Discrete Math Quiz: Negating Quantified Statements
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Negating Quantified StatementsQuestion 1 of 20

Let P(x,y)P(x,y) be the predicate "xx divides yy" where xx and yy are positive integers. Which of the following is the negation of the statement "∃x∀y(P(x,y)→y≥2x)\exists x \forall y (P(x,y) \rightarrow y \geq 2x)"?

∀x∃y(P(x,y)∧y<2x)\forall x \exists y (P(x,y) \wedge y < 2x)
∃x∀y(P(x,y)∧y<2x)\exists x \forall y (P(x,y) \wedge y < 2x)
∀x∃y(¬P(x,y)∨y<2x)\forall x \exists y (\neg P(x,y) \vee y < 2x)
∃x∃y(P(x,y)∧y<2x)\exists x \exists y (P(x,y) \wedge y < 2x)
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Discrete Math Quiz

Discrete Math Quiz: Negating Quantified Statements

Practice Negating Quantified Statements in Discrete Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Negating Quantified Statements, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let P(x,y)P(x,y) be the predicate "xx divides yy" where xx and yy are positive integers. Which of the following is the negation of the statement "∃x∀y(P(x,y)→y≥2x)\exists x \forall y (P(x,y) \rightarrow y \geq 2x)"?

  1. ∀x∃y(P(x,y)∧y<2x)\forall x \exists y (P(x,y) \wedge y < 2x) (correct answer)
  2. ∃x∀y(P(x,y)∧y<2x)\exists x \forall y (P(x,y) \wedge y < 2x)
  3. ∀x∃y(¬P(x,y)∨y<2x)\forall x \exists y (\neg P(x,y) \vee y < 2x)
  4. ∃x∃y(P(x,y)∧y<2x)\exists x \exists y (P(x,y) \wedge y < 2x)
Explanation: The original statement is ∃x∀y(P(x,y)→y≥2x)\exists x \forall y (P(x,y) \rightarrow y \geq 2x). To negate this, we apply ¬(∃x∀yQ(x,y))≡∀x¬(∀yQ(x,y))≡∀x∃y¬Q(x,y)\neg(\exists x \forall y Q(x,y)) \equiv \forall x \neg(\forall y Q(x,y)) \equiv \forall x \exists y \neg Q(x,y). Here, Q(x,y)=P(x,y)→y≥2xQ(x,y) = P(x,y) \rightarrow y \geq 2x, so ¬Q(x,y)=¬(P(x,y)→y≥2x)=P(x,y)∧y<2x\neg Q(x,y) = \neg(P(x,y) \rightarrow y \geq 2x) = P(x,y) \wedge y < 2x. Therefore, the negation is ∀x∃y(P(x,y)∧y<2x)\forall x \exists y (P(x,y) \wedge y < 2x). Choice B fails to negate the existential quantifier. Choice C incorrectly negates the implication. Choice D fails to negate both quantifiers properly.

Question 2

A computer scientist claims: "For any algorithm AA, if AA runs in polynomial time, then there exists an input size nn such that for every input of size nn, the algorithm terminates within n3n^3 steps." Which statement correctly negates this claim?

  1. For every algorithm, if it does not run in polynomial time, then for every input size nn, there exists an input of size nn requiring more than n3n^3 steps.
  2. For every algorithm that runs in polynomial time, there exists an input size nn such that every input of size nn requires more than n3n^3 steps.
  3. There exists an algorithm that does not run in polynomial time, and for some input size nn, every input of size nn terminates within n3n^3 steps.
  4. There exists an algorithm that runs in polynomial time, but for every input size nn, there exists an input of size nn requiring more than n3n^3 steps. (correct answer)
Explanation: When you encounter logical negation problems, you need to carefully transform quantifiers and flip the final condition. The original claim has the structure: "For any algorithm A, if A runs in polynomial time, then [there exists an n such that for every input of size n, the algorithm terminates within n³ steps]." To negate this statement, you need an algorithm that runs in polynomial time but fails the conclusion. The negation becomes: "There exists a polynomial-time algorithm such that for every input size n, there exists an input of size n requiring more than n³ steps." This means no matter what input size you pick, you can always find at least one input of that size that takes longer than n³ steps. Option D captures this negation perfectly - it identifies a polynomial-time algorithm that consistently has some input requiring more than n³ steps at every input size. Option A incorrectly negates the hypothesis (focusing on non-polynomial algorithms) rather than the entire implication. Option B keeps "every algorithm" when the negation should have "there exists an algorithm," and it also mishandles the quantifier structure. Option C discusses non-polynomial algorithms and has the final condition backwards - it says inputs terminate within n³ steps rather than requiring more. The key strategy for logical negation is systematically flipping quantifiers (∀ becomes ∃ and vice versa) and negating the final condition. Always work from the inside out, and remember that negating "if P then Q" gives you "P and not Q."

