Proof of the open interval $(-1,1)$ with respect to $*$ is a group [duplicate]
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Proving a set is an abelian group.
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Hi,so I just can't get my head through the solution for 4th problem.I successfully proved associative law and when I wrote an equation for finding identity element I did get the same thing,but then I don't understand how they get $e(1 - x^2) = 0$. I always get $x + e = x^2e + x$.
I'd appreciate if someone could explain. I understand that logically speaking, it should be $0$.
abstract-algebra algebra-precalculus
asked Sep 11 at 3:32
Cat
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marked as duplicate by Arnaud D., Lord Shark the Unknown abstract-algebra
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This question already has an answer here:
Proving a set is an abelian group.
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Hi,so I just can't get my head through the solution for 4th problem.I successfully proved associative law and when I wrote an equation for finding identity element I did get the same thing,but then I don't understand how they get $e(1 - x^2) = 0$. I always get $x + e = x^2e + x$.
I'd appreciate if someone could explain. I understand that logically speaking, it should be $0$.
abstract-algebra algebra-precalculus
asked Sep 11 at 3:32
Cat
184
marked as duplicate by Arnaud D., Lord Shark the Unknown abstract-algebra
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Sep 11 at 13:39
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up vote
3
down vote
favorite
This question already has an answer here:
Proving a set is an abelian group.
4 answers
Hi,so I just can't get my head through the solution for 4th problem.I successfully proved associative law and when I wrote an equation for finding identity element I did get the same thing,but then I don't understand how they get $e(1 - x^2) = 0$. I always get $x + e = x^2e + x$.
I'd appreciate if someone could explain. I understand that logically speaking, it should be $0$.
abstract-algebra algebra-precalculus
asked Sep 11 at 3:32
Cat
184
This question already has an answer here:
Proving a set is an abelian group.
4 answers
Hi,so I just can't get my head through the solution for 4th problem.I successfully proved associative law and when I wrote an equation for finding identity element I did get the same thing,but then I don't understand how they get $e(1 - x^2) = 0$. I always get $x + e = x^2e + x$.
I'd appreciate if someone could explain. I understand that logically speaking, it should be $0$.
This question already has an answer here:
Proving a set is an abelian group.
4 answers
abstract-algebra algebra-precalculus
abstract-algebra algebra-precalculus
asked Sep 11 at 3:32
Cat
184
asked Sep 11 at 3:32
Cat
184
edited Sep 11 at 12:45
asked Sep 11 at 3:32
Cat
184
asked Sep 11 at 3:32
Cat
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asked Sep 11 at 3:32
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Sep 11 at 13:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Let $e$ be the identity element. Then
$$x*e=x=e*x=fracx+exe+1$$ so, $x(xe+1)=x+e$ and so $x^2e+x=x+e$ which implies $$e-x^2e=0$$ that is, $$e(1-x^2)=0$$ and so $e=0$
Verification: $$x*0=fracx+0x.0+1=x=0*x$$
Added: To find the inverse of an arbitrary $x$, assume $y$ is the inverse of $x$. That is, $$x*y=0=fracx+yxy+1$$
Therefore, numerator must be zero and so $y=-x$
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
thank you soo much!
– Cat
Sep 11 at 13:03
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
add a comment |
up vote
7
down vote
Here is an alternative solution. I'm sure not everyone will like it.
Observe that
$$tanh(a+b)=fractanh a+tanh b1+tanh atanh b.$$
Therefore
$$x*y=tanhleft(tanh^-1x+tanh^-1yright).tag*$$
From (*) one can read off associativity easily:
beginalign
x*(y*z)&=tanh(tanh^-1x+tanh^-1(y*z))\
&=tanh(tanh^-1x+(tanh^-1y+tanh^-1z))\
&=cdots=(x*y)*z.
endalign
Since $0$ is the additive identity and $tanh 0=0$ then $0$
is the identity for $*$. Also
$$x*(-x)=tanh(tanh^-1x+tanh^-1(-x))
=tanh(tanh^-1x-tanh^-1x)=tanh0=0$$
etc.
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Let $e$ be the identity element. Then
$$x*e=x=e*x=fracx+exe+1$$ so, $x(xe+1)=x+e$ and so $x^2e+x=x+e$ which implies $$e-x^2e=0$$ that is, $$e(1-x^2)=0$$ and so $e=0$
Verification: $$x*0=fracx+0x.0+1=x=0*x$$
Added: To find the inverse of an arbitrary $x$, assume $y$ is the inverse of $x$. That is, $$x*y=0=fracx+yxy+1$$
Therefore, numerator must be zero and so $y=-x$
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
thank you soo much!
– Cat
Sep 11 at 13:03
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
add a comment |
up vote
6
down vote
accepted
Let $e$ be the identity element. Then
$$x*e=x=e*x=fracx+exe+1$$ so, $x(xe+1)=x+e$ and so $x^2e+x=x+e$ which implies $$e-x^2e=0$$ that is, $$e(1-x^2)=0$$ and so $e=0$
Verification: $$x*0=fracx+0x.0+1=x=0*x$$
Added: To find the inverse of an arbitrary $x$, assume $y$ is the inverse of $x$. That is, $$x*y=0=fracx+yxy+1$$
Therefore, numerator must be zero and so $y=-x$
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
thank you soo much!
