Mathematical-Reasoning · Unit 1 · Video 4 · Interactive Practice

Using Conditional Statements: Counterexamples, Theorems and Closure Properties

IKey Formulas

Tool	Statement	Note
The one false case	$P \to Q$ is false only when $P$ is true and $Q$ is false	All three other rows are true
Disproving $P \to Q$	One case with $P$ true and $Q$ false	A single counterexample suffices
Using a true theorem $P \to Q$	$P$ true $\Rightarrow$ $Q$ true; $\ Q$ false $\Rightarrow$ $P$ false	$P$ false or $Q$ true: no conclusion
Closure of $\mathbb{Q}$	$\frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}, \quad \frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$	$a, b, c, d \in \mathbb{Z}$ , $b \neq 0$ , $d \neq 0$

Key Insight: One example can disprove a conditional, but no number of examples can prove one: showing a set is closed needs an argument that covers every case.

IIUsing a Theorem — Which Facts Give a Conclusion?

Given only “if $f$ is differentiable at $a$ , then $f$ is continuous at $a$ ”, which known facts force a conclusion?

Function

Theorem gives

IIICounterexamples to Closure

A closure property is a conditional, so one pair with a true hypothesis and a false conclusion disproves it.

Set Operation

💡 Challenge: find a counterexample showing $\mathbb{Z}$ is not closed under division, then hunt for one against $\mathbb{Z}$ under multiplication. None exists for multiplication, and no amount of searching can establish that; only an argument like the one below can.

IVProving $\mathbb{Q}$ Is Closed

One general argument covers every pair of rationals; any particular pair is just one instance of it.

Operation a 3 b 4 c −2 d 5

Step 1 · Givens — $\frac{a}{b}, \frac{c}{d} \in \mathbb{Q}$ with $a, b, c, d \in \mathbb{Z}$ , $b \neq 0$ , $d \neq 0$

Step 2 · Combine the fractions

\frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}

Step 3 · Numerator and denominator are integers

$ac \in \mathbb{Z}$ and $bd \in \mathbb{Z}$ , because $\mathbb{Z}$ is closed under multiplication.

Step 4 · Nonzero denominator

$b \neq 0$ and $d \neq 0 \Rightarrow bd \neq 0$ , so $\frac{ac}{bd}$ is a quotient of integers with nonzero denominator: $\frac{ac}{bd} \in \mathbb{Q}$ .

VQuiz Questions

Problem 1 · The One False Case

Given: Ed has exactly $35 in his wallet. Which conditional is false?

✅ Correct! The hypothesis is true (he has exactly $35) and the conclusion is false (35 is less than 40): the one row where a conditional is false.

❌ Not quite. Ed does not have $70, so the hypothesis is false, and a conditional with a false hypothesis is true whatever the conclusion.

❌ Not quite. Both parts are true here (35 is at least 30), so the conditional is true.

Show solution

Evaluate hypothesis $P$ and conclusion $Q$ for each statement, knowing Ed holds exactly $35:

(a) $P$ true, $Q$ true (35 ≥ 30): true.
(b) $P$ false, $Q$ false (35 < 60): true.
(c) $P$ true, $Q$ false (35 < 40): false.
(d) $P$ false, $Q$ true (35 ≥ 20): true.

Only (c) lands in the row $P$ true, $Q$ false, so it is the false conditional.

Problem 2 · The Theorem Runs One Way

Given: the theorem “if $f$ is differentiable at $a$ , then $f$ is continuous at $a$ ”, and the fact that a function $g$ is continuous at $3$ . Using only this theorem, what can you conclude?

✅ Correct! A true conclusion fits both row 1 (

P

true) and row 3 (

P

false), so the hypothesis could go either way.

❌ Close, but that runs the theorem backwards.

f(x) = |x|

is continuous at

0

but not differentiable there: continuity does not give differentiability.

❌ Not quite.

f(x) = \sin x

is continuous at

0

and differentiable there, so continuity cannot rule differentiability out either.

❌ Not quite. The theorem is true for every function; for

g

it simply lands in a row that gives no information.