Question 3

Consider the mathematical statement: "∃n∈Z+∀k∈Z+∃m∈Z+(k∣nm→k≤n)\exists n \in \mathbb{Z}^+ \forall k \in \mathbb{Z}^+ \exists m \in \mathbb{Z}^+ (k | n^m \rightarrow k \leq n)" where k∣nmk | n^m means "kk divides nmn^m". Which is the correct negation?

  1. ∀n∈Z+∀k∈Z+∃m∈Z+(k∣nm∧k>n)\forall n \in \mathbb{Z}^+ \forall k \in \mathbb{Z}^+ \exists m \in \mathbb{Z}^+ (k | n^m \wedge k > n)
  2. ∃n∈Z+∃k∈Z+∀m∈Z+(k∤nm∨k>n)\exists n \in \mathbb{Z}^+ \exists k \in \mathbb{Z}^+ \forall m \in \mathbb{Z}^+ (k \nmid n^m \vee k > n)
  3. ∀n∈Z+∃k∈Z+∀m∈Z+(k∣nm∧k>n)\forall n \in \mathbb{Z}^+ \exists k \in \mathbb{Z}^+ \forall m \in \mathbb{Z}^+ (k | n^m \wedge k > n) (correct answer)
  4. ∃n∈Z+∀k∈Z+∀m∈Z+(k∤nm∧k>n)\exists n \in \mathbb{Z}^+ \forall k \in \mathbb{Z}^+ \forall m \in \mathbb{Z}^+ (k \nmid n^m \wedge k > n)
Explanation: When you encounter logical negation problems with multiple quantifiers, you need to systematically apply De Morgan's laws and flip each quantifier while negating the final statement. The original statement is ∃n∈Z+∀k∈Z+∃m∈Z+(k∣nm→k≤n)\exists n \in \mathbb{Z}^+ \forall k \in \mathbb{Z}^+ \exists m \in \mathbb{Z}^+ (k | n^m \rightarrow k \leq n). To negate this properly, you flip each quantifier from left to right: ∃\exists becomes ∀\forall, ∀\forall becomes ∃\exists, and ∃\exists becomes ∀\forall. This gives us ∀n∈Z+∃k∈Z+∀m∈Z+[negation of (k∣nm→k≤n)]\forall n \in \mathbb{Z}^+ \exists k \in \mathbb{Z}^+ \forall m \in \mathbb{Z}^+ [\text{negation of } (k | n^m \rightarrow k \leq n)]. Next, you need to negate the implication k∣nm→k≤nk | n^m \rightarrow k \leq n. Since ¬(P→Q)≡P∧¬Q\neg(P \rightarrow Q) \equiv P \wedge \neg Q, this becomes k∣nm∧k>nk | n^m \wedge k > n. Therefore, the complete negation is ∀n∈Z+∃k∈Z+∀m∈Z+(k∣nm∧k>n)\forall n \in \mathbb{Z}^+ \exists k \in \mathbb{Z}^+ \forall m \in \mathbb{Z}^+ (k | n^m \wedge k > n), which matches choice C. Choice A fails to flip the first quantifier (keeps ∃n\exists n instead of changing to ∀n\forall n). Choice B incorrectly flips the second quantifier to ∃\exists when it should stay ∃\exists, and uses ∨\vee instead of ∧\wedge in the negated implication. Choice D keeps the first quantifier as ∃\exists instead of flipping it to ∀\forall. Remember: when negating nested quantifiers, work systematically from left to right, flipping each one, then carefully negate the final logical statement using the proper equivalences for implications.

Question 4

A software engineer claims: "Every function that is recursive has the property that for some input size nn, all inputs of that size cause the function to make at most log⁡n\log n recursive calls." If this claim is false, what must be true?

