Show that the operation $*$ defined on $\{0,1,2,3,4\}$ by $a*b=a imes b\pmod5$ is a binary operation. Test whether it is associative and commutative. Test whether the identity exists. If it exists, investigate about the inverse for each element.

Question

Show that the operation $*$ defined on $\{0,1,2,3,4\}$ by $a*b=a	imes b\pmod5$ is a binary operation. Test whether it is associative and commutative. Test whether the identity exists. If it exists, investigate about the inverse for each element.

Accepted Answer

>

> [!TLDR]
> $a*b=ab\bmod5$ on $\{0,1,2,3,4\}$ is closed, commutative and associative, has identity $1$, and every nonzero element has an inverse; $0$ has none.
>
> **Binary operation (closure):** For any $a,b\in\{0,1,2,3,4\}$, $ab\bmod5\in\{0,1,2,3,4\}$ by definition of remainder mod $5$. So $*$ is a well-defined binary operation on the set.
>
> **Commutative:** $a*b=ab\bmod5=ba\bmod5=b*a$, since ordinary multiplication of integers is commutative. So $*$ is commutative.
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> **Associative:** $(a*b)*c=(ab\bmod5)*c\equiv(ab)c\equiv a(bc)\pmod5\equiv a*(b*c)$, since modular arithmetic respects multiplication (working mod $5$ throughout doesn't change the final remainder of a triple product). So $*$ is associative.
>
> **Identity:** We need $e$ with $a*e=a$ for all $a$, i.e. $ae\equiv a\pmod5$. Take $e=1$: $a	imes1=a$ for every $a\in\{0,\dots,4\}$. So the identity element is $e=1$.
>
> **Inverses:** need $b$ with $a*b=1$, i.e. $ab\equiv1\pmod5$:
> …

Q.Show that the operation $$ defined on $\{0,1,2,3,4\}$ by $ab=a\times b\pmod5$ is a binary operation. Test whether it is associative and commutative. Test whether the identity exists. If it exists, investigate about the inverse for each element.

Concept understanding — Binary Operations

Binary Operations

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