§2.3 Exercises

Exercise 2.3.1

Find all the units in:

• (a) \(\mathbb{Z}_7\)
• (b) \(\mathbb{Z}_8\)
• (c) \(\mathbb{Z}_9\)
• (d) \(\mathbb{Z}_{10}\)

Exercise 2.3.2

Find all the zero divisors in:

• (a) \(\mathbb{Z}_7\)
• (b) \(\mathbb{Z}_8\)
• (c) \(\mathbb{Z}_9\)
• (d) \(\mathbb{Z}_{10}\)

Exercise 2.3.3

Based on Exercises 1 and 2, make a conjecture about units and zero divisors in \(\mathbb{Z}_n\).

Exercise 2.3.4

How many solutions does the equation \(6x = 4\) have in:

• (a) \(\mathbb{Z}_7\)?
• (b) \(\mathbb{Z}_8\)?
• (c) \(\mathbb{Z}_9\)?
• (d) \(\mathbb{Z}_{10}\)?

Exercise 2.3.5

If \(a\) is a unit and \(b\) is a zero divisor in \(\mathbb{Z}_n\), show that \(ab\) is a zero divisor.

Exercise 2.3.6

If \(n\) is composite, prove that there is at least one zero divisor in \(\mathbb{Z}_n\). (See Exercise 2.)

Exercise 2.3.7

Without using Theorem 2.8, prove that if \(p\) is prime and \(ab = 0\) in \(\mathbb{Z}_p\), then \(a = 0\) or \(b = 0\). Hint: Theorem 1.8.

Exercise 2.3.8

(a) Give three examples of equations of the form \(ax = b\) in \(\mathbb{Z}_{12}\) that have no nonzero solutions.

(b) For each of the equations in part (a), does the equation \(ax = 0\) have a nonzero solution?

Exercise 2.3.9

(a) If \(a\) is a unit in \(\mathbb{Z}_n\), prove that \(a\) is not a zero divisor.

(b) If \(a\) is a zero divisor in \(\mathbb{Z}_n\), prove that \(a\) is not a unit. Hint: Think contrapositive in part (a).

Exercise 2.3.10

Prove that every nonzero element of \(\mathbb{Z}_n\) is either a unit or a zero divisor, but not both. Hint: Exercise 9 provides the proof of "not both."

Exercise 2.3.11

Without using Exercises 13 and 14, prove: If \(a, b \in \mathbb{Z}_n\) and \(a\) is a unit, then the equation \(ax = b\) has a unique solution in \(\mathbb{Z}_n\). Note: You must find a solution for the equation and show that this solution is the only one.

Exercise 2.3.12

Let \(a, b, n\) be integers with \(n > 1\) and let \(d = (a, n)\). If the equation \([a]x = [b]\) has a solution in \(\mathbb{Z}_n\), prove that \(d \mid b\). Hint: If \(x = [r]\) is a solution, then \([ar] = [b]\) so that \(ar - b = kn\) for some integer \(k\).

Exercise 2.3.13

Let \(a, b, n\) be integers with \(n > 1\). Let \(d = (a, n)\) and assume \(d \mid b\). Prove that the equation \([a]x = [b]\) has a solution in \(\mathbb{Z}_n\) as follows:

• (a) Explain why there are integers \(u, v, a_1, b_1, n_1\) such that \(au + nv = d\), \(a = da_1\), \(b = db_1\), \(n = dn_1\).
• (b) Show that each of \[ [ub], [ub + n_1], [ub + 2n_1], [ub + 3n_1], \dots, [ub + (d - 1)n_1] \] is a solution of \([a]x = [b]\).

Exercise 2.3.14

Let \(a, b, n\) be integers with \(n > 1\). Let \(d = (a, n)\) and assume \(d \mid b\). Prove that the equation \([a]x = [b]\) has distinct solutions in \(\mathbb{Z}_n\) as follows:

• (a) Show that the solutions listed in Exercise 13 (b) are all distinct. Hint: \([r] = [s]\) if and only if \(n_1 \mid (r - ub_1)\).
• (b) If \(x = [r]\) is any solution of \([a]x = [b]\), show that \([r] = [ub_1 + kn_1]\) for some integer \(k\) with \(0 \leq k \leq d - 1\). Hint: \([ar] = [aub_1] = [0]\) (Why?), so that \(n_1(a(r - ub_1))\). Show that \(n_1 \mid (a(r - ub_1))\) and use Theorem 1.4 to show that \(n_1 \mid (r - ub_1)\).

Exercise 2.3.15

Use Exercise 13 to solve the following equations:

• (a) \(15x = 9\) in \(\mathbb{Z}_8\)
• (b) \(25x = 10\) in \(\mathbb{Z}_{65}\)

Exercise 2.3.16

If \(a \neq 0\) and \(b\) are elements of \(\mathbb{Z}_n\) and \(ax = b\) has no solutions in \(\mathbb{Z}_n\), prove that \(a\) is a zero divisor.

Exercise 2.3.17

Prove that the product of two units in \(\mathbb{Z}_n\) is also a unit.

Exercise 2.3.18

The usual ordering of \(\mathbb{Z}\) by \(<\) is transitive and behaves nicely with respect to addition. Show that there is no ordering of \(\mathbb{Z}_n\) such that:

• (i) if \(a < b\) and \(b < c\), then \(a < c\);
• (ii) if \(a < b\), then \(a + c < b + c\) for every \(c\) in \(\mathbb{Z}_n\).

Hint: If there is such an ordering with \(0 < 1\), then adding 1 repeatedly to both sides shows that \(0 < 1 < 2 < \cdots < n - 1\) by (ii). Thus \(0 < n=0\), a contradiction. Make a similar argument when \(1 < 0\).