  1. There exists a non-recursive function such that for some input size nn, all inputs of size nn cause at most log⁡n\log n recursive calls.
  2. For every recursive function, there exists an input size nn such that all inputs of size nn cause more than log⁡n\log n recursive calls.
  3. There exists a recursive function such that for every input size nn, there exists an input of size nn causing more than log⁡n\log n recursive calls. (correct answer)
  4. Every recursive function has the property that for all input sizes nn, there exists an input of size nn causing more than log⁡n\log n recursive calls.
Explanation: When you encounter a question about negating logical statements, you need to carefully apply the rules for negating quantifiers and logical connectives. The original claim uses universal quantification ("every function") combined with existential quantification ("for some input size"). The claim states: "Every recursive function has the property that for some input size nn, all inputs of that size cause at most log⁡n\log n recursive calls." In logical form, this is: ∀(recursive functions) ∃n ∀(inputs of size n)[calls ≤ log⁡n\log n]. To negate this statement, you apply De Morgan's laws and flip the quantifiers: ∃(recursive function) ∀n ∃(input of size n)[calls > log⁡n\log n]. This translates to: "There exists a recursive function such that for every input size nn, there exists an input of size nn causing more than log⁡n\log n recursive calls." This matches answer choice C exactly. Answer A is wrong because it discusses non-recursive functions, which are irrelevant to negating a claim about recursive functions. Answer B incorrectly negates the statement by changing "some input size" to "an input size" while keeping "all inputs," creating a different logical structure. Answer D uses universal quantification for functions ("every recursive function") instead of existential, making it far too strong a claim. Strategy tip: When negating complex logical statements, work systematically through each quantifier from left to right, flipping ∀ to ∃ and vice versa, while also negating the final condition. Practice translating between English and logical notation to avoid common quantifier mistakes.

Question 5

A network administrator claims: "In every computer network, if the network uses TCP protocol, then there exists a router such that all data packets passing through that router are delivered without errors." Assuming this claim is incorrect, which scenario demonstrates its falsehood?

  1. In every computer network using TCP protocol, there exists a router where all data packets are delivered with errors.
  2. There exists a computer network using TCP protocol where every router has at least one data packet that is delivered with errors. (correct answer)
  3. There exists a computer network not using TCP protocol where every router delivers all data packets without errors.
  4. Every computer network using TCP protocol has the property that all routers deliver some data packets with errors.
Explanation: When you encounter logical statements and their negations, the key is carefully translating between everyday language and formal logic structure. The administrator's claim has the logical form: "For all networks, if TCP is used, then there exists a router where all packets are delivered without errors." To demonstrate this claim is false, you need to find a counterexample - a specific scenario where the claim fails. The original statement says that in ANY TCP network, you can always find AT LEAST ONE router that delivers all packets perfectly. To disprove this, you need to show there exists a TCP network where NO such perfect router exists - meaning every router in that network has at least some packet errors. Option B correctly provides this counterexample: "There exists a computer network using TCP protocol where every router has at least one data packet that is delivered with errors." This directly contradicts the claim by showing a TCP network where you cannot find even one error-free router. Option A incorrectly negates the statement by claiming ALL routers have ALL packets with errors - this is too strong and misrepresents the logical structure. Option C discusses non-TCP networks, which is irrelevant since the original claim only applies to TCP networks. Option D makes a universal claim about ALL TCP networks having errors, but the administrator's claim could still be true if each network had one perfect router alongside the imperfect ones. Remember: when disproving "for all X, there exists Y" statements, look for "there exists X where no such Y exists."

Question 6

Consider the statement about sequences: "∀{an}∃N∈N∀n≥N(an>0→∃m>n such that am<an)\forall \{a_n\} \exists N \in \mathbb{N} \forall n \geq N (a_n > 0 \rightarrow \exists m > n \text{ such that } a_m < a_n)" where {an}\{a_n\} denotes a sequence of real numbers. What is the correct negation?

