Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2?

Solution 1:

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{x\left(\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*}

Solution 2:

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:

$$\begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\\ &= f^{-1}\left(f\left(f^{-1}(f(x)f(y))\right)f(z)\right)\\ &= f^{-1}(f(x)f(y)f(z)) \end{align}$$

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:

$$xy = e^{\log x + \log y}$$

Solution 3:

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.