Associativity of $x*y := \frac{xy}{x+y+1}$

Question

In order to show something is associative one must show that $(x*y)*z$ = $x*(y*z)$. I want to show that $x * y = \frac{xy}{x+y+1}$ is associative.

This is for self-study (I'm learning algebra over the summer) and need help finishing the proof.

Below are my (hopefully not incorrect) steps,

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

Applying the binary operation * I obtain

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

Here is where my first question comes:

Question 1: Do I multiple x (in numerator) using the definition of *, or do I multiply x the "normal" way? If I multiply x the "normal" way, then why do I not use the * definition recursively for each application of *?

Assuming I do multiplication the "normal" way I obtain,

3) $\frac{\left(\frac{xyz}{y+z+1}\right)}{x+\frac{yz}{y+z+1}+ 1}$

I know in order to add fractions they must have the same denominators. This means the denominator for each term needs to be y+z+1.

So this brings me to

4) $\frac{\left(\frac{xyz}{y+z+1}\right)}{{y+z+1}(x+\frac{yz}{y+z+1}+ 1)}$

Here is where I become and more uncertain...

Question 2: Again, do I use the * definition of multiplication or do I use the "normal" definition of multiplication for multiplying the y+z+1 across and why? Again, assuming I use "normal" multiplication I obtain

5) $\frac{\left(\frac{xyz}{y+z+1}\right)}{\frac{xy+xz+x+yz+y+z+1)}{y+z+1}}$

From here I'm suck and not sure how to reduce this any further.

Question 3: How do I proceed from step 5) to the conclusion? I believe once I understand this, I will be able to show $(x*y)*z$,I would like to see the rest of the proof from where I am stuck.

Thank you very much in advance for the help.

Disclaimer: A similar question has been asked & answered in this thread, but I have different questions than the ones they posed.

Answer 1

The operation is $ x*y = \dfrac{xy}{x+y+1}$, not $\dfrac{x*y}{x+y+1}$.

So $$(x*y)*z = \dfrac{\dfrac{xy}{x+y+1} z}{\dfrac{xy}{x+y+1} + z + 1} = \dfrac{xyz}{xy + xz + yz + x + y + z + 1}$$ and similarly $$ x*(yz) = \dfrac{x\dfrac{yz}{y+z+1}}{x+\dfrac{yz}{y+z+1}+1} = \dfrac{xyz}{xy + xz + yz + x + y + z + 1}$$

Answer 2

If your expression contains a $*$ in the numerator, you must use the definition of $*$ there, not $\cdot$. However, applying the definition pf $*$ should actuall yget you rid of (one) $*$ each time.

So you have to show that $$x*(y*z)=x*\frac{yz}{y+z+1}=\frac{x\cdot \frac{yz}{y+z+1}}{x+\frac{yz}{y+z+1}+1} $$ is the same as $$(x*y)*z=\frac{xy}{x+y+1}*z=\frac{\frac{xy}{x+y+1}\cdot z}{\frac{xy}{x+y+1}+z+1} $$ To simplify both expressions, expand the fraction with $y+z+1$ (or $x+y+1$)

Answer 3

To answer your first question, we could write the $*$ operation using notation that might be a bit more familiar: $$x * y = \frac{xy}{x+y+1} = f(x,y).$$ The numerator represents the usual multiplication of $x$ and $y$, not the operation $*$. So $$x * (y * z) = f(x,f(y,z)),$$ for example, and $$(x * y) * z = f(f(x,y),z).$$ That said, this notation is a little more verbose, and you can see it's not necessary to write so many $f$s. So we will stick to the original notation.