  1. ∃{an}∃N∈N∀n≥N(an≤0∨∀m>n,am≥an)\exists \{a_n\} \exists N \in \mathbb{N} \forall n \geq N (a_n \leq 0 \vee \forall m > n, a_m \geq a_n)
  2. ∃{an}∀N∈N∃n≥N(an>0∧∀m>n,am≥an)\exists \{a_n\} \forall N \in \mathbb{N} \exists n \geq N (a_n > 0 \wedge \forall m > n, a_m \geq a_n) (correct answer)
  3. ∀{an}∀N∈N∃n≥N(an>0∧∀m>n,am≥an)\forall \{a_n\} \forall N \in \mathbb{N} \exists n \geq N (a_n > 0 \wedge \forall m > n, a_m \geq a_n)
  4. ∃{an}∀N∈N∀n≥N(an≤0∨∃m>n such that am≥an)\exists \{a_n\} \forall N \in \mathbb{N} \forall n \geq N (a_n \leq 0 \vee \exists m > n \text{ such that } a_m \geq a_n)
Explanation: When you encounter logical negation problems involving multiple quantifiers, you must systematically flip each quantifier and negate the final statement. The key is working from left to right through the quantifier chain. The original statement says: "For all sequences, there exists some N where for all n ≥ N, if a term is positive, then there's a later term that's smaller." To negate this complex statement, you flip ∀\forall to ∃\exists, ∃\exists to ∀\forall, then ∀\forall to ∃\exists, and finally negate the implication an>0→∃m>n such that am<ana_n > 0 \rightarrow \exists m > n \text{ such that } a_m < a_n. The negation of an implication P→QP \rightarrow Q is P∧¬QP \wedge \neg Q. So an>0→∃m>n such that am<ana_n > 0 \rightarrow \exists m > n \text{ such that } a_m < a_n becomes an>0∧∀m>n,am≥ana_n > 0 \wedge \forall m > n, a_m \geq a_n. This gives us answer B: there exists a sequence where for every N, we can find some n ≥ N such that the term is positive but no later term is smaller. Option A incorrectly keeps ∀N\forall N as ∃N\exists N instead of flipping it. Option C fails to change the initial ∀{an}\forall \{a_n\} to ∃{an}\exists \{a_n\}. Option D incorrectly negates the implication as a disjunction an≤0∨…a_n \leq 0 \vee \ldots and changes << to ≥\geq instead of properly negating the existential statement. Remember: when negating nested quantifiers, flip each one systematically from left to right, then carefully negate the final logical statement using De Morgan's laws and implication negation rules.

Question 7

A database administrator states: "For every table in the database, if the table has a primary key, then there exists a query that can retrieve every record from that table in constant time." Assuming this claim is false, which scenario must exist?

  1. For every table without a primary key, there exists a query that can retrieve all records from that table in constant time.
  2. Every table with a primary key has at least one query that takes more than constant time to retrieve all records from that table.
  3. There is a table without a primary key such that no query can retrieve all records from that table in constant time.
  4. There is a table with a primary key such that every query takes more than constant time to retrieve all records from that table. (correct answer)
Explanation: When you encounter questions about logical statements and their negations, you need to carefully analyze what it means for a conditional statement to be false. The administrator's claim is a universal statement: "For every table in the database, if the table has a primary key, then there exists a query that can retrieve every record from that table in constant time." To prove this statement false, you need to find just one counterexample - a single table that has a primary key but for which no query can retrieve all records in constant time. This means every possible query for that table takes more than constant time, which is exactly what answer D describes. Let's examine why the other options are incorrect. Option A discusses tables without primary keys, but the original claim only makes assertions about tables that do have primary keys, so this is irrelevant to disproving the claim. Option B states that every table with a primary key has "at least one" slow query, but this doesn't contradict the original claim - the administrator only claimed that there exists some fast query, not that all queries are fast. Option C again focuses on tables without primary keys, which doesn't address the administrator's specific claim about tables with primary keys. Remember that to disprove a "for all" statement of the form "for all X, if P(X) then Q(X)," you only need one example where P(X) is true but Q(X) is false. Look for counterexamples that directly contradict the specific conditions mentioned in the original claim.

Question 8

A mathematician states: "For every prime number pp, there exists a positive integer kk such that for all integers aa not divisible by pp, we have ak≡1(modp)a^k \equiv 1 \pmod{p}." Which statement represents the negation of this claim?

