Commutativity

Let $*_1\colon \Z\times \Z\longrightarrow \Z$ be the operation such that $\forall a,b\in\Z, a*_1b:=a-b$ .

Is $*_1$ commutative ?

Let $*_2\colon \Z\times \Z\longrightarrow \R$ be the operation such that $\forall a,b\in\Z, a*_2b:=2^{a-b}$ .

Is $*_2$ commutative ?

Let $*_3\colon \mathbb{Q}^*\times\mathbb{Q}\longrightarrow \Q$ be the operation such that $\forall a,b\in\mathbb{Q}^*, a*_3b:=\frac{a}{b}+\frac{b}{a}$ .

Is $*_3$ commutative ?

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Solution by Sequoia

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Take $a=0$ and $b=1$ , then $a*_1b=a-b=0-1=-1$ but $b*_1a=b-a=1-0=1\neq a*_1b$ so $*_1$ is not commutative.
Take $a=1$ and $b=0$ , then $a*_2b=2^{a-b}=\frac{1}{2}$ but $b*_1a=2^{b-a}=2\neq a*_2b$ so $*_2$ is not commutative.
This operation is commutative. Indeed, if one consider $a,b\in\mathbb{Q}^*$ , then one gets:

$b*_3a=\frac{b}{a}+\frac{a}{b}=\frac{a}{b}+\frac{b}{a}=a*_3b,$ which is exactly what we wanted !

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