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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:

$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} $$

$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) =\frac{x\left(\frac{yz}{y+z+1}\right)}{x+\frac{yz}{y+z+1} + 1} $$

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.

TMM
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bjsdaiyu
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3 Answers3

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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{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*}

Rebecca J. Stones
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4

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:

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

Thomas Andrews
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1

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.

heropup
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