  1. For every prime number pp, there exists a positive integer kk such that there exists an integer aa not divisible by pp with ak≢1(modp)a^k \not\equiv 1 \pmod{p}.
  2. There exists a prime number pp such that for every positive integer kk, there exists an integer aa not divisible by pp with ak≢1(modp)a^k \not\equiv 1 \pmod{p}. (correct answer)
  3. There exists a prime number pp and a positive integer kk such that for all integers aa divisible by pp, we have ak≢1(modp)a^k \not\equiv 1 \pmod{p}.
  4. For every prime number pp and every positive integer kk, there exists an integer aa not divisible by pp with ak≢1(modp)a^k \not\equiv 1 \pmod{p}.
Explanation: When you encounter logical statements with multiple quantifiers, the key skill is systematically negating each quantifier while preserving the logical structure. The original statement has the form: "For every pp (∀p), there exists kk (∃k) such that for all aa (∀a), property P holds." To negate this statement, you must flip each quantifier in sequence: ∀ becomes ∃, ∃ becomes ∀, and ∀ becomes ∃, while negating only the final condition. This gives us: "There exists a prime pp such that for every positive integer kk, there exists an integer aa not divisible by pp with ak≢1(modp)a^k \not\equiv 1 \pmod{p}." This matches choice B exactly. Choice A incorrectly keeps the first quantifier as "for every" instead of changing it to "there exists." Choice C introduces an irrelevant change by considering integers divisible by pp rather than those not divisible by pp, which completely alters the mathematical meaning. Choice D fails to properly negate the middle quantifier—it should change "there exists kk" to "for every kk," but instead writes "for every pp and every kk." The key strategy for quantifier negation problems is to work left-to-right through the statement, systematically flipping each quantifier (∀↔∃) while keeping the same variables and conditions until you reach the final statement, which you negate. Practice this mechanical approach—it prevents the logical errors that create most wrong answer choices.

Question 9

Consider the statement about graphs: "∃G∀v∈V(G)∃u∈V(G)(deg⁡(v)≥3→d(u,v)=1)\exists G \forall v \in V(G) \exists u \in V(G) (\deg(v) \geq 3 \rightarrow d(u,v) = 1)" where V(G)V(G) is the vertex set of graph GG, deg⁡(v)\deg(v) is the degree of vertex vv, and d(u,v)d(u,v) is the distance between vertices uu and vv. What is its negation?

  1. ∀G∀v∈V(G)∃u∈V(G)(deg⁡(v)≥3∧d(u,v)≠1)\forall G \forall v \in V(G) \exists u \in V(G) (\deg(v) \geq 3 \wedge d(u,v) \neq 1)
  2. ∃G∃v∈V(G)∀u∈V(G)(deg⁡(v)<3∨d(u,v)≠1)\exists G \exists v \in V(G) \forall u \in V(G) (\deg(v) < 3 \vee d(u,v) \neq 1)
  3. ∀G∃v∈V(G)∀u∈V(G)(deg⁡(v)≥3∧d(u,v)≠1)\forall G \exists v \in V(G) \forall u \in V(G) (\deg(v) \geq 3 \wedge d(u,v) \neq 1) (correct answer)
  4. ∃G∀v∈V(G)∀u∈V(G)(deg⁡(v)<3∧d(u,v)≠1)\exists G \forall v \in V(G) \forall u \in V(G) (\deg(v) < 3 \wedge d(u,v) \neq 1)
Explanation: When negating complex logical statements with multiple quantifiers and implications, you must systematically apply negation rules while being careful about the order of operations. To negate the original statement, work from outside to inside. The statement has the structure ∃G∀v∃u(P→Q)\exists G \forall v \exists u (P \rightarrow Q) where PP is "deg⁡(v)≥3\deg(v) \geq 3" and QQ is "d(u,v)=1d(u,v) = 1". First, flip the quantifiers: ∃G\exists G becomes ∀G\forall G, ∀v\forall v becomes ∃v\exists v, and ∃u\exists u becomes ∀u\forall u. Next, negate the implication P→QP \rightarrow Q. Since ¬(P→Q)≡P∧¬Q\neg(P \rightarrow Q) \equiv P \wedge \neg Q, we get "deg⁡(v)≥3∧d(u,v)≠1\deg(v) \geq 3 \wedge d(u,v) \neq 1". This gives us ∀G∃v∈V(G)∀u∈V(G)(deg⁡(v)≥3∧d(u,v)≠1)\forall G \exists v \in V(G) \forall u \in V(G) (\deg(v) \geq 3 \wedge d(u,v) \neq 1), which is answer C. Answer A incorrectly keeps ∀G\forall G and ∀v\forall v but should have ∃v\exists v. Answer B fails to flip the first quantifier (keeps ∃G\exists G instead of ∀G\forall G) and incorrectly negates the implication as a disjunction rather than a conjunction. Answer D correctly flips quantifiers but wrongly negates deg⁡(v)≥3\deg(v) \geq 3 to deg⁡(v)<3\deg(v) < 3 when it should remain deg⁡(v)≥3\deg(v) \geq 3 in the negated implication. Remember: when negating implications within quantified statements, the implication P→QP \rightarrow Q becomes P∧¬QP \wedge \neg Q, not ¬P∧¬Q\neg P \wedge \neg Q. Practice this pattern—it's a common source of errors.