– Cat
Sep 11 at 13:03
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Let $e$ be the identity element. Then
$$x*e=x=e*x=fracx+exe+1$$ so, $x(xe+1)=x+e$ and so $x^2e+x=x+e$ which implies $$e-x^2e=0$$ that is, $$e(1-x^2)=0$$ and so $e=0$
Verification: $$x*0=fracx+0x.0+1=x=0*x$$
Added: To find the inverse of an arbitrary $x$, assume $y$ is the inverse of $x$. That is, $$x*y=0=fracx+yxy+1$$
Therefore, numerator must be zero and so $y=-x$
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
Let $e$ be the identity element. Then
$$x*e=x=e*x=fracx+exe+1$$ so, $x(xe+1)=x+e$ and so $x^2e+x=x+e$ which implies $$e-x^2e=0$$ that is, $$e(1-x^2)=0$$ and so $e=0$
Verification: $$x*0=fracx+0x.0+1=x=0*x$$
Added: To find the inverse of an arbitrary $x$, assume $y$ is the inverse of $x$. That is, $$x*y=0=fracx+yxy+1$$
Therefore, numerator must be zero and so $y=-x$
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
edited Sep 11 at 4:14
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
answered Sep 11 at 4:08
Chinnapparaj R
4,338725
4,338725
thank you soo much!
– Cat
Sep 11 at 13:03
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
add a comment |
thank you soo much!
– Cat
Sep 11 at 13:03
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
thank you soo much!
– Cat
Sep 11 at 13:03
thank you soo much!
– Cat
Sep 11 at 13:03
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
You are welcome!
– Chinnapparaj R
Sep 12 at 4:36
add a comment |
up vote
7
down vote
Here is an alternative solution. I'm sure not everyone will like it.
Observe that
$$tanh(a+b)=fractanh a+tanh b1+tanh atanh b.$$
Therefore
$$x*y=tanhleft(tanh^-1x+tanh^-1yright).tag*$$
From (*) one can read off associativity easily:
beginalign
x*(y*z)&=tanh(tanh^-1x+tanh^-1(y*z))\
&=tanh(tanh^-1x+(tanh^-1y+tanh^-1z))\
&=cdots=(x*y)*z.
endalign
Since $0$ is the additive identity and $tanh 0=0$ then $0$
is the identity for $*$. Also
$$x*(-x)=tanh(tanh^-1x+tanh^-1(-x))
=tanh(tanh^-1x-tanh^-1x)=tanh0=0$$
etc.
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
add a comment |
up vote
7
down vote
Here is an alternative solution. I'm sure not everyone will like it.
Observe that
$$tanh(a+b)=fractanh a+tanh b1+tanh atanh b.$$
Therefore
$$x*y=tanhleft(tanh^-1x+tanh^-1yright).tag*$$
From (*) one can read off associativity easily:
beginalign
x*(y*z)&=tanh(tanh^-1x+tanh^-1(y*z))\
&=tanh(tanh^-1x+(tanh^-1y+tanh^-1z))\
&=cdots=(x*y)*z.
endalign
Since $0$ is the additive identity and $tanh 0=0$ then $0$
is the identity for $*$. Also
$$x*(-x)=tanh(tanh^-1x+tanh^-1(-x))
=tanh(tanh^-1x-tanh^-1x)=tanh0=0$$
etc.
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
add a comment |
up vote
7
down vote
up vote
7
down vote
Here is an alternative solution. I'm sure not everyone will like it.
Observe that
$$tanh(a+b)=fractanh a+tanh b1+tanh atanh b.$$
Therefore
$$x*y=tanhleft(tanh^-1x+tanh^-1yright).tag*$$
From (*) one can read off associativity easily:
beginalign
x*(y*z)&=tanh(tanh^-1x+tanh^-1(y*z))\
&=tanh(tanh^-1x+(tanh^-1y+tanh^-1z))\
&=cdots=(x*y)*z.
endalign
Since $0$ is the additive identity and $tanh 0=0$ then $0$
is the identity for $*$. Also
$$x*(-x)=tanh(tanh^-1x+tanh^-1(-x))
=tanh(tanh^-1x-tanh^-1x)=tanh0=0$$
etc.
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
Here is an alternative solution. I'm sure not everyone will like it.
Observe that
$$tanh(a+b)=fractanh a+tanh b1+tanh atanh b.$$
Therefore
$$x*y=tanhleft(tanh^-1x+tanh^-1yright).tag*$$
From (*) one can read off associativity easily:
beginalign
x*(y*z)&=tanh(tanh^-1x+tanh^-1(y*z))\
&=tanh(tanh^-1x+(tanh^-1y+tanh^-1z))\
&=cdots=(x*y)*z.
endalign
Since $0$ is the additive identity and $tanh 0=0$ then $0$
is the identity for $*$. Also
$$x*(-x)=tanh(tanh^-1x+tanh^-1(-x))
=tanh(tanh^-1x-tanh^-1x)=tanh0=0$$
etc.
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
answered Sep 11 at 4:36
Lord Shark the Unknown
95.7k957125
95.7k957125
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
add a comment |
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
Lol I didn't expected a trigonometric approach. Very creative!! :)
– manooooh
Sep 11 at 4:48
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
ike it! Shark it! :+)
– mrs
Sep 11 at 5:53
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
couldn't understood this now, but i hope i will someday thanks !
– Cat
Sep 11 at 13:04
add a comment |
