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Home , Additive identity, Additive inverse, Identity element, Zero element

(c)Karen E. Smith 2017 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License.

Math 412. Worksheet on Properties of Zn. Professor Karen E. Smith

NOTATION: The notation ZN denotes the of congruence classes modulo N.

Recall that there is a natural on Zn defined by [a]N + [b]N = [a + b]N , and a natural on Zn defined by [a]N × [b]N = [ab]N .

These two operations (addition and multiplication) give the set Zn the structure of a commuta- tive . This means that

(1) The addition is associative, commutative, has an [0]N , and every ele- ment of ZN has an . (2) The multiplication is associative, commutative, has an [1]N . (3) Multiplication distributes over addition: [a]N ([b]N + [c]N ) = [a]N [b]N + [a]N [c]N for all a, b, c, N.

THEOREM: The class [a]N has a in ZN if and only if gcd(a, N) = 1.

A.WARM-UP (1) Discuss the addition and multiplication with your team so that everyone understands. Remember, the elements of ZN are sets so it is non-trivial! (2) What is the element of Z6? Explain and verify. (3) What is the multiplicative identity for Z6 ? Explain and verify. (4) Compute [7]11 + [4]11. What is the additive inverse of [4]11? (5) Compute [4]11 × [3]11. What is the multiplicative inverse of [3]11. (6) In Z6, verify the of addition in a special case by computing ([63]6 + [−184]6) + [4]6 and [63]6 + ([−184]6 + [4]6). (7) Verify the in Z6 in a special case by computing both [6001]6([49]6 + [−18]6) and [6001]6[49]6 + [6001]6[−18]6. (8) Compute [−60052]6 + [4]6; now find the additive inverse of [−60052]6. (9) Compute [688]15 × [37]15. Now find the multiplicative inverse of [688]15.

(1) To add two congruence classes [a]N and [b]N , we take any element of each, and add those. The sum class is the congruence class containing that . Since a is the most obvious element of [a]N and b is the most obvious element of [b]N , we can write the formula [a]N + [b]N = [a + b]N . We can also define the sum as the set of all x + y where x ∈ [a]N and y ∈ [b]N . Similar for multiplication, using times instead of plus. (2) The additive identity (or ZERO) is [0]6, since [0]6 + [a]6 = [0 + a]6 = [a]6 for all [a]6 ∈ Z6. (3) The multiplicative identity is [1]6 since [1]6 × [a]6 = [1 × a]6 = [a]6 for all l [a]6 ∈ Z6. (4) [7]11 + [4]11 = 0, so the additive inverse of [4]11 is [7]11. (5) [4]11 × [3]11 = [1]11, so the multiplicative inverse of [3]11 is [4]11. (6) Easy: ([3] + [−4]) + [4] = [−1] + [4] = 3 and [3] + ([−4]) + [4]) = [3] as well. (7) Easy: [1]6([1]6 + [0]6) = [1] and [1]6[1]6 + [1]6[0]6 = [1]. (8) [−60052]6 + [4]6 = [−60048]6 = [0]; by definition, the additive inverse of [−60052]6 is [4]. (9) Compute [688]15 ×[37]15 = [88]15 ×[7]15 = [−2]15 ×[7]15 = [−14]15 = [1]15. Sothe multiplicative inverse of [688]15 is [37]15 or [7]15.

B.EASY PROOFSAND COMPUTATIONS (1) Prove that in Zn, the additive inverse of [a]n is [−a]n. (2) How can we define in Zn using (1)? Is this the same as defining [a]N −[b]N = [a − b]N ? (3) Which elements of Z6 have multiplicative inverses? For each that does, find it. (4) Explain the relationship between the statements “[a]x = 1 has a solution in Zn” and “[a] has a multiplicative inverse in Zn”.

(1) Check: [a]n + [−a]n = [a + −a]n = [0]n, which is the of Zn. By definition, we just showed that −[a] (a notation for the additive inverse of [a]) is [−a]. (2) [a]n − [b]n can be interpreted as [a] PLUS the additive inverse of [b]. So [a]n − [b]n = [a] + [−b] = [a − b] (I am dropping the subscript n when it is supposed to be clear from context..) (3) We can test each of the six elements [0], [1], [2], [3], [4].[5]. Using the Theorem, only [1]6 and [5]6 have multiplicative inverses. The multiplicative inverse of [1] is [1]. The multiplicative inverse of [5] is easier to find if we represent this class by [−1]. Then it’s easy: since [−1][−1] = [1], we see that [−1] is its own inverse. (4) The two statements are equivalent. Yet another equivalent statement is “[a] is a in Zn”.

