Question

Prove or disprove: If$\left[a\right]\odot \left[b\right]=\left[a\right]\odot \left[c\right]$and$\left[a\right]\ne \left[0\right]$in${\mathrm{\mathbb{Z}}}_n$, then $\left[b\right]=\left[c\right]$.

Short answer

The given result is false. It is proved that when$\left[a\right]\odot \left[b\right]=\left[a\right]\odot \left[c\right]$ and$\left[a\right]\ne \left[0\right]$ in ${\mathrm{\mathbb{Z}}}_n$, then $\left[b\right]=\left[c\right]$.

Step-by-step solution

Definition of multiplication ℤn

According to the definition of multiplication in ${\mathrm{\mathbb{Z}}}_n$, we can say that role="math" localid="1646390066621" width="109" height="21">a⊙c=ac
.

Prove or disprove: If a⊙b=a⊙c and a≠0 in ℤn, then b=c 

Let us prove or disprove the result by considering the example.

For example, assume that$\left[a\right]=\left[6\right],\left[b\right]=\left[4\right]$and$\left[c\right]=\left[6\right]$ belong in ${\mathrm{\mathbb{Z}}}_n$. Then we have

.role="math" localid="1646391031301" $\begin{array}{rcl}\left[6\right]\odot \left[4\right]& =& \left[6\right]\odot \left[6\right]\\ \left[6·6\right]& =& \left[6·4\right]\\ \left[0\right]& =& \left[0\right]\\ & & \end{array}$

This implies that $\left[6\right]\odot \left[4\right]=\left[6\right]\odot \left[6\right]$ in ${\mathrm{\mathbb{Z}}}_n$.

We can observe that$\left[6\right]\ne \left[0\right]$ but$\left[4\right]\ne \left[6\right]$ in _{${\mathrm{\mathbb{Z}}}_n$}. Then the given result is false.

Hence,it is proved that when$\left[a\right]\odot \left[b\right]=\left[a\right]\odot \left[c\right]$ and$\left[a\right]\ne \left[0\right]$ in ${\mathrm{\mathbb{Z}}}_n$, then .