Question 10

Consider the statement: "For every integer mm, if mm is odd, then there exists an integer kk such that m3−m=8km^3 - m = 8k." Which statement below is logically equivalent to the negation of this statement?

  1. For every integer mm, if mm is odd, then for every integer kk, we have m3−m≠8km^3 - m \neq 8k.
  2. For every odd integer mm, there exists an integer kk such that m3−m≠8km^3 - m \neq 8k.
  3. There exists an integer mm such that mm is odd and for every integer kk, we have m3−m≠8km^3 - m \neq 8k.
  4. There exists an odd integer mm such that for every integer kk, we have m3−m≠8km^3 - m \neq 8k. (correct answer)
Explanation: When you encounter questions about negating logical statements with quantifiers, remember that negation flips universal quantifiers (∀) to existential quantifiers (∃) and vice versa, while also negating the final conclusion. The original statement has the structure: "For every integer mm, if mm is odd, then there exists an integer kk such that m3−m=8km^3 - m = 8k." In logical notation, this is ∀m:odd(m)→∃k:m3−m=8k\forall m: \text{odd}(m) \rightarrow \exists k: m^3 - m = 8k. To negate this properly, you need to find a counterexample that makes the original statement false. This means finding an odd integer mm where the conclusion fails—that is, where no integer kk satisfies m3−m=8km^3 - m = 8k. The negation becomes: "There exists an odd integer mm such that for every integer kk, we have m3−m≠8km^3 - m \neq 8k." Option D captures this structure perfectly. Option A incorrectly keeps the universal quantifier "for every integer mm" instead of changing it to existential. Option B makes a similar error by maintaining "for every odd integer mm" rather than "there exists." Option C separates the conditions awkwardly—it says "there exists an integer mm such that mm is odd" instead of the more direct "there exists an odd integer mm," though this is logically equivalent to D, making D the more natural choice. When negating complex logical statements, work systematically: flip each quantifier, then negate the final predicate. Practice identifying the logical structure before attempting the negation.

Question 11

Let DD be a domain and P(x,y)P(x,y) a binary predicate. Consider the statement: "For every element xx in DD, if there exists an element yy in DD such that P(x,y)P(x,y), then for all elements zz in DD, we have P(x,z)P(x,z)." Which choice below represents the negation of this statement?

  1. For every element xx in DD, there exists an element yy in DD such that P(x,y)P(x,y) and there exists an element zz in DD such that ¬P(x,z)\neg P(x,z).
  2. There exists an element xx in DD such that there exists an element yy in DD with P(x,y)P(x,y), and there exists an element zz in DD with ¬P(x,z)\neg P(x,z). (correct answer)
  3. There exists an element xx in DD such that for every element yy in DD, if P(x,y)P(x,y), then there exists an element zz in DD with ¬P(x,z)\neg P(x,z).
  4. For every element xx in DD, if there exists an element yy in DD such that P(x,y)P(x,y), then there exists an element zz in DD such that ¬P(x,z)\neg P(x,z).
Explanation: When you encounter negation problems in discrete math, you need to systematically apply De Morgan's laws and the rules for negating quantifiers. The key insight is that negating "for all" gives "there exists" and vice versa, while negating an implication A→BA \rightarrow B gives A∧¬BA \land \neg B. Let's first translate the original statement into logical form: ∀x∈D:(∃y∈D:P(x,y))→(∀z∈D:P(x,z))\forall x \in D: (\exists y \in D: P(x,y)) \rightarrow (\forall z \in D: P(x,z)). To negate this, we need to find when this implication fails for at least one xx. The negation becomes: ∃x∈D:¬[(∃y∈D:P(x,y))→(∀z∈D:P(x,z))]\exists x \in D: \neg[(\exists y \in D: P(x,y)) \rightarrow (\forall z \in D: P(x,z))]. Using the rule that ¬(A→B)≡A∧¬B\neg(A \rightarrow B) \equiv A \land \neg B, this simplifies to: ∃x∈D:(∃y∈D:P(x,y))∧¬(∀z∈D:P(x,z))\exists x \in D: (\exists y \in D: P(x,y)) \land \neg(\forall z \in D: P(x,z)). Finally, ¬(∀z∈D:P(x,z))≡∃z∈D:¬P(x,z)\neg(\forall z \in D: P(x,z)) \equiv \exists z \in D: \neg P(x,z), giving us choice B. Choice A incorrectly keeps the universal quantifier ∀x\forall x instead of changing it to ∃x\exists x. Choice C misunderstands the structure by introducing an unnecessary conditional within the negation. Choice D fails to negate the initial universal quantifier and creates a weaker statement than the true negation. Remember: when negating complex logical statements, work systematically from the outside in, flipping quantifiers and applying De Morgan's laws at each step. Practice with the standard negation patterns until they become automatic.

