Question

Showthat\({{\rm{Z}}_m}\)with multiplication modulo m, where\({\rm{m}} \ge {\rm{2}}\)is an integer, satisfies the closure, associative, and commutativity properties, and\({\rm{1}}\)is a multiplicative identity.

Short answer

The multiplication modulo m satisfies the closure, associative and commutative properties and \(1\)is a multiplicative identity.

Step-by-step solution

Step 1

DEFINITIONS

Theorem \(5\)- Let m be a positive integer. If \(a \equiv b(\bmod {\rm{ m)and c}} \equiv {\rm{d(mod m)}}\), then \(a + c \equiv b + d(\bmod {\rm{ m)}}\)and \({\rm{ac}} \equiv b{\rm{d(mod m)}}\).

Corollary \({\rm{2}}\)- Let m be a positive integer and let a and b be integers.

\(\begin{array}{l}{\rm{(a + b)mod m = ((a mod m) + (b mod m)mod m}}\\{\rm{(ab)mod m = ((a mod m)(b mod m))mod m}}\end{array}\)

CLOSURE:

Given: \({\rm{a}} \in {{\rm{Z}}_m}{\rm{ and b}} \in {{\rm{Z}}_m}{\rm{ and m}} \ge {\rm{2}}\)

To proof: \({\rm{a}} \cdot {}_m{\rm{b}} \in {{\rm{Z}}_m}\)

PROOF

By definition of the multiplication modulo m:

\({\rm{a}} \cdot {}_m{\rm{b = (a}} \cdot {\rm{b)mod m = ab mod m}}\)

Since \({\rm{ab mod m}} \in {{\rm{Z}}_m}\), we then also known that \({\rm{a}} \cdot {}_m{\rm{b}} \in {{\rm{Z}}_m}\).

Step 2

ASSOCIATIVE

Given: \({\rm{a}} \in {{\rm{Z}}_m}\)and \({\rm{b}} \in {{\rm{Z}}_m}\) and \({\rm{m}} \ge {\rm{2}}\).

To proof: \({\rm{(a}} \cdot {}_m{\rm{b)}} \cdot {}_mc = a \cdot {}_m(b \cdot {}_mc)\)

PROOF-

By definition of the multiplication modulo m:

\({\rm{(a}} \cdot {}_mb) \cdot {}_mc\)

\(\begin{array}{l} = ((a \cdot b)\bmod {\rm{ m)}} \cdot {}_mc\\ = (((a \cdot b)\bmod {\rm{ m)}} \cdot {\rm{c)mod m}}\end{array}\)

By corollary \({\rm{(a}} \cdot {}_mb) \cdot {}_mc2\):

\( = ((a \cdot b) \cdot c)\bmod {\rm{ m}}\)

Using the associative property of multiplication in Z:

\( = (a \cdot (b \cdot c))\bmod {\rm{ m}}\)

By corollary \(2\):

\( = (a \cdot (b \cdot c{\rm{ }}\bmod {\rm{ m))mod m}}\)

By definition of the multiplication modulo m:

\( = (a \cdot {}_m(b \cdot c{\rm{ }}\bmod {\rm{ m))}}\)

\( = a \cdot {}_m(b \cdot {}_mc))\).

Step 3

COMMUTATIVE

Given: \({\rm{a}} \in {{\rm{Z}}_m}\)and \(b \in {{\rm{Z}}_m}\) and \({\rm{m}} \ge {\rm{2}}\).

To proof: \({\rm{a}} \cdot {}_mb = b \cdot {}_ma\)

PROOF

By definition of the multiplication modulo m:

\({\rm{a}} \cdot {}_mb\)

\( = (a \cdot b)\bmod {\rm{ m}}\)

Using the commutative property of multiplication in Z:

\( = (b \cdot a)\bmod {\rm{ m}}\)

By definition of the multiplication modulo m:

\( = b \cdot {}_ma\)

Step 4

MULTIPLICATIVE IDENTITY

Given: \({\rm{a}} \in {{\rm{Z}}_m}\)and \({\rm{m}} \ge {\rm{2}}\).

To proof: \({\rm{a}} \cdot {}_m1 = 1 \cdot {}_ma = a\)

PROOF

By definition of the multiplication modulo m:

\({\rm{a}} \cdot {}_m1 = (a \cdot 1)mod{\rm{ m}} = a\bmod {\rm{ m = a}}\)

Note: The last equality holds, because \({\rm{a}} \in {{\rm{Z}}_m}\).

Thus we have shown that \(1\)is the identity element for multiplication modulo m.