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I am given the following exercice.

Is the operation $x ∆ y = 2xy - 3x + 3y$ Associative on the set of real number?

In order to do this, one must check that

$(x ∆ y) ∆ z = x ∆ (y ∆ z)$

So answer provided is:

$( x ∆ y ) ∆ z =(2xy - 3x + 3y) ∆ z = 4xyz - 6xz + 6yz - 6xy + 9x - 9y + 3z$

$x ∆ (y ∆ z) = x ∆ (2xy - 3x + 3z) = 4xyz - 6xy + 6xz - 3x + 6yz - 9y + 9z$

This shows that both sides aren't the same, thus the operation is not associative.

However, I have no idea how this is done. I don't understand where to value of z if being defined or how it is calculated.

Any help would help, I'm very stuck. Thanks

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    Your question is weird. In order to show that that operation is associative you have to prove that $(x ∆ y) ∆ z = x ∆ (y ∆ z)$ for all possible choices of real numbers $x$, $y$ and $z$, and to show that it is not it is enough for you to find three real numbers such that equality does not hold. Can you find such three numbers? – 2017-01-10
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    In the first "equation" first we calculate $(x \Delta y)$ i.e. $(2xy-3x+3y)$. Then call it $k$ and calculate $(k \Delta z)$ i.e. $2(2xy-3x+3y)z-3(2xy-3x+3y)+3z$. – 2017-01-10
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    $x,y,z \in \mathbb R$; to check if $\Delta$ is associative, it must be associative $\forall x,y,z \in \mathbb R$, thus $z$ is a generic element of your set, in this case $\mathbb R$ – 2017-01-10
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    @MauroALLEGRANZA Where is z being defined. If k is the expression and z a variable, wouldn't it just be $(2xy - 3x + 3y) Δ z $? – 2017-01-10
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    @MauroALLEGRANZA How would I go about doing that? I'm really lost with this. To me it would just be $2xyz−3xz+3yz$. Prehaps you could write an explination as an answers instead of in comments. Thank you! – 2017-01-10
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    @MauroALLEGRANZA I understand that the calculations work. I wonder why replace $xy$ with $xΔy$ while also being able to replace $x$ with $xΔy$. – 2017-01-10
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    @MauroALLEGRANZA Please do answer. I am stuck on this question which seems simple but I can't seem to understand it. It's incredibly frustrating. – 2017-01-10

1 Answers 1

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First "equation": $(x \Delta y) \Delta z$

First we have to calculate $(xΔy)$ i.e. $(2xy−3x+3y)$ : call it $k$.

Then we have to calculate $(kΔz)$ i.e. $(2kz−3k+3z)$.

Finally, we have to "plug in" $(2xy−3x+3y)$ in place of $k$ in the previous expression to get :

$2(2xy−3x+3y)z-3(2xy−3x+3y)+3z$.

This in turn is :

$4xyz-6xz+6yz-6xy+9x-9y+3z$.

Second "equation": $x \Delta (y \Delta z)$.

First calculate $(yΔz)$ i.e. $(2yz−3y+3z)$ : call it $k$.

Then calculate $(xΔk)$ and complete it in the same way.