Question 12

Let R(x,y,z)R(x,y,z) be a ternary relation. Consider the statement: "For every xx, there exist yy and zz such that if y≠zy \neq z, then R(x,y,z)R(x,y,z)." Which of the following is the correct negation?

  1. There exists xx such that for all yy and zz, if y≠zy \neq z, then ¬R(x,y,z)\neg R(x,y,z).
  2. There exists xx such that for all yy and zz, we have y≠zy \neq z and ¬R(x,y,z)\neg R(x,y,z). (correct answer)
  3. For every xx, there exist yy and zz such that y≠zy \neq z and ¬R(x,y,z)\neg R(x,y,z).
  4. There exists xx such that for all yy and zz with y=zy = z, we have ¬R(x,y,z)\neg R(x,y,z).
Explanation: When negating complex logical statements with multiple quantifiers and implications, you must systematically apply De Morgan's laws and the rules for negating quantifiers and conditionals. The original statement is: "For every xx, there exist yy and zz such that if y≠zy \neq z, then R(x,y,z)R(x,y,z)." In logical notation: ∀x∃y∃z(y≠z→R(x,y,z))\forall x \exists y \exists z (y \neq z \rightarrow R(x,y,z)). To negate this, we apply the rules step by step: ∀x\forall x becomes ∃x\exists x, and ∃y∃z\exists y \exists z becomes ∀y∀z\forall y \forall z. Next, we must negate the implication y≠z→R(x,y,z)y \neq z \rightarrow R(x,y,z). The negation of P→QP \rightarrow Q is P∧¬QP \wedge \neg Q. So (y≠z→R(x,y,z))(y \neq z \rightarrow R(x,y,z)) becomes y≠z∧¬R(x,y,z)y \neq z \wedge \neg R(x,y,z). The complete negation is: "There exists xx such that for all yy and zz, we have y≠zy \neq z and ¬R(x,y,z)\neg R(x,y,z)," which matches answer B. Answer A incorrectly keeps the implication structure instead of converting it to a conjunction. Answer C fails to change the universal quantifier on xx to existential. Answer D completely mishandles the implication negation by changing the condition to y=zy = z instead of properly negating y≠z→R(x,y,z)y \neq z \rightarrow R(x,y,z). Remember: when negating implications, ¬(P→Q)≡P∧¬Q\neg(P \rightarrow Q) \equiv P \wedge \neg Q. The antecedent stays the same while only the consequent gets negated, then they're connected by "and" rather than "if-then."

Question 13

Consider the statement: "For every positive integer nn, there exists a prime number pp such that p>n2p > n^2 and pp divides n!+1n! + 1." Which of the following is the correct negation of this statement?

  1. There exists a positive integer nn such that for every prime number pp, either p≤n2p \leq n^2 or pp does not divide n!+1n! + 1. (correct answer)
  2. For every positive integer nn, there exists a prime number pp such that p≤n2p \leq n^2 and pp does not divide n!+1n! + 1.
  3. There exists a positive integer nn such that for every prime number pp, if p>n2p > n^2, then pp does not divide n!+1n! + 1.
  4. For every positive integer nn and every prime number pp, either p≤n2p \leq n^2 or pp does not divide n!+1n! + 1.
Explanation: The original statement has the form ∀n∃p(P(n,p)∧Q(n,p))\forall n \exists p (P(n,p) \land Q(n,p)) where P(n,p)P(n,p) is "p>n2p > n^2" and Q(n,p)Q(n,p) is "pp divides n!+1n! + 1". The negation is ∃n∀p¬(P(n,p)∧Q(n,p))\exists n \forall p \neg(P(n,p) \land Q(n,p)), which becomes ∃n∀p(¬P(n,p)∨¬Q(n,p))\exists n \forall p (\neg P(n,p) \lor \neg Q(n,p)), or "there