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Let $*:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}$ on the integers by the formula $x*y=x$ for any $x,y\in\mathbb{Z}$. Decide whether $*$ is associative and/or commutative.

To the best of my understanding this means I'm taking a Cartesian product $(x,y)$ and sending it to $x$. I think this would be both associative and commutative since it appears this only happens when the pair is $(x,0)$ with $(0,0)$.

Associativity means $(x*y)*z=x*(y*z)$. By this, if $x=(x,0), y=(0,0)$, then $z$ must also be $(0,0)$ by necessity which clearly is associative.

Commutativity means $x*y=y*x$. If $x=(x,0)$, then $y$ must be $(0,0)$. This is clearly commutative.

I'm struggling to put this into formal terminology. Are my thoughts correct? If not, what adjustments should I consider?

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    $\begingroup$ $*$ is just some operation just as multiplication, addition are operations. When we write $3+4$, we don't usually think of a map from the Cartesian product $(3,4)$ to $7$, we just think of $3+4$. Same here. $\endgroup$ – copper.hat Jun 21 at 20:27
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    $\begingroup$ You can never prove a general property with an example $\endgroup$ – Miguel Jun 21 at 20:28
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    $\begingroup$ One way to approach it as just 'string' replacement. When you see $x * y$ replace it by $x$. Then $x * (y * z) = x * (y) = x * y = x $. $\endgroup$ – copper.hat Jun 21 at 20:29
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    $\begingroup$ I don't understand "this only happens when the pair is $(x,0)$ with $(0,0)$." What does "this"mean here? I hope it doesn't refer to sending $(x,y)$ to $x$, since the problem says this happens "for any $x,y\in\mathbb Z$." Also "a Cartesian product $(x,y)$" doesn't make sense because $(x,y)$ is not a Cartesian product; it's one element of a Cartesian product. $\endgroup$ – Andreas Blass Jun 21 at 20:30
  • $\begingroup$ Thanks, I misread the definition of $*$, all your responses make sense. $\endgroup$ – SprNtndoChlmrs Jun 22 at 2:24
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It is associative, because for all $x,y,z\in\mathbb{Z}$, you have that $$x\star (y\star z)=x\star y=x=(x\star y)\star z$$ It is not commutative, for example because $$0\star 1=0\neq 1=1\star 0$$

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    $\begingroup$ Last line starts with $0*1$ $\endgroup$ – Miguel Jun 21 at 20:27
  • $\begingroup$ Much appreciated. I'm slightly new to problems such as these, when I see general elements sometimes I talk myself out of the solution. $\endgroup$ – SprNtndoChlmrs Jun 22 at 2:25

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