C.SOLVE THE FOLLOWING EQUATIONS FOR x IN Z7: (1) [5]x = 1. (2) [6]x = 1. Solve this one by finding a convenient representative for the class [6]. (3) [12]x + [703] = [4]

(1) x = [3] (2) x = [6] (3) x = [3]

D.USE THE FACT THAT 688 × 37 − 1697 × 15 = 1 to find: (1) The multiplicative inverse of [1697] in Z688. (2) A solution to [1697]x = 1 in Z688. (3) A solution to [1697]x = 2 in Z688. (4) The multiplicative inverse of [1697] in Z37. (5) A solution to [1697]x + [4] = [3] in Z37.

(1) We can write the equation as 688 × 37 = 1697 × (−15) − 1, which means that 1697 × (−15) =∼ 1 mod 688, so [1697] × [−15)] = [1] in Z688. So the mutiplicative inverse of [1697] in Z688 is [−15]688. (2) x = [−15]688. (3) Multiply both sides by the inverse of [1697] to fet x = [−30]. (4) The inverse of [1697] in Z37 is [−15]37, which is a completely different set than [−15]688 (though they both contain -15). (5) x = [15]37.

E.PROOFOFTHE THEOREM: The class [a]N has a multiplicative inverse in ZN if and only if gcd(a, N) = 1. We will use the Theorem 1.2 from last week: For a, b ∈ Z, the gcd (a, b) is the smallest positive integer that is a Z- of a and b. (1) Suppose that ab + Ny = 1. Prove that [a][b] = [1] in ZN . (2) Use (2) and Theorem 1.2 to prove that if gcd(a, N) = 1, then [a]N has a multiplicative inverse in ZN . (3) Suppose that [b] is the multiplicative inverse of [a] in ZN . Show that 1 is a linear combination of a and N. (4) Show that if [a] has a multiplicative inverse in ZN , then gcd(a, N) = 1. [Hint: translate a statement about classes in ZN to a statement about integers; then use Theorem 1.2] (5) Prove the Theorem.

See the proof of Theorem 2.9 on page 40.

F. FACT: 863 is prime. (1) To verify 863 is prime, explain why we only need to check whether any prime less than 30 divide it. (2) Does the class [430] in Z863 have a multiplicative inverse? Explain without computing it. (3) What is the smallest positive integer you can write as a Z-linear combination of 863 and 430? Explain without computing anything. 1 (4) Find explicit x, y in Z so that 430x + 863y = 1. (5) Compute the multiplicative inverse of [430] in Z863. (6) Solve for x in the ring Z863 : [430]x − [860] = [4]. √ (1) We can check n√is prime√ by checking that all prime less that n are not factors (Theorem 1.10 in the text). Here 863 < 900 = 30. (2) Yes, all non-zero elements do. Because we can represent each element of Z863 by [a]863 where 0 ≤ a < 863 (division algorithm). Note that (a, 863) = 1 except when a = 0. (3) Because 863 is prime, (430, 863) = 1. So 1 is the smallest positive integer combination of 430 and 863. (4) We use the Euclidean algorithm: 863 = 2 × (430) + 3. Then 430 = 143 × 3 + 1. So 1 = 430 − 143 × 3 = 430 − 143(863 − 2 × 430) = (−143)(863) + (287)(430). (5) The multiplicative inverse of [430] is [287] in Z863. (6) [430]x − [860] = [4] is [430]x + [−860] = [4], which is [430]x + [3] = [4]. Subtracting [3] from both sides: [430]x = [1]. So the answer is [287] in Z863.

k BONUS: Suppose there exists non-zero [x] ∈ Zn is such that [x] = [0] for some k > 0. What can you say about the decomposition of n into its prime factors?

1Hint: Use the Euclidean algorithm backwards, as we did on a worksheet the first week of class.