Mathematical-Reasoning · Unit 7 · Video 2 · Interactive Practice

Writing a Proof by Contrapositive: ab = 0 and De Morgan

IKey Formulas

Formula	Name	Role in the proofs
$P \to Q \equiv \neg Q \to \neg P$	Contrapositive	Assume $\neg Q$ and deduce $\neg P$ : a direct proof of the contrapositive
$\neg(P \wedge Q) \equiv \neg P \vee \neg Q$	De Morgan's Law	$\neg(a \ne 0 \wedge b \ne 0) \equiv (a = 0) \vee (b = 0)$
$X \to (Y \vee Z) \equiv (X \wedge \neg Y) \to Z$	Conditional with a disjunctive conclusion	Supplies the extra assumption $a \ne 0$
$(ab = 0 \wedge a \ne 0) \to b = 0$	The statement actually proved	Multiply both sides of $ab = 0$ by $\tfrac{1}{a}$

Key Insight: To prove a statement, prove a logically equivalent one instead, and say in the first lines which statement you are proving.

IINegating a Conjunction

The negation of $a \ne 0 \wedge b \ne 0$ holds on both axes, not only at the origin.

Shade

IIIRewriting $X \to (Y \vee Z)$

Which rewrites of $X \to (Y \vee Z)$ agree with it on all eight rows of a truth table?

Rewrite

IVWriting the Proof Line by Line

Each line of a proof by contrapositive announces the method or states why the assertion is true.

Proof Line 1

VQuiz Questions

Problem 1 · Form the Contrapositive

Given: for all integers $m$ and $n$ , if $mn$ is even, then $m$ is even or $n$ is even. Which statement is its contrapositive?

✅ Correct! The conclusion is a disjunction, so De Morgan's Law turns its negation into “

m

is odd and

n

is odd”, and the negation of “

mn

is even” is “

mn

is odd”.

❌ Close, but… each part was negated and the “or” was kept. De Morgan's Law gives

\neg(A \vee B) \equiv \neg A \wedge \neg B

, so the “or” must become “and”.

❌ Not quite. This swaps hypothesis and conclusion without negating them: it is the converse, which is not logically equivalent to the original statement.

❌ Not quite. The new hypothesis is right, but the new conclusion must be the negation of the old hypothesis: “

mn

is odd”, not “

mn

is even”.

Show solution

Write the statement as $P \to Q$ with $P$ : “ $mn$ is even” and $Q$ : “ $m$ is even or $n$ is even”. The contrapositive is $\neg Q \to \neg P$ .

By De Morgan's Law, $\neg Q$ is “ $m$ is not even and $n$ is not even”; for integers, not even means odd, so $\neg Q$ is “ $m$ is odd and $n$ is odd”. Likewise $\neg P$ is “ $mn$ is odd”.

The contrapositive: for all integers $m$ and $n$ , if $m$ is odd and $n$ is odd, then $mn$ is odd, the result about products of odd integers proved earlier in the course.

Problem 2 · The Tempting Wrong Negation

Given: a student writes the contrapositive of “for all $a, b \in \mathbb{R}$ , if $a \ne 0$ and $b \ne 0$ , then $ab \ne 0$ ” as “for all $a, b \in \mathbb{R}$ , if $ab = 0$ , then $a = 0$ and $b = 0$ ”. Which pair $(a, b)$ is a counterexample to the student's statement?

✅ Correct! Here

ab = 0

is true but “

a = 0

and

b = 0

” is false, because

b = 5

. The student kept “and” where De Morgan's Law requires “or”.

❌ Not quite. At

(0, 0)

both the hypothesis

ab = 0

and the conclusion “

a = 0

and

b = 0

” are true, so the conditional is true there.

❌ Not quite. For this pair

ab \ne 0

, so the hypothesis is false and the conditional is true. A counterexample needs a true hypothesis and a false conclusion.

Show solution

A counterexample to $X \to W$ makes $X$ true and $W$ false. Here $X$ is $ab = 0$ and $W$ is $(a = 0) \wedge (b = 0)$ .

$(0, 0)$ : $ab = 0$ and $W$ is true, so the conditional is true.
$(0, 5)$ : $ab = 0$ but $b \ne 0$ , so $W$ is false: a counterexample.
$(3, -3)$ : $ab = -9 \ne 0$ , so the hypothesis is false.
$(\tfrac{1}{2}, 2)$ : $ab = 1 \ne 0$ , so the hypothesis is false.

The correct contrapositive uses De Morgan's Law: $\neg(a \ne 0 \wedge b \ne 0) \equiv (a = 0) \vee (b = 0)$ , giving “if $ab = 0$ , then $a = 0$ or $b = 0$ ”, which $(0, 5)$ satisfies.

Problem 3 · Rewrite, Then Prove

Given: for all real numbers $x$ , if $x^2 = 9$ , then $x = 3$ or $x = -3$ . Use $X \to (Y \vee Z) \equiv (X \wedge \neg Y) \to Z$ with $Y$ : $x = 3$ .

Which statement is logically equivalent to the given one?

