Contrapositive with Quantifiers, Sets, and Limits

Reviewed by Alva Benedict B.

Alva Benedict B., PhD

College Expert

Review Board Member

Alva Benedict B. is an experienced mathematician and math content developer with over 15 years of teaching and tutoring experience across high school, undergraduate, and test prep levels. He specializes in Algebra, Calculus, and Statistics, and holds advanced academic training in Mathematics with extensive expertise in LaTeX-based math content development.

, PhD

By Thames

Thames

Community Contributor

Quizzes Created: 8156 | Total Attempts: 9,588,805

| Questions: 15 | Updated: Jan 28, 2026

Share on Facebook Share on Twitter Share on Whatsapp Share on Pinterest Share on Email Copy to Clipboard Embed on your website

Question 1 / 16

🏆 Rank #-- ▾

0 %

0/100

Score 0/100

1) Give the contrapositive of "If x ∈ A ∩ B, then x ∈ A":

If x ∉ A, then x ∉ A ∩ B

If x ∈ A, then x ∈ A ∩ B

If x ∉ A ∩ B, then x ∉ A

If x ∈ A ∩ B, then x ∉ A

The contrapositive negates the conclusion and then implies the negation of the hypothesis. The negation of “x ∈ A” is “x ∉ A,” and this implies “x ∉ A ∩ B,” since membership in the intersection requires membership in A.

Explanation

Submit

About This Quiz

Contrapositive With Quantifiers, Sets, And Limits - Quiz

Ready to level up your contrapositive skills in more abstract math? In this quiz, you’ll apply contraposition to statements about sets, intersections, divisibility, factorial inequalities, limits, and ε–δ style definitions. You’ll practice negating “for all” and “there exists,” working with conditions like A ⊆ B ∪ C, and handling precise... see morestatements about sequences and limits. Along the way, you’ll see how carefully negating conclusions like “x = 0 and y = 0” or “a divides c” leads to a clearer proof strategy. By the end, you’ll be more comfortable using contrapositive proofs in real analysis, number theory, and set theory.
see less

2)

What first name or nickname would you like us to use?

You may optionally provide this to label your report, leaderboard, or certificate.

2) To prove "If f(x) = 0 has no solution, then f(x) > 0 for all x" by contrapositive:

Assume f(x) ≤ 0 for some x

Assume f(x) = 0 has a solution

Assume f(x) < 0 for all x

Assume f(x) ≠ 0 for all x

Negating “f(x) > 0 for all x” yields “there exists x such that f(x) ≤ 0.” This is the assumption used in the contrapositive proof.

Explanation

Negating “f(x) > 0 for all x” yields “there exists x such that f(x) ≤ 0.” This is the assumption used in the contrapositive proof.

Submit

3) The contrapositive of "If ∀x P(x), then ∃x Q(x)" is:

If ∀x ¬Q(x), then ∃x ¬P(x)

If ∃x ¬Q(x), then ∀x ¬P(x)

If ∃x Q(x), then ¬∀x P(x)

If ∀x Q(x), then ∃x P(x)

The contrapositive of an implication "If R, then S" is formed by negating both components and reversing their order, resulting in "If not S, then not R". Here, R is "for all x, P(x)" and S is "there exists x such that Q(x)". Negating “∃x Q(x)” gives “∀x ¬Q(x).” Negating “∀x P(x)” gives “∃x ¬P(x).” Therefore the contrapositive matches option A.

Explanation

Submit

4) For "If a divides bc and gcd(a,b)=1, then a divides c", the contrapositive assumes:

A does not divide c

Gcd(a,b) ≠ 1

A does not divide bc

Both A and B

The conclusion is “a divides c,” whose negation is “a does not divide c.” This is the assumption used for the contrapositive. The condition gcd(a,b)=1 remains unchanged.

Explanation

The conclusion is “a divides c,” whose negation is “a does not divide c.” This is the assumption used for the contrapositive. The condition gcd(a,b)=1 remains unchanged.

Submit

5) The contrapositive of "If n! > 2ⁿ, then n ≥ 4" is:

If n < 4, then n! ≤ 2ⁿ

If n! ≤ 2ⁿ, then n < 4

If n ≤ 3, then n! ≤ 2ⁿ

Both A and C

Negating “n ≥ 4” yields “n 2ⁿ” gives “n! ≤ 2ⁿ.” So the contrapositive is “If n

Explanation

Negating “n ≥ 4” yields “n 2ⁿ” gives “n! ≤ 2ⁿ.” So the contrapositive is “If n

Submit

6) To prove "If A ⊆ B ∪ C and A ∩ B = ∅, then A ⊆ C" by contrapositive:

Assume A ⊄ C

Assume A ⊆ B ∪ C

Assume A ∩ B ≠ ∅

Assume A ⊆ C

The conclusion is “A ⊆ C,” so its negation is “A ⊄ C.” This is the starting point of the contrapositive proof. The other conditions remain part of the implication’s hypothesis.

