure, identity, inverse, assoc, and comm.
For closure I want to prove using contrapositive. So I know I have to start with assuming that $a*b$ is not in S and show that $a$ is not in S or $b$ is not in $S$. But I'm really confused with $S = mathbbR setminus -1$. Because I know I can say $a*b = a+b+ab$ if it was in S but I'm assuming it is not.
abstract-algebra proof-writing proof-explanation abelian-groups
edited Sep 2 at 5:09
Guido A.
4,728727
asked Sep 2 at 5:08
Rose
1348
I need clarification of what it means when they say Let $S = mathbbR setminus -1$. Does this mean all the real numbers except $-1$.
Problem: Let $S = mathbbR setminus -1$ and define a binary operation on $S$ by $a*b = a + b + ab$. Prove that $(S,*)$ is an abelian group.
What I know: I know I need to show closure, identity, inverse, assoc, and comm.
For closure I want to prove using contrapositive. So I know I have to start with assuming that $a*b$ is not in S and show that $a$ is not in S or $b$ is not in $S$. But I'm really confused with $S = mathbbR setminus -1$. Because I know I can say $a*b = a+b+ab$ if it was in S but I'm assuming it is not.
abstract-algebra proof-writing proof-explanation abelian-groups
abstract-algebra proof-writing proof-explanation abelian-groups
edited Sep 2 at 5:09
Guido A.
4,728727
asked Sep 2 at 5:08
Rose
1348
edited Sep 2 at 5:09
Guido A.
4,728727
asked Sep 2 at 5:08
Rose
1348
edited Sep 2 at 5:09
Guido A.
4,728727
edited Sep 2 at 5:09
Guido A.
4,728727
edited Sep 2 at 5:09
Guido A.
4,728727
4,728727
asked Sep 2 at 5:08
Rose
1348
asked Sep 2 at 5:08
Rose
1348
asked Sep 2 at 5:08
Rose
1348
1348
You know that the reals are closed by sum and multiplication, and so $a * b$ will always lie on $mathbbR$. Thus, $a * b not in S$ can only occur when $a * b = -1$.
– Guido A.
Sep 2 at 5:11
2
Note that $ab+a+b=(a+1)(b+1)-1$, and consider some of the identities related to $0$ in $Bbb R$...
– abiessu
Sep 2 at 5:12
Your question is already has an answer here
– Chinnapparaj R
Sep 2 at 5:21
Have you tried calculating a few things with this operation? Like what do you get for $0*0$, $1*1$, or $-2*-2$? What about $-2*0$ or $10*1$? Just so a few random ones. Then you can try some general stuff, like $0*x$. Familiarise yourself with the operation. Only once you've done that should you try to prove general properties like associativity and commutativity. It will probably be a lot easier by then.
– Arthur
Sep 2 at 5:21
Here's a hint: if $a, b in mathbbR setminus -1$, then $(a + 1)(b + 1) neq 0$.
– Theo Bendit
Sep 2 at 5:32
add a comment |Â
You know that the reals are closed by sum and multiplication, and so $a * b$ will always lie on $mathbbR$. Thus, $a * b not in S$ can only occur when $a * b = -1$.
– Guido A.
Sep 2 at 5:11
2
Note that $ab+a+b=(a+1)(b+1)-1$, and consider some of the identities related to $0$ in $Bbb R$...
– abiessu
Sep 2 at 5:12
Your question is already has an answer here
– Chinnapparaj R
Sep 2 at 5:21
Have you tried calculating a few things with this operation? Like what do you get for $0*0$, $1*1$, or $-2*-2$? What about $-2*0$ or $10*1$? Just so a few random ones. Then you can try some general stuff, like $0*x$. Familiarise yourself with the operation. Only once you've done that should you try to prove general properties like associativity and commutativity. It will probably be a lot easier by then.
– Arthur
Sep 2 at 5:21
Here's a hint: if $a, b in mathbbR setminus -1$, then $(a + 1)(b + 1) neq 0$.
– Theo Bendit
Sep 2 at 5:32
You know that the reals are closed by sum and multiplication, and so $a * b$ will always lie on $mathbbR$. Thus, $a * b not in S$ can only occur when $a * b = -1$.
– Guido A.
Sep 2 at 5:11
You know that the reals are closed by sum and multiplication, and so $a * b$ will always lie on $mathbbR$. Thus, $a * b not in S$ can only occur when $a * b = -1$.
– Guido A.
Sep 2 at 5:11
2
2
Note that $ab+a+b=(a+1)(b+1)-1$, and consider some of the identities related to $0$ in $Bbb R$...
– abiessu
Sep 2 at 5:12
Note that $ab+a+b=(a+1)(b+1)-1$, and consider some of the identities related to $0$ in $Bbb R$...
– abiessu
Sep 2 at 5:12
Your question is already has an answer here
– Chinnapparaj R
Sep 2 at 5:21
Your question is already has an answer here
– Chinnapparaj R
Sep 2 at 5:21
Have you tried calculating a few things with this operation? Like what do you get for $0*0$, $1*1$, or $-2*-2$? What about $-2*0$ or $10*1$? Just so a few random ones. Then you can try some general stuff, like $0*x$. Familiarise yourself with the operation. Only once you've done that should you try to prove general properties like associativity and commutativity. It will probably be a lot easier by then.
– Arthur
Sep 2 at 5:21
Have you tried calculating a few things with this operation? Like what do you get for $0*0$, $1*1$, or $-2*-2$? What about $-2*0$ or $10*1$? Just so a few random ones. Then you can try some general stuff, like $0*x$. Familiarise yourself with the operation. Only once you've done that should you try to prove general properties like associativity and commutativity. It will probably be a lot easier by then.
– Arthur
Sep 2 at 5:21
Here's a hint: if $a, b in mathbbR setminus -1$, then $(a + 1)(b + 1) neq 0$.
– Theo Bendit
Sep 2 at 5:32
Here's a hint: if $a, b in mathbbR setminus -1$, then $(a + 1)(b + 1) neq 0$.
– Theo Bendit
Sep 2 at 5:32
add a comment |Â
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You know that the reals are closed by sum and multiplication, and so $a * b$ will always lie on $mathbbR$. Thus, $a * b not in S$ can only occur when $a * b = -1$.
– Guido A.
Sep 2 at 5:11
2
Note that $ab+a+b=(a+1)(b+1)-1$, and consider some of the identities related to $0$ in $Bbb R$...
– abiessu
Sep 2 at 5:12
Your question is already has an answer here
– Chinnapparaj R
Sep 2 at 5:21
Have you tried calculating a few things with this operation? Like what do you get for $0*0$, $1*1$, or $-2*-2$? What about $-2*0$ or $10*1$? Just so a few random ones. Then you can try some general stuff, like $0*x$. Familiarise yourself with the operation. Only once you've done that should you try to prove general properties like associativity and commutativity. It will probably be a lot easier by then.
– Arthur
Sep 2 at 5:21
Here's a hint: if $a, b in mathbbR setminus -1$, then $(a + 1)(b + 1) neq 0$.
– Theo Bendit
Sep 2 at 5:32