Question

Compute the following products.

1. ${\left(\left[a\right]\oplus \left[b\right]\right)}^2$in${\mathrm{\mathbb{Z}}}_2$
2. ${\left(\left[a\right]\oplus \left[b\right]\right)}^3$in ${\mathrm{\mathbb{Z}}}_3$ [Hint: Exercise 11(a) may be helpful.]
3. ${\left(\left[a\right]\oplus \left[b\right]\right)}^5$in ${\mathrm{\mathbb{Z}}}_5$ [Hint: See Exercise 11(c).]
4. Based on the results of parts (a)-(c), what do you think${\left(\left[a\right]\oplus \left[b\right]\right)}^7$is equal to in${\mathrm{\mathbb{Z}}}_7$?

Short answer

1. ${\left(\left[a\right]\oplus \left[b\right]\right)}^2={\left[a\right]}^2\oplus {\left[b\right]}^2$
   
2. ${\left(\left[a\right]\oplus \left[b\right]\right)}^3={\left[a\right]}^3\oplus {\left[b\right]}^3$
   
3. role="math" localid="1646398251102" ${\left(\left[a\right]\oplus \left[b\right]\right)}^5={\left[a\right]}^5\oplus {\left[b\right]}^5$
   
4. ${\left(\left[a\right]\oplus \left[b\right]\right)}^7={\left[a\right]}^7\oplus {\left[b\right]}^7$
   

Step-by-step solution

Find the product a⊕b2 in ℤn 

Simplify the given expression asfollows:

$\begin{array}{rcl}{\left(\left[a\right]\oplus \left[b\right]\right)}^2& =& {\left[a\right]}^2\oplus 2\left[a\right]\odot \left[b\right]\oplus {\left[b\right]}^2\\ & =& {\left[a\right]}^2\oplus 2\left[ab\right]\oplus {\left[b\right]}^2\\ & =& {\left[a\right]}^2\oplus 0\oplus {\left[b\right]}^2\\ & =& {\left[a\right]}^2\oplus {\left[b\right]}^2\end{array}$

Thus,${\left(\left[a\right]\oplus \left[b\right]\right)}^2={\left[a\right]}^2\oplus {\left[b\right]}^2$ .

Find the product  a⊕b3 in ℤ3

Simplify the given expression as follows:$\begin{array}{rcl}{\left(\left[a\right]\oplus \left[b\right]\right)}^3& =& {\left[a\right]}^3\oplus {\left[b\right]}^3\oplus 3{\left[a\right]}^2\odot \left[b\right]\oplus 3\left[a\right]\odot {\left[b\right]}^2\\ & =& {\left[a\right]}^3\oplus {\left[b\right]}^3\oplus 3\left[a^{2}b\right]\oplus 3\left[a{b}^2\right]\\ & =& {\left[a\right]}^3\oplus {\left[b\right]}^3\oplus 0\oplus 0\\ & =& {\left[a\right]}^3\oplus {\left[b\right]}^3\\ & & \end{array}$

Thus,${\left(\left[a\right]\oplus \left[b\right]\right)}^3={\left[a\right]}^3\oplus {\left[b\right]}^3$

Find the product a⊕b5 in ℤ5

Simplify the given expression as follows:

$\begin{array}{rcl}{\left(\left[a\right]\oplus \left[b\right]\right)}^5& =& {\left[a\right]}^5\oplus {\left[b\right]}^5\oplus 5\left({\left[a\right]}^4\odot \left[b\right]\oplus 2{\left[a\right]}^3\odot {\left[b\right]}^2\oplus 2{\left[a\right]}^2\odot {\left[b\right]}^3\oplus \left[a\right]4\odot {\left[b\right]}^4\right)\\ & =& {\left[a\right]}^5\oplus {\left[b\right]}^5\oplus 0\\ & =& {\left[a\right]}^5\oplus {\left[b\right]}^5\end{array}$

Thus, $\begin{array}{rcl}{\left(\left[a\right]\oplus \left[b\right]\right)}^5& =& {\left[a\right]}^5\oplus {\left[b\right]}^5\end{array}$

Find the product  a+b7 in ℤ7

Based on (a)and(c), we can say that$\begin{array}{rcl}{\left(\left[a\right]\oplus \left[b\right]\right)}^7& =& {\left[a\right]}^7\oplus {\left[b\right]}^7\end{array}$ in ${\mathrm{\mathbb{Z}}}_7$.