Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2?
I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) = frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) =frac{xleft(frac{yz}{y+z+1}right)}{x+frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative. This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something. Sorry for bothering people with such a basic question.
abstract-algebra
add a comment |
I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) = frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) =frac{xleft(frac{yz}{y+z+1}right)}{x+frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative. This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something. Sorry for bothering people with such a basic question.
abstract-algebra
1
On which set is $*$ defined?
– Rebecca J. Stones
Jan 26 '14 at 0:35
There is no cancellation. It is more pleasant to work with numbers, like $1,1,2$.
– André Nicolas
Jan 26 '14 at 1:36
@Rebecca the set is $x > 0$. $o(1,o(1,2))$ and $o(o(1,1),2)$ ("o" being the operator function) give the same value of 1/5... sure looks like there's cancellation to me.
– bjsdaiyu
Jan 27 '14 at 1:37
add a comment |
I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) = frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) =frac{xleft(frac{yz}{y+z+1}right)}{x+frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative. This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something. Sorry for bothering people with such a basic question.
abstract-algebra
I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) = frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = left(frac{xy}{x+y+1}right) * z = frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * left(frac{yz}{y+z+1}right) =frac{xleft(frac{yz}{y+z+1}right)}{x+frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative. This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something. Sorry for bothering people with such a basic question.
abstract-algebra
abstract-algebra
edited Jan 25 '14 at 23:54
TMM
9,11032848
9,11032848
asked Jan 25 '14 at 23:45
bjsdaiyu
283
283
1
On which set is $*$ defined?
– Rebecca J. Stones
Jan 26 '14 at 0:35
There is no cancellation. It is more pleasant to work with numbers, like $1,1,2$.
– André Nicolas
Jan 26 '14 at 1:36
@Rebecca the set is $x > 0$. $o(1,o(1,2))$ and $o(o(1,1),2)$ ("o" being the operator function) give the same value of 1/5... sure looks like there's cancellation to me.
– bjsdaiyu
Jan 27 '14 at 1:37
add a comment |
1
On which set is $*$ defined?
– Rebecca J. Stones
Jan 26 '14 at 0:35
There is no cancellation. It is more pleasant to work with numbers, like $1,1,2$.
– André Nicolas
Jan 26 '14 at 1:36
@Rebecca the set is $x > 0$. $o(1,o(1,2))$ and $o(o(1,1),2)$ ("o" being the operator function) give the same value of 1/5... sure looks like there's cancellation to me.
– bjsdaiyu
Jan 27 '14 at 1:37
1
1
On which set is $*$ defined?
– Rebecca J. Stones
Jan 26 '14 at 0:35
On which set is $*$ defined?
– Rebecca J. Stones
Jan 26 '14 at 0:35
There is no cancellation. It is more pleasant to work with numbers, like $1,1,2$.
– André Nicolas
Jan 26 '14 at 1:36
There is no cancellation. It is more pleasant to work with numbers, like $1,1,2$.
– André Nicolas
Jan 26 '14 at 1:36
@Rebecca the set is $x > 0$. $o(1,o(1,2))$ and $o(o(1,1),2)$ ("o" being the operator function) give the same value of 1/5... sure looks like there's cancellation to me.
– bjsdaiyu
Jan 27 '14 at 1:37
@Rebecca the set is $x > 0$. $o(1,o(1,2))$ and $o(o(1,1),2)$ ("o" being the operator function) give the same value of 1/5... sure looks like there's cancellation to me.