Explanation

The conclusion is “A ⊆ C,” so its negation is “A ⊄ C.” This is the starting point of the contrapositive proof. The other conditions remain part of the implication’s hypothesis.

Submit

7) Negation of "For all ε > 0, there exists δ > 0 such that P(ε,δ)":

There exists ε > 0 such that for all δ > 0, ¬P(ε,δ)

For all ε > 0, there exists δ > 0 such that ¬P(ε,δ)

There exists ε ≤ 0 such that for all δ > 0, ¬P(ε,δ)

For all ε > 0, for all δ > 0, ¬P(ε,δ)

The negation of ∀ε ∃δ P(ε,δ) is ∃ε ∀δ ¬P(ε,δ), preserving the constraints ε > 0 and δ > 0.

Explanation

The negation of ∀ε ∃δ P(ε,δ) is ∃ε ∀δ ¬P(ε,δ), preserving the constraints ε > 0 and δ > 0.

Submit

8) Why is contrapositive useful for "If f is differentiable at a, then f is continuous at a"?

Differentiable is easier to work with

Discontinuity has a clearer definition

Contrapositive is always easier

Continuity is harder to verify

Showing discontinuity (the negation of continuity) is often easier because you can demonstrate a failure of the ε-δ condition. This makes the contrapositive method effective.

Explanation

Showing discontinuity (the negation of continuity) is often easier because you can demonstrate a failure of the ε-δ condition. This makes the contrapositive method effective.

Submit

9) The contrapositive of "If lim(aₙ) = L and lim(bₙ) = M, then lim(aₙ + bₙ) = L + M" is:

If lim(aₙ + bₙ) ≠ L + M, then lim(aₙ) ≠ L or lim(bₙ) ≠ M

If lim(aₙ) ≠ L or lim(bₙ) ≠ M, then lim(aₙ + bₙ) ≠ L + M

If lim(aₙ + bₙ) = L + M, then lim(aₙ) = L and lim(bₙ) = M

If lim(aₙ + bₙ) ≠ L + M, then lim(aₙ) ≠ L and lim(bₙ) ≠ M

Negating the conclusion “= L + M” yields “≠ L + M.” Negating the conjunction “lim(aₙ)=L AND lim(bₙ)=M” gives “lim(aₙ) ≠ L OR lim(bₙ) ≠ M.”

Explanation

Negating the conclusion “= L + M” yields “≠ L + M.” Negating the conjunction “lim(aₙ)=L AND lim(bₙ)=M” gives “lim(aₙ) ≠ L OR lim(bₙ) ≠ M.”

Submit

10) Proving the contrapositive is valid because of logical equivalence.

True

False

Cannot determine from assumption alone

X + y ≠ 10

A statement and its contrapositive always share the same truth value. Proving the contrapositive therefore proves the original implication.

Explanation

A statement and its contrapositive always share the same truth value. Proving the contrapositive therefore proves the original implication.

Submit

11) For "If x² + y² = 0, then x = 0 and y = 0", the contrapositive assumes:

X ≠ 0 or y ≠ 0

X ≠ 0 and y ≠ 0

X² + y² ≠ 0

X = 0 or y = 0

Negating “x = 0 and y = 0” yields “x ≠ 0 or y ≠ 0.” This is the assumption used in the contrapositive proof.

Explanation

Negating “x = 0 and y = 0” yields “x ≠ 0 or y ≠ 0.” This is the assumption used in the contrapositive proof.

Submit

12) To prove "If x is irrational, then √x is irrational" by contrapositive, which assumption should you start with?

Assume √x is rational

Assume x is irrational

Assume √x is irrational

Assume x is rational

The contrapositive of "If P, then Q" is "If not Q, then not P." Here, P is "x is irrational" and Q is "√x is irrational." The negation of Q is "√x is rational," and the negation of P is "x is rational." Thus, the contrapositive is "If √x is rational, then x is rational." To prove this, you must start by assuming "√x is rational" (not Q) and derive "x is rational" (not P). Choice B is correct. Choice A would start a direct proof, not contrapositive. Choice D is the conclusion, not the assumption.

Explanation

Submit

13) When is direct proof preferred over contrapositive?

When P and Q are simpler than ¬P and ¬Q

When contrapositive is false

When the statement is biconditional

When Q implies P

Direct proof is more efficient when the original statements are simpler to work with than their negations.

Explanation

Direct proof is more efficient when the original statements are simpler to work with than their negations.

Submit

14) A student tries to prove “If a sequence converges, then it is bounded” by contrapositive. Which shows misunderstanding?

Assuming the sequence converges and trying to prove it is unbounded

Trying to show the sequence does not converge

Assuming the sequence is unbounded

Showing that if the sequence is not bounded, then it cannot converge

To use the contrapositive, we must assume the negation of the conclusion, which is "the sequence is not bounded." Then we aim to show the negation of the hypothesis, which is “the sequence does not converge.” Assuming the original hypothesis (“sequence converges”) and trying to derive the negation of the conclusion violates the contrapositive structure and mixes direct proof with inverse reasoning, making the argument invalid.

Explanation

Submit