Mathematical-Reasoning · Unit 29 · Video 4 · Interactive Practice

Zero Divisors in ℤ₆: When the Zero Product Property Fails

IKey Formulas

Formula	Meaning
$\forall x, y \in \mathbb{Z},\ xy = 0 \rightarrow (x = 0 \vee y = 0)$	the zero product property in the integers
$\forall x, y \in \mathbb{Z},\ (x \ne 0 \wedge y \ne 0) \rightarrow xy \ne 0$	its contrapositive: a product of nonzero integers is nonzero
$[x] = [0]$ in $\mathbb{Z}_n \iff n \mid x$	when a class is the zero class
$[2] \odot [3] = [6] = [0]$ in $\mathbb{Z}_6$	a counterexample: two nonzero classes with product $[0]$

Key Insight: A property of the integers does not automatically carry over to $\mathbb{Z}_n$ ; for a small modulus the multiplication table settles it. In $\mathbb{Z}_6$ , $6 \mid 2 \cdot 3$ although $6$ divides neither $2$ nor $3$ .

IIThe Nonzero Block of ℤₙ

The zero product property holds in $\mathbb{Z}_n$ exactly when $[0]$ never appears among products of nonzero classes.

n 6

IIIFrom the Property to Its Contrapositive

The contrapositive tells us exactly which table entries to inspect.

Step 1

Step 1 — The zero product property

\forall x, y \in \mathbb{Z}, \quad xy = 0 \ \rightarrow\ (x = 0 \ \vee\ y = 0)

Step 2 — Negate the conclusion (De Morgan: $\vee$ becomes $\wedge$ )

\neg(x = 0 \ \vee\ y = 0) \ \equiv\ x \ne 0 \ \wedge\ y \ne 0

Step 3 — Negate the hypothesis

\neg(xy = 0) \ \equiv\ xy \ne 0

Step 4 — The contrapositive

\forall x, y \in \mathbb{Z}, \quad (x \ne 0 \ \wedge\ y \ne 0) \ \rightarrow\ xy \ne 0

Step 5 — The same statement in $\mathbb{Z}_6$ fails

[2] \ne [0], \quad [3] \ne [0], \quad [2] \odot [3] = [6] = [0]

IVQuiz Questions

Problem 1 · A Product in ℤ₆

In $\mathbb{Z}_6$ , compute $[5] \odot [2]$ and write it as $[r]$ with $0 \le r \le 5$ .

✅ Correct!

5 \cdot 2 = 10 = 6 \cdot 1 + 4

❌ Not quite.

10

is not a multiple of

6

, so the product is not

[0]

❌ Not quite.

[5] \oplus [2] = [7] = [1]

is the sum;

\odot

multiplies.

❌ Not quite. Multiply

5 \cdot 2

, then keep the remainder on division by

6

Show solution

[5] \odot [2] = [10], \qquad 10 = 6 \cdot 1 + 4 \ \Rightarrow\ [5] \odot [2] = [4].

Problem 2 · Writing the Contrapositive

Which statement is the contrapositive of: for all integers $x$ and $y$ , if $xy = 0$ , then $x = 0$ or $y = 0$ ?

✅ Correct! Negate both parts and swap them; by De Morgan, the negation of "

x = 0

y = 0

" is "

x \ne 0

and

y \ne 0

❌ Close, but… negating an "or" gives an "and" (De Morgan's law). This statement is false:

x = 0

y = 5

gives

xy = 0

❌ Not quite. This negates the parts without swapping them — the inverse, which is equivalent to the converse, not to the original.

❌ Not quite. This runs from the conclusion back to the hypothesis without negating anything, so it is a form of the converse. The contrapositive negates both parts.

Show solution

The contrapositive of $P \rightarrow Q$ is $\neg Q \rightarrow \neg P$ . Here $P$ is $xy = 0$ and $Q$ is $x = 0 \vee y = 0$ , so

\neg Q:\ x \ne 0 \wedge y \ne 0, \qquad \neg P:\ xy \ne 0.

The contrapositive reads: for all integers $x$ and $y$ , if $x \ne 0$ and $y \ne 0$ , then $xy \ne 0$ . It is logically equivalent to the original.

Problem 3 · Which ℤₙ Keeps the Property?

In which of these does the zero product property hold for all classes?

✅ Correct! No product

xy

with

1 \le x, y \le 6

is a multiple of

7

, so

[0]

never appears in the nonzero block.

❌ Not quite.

[2] \odot [2] = [4] = [0]

\mathbb{Z}_4

❌ Not quite.

[2] \odot [4] = [8] = [0]

\mathbb{Z}_8

❌ Not quite.

[3] \odot [3] = [9] = [0]

\mathbb{Z}_9

Show solution

Each composite modulus gives a counterexample: $2 \cdot 2 = 4$ , $2 \cdot 4 = 8$ , $3 \cdot 3 = 9$ . For $\mathbb{Z}_7$ , checking the $36$ products $xy$ with $1 \le x, y \le 6$ shows none is divisible by $7$ , so the nonzero block of the multiplication table contains no $[0]$ and the contrapositive — hence the property — holds.

Problem 4 · Partners of [5] in ℤ₁₀

In $\mathbb{Z}_{10}$ , which nonzero classes $[y]$ satisfy $[5] \odot [y] = [0]$ ?

✅ Correct!

10 \mid 5y

exactly when

y

is even, giving

[2], [4], [6], [8]

among the nonzero classes.

❌ Not quite.

[2]

works, but so does

[4]

5 \cdot 4 = 20 \equiv 0 \pmod{10}

❌ Not quite.

5 \cdot 5 = 25 = 10 \cdot 2 + 5

, so

[5] \odot [5] = [5]

❌ Not quite.

5 \cdot 2 = 10

, so

[5] \odot [2] = [0]

Show solution

$[5] \odot [y] = [5y]$ , and $[5y] = [0]$ exactly when $10 \mid 5y$ , that is, when $2 \mid y$ . Among $y = 1, \ldots, 9$ :

5 \cdot 2 = 10, \quad 5 \cdot 4 = 20, \quad 5 \cdot 6 = 30, \quad 5 \cdot 8 = 40

are multiples of $10$ , while $5y$ for odd $y$ leaves remainder $5$ . So $[y] \in \{[2], [4], [6], [8]\}$ .

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