Determine the identity element and inverse elements of $3$ and $-2$.
$begingroup$
Determine the identity element and inverse elements of $3$ and $-2$.
Given an algebraic structure $(G,*)$ with $G=R-{1}$, the set of real numbers without the unit number and $* $ stands for binary operation defined by $a* b=a+b-ab$ for all $a,b in G$.
My Attempt:
Let $e$ be the identity element.
$$a* e=a+e-ae$$
$$a=a+e-a$$
$$e=a$$
What is this giving? I don't understand this.
group-theory
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
$endgroup$
add a comment |
$begingroup$
Determine the identity element and inverse elements of $3$ and $-2$.
Given an algebraic structure $(G,*)$ with $G=R-{1}$, the set of real numbers without the unit number and $* $ stands for binary operation defined by $a* b=a+b-ab$ for all $a,b in G$.
My Attempt:
Let $e$ be the identity element.
$$a* e=a+e-ae$$
$$a=a+e-a$$
$$e=a$$
What is this giving? I don't understand this.
group-theory
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
$endgroup$
$begingroup$
Start with $e=e*e=e+e-e^2.$ Since $1$ is excluded, you'll have only one option for $e.$ Then it is not difficult to continue.
$endgroup$
– user376343
Jan 17 at 9:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 28 at 12:56
add a comment |
$begingroup$
Determine the identity element and inverse elements of $3$ and $-2$.
Given an algebraic structure $(G,*)$ with $G=R-{1}$, the set of real numbers without the unit number and $* $ stands for binary operation defined by $a* b=a+b-ab$ for all $a,b in G$.
My Attempt:
Let $e$ be the identity element.
$$a* e=a+e-ae$$
$$a=a+e-a$$
$$e=a$$
What is this giving? I don't understand this.
group-theory
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
$endgroup$
Determine the identity element and inverse elements of $3$ and $-2$.
Given an algebraic structure $(G,*)$ with $G=R-{1}$, the set of real numbers without the unit number and $* $ stands for binary operation defined by $a* b=a+b-ab$ for all $a,b in G$.
My Attempt:
Let $e$ be the identity element.
$$a* e=a+e-ae$$
$$a=a+e-a$$
$$e=a$$
What is this giving? I don't understand this.
group-theory
group-theory
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
asked Jan 17 at 9:22
blue_eyed_...blue_eyed_...
3,25921649
3,25921649
$begingroup$
Start with $e=e*e=e+e-e^2.$ Since $1$ is excluded, you'll have only one option for $e.$ Then it is not difficult to continue.
$endgroup$
– user376343
Jan 17 at 9:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 28 at 12:56
add a comment |
$begingroup$
Start with $e=e*e=e+e-e^2.$ Since $1$ is excluded, you'll have only one option for $e.$ Then it is not difficult to continue.
$endgroup$
– user376343
Jan 17 at 9:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 28 at 12:56
$begingroup$
Start with $e=e*e=e+e-e^2.$ Since $1$ is excluded, you'll have only one option for $e.$ Then it is not difficult to continue.
$endgroup$
– user376343
Jan 17 at 9:50
$begingroup$
Start with $e=e*e=e+e-e^2.$ Since $1$ is excluded, you'll have only one option for $e.$ Then it is not difficult to continue.
$endgroup$
– user376343
Jan 17 at 9:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 28 at 12:56
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 28 at 12:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You made a very simple mistake:
$$a+e-aneq a+e-ae$$
The product $ae$ is not $a*e$, it is $acdot e$, where $cdot$ means standard multiplication in $mathbb R$. There is no reason why $ae$ should equal $a$. So your second line in the equation should read
$$a=a+e-ae$$
and you can continue from there.
answered Jan 17 at 9:27
5xum5xum
90.6k394161
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You made a very simple mistake:
$$a+e-aneq a+e-ae$$
The product $ae$ is not $a*e$, it is $acdot e$, where $cdot$ means standard multiplication in $mathbb R$. There is no reason why $ae$ should equal $a$. So your second line in the equation should read
$$a=a+e-ae$$
and you can continue from there.
answered Jan 17 at 9:27
5xum5xum
90.6k394161
$endgroup$
add a comment |
$begingroup$
You made a very simple mistake:
$$a+e-aneq a+e-ae$$
The product $ae$ is not $a*e$, it is $acdot e$, where $cdot$ means standard multiplication in $mathbb R$. There is no reason why $ae$ should equal $a$. So your second line in the equation should read
$$a=a+e-ae$$
and you can continue from there.
answered Jan 17 at 9:27
5xum5xum
90.6k394161
$endgroup$
add a comment |
$begingroup$
You made a very simple mistake:
$$a+e-aneq a+e-ae$$
The product $ae$ is not $a*e$, it is $acdot e$, where $cdot$ means standard multiplication in $mathbb R$. There is no reason why $ae$ should equal $a$. So your second line in the equation should read
$$a=a+e-ae$$
and you can continue from there.
answered Jan 17 at 9:27
5xum5xum
90.6k394161
$endgroup$
You made a very simple mistake:
$$a+e-aneq a+e-ae$$
The product $ae$ is not $a*e$, it is $acdot e$, where $cdot$ means standard multiplication in $mathbb R$. There is no reason why $ae$ should equal $a$. So your second line in the equation should read
$$a=a+e-ae$$
and you can continue from there.
answered Jan 17 at 9:27
5xum5xum
90.6k394161
answered Jan 17 at 9:27
5xum5xum
90.6k394161
answered Jan 17 at 9:27
5xum5xum
90.6k394161
answered Jan 17 at 9:27
5xum5xum
90.6k394161
90.6k394161
add a comment |
add a comment |
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$begingroup$
Start with $e=e*e=e+e-e^2.$ Since $1$ is excluded, you'll have only one option for $e.$ Then it is not difficult to continue.
$endgroup$
– user376343
Jan 17 at 9:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 28 at 12:56