After writing $x^2 - 9 = (x - 3)(x + 3) = 0$ , which step yields $x = -3$ ?

✅ Correct! The rewrite adds the assumption

x \ne 3

, which makes the factor

x - 3

nonzero, and the proposition

(ab = 0 \wedge a \ne 0) \to b = 0

finishes the proof.

❌ Not quite. The rule adds

\neg Y

, the negation of

x = 3

, to the hypothesis, not

Y

itself.

❌ Close, but… the hypothesis is right, but the conclusion of

(X \wedge \neg Y) \to Z

Z

itself:

x = -3

, not its negation.

❌ Not quite. This changes “or” into “and” in the conclusion; no real number is both

3

and

-3

, so this statement fails at

x = 3

❌ Check the rewrite. In

(X \wedge \neg Y) \to Z

, the hypothesis gains

x \ne 3

and the conclusion is

x = -3

❌ Not quite. A product equal to zero forces at least one factor to be zero, not both: “

a = 0

b = 0

”.

❌ Not quite. The known nonzero factor is

x - 3

, from the assumption

x \ne 3

. Nothing so far says

x + 3 \ne 0

❌ Not quite. This step drops

x = -3

, and the assumption

x \ne 3

already rules out

x = 3

❌ Check the last step. Use the assumption

x \ne 3

to name the nonzero factor.

Show solution

Rewrite. With $X$ : $x^2 = 9$ , $Y$ : $x = 3$ , $Z$ : $x = -3$ , the statement $X \to (Y \vee Z)$ is logically equivalent to $(X \wedge \neg Y) \to Z$ : for all real $x$ , if $x^2 = 9$ and $x \ne 3$ , then $x = -3$ .

Proof. We prove this logically equivalent statement. Let $x \in \mathbb{R}$ with $x^2 = 9$ and $x \ne 3$ . Then

x^2 - 9 = 0, \qquad (x - 3)(x + 3) = 0.

Since $x \ne 3$ , the factor $x - 3$ is nonzero. By the proposition proved in the video (if $ab = 0$ and $a \ne 0$ , then $b = 0$ ), with $a = x - 3$ and $b = x + 3$ , we get $x + 3 = 0$ , so $x = -3$ . $\blacksquare$

Problem 4 · Assume the Other Factor Is Nonzero

Given: to prove “for all $a, b \in \mathbb{R}$ , if $ab = 0$ , then $a = 0$ or $b = 0$ ”, a classmate assumes $b \ne 0$ instead of $a \ne 0$ .

Which statement does the classmate prove?

Which step finishes the classmate's proof?

✅ Correct! Since

Y \vee Z \equiv Z \vee Y

, the rule also gives

(X \wedge \neg Z) \to Y

; then

b \ne 0

makes

\tfrac{1}{b}

exist, and

(ab)\cdot\tfrac{1}{b} = a\,(b \cdot \tfrac{1}{b}) = a = 0

❌ Not quite. The assumption is

b \ne 0

, the negation of

b = 0

; the hypothesis must gain

\neg Z

❌ Close, but… the conclusion stays the remaining disjunct,

a = 0

, not its negation.

❌ Not quite. This negates both the hypothesis

ab = 0

and the conclusion

a = 0

. The hypothesis

ab = 0

is kept as it is, only

b \ne 0

is added to it, and the conclusion stays

a = 0

❌ Check the statement. Swap the roles of

Y

and

Z

(X \wedge \neg Y) \to Z

❌ Not quite. Nothing says

a \ne 0

; in fact the goal is

a = 0

, so

\tfrac{1}{a}

need not exist.

❌ Not quite. Here

ab = 0

, and division by

0

is undefined.

❌ Not quite. This gives

ab + b = b

, which does not isolate

a

❌ Check the step. Only the factor assumed nonzero has a reciprocal you can use.

Show solution

With $X$ : $ab = 0$ , $Y$ : $a = 0$ , $Z$ : $b = 0$ , the disjunction $Y \vee Z$ is logically equivalent to $Z \vee Y$ , so

X \to (Z \vee Y) \equiv (X \wedge \neg Z) \to Y.

The classmate proves: if $ab = 0$ and $b \ne 0$ , then $a = 0$ .

Let $a, b \in \mathbb{R}$ with $ab = 0$ and $b \ne 0$ . Since $b \ne 0$ , the number $\tfrac{1}{b}$ exists. Multiplying both sides by it:

(ab)\cdot\tfrac{1}{b} = 0 \cdot \tfrac{1}{b}, \qquad a\left(b \cdot \tfrac{1}{b}\right) = 0, \qquad a \cdot 1 = 0, \qquad a = 0.

The second equation uses the associative property and $0 \cdot \tfrac{1}{b} = 0$ . Either choice of nonzero factor gives a valid proof. $\blacksquare$

Solved: 0 / 4

Writing a Proof by Contrapositive: ab = 0 and De Morgan

IKey Formulas

IINegating a Conjunction

IIIRewriting X→(Y∨Z)X \to (Y \vee Z)X→(Y∨Z)

IVWriting the Proof Line by Line

VQuiz Questions

IIIRewriting $X \to (Y \vee Z)$