– bjsdaiyu
Jan 27 '14 at 1:37
add a comment |
3 Answers
3
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oldest
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As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
begin{align*}
(x * y) * z &= left(frac{xy}{x+y+1}right) * z \
&= frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} \
&= frac{left(frac{xyz}{x+y+1}right)}{frac{xy+(z+1)(x+y+1)}{x+y+1}} \
&= frac{xyz}{xy+(z+1)(x+y+1)} \
&= frac{xyz}{xy+xz+yz+x+y+z+1} \
&= frac{xyz}{yz+(x+1)(y+z+1)} \
&= frac{left(frac{xyz}{y+z+1}right)}{frac{yz+(x+1)(y+z+1)}{y+z+1}} \
&= frac{xleft(frac{yz}{y+z+1}right)}{x + frac{yz}{y+z+1} + 1} \
&= x*left(frac{yz}{y+z+1}right) \
&= x*(y*z).
end{align*}
add a comment |
Let $f(x)=1+frac 1x$. Then $f^{-1}(z)=frac{1}{z-1}$ and show that:
$$x*y = f^{-1}(f(x) f(y))$$
This means that:
$$begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\
&= f^{-1}left(fleft(f^{-1}(f(x)f(y))right)f(z)right)\
&= f^{-1}(f(x)f(y)f(z))
end{align}$$
Since $*$ is commutative, we get:
$$begin{align}x*(y*z)&= (y*z)*x\
&= f^{-1}(f(y)f(z)f(x))
end{align}$$
Clearly, these two are equal.
This is true in general for any invertible function $f$ if we define $*$ in this way.
Most easily defined associative operators on the real numbers are of this form. Even addition can be defined this way in terms of multiplication, with $f(x)=e^x$ and $f^{-1}(x)=log x$, then:
$$x+y = log(e^xe^y)$$
Alternatively, of course, you can define positive multiplication in terms of addition:
$$xy = e^{log x + log y}$$
add a comment |
Define $a = x^{-1}$, $b = y^{-1}$, and $c = z^{-1}$. Furthermore, let $d = (x*y)^{-1}$. Then it is easy to see that $$d = (1+a)(1+b) - 1.$$ Then $$((x*y)*z)^{-1} = (d^{-1}*c^{-1})^{-1} = (1+d)(1+c)-1 = (1+a)(1+b)(1+c)-1.$$ Being that this is a symmetric function in $a,b,c$, it immediately follows that $*$ is associative.
add a comment |
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3 Answers
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3 Answers
3
active
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As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
begin{align*}
(x * y) * z &= left(frac{xy}{x+y+1}right) * z \
&= frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} \
&= frac{left(frac{xyz}{x+y+1}right)}{frac{xy+(z+1)(x+y+1)}{x+y+1}} \
&= frac{xyz}{xy+(z+1)(x+y+1)} \
&= frac{xyz}{xy+xz+yz+x+y+z+1} \
&= frac{xyz}{yz+(x+1)(y+z+1)} \
&= frac{left(frac{xyz}{y+z+1}right)}{frac{yz+(x+1)(y+z+1)}{y+z+1}} \
&= frac{xleft(frac{yz}{y+z+1}right)}{x + frac{yz}{y+z+1} + 1} \
&= x*left(frac{yz}{y+z+1}right) \
&= x*(y*z).
end{align*}
add a comment |
As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
begin{align*}
(x * y) * z &= left(frac{xy}{x+y+1}right) * z \
&= frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} \
&= frac{left(frac{xyz}{x+y+1}right)}{frac{xy+(z+1)(x+y+1)}{x+y+1}} \
&= frac{xyz}{xy+(z+1)(x+y+1)} \
&= frac{xyz}{xy+xz+yz+x+y+z+1} \
&= frac{xyz}{yz+(x+1)(y+z+1)} \
&= frac{left(frac{xyz}{y+z+1}right)}{frac{yz+(x+1)(y+z+1)}{y+z+1}} \
&= frac{xleft(frac{yz}{y+z+1}right)}{x + frac{yz}{y+z+1} + 1} \
&= x*left(frac{yz}{y+z+1}right) \
&= x*(y*z).
end{align*}
add a comment |
As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
begin{align*}
(x * y) * z &= left(frac{xy}{x+y+1}right) * z \
&= frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} \
&= frac{left(frac{xyz}{x+y+1}right)}{frac{xy+(z+1)(x+y+1)}{x+y+1}} \
&= frac{xyz}{xy+(z+1)(x+y+1)} \
&= frac{xyz}{xy+xz+yz+x+y+z+1} \
&= frac{xyz}{yz+(x+1)(y+z+1)} \
&= frac{left(frac{xyz}{y+z+1}right)}{frac{yz+(x+1)(y+z+1)}{y+z+1}} \
&= frac{xleft(frac{yz}{y+z+1}right)}{x + frac{yz}{y+z+1} + 1} \
&= x*left(frac{yz}{y+z+1}right) \
&= x*(y*z).