Show solution

Let $P$ : “ $g$ is differentiable at $3$ ” and $Q$ : “ $g$ is continuous at $3$ ”. We know $Q$ is true and $P \to Q$ is true.

The rows of the truth table with $Q$ true are $P$ true, $Q$ true (conditional true) and $P$ false, $Q$ true (conditional true). Both are consistent with the theorem, so $P$ may be true or false.

Examples confirm both are possible: $\sin x$ at $0$ (differentiable) and $|x|$ at $0$ (not differentiable). The theorem gives no conclusion.

Problem 3 · Choosing a Valid Counterexample

Given: two closure claims, each a conditional. Pick the pair that is a genuine counterexample to each.

If $x$ and $y$ are irrational, then $x \cdot y$ is irrational.

If $x, y \in \mathbb{N}$ , then $x \div y \in \mathbb{N}$ .

✅ Correct!

\sqrt{2}

is irrational but

\sqrt{2}\cdot\sqrt{2} = 2

is rational, and

1, 2 \in \mathbb{N}

but

\tfrac{1}{2} \notin \mathbb{N}

: each has a true hypothesis and a false conclusion.

❌ Not a counterexample. The product here is irrational, so the conclusion is true; a counterexample needs the conclusion to fail.

❌ Not a counterexample.

2

and

3

are rational, so the hypothesis is false and the pair says nothing about the claim.

❌ Not quite. Look for two irrationals whose product is rational.

❌ Not a counterexample.

2 \in \mathbb{N}

, so the conclusion is true for this pair.

❌ Not a counterexample. One of the inputs is not a natural number (

-4 \notin \mathbb{N}

0 \notin \mathbb{N}

), so the hypothesis is false.

❌ Not quite. Both inputs must be natural numbers and the quotient must not be.

Show solution

A counterexample to $P \to Q$ must make $P$ true and $Q$ false.

Irrationals under $\times$ : take $x = y = \sqrt{2}$ . Both are irrational (hypothesis true), and $\sqrt{2}\cdot\sqrt{2} = 2 = \tfrac{2}{1}$ is rational (conclusion false). The products $\pi^2$ and $\sqrt{6}$ are irrational, and $2, 3$ fail the hypothesis.

$\mathbb{N}$ under $\div$ : take $x = 1$ , $y = 2$ . Both are natural numbers, and $1 \div 2 = \tfrac{1}{2}$ is not. The pair $6, 3$ gives $2 \in \mathbb{N}$ ; the pairs with $-4$ or $0$ fail the hypothesis, since in this course $0 \notin \mathbb{N}$ .

Problem 4 · Proving a Closure Property

Call an integer even when it equals $2k$ for some integer $k$ . Which argument proves that the set of even integers is closed under addition?

✅ Correct! Arbitrary even

x

and

y

are handled at once, and the sum is shown to be twice an integer.

❌ Not quite. Examples can support a guess but never cover every pair; closure needs a general argument.

❌ Close, but… this shows

x + y

is an integer, not that it is even; the conclusion must be

x + y = 2k

for some integer

k

❌ Not quite. Using one

m

for both forces

x = y

, so only pairs of equal numbers are covered.

Show solution

The claim is the conditional: if $x$ and $y$ are even integers, then $x + y$ is an even integer.

Assume the hypothesis: $x = 2m$ , $y = 2n$ for some integers $m, n$ (separate letters, since $x$ and $y$ are arbitrary). Then

x + y = 2m + 2n = 2(m + n).

Since $\mathbb{Z}$ is closed under addition, $m + n \in \mathbb{Z}$ , so $x + y$ is $2$ times an integer: it is even. The argument works for every pair, exactly as the $a, b, c, d$ argument did for $\mathbb{Q}$ .

Solved: 0 / 4

Using Conditional Statements: Counterexamples, Theorems and Closure Properties

IKey Formulas

IIUsing a Theorem — Which Facts Give a Conclusion?

IIICounterexamples to Closure

IVProving Q\mathbb{Q}Q Is Closed

VQuiz Questions

IVProving $\mathbb{Q}$ Is Closed