How to prove $A=R-{-1}$ and $a*b = a+b+ab $ is a binary operation?
$begingroup$
$A=R-{-1}$ and $a*b = a+b+ab $
- Show that * is a binary operation on A
- Show that * is associative
- Show that there is an identity element in A for *
- Show that every element in A has an inverse with respect to *
I'm new to this lesson. All I know to prove is associativity,
let $a,b,cin A$
Consider:$$(a*b)*c=(a+b+ab)*(c)$$
$$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
$$(a*b)*c=a+b+c+ab+ac+bc+abc$$
Now consider:
$$a*(b*c)=a*(b+c+bc)$$
$$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
$$a*(b*c)=a+b+c+ab+ab+ac+abc$$
Since we get the same result it is associative.
How to prove part 1? and How do I prove that identity exist? Thanks in advance!
P.S: also I'm curious to know why -1 is omitted in the set?
binary-operations
asked Dec 15 '18 at 5:57
emilemil
431410
$endgroup$
add a comment |
$begingroup$
$A=R-{-1}$ and $a*b = a+b+ab $
- Show that * is a binary operation on A
- Show that * is associative
- Show that there is an identity element in A for *
- Show that every element in A has an inverse with respect to *
I'm new to this lesson. All I know to prove is associativity,
let $a,b,cin A$
Consider:$$(a*b)*c=(a+b+ab)*(c)$$
$$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
$$(a*b)*c=a+b+c+ab+ac+bc+abc$$
Now consider:
$$a*(b*c)=a*(b+c+bc)$$
$$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
$$a*(b*c)=a+b+c+ab+ab+ac+abc$$
Since we get the same result it is associative.
How to prove part 1? and How do I prove that identity exist? Thanks in advance!
P.S: also I'm curious to know why -1 is omitted in the set?
binary-operations
asked Dec 15 '18 at 5:57
emilemil
431410
$endgroup$
add a comment |
$begingroup$
$A=R-{-1}$ and $a*b = a+b+ab $
- Show that * is a binary operation on A
- Show that * is associative
- Show that there is an identity element in A for *
- Show that every element in A has an inverse with respect to *
I'm new to this lesson. All I know to prove is associativity,
let $a,b,cin A$
Consider:$$(a*b)*c=(a+b+ab)*(c)$$
$$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
$$(a*b)*c=a+b+c+ab+ac+bc+abc$$
Now consider:
$$a*(b*c)=a*(b+c+bc)$$
$$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
$$a*(b*c)=a+b+c+ab+ab+ac+abc$$
Since we get the same result it is associative.
How to prove part 1? and How do I prove that identity exist? Thanks in advance!
P.S: also I'm curious to know why -1 is omitted in the set?
binary-operations
asked Dec 15 '18 at 5:57
emilemil
431410
$endgroup$
$A=R-{-1}$ and $a*b = a+b+ab $
- Show that * is a binary operation on A
- Show that * is associative
- Show that there is an identity element in A for *
- Show that every element in A has an inverse with respect to *
I'm new to this lesson. All I know to prove is associativity,
let $a,b,cin A$
Consider:$$(a*b)*c=(a+b+ab)*(c)$$
$$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$
$$(a*b)*c=a+b+c+ab+ac+bc+abc$$
Now consider:
$$a*(b*c)=a*(b+c+bc)$$
$$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$
$$a*(b*c)=a+b+c+ab+ab+ac+abc$$
Since we get the same result it is associative.
How to prove part 1? and How do I prove that identity exist? Thanks in advance!
P.S: also I'm curious to know why -1 is omitted in the set?
binary-operations
binary-operations
asked Dec 15 '18 at 5:57
emilemil
431410
asked Dec 15 '18 at 5:57
emilemil
431410
asked Dec 15 '18 at 5:57
emilemil
431410
asked Dec 15 '18 at 5:57
emilemil
431410
asked Dec 15 '18 at 5:57
emilemil
431410
431410
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
$$a*b+1=a+b+ab+1ne0.$$
For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
needs is that $a*e=a$, that is
$$a+e+ae=a.$$
Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?
Once you have done that, do inverses. You then need to solve $a*b=e$ for
$b$ in terms of $a$, that is
$$a+b+ab=e$$
where you now know $e$.
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
1
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
1
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
|
show 2 more comments
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$begingroup$
For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
$$a*b+1=a+b+ab+1ne0.$$
For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
needs is that $a*e=a$, that is
$$a+e+ae=a.$$
Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?
Once you have done that, do inverses. You then need to solve $a*b=e$ for
$b$ in terms of $a$, that is
$$a+b+ab=e$$
where you now know $e$.
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
1
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
1
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
|
show 2 more comments
$begingroup$
For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
$$a*b+1=a+b+ab+1ne0.$$
For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
needs is that $a*e=a$, that is
$$a+e+ae=a.$$
Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?
Once you have done that, do inverses. You then need to solve $a*b=e$ for
$b$ in terms of $a$, that is
$$a+b+ab=e$$
where you now know $e$.
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
1
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
1
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
|
show 2 more comments
$begingroup$
For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
$$a*b+1=a+b+ab+1ne0.$$
For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
needs is that $a*e=a$, that is
$$a+e+ae=a.$$
Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?
Once you have done that, do inverses. You then need to solve $a*b=e$ for
$b$ in terms of $a$, that is
$$a+b+ab=e$$
where you now know $e$.
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
For the binary operation, you need to prove that $a*bne-1$ iff $a$, $bne-1$, that is
$$a*b+1=a+b+ab+1ne0.$$
For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one
needs is that $a*e=a$, that is
$$a+e+ae=a.$$
Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?
Once you have done that, do inverses. You then need to solve $a*b=e$ for
$b$ in terms of $a$, that is
$$a+b+ab=e$$
where you now know $e$.
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
edited Dec 15 '18 at 6:46
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 15 '18 at 6:02
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
1
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
1
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
|
show 2 more comments
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
1
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
1
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
It is very clear after your hint. so identity would be $e=0$. Thanks a lot.
$endgroup$
– emil
Dec 15 '18 at 6:11
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
For the binary proof operation how can we prove that $a*bneq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it?
$endgroup$
– emil
Dec 15 '18 at 6:44
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
$begingroup$
@emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation.
$endgroup$
– bof
Dec 15 '18 at 6:55
1
1
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
$begingroup$
@emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $ane-1$ I can divide by $1+a$ and get $$b=frac{-1-a}{1+a}=frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$.
$endgroup$
– bof
Dec 15 '18 at 7:14
1
1
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
$begingroup$
@emil Maybe everything will be easier if you note that the definition of $a*b$ can be rewritten as $$a*b=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$.
$endgroup$
– bof
Dec 15 '18 at 7:18
|
show 2 more comments
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