# Which elements of the ring #ZZ_14# are invertible?

##### 1 Answer

#### Explanation:

**Ring**

A *ring* is a set

#R# is an abelian group under addition:

#AA a, b, c in R, (a+b)+c = a+(b+c)" "# (associativity)

#AA a, b in R, a+b = b+a" "# (commutativity)

#EE 0 in R : AA a in R, a + 0 = 0 + a = a" "# (identity)

#AA a in R, EE (-a) in R : a + (-a) = (-a) + a = 0" "# (inverse)

- The non-zero elements of
#R# are a monoid under multiplication:

#AA a, b, c in R "\" {0}, (axxb)xxc = axx(bxxc)" "# (associativity)

#EE 1 in R : AA a in R "\" {0}, a xx 1 = 1 xx a = a" "# (identity)

- Multiplication is (left and right) distributive over addition:

#AA a, b, c in R, a xx (b+c) = (a xx b) + (a xx c)" "# (left distributivity)

#AA a, b, c in R, (a+b) xx c = (a xx c) + (b xx c)" "# (right distributivity)

**Invertible elements and zero divisors**

An element *invertible* if there is some element

#a xx a^(-1) = a^(-1) xx a = 1#

An element *zero divisor* if there is some element

#a xx b = 0#

Note that any zero divisor is not invertible:

Suppose

Then:

#b = 1 xx b = (a^(-1) xx a) xx b = a^(-1) xx (a xx b) = a^(-1) xx 0 = 0#

Any multiple of a zero divisor is also a zero divisor:

If

#AA c in R, (c xx a) xx b = c xx (a xx b) = c xx 0 = 0#

Addition and multiplication of integers modulo

- Multiplication is commutative:

#AA a, b in R, a xx b = b xx a" "# (commutativity)

This property makes *commutative ring*

However, note that

#2 xx 7 -= 0" "# modulo#14#

So we have seen that

Hence the following numbers are not invertible in

#0, 2, 4, 6, 7, 8, 10, 12#

That leaves:

#1, 3, 5, 9, 11, 13#

To confirm, note that:

#3 xx 5 = 15 -= 1" "# modulo#14#

#9 xx 11 = 99 -= 1" "# modulo#14#

#13 xx 13 = 169 -= 1" "# modulo#14#

So these elements are all invertible.