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Chapter 14: Problem 12

Find the multiplicative inverse of each element in \(\mathbf{Z}_{11}, \mathbf{Z}_{13}\), and \(\mathbf{Z}_{17}\).

Short Answer

Expert verified
The multiplicative inverses for each element in the respective sets are as follows: In \mathbf{Z}_{11},: 1 ↔ 1, 2 ↔ 6, 3 ↔ 4, 4 ↔ 3, 5 ↔ 9, 6 ↔ 2, 7 ↔ 8, 8 ↔ 7, 9 ↔ 5, 10 ↔ 10. In \mathbf{Z}_{13},: 1 ↔ 1, 2 ↔ 7, 3 ↔ 9, 4 ↔ 10, 5 ↔ 8, 6 ↔ 11, 7 ↔ 2, 8 ↔ 5, 9 ↔ 3, 10 ↔ 4, 11 ↔ 6, 12 ↔ 12. Finally, in \mathbf{Z}_{17},: 1 ↔ 1, 2 ↔ 9, 3 ↔ 6, 4 ↔ 13, 5 ↔ 7, 6 ↔ 3, 7 ↔ 5, 8 ↔ 15, 9 ↔ 2, 10 ↔ 12, 11 ↔ 16, 12 ↔ 10, 13 ↔ 4, 14 ↔ 11, 15 ↔ 8, 16 ↔ 14.

Step by step solution

01

Find the Numbers for Each Set

First, identify the elements within each set. The sets are: \(\mathbf{Z}_{11}, \mathbf{Z}_{13},\text{ and } \mathbf{Z}_{17}\). These are finite integer sets, from 1 to \(n-1\), where \(n\) is the subscript value. Therefore, \(\mathbf{Z}_{11} = \{1, 2, 3, ..., 10\}, \mathbf{Z}_{13} = \{1, 2, 3, ..., 12\}, \text{ and } \mathbf{Z}_{17} = \{1, 2, 3, ...,16\}\).
02

Identify Coprime Numbers

We need to identify the numbers which are coprime to the modulus. In these cases, since 11, 13 and 17 are prime numbers, all numbers less than these are co-prime to them. Thus all numbers in each set satisfy this condition.
03

Find Multiplicative Inverse for Each Number in \(\mathbf{Z}_{11}\)

We apply the concept of multiplicative inverse for each number \(i\) in the formed set \(\mathbf{Z}_{11} = \{1, 2, 3, ..., 10\}\), and find \(j\) such that \((i \times j) \% 11 = 1\). Doing so, we obtain the equivalence: 1 ↔ 1, 2 ↔ 6, 3 ↔ 4, 4 ↔ 3, 5 ↔ 9, 6 ↔ 2, 7 ↔ 8, 8 ↔ 7, 9 ↔ 5, 10 ↔ 10.
04

Find Multiplicative Inverse for Each Number in \(\mathbf{Z}_{13}\)

Now we find the multiplicative inverse for each number \(i\) in the set \(\mathbf{Z}_{13} = \{1, 2, 3, ..., 12\}\), and identify \(j\) such that \((i \times j) \% 13 = 1\). We obtain the equivalence: 1 ↔ 1, 2 ↔ 7, 3 ↔ 9, 4 ↔ 10, 5 ↔ 8, 6 ↔ 11, 7 ↔ 2, 8 ↔ 5, 9 ↔ 3, 10 ↔ 4, 11 ↔ 6, 12 ↔ 12.
05

Find Multiplicative Inverse for Each Number in \(\mathbf{Z}_{17}\)

Lastly, for each number \(i\) in the set \(\mathbf{Z}_{17} = \{1, 2, 3,...,16\}\), we find \(j\) such that \((i \times j) \% 17 = 1\). We get: 1 ↔ 1, 2 ↔ 9, 3 ↔ 6, 4 ↔ 13, 5 ↔ 7, 6 ↔ 3, 7 ↔ 5, 8 ↔ 15, 9 ↔ 2, 10 ↔ 12, 11 ↔ 16, 12 ↔ 10, 13 ↔ 4, 14 ↔ 11, 15 ↔ 8, 16 ↔ 14.

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Most popular questions from this chapter

\(R\) is a field, how many ideals does \(R\) have?

Prove that for all integers \(n\) exactly one of \(n, 2 n-1\), and \(2 n+1\) is divisible by 3 .

a) For ring \((R,+, \cdot)\) and each \(a \in R\), we define \(a^{1}=a\), and \(a^{n+1}=a^{n} a\), for all \(n \in \mathbf{Z}^{+}\). Prove that for all \(m, n\) \(\in \mathbf{Z}^{+},\left(a^{m}\right)\left(a^{n}\right)=a^{m+n}\) and \(\left(a^{m}\right)^{n}=a^{m n} .\) b) Can you suggest how we might define \(a^{0}\) or \(a^{-n}, n\) \(\in \mathbf{Z}^{+}\), including any necessary conditions that \(R\) must satisfy for these definitions to make sense?

If \(a, b\), and \(c\) are any elements in a ring \((R,+, \cdot)\), prove that (a) \(a(b-c)=a b-(a c)=a b-a c\) and (b) \((b-c) a=\) \(b a-(c a)=b a-c a\).

Prove that in any list of \(n\) consecutive integers, one of the integers is divisible by \(n\).
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