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Now, I think the best way to proceed is to observe $$x * y = \frac{xy}{x+y+1} = \frac{xy}{-xy + xy + x + y + 1} = \frac{xy}{(x+1)(y+1)-xy} = \frac{1}{(1+1/x)(1+1/y) - 1}.$$ This suggests to us to let $a = 1 + 1/x$, $b = 1 + 1/y$, $c = 1 + 1/z$, so that $$x * (y * z) = x * \frac{1}{bc-1} = \frac{1}{a\left(1 + \frac{1}{\frac{1}{bc - 1}}\right) - 1} = \frac{1}{abc-1}.$$ Now the symmetry is evident, and it is straightforward to compute $$(x * y) * z = \frac{1}{ab-1} * z = \frac{1}{abc-1}.$$

Answer 4

Too long for a comment:

@heropup: Your formula for $x*y$ shows that the $*$ operation is the one obtained by transport of structure of the usual multiplication. Indeed, let $$ f\colon \mathbf R\setminus\{0\}\rightarrow \mathbf R\setminus\{1\} $$ be the map defined by $$ f(x)=1+\frac1x. $$ Then $f$ is a bijection and its inverse map is given by $$ f^{-1}(y)=\frac1{y-1} $$ for all $y\in\mathbf R\setminus\{1\}$. We then indeed have, as per your formula, $$ x*y= \frac{1}{(1+\frac1x)(1+\frac1y)-1}=f^{-1}(f(x)\cdot f(y)), $$ where $\cdot$ is usual multiplication on $\mathbf R\setminus\{1\}$. This is what one calls an operation obtained by transport of structure: if $f\colon E\rightarrow F$ is a bijection between two sets, and $\cdot$ is a binary operation on $F$, then one has an induced binary operation $*$ on $E$ defined by $$ x*y=f^{-1}(f(x)\cdot f(y)) $$ for all $x,y\in E$. Its a kind of ping-pong principle, you map the elements $x$ and $y$ of $E$ in $F$ where you can multiply them, and then map the result back. The operation $*$ is the unique operation on $E$ for which $$ f(x*y)=f(x)\cdot f(y) $$ for all $x,y\in E$. The binary operation $*$ obtained by transport of structure has all properties that the binary operation $\cdot$ has: if $\cdot$ is associative then $*$ is so, if $\cdot$ has a neutral element then $*$ has one, and in that case, if inverses exist for $\cdot$ then inverses exists for $*$, if $\cdot$ has an absorbing element then $*$ has one, if $\cdot$ is commutative then $*$ is so, and so on.

Now, returning to your specific example and the specific bijection $f$, the usual multiplication $\cdot$ on $\mathbf R\setminus\{1\}$ is not an operation on $\mathbf R\setminus\{1\}$. Indeed, the product of two real numbers different from $1$ can very well be equal to $1$, which is not an element of the set $\mathbf R\setminus\{1\}$. Therefore, the formula above only holds for $x,y\in \mathbf R\setminus\{0\}$ for which $f(x)\cdot f(y)\neq1$. That is probably why your operation $*$ has been restricted to the positive real numbers, which corresponds exactly to the subset $(1,+\infty)$ of $\mathbf R\setminus\{1\}$ where $\cdot$ is a binary operation.

In any case, it is now clear that $*$ is associative since $\cdot$ is so, but much more can be derived now: for example, since the product $\cdot$ is a group law on $\mathbf R\setminus\{0\}$, it would be natural to consider $f$ as a map $$ f\colon (\mathbf R\cup\{\infty\})\setminus\{0\}\rightarrow \mathbf R\setminus\{0\} $$ still defined by $f(x)=1+\frac1x$ for $x\in\mathbf R\setminus\{0\}$, and with $f(\infty)=1$. Then $f$ is again a bijection. Since $\cdot$ is a commutative group structure on the codomain of $f$, one deduces that $*$ defines a group structure on $(\mathbf R\cup\{\infty\})\setminus\{0\}$! The neutral element is the element $\infty$. The inverse $x^{*-1}$ of $x$ with respect to $*$ is $$ x^{*-1}=f^{-1}(f(x)^{\cdot-1})= \frac1{\frac1{1+\frac1x}-1}, $$ where $x^{\cdot-1}$ is the usual inverse $\frac1x$.

The moral is that transport of structure is a wonderful machine to create exercises. Pick your favorite group $G$, take any bijection $f\colon E\rightarrow G$, and consider the binary operation on $E$ obtained by transport of structure. Such an operation can be really weird looking, but still you can be sure that it defines a group structure on your set $E$!