end{align*}
As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
begin{align*}
(x * y) * z &= left(frac{xy}{x+y+1}right) * z \
&= frac{left(frac{xy}{x+y+1}right)z}{frac{xy}{x+y+1} + z + 1} \
&= frac{left(frac{xyz}{x+y+1}right)}{frac{xy+(z+1)(x+y+1)}{x+y+1}} \
&= frac{xyz}{xy+(z+1)(x+y+1)} \
&= frac{xyz}{xy+xz+yz+x+y+z+1} \
&= frac{xyz}{yz+(x+1)(y+z+1)} \
&= frac{left(frac{xyz}{y+z+1}right)}{frac{yz+(x+1)(y+z+1)}{y+z+1}} \
&= frac{xleft(frac{yz}{y+z+1}right)}{x + frac{yz}{y+z+1} + 1} \
&= x*left(frac{yz}{y+z+1}right) \
&= x*(y*z).
end{align*}
answered Jan 27 '14 at 2:25
Rebecca J. Stones
20.9k22781
20.9k22781
add a comment |
add a comment |
Let $f(x)=1+frac 1x$. Then $f^{-1}(z)=frac{1}{z-1}$ and show that:
$$x*y = f^{-1}(f(x) f(y))$$
This means that:
$$begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\
&= f^{-1}left(fleft(f^{-1}(f(x)f(y))right)f(z)right)\
&= f^{-1}(f(x)f(y)f(z))
end{align}$$
Since $*$ is commutative, we get:
$$begin{align}x*(y*z)&= (y*z)*x\
&= f^{-1}(f(y)f(z)f(x))
end{align}$$
Clearly, these two are equal.
This is true in general for any invertible function $f$ if we define $*$ in this way.
Most easily defined associative operators on the real numbers are of this form. Even addition can be defined this way in terms of multiplication, with $f(x)=e^x$ and $f^{-1}(x)=log x$, then:
$$x+y = log(e^xe^y)$$
Alternatively, of course, you can define positive multiplication in terms of addition:
$$xy = e^{log x + log y}$$
add a comment |
Let $f(x)=1+frac 1x$. Then $f^{-1}(z)=frac{1}{z-1}$ and show that:
$$x*y = f^{-1}(f(x) f(y))$$
This means that:
$$begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\
&= f^{-1}left(fleft(f^{-1}(f(x)f(y))right)f(z)right)\
&= f^{-1}(f(x)f(y)f(z))
end{align}$$
Since $*$ is commutative, we get:
$$begin{align}x*(y*z)&= (y*z)*x\
&= f^{-1}(f(y)f(z)f(x))
end{align}$$
Clearly, these two are equal.
This is true in general for any invertible function $f$ if we define $*$ in this way.
Most easily defined associative operators on the real numbers are of this form. Even addition can be defined this way in terms of multiplication, with $f(x)=e^x$ and $f^{-1}(x)=log x$, then:
$$x+y = log(e^xe^y)$$
Alternatively, of course, you can define positive multiplication in terms of addition:
$$xy = e^{log x + log y}$$
add a comment |
Let $f(x)=1+frac 1x$. Then $f^{-1}(z)=frac{1}{z-1}$ and show that:
$$x*y = f^{-1}(f(x) f(y))$$
This means that:
$$begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\
&= f^{-1}left(fleft(f^{-1}(f(x)f(y))right)f(z)right)\
&= f^{-1}(f(x)f(y)f(z))
end{align}$$
Since $*$ is commutative, we get:
$$begin{align}x*(y*z)&= (y*z)*x\
&= f^{-1}(f(y)f(z)f(x))
end{align}$$
Clearly, these two are equal.
This is true in general for any invertible function $f$ if we define $*$ in this way.
