The question I am needing to answer is highlighted in yellow. Everything italicized is what I have come up with. I would like to make sure that I am on the right track/doing it right or if there are additional explanations/items missing from my answer.

B. Given the set of integers mod m denoted Z. the elements of Z. are denoted [x]., where x is an integer from 0 to m - 1. Each element [x]. is an equivalence class of integers that has the same integer remainder as x when divided by m. Consider, for example, Zz = {[0]7, [1],, [217, [3]7, [417, [517, [6],]. The element [5], represents the infinite set of integers of the form 5 plus an integer multiple of 7. That is, [5],= {. ..-9, -2, 5, 12, 19, 26, . ..], or, more formally, [5], = (y: y = 5 + 7q for some integer q]. Modular addition, +, is defined on the set Zo in terms of integer addition as follows: [alm + [b]m = [a + bl.. Modular multiplication, *, is defined on the set Z. in terms of integer multiplication as follows: [a]m * [b]m = [a * b]m. The set Za with these operations is a ring for any integer m. 1. Choose a two-digit even integer for m and, using a counterexample, explain why Zo with the operations of + and * is not an integral domain. Let Zm = Z12, which is an integral domain if it is a ring for any ab = 0, meaning either a = 0 or b = 0. Let Z12=([0]12, [1]12, [2)12, [3]12, [4]12, [5]12, [6]12, [712, [8]12, [9]12, [10]12, [11]12} Let [al12 E Z12 and let [b]12 E Z12 [al12 * [bl12= [a * bl12 where a=2 and b=6. [2)12 * [6)12 = [2 * 6]12 = [12)12 = [0]12- Since [212 and [6)12 are zero divisors, Z12 is not an integral domain