Most easily defined associative operators on the real numbers are of this form. Even addition can be defined this way in terms of multiplication, with $f(x)=e^x$ and $f^{-1}(x)=log x$, then:
$$x+y = log(e^xe^y)$$
Alternatively, of course, you can define positive multiplication in terms of addition:
$$xy = e^{log x + log y}$$
Let $f(x)=1+frac 1x$. Then $f^{-1}(z)=frac{1}{z-1}$ and show that:
$$x*y = f^{-1}(f(x) f(y))$$
This means that:
$$begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\
&= f^{-1}left(fleft(f^{-1}(f(x)f(y))right)f(z)right)\
&= f^{-1}(f(x)f(y)f(z))
end{align}$$
Since $*$ is commutative, we get:
$$begin{align}x*(y*z)&= (y*z)*x\
&= f^{-1}(f(y)f(z)f(x))
end{align}$$
Clearly, these two are equal.
This is true in general for any invertible function $f$ if we define $*$ in this way.
Most easily defined associative operators on the real numbers are of this form. Even addition can be defined this way in terms of multiplication, with $f(x)=e^x$ and $f^{-1}(x)=log x$, then:
$$x+y = log(e^xe^y)$$
Alternatively, of course, you can define positive multiplication in terms of addition:
$$xy = e^{log x + log y}$$
edited Aug 23 '14 at 18:43
answered Aug 23 '14 at 17:59
Thomas Andrews
130k11146297
130k11146297
add a comment |
add a comment |
Define $a = x^{-1}$, $b = y^{-1}$, and $c = z^{-1}$. Furthermore, let $d = (x*y)^{-1}$. Then it is easy to see that $$d = (1+a)(1+b) - 1.$$ Then $$((x*y)*z)^{-1} = (d^{-1}*c^{-1})^{-1} = (1+d)(1+c)-1 = (1+a)(1+b)(1+c)-1.$$ Being that this is a symmetric function in $a,b,c$, it immediately follows that $*$ is associative.
add a comment |
Define $a = x^{-1}$, $b = y^{-1}$, and $c = z^{-1}$. Furthermore, let $d = (x*y)^{-1}$. Then it is easy to see that $$d = (1+a)(1+b) - 1.$$ Then $$((x*y)*z)^{-1} = (d^{-1}*c^{-1})^{-1} = (1+d)(1+c)-1 = (1+a)(1+b)(1+c)-1.$$ Being that this is a symmetric function in $a,b,c$, it immediately follows that $*$ is associative.
add a comment |
Define $a = x^{-1}$, $b = y^{-1}$, and $c = z^{-1}$. Furthermore, let $d = (x*y)^{-1}$. Then it is easy to see that $$d = (1+a)(1+b) - 1.$$ Then $$((x*y)*z)^{-1} = (d^{-1}*c^{-1})^{-1} = (1+d)(1+c)-1 = (1+a)(1+b)(1+c)-1.$$ Being that this is a symmetric function in $a,b,c$, it immediately follows that $*$ is associative.
Define $a = x^{-1}$, $b = y^{-1}$, and $c = z^{-1}$. Furthermore, let $d = (x*y)^{-1}$. Then it is easy to see that $$d = (1+a)(1+b) - 1.$$ Then $$((x*y)*z)^{-1} = (d^{-1}*c^{-1})^{-1} = (1+d)(1+c)-1 = (1+a)(1+b)(1+c)-1.$$ Being that this is a symmetric function in $a,b,c$, it immediately follows that $*$ is associative.
answered Jan 27 '14 at 2:55
heropup
62.5k66099
62.5k66099
add a comment |
add a comment |
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1
On which set is $*$ defined?
– Rebecca J. Stones
Jan 26 '14 at 0:35
There is no cancellation. It is more pleasant to work with numbers, like $1,1,2$.
– André Nicolas
Jan 26 '14 at 1:36
@Rebecca the set is $x > 0$. $o(1,o(1,2))$ and $o(o(1,1),2)$ ("o" being the operator function) give the same value of 1/5... sure looks like there's cancellation to me.
– bjsdaiyu
Jan 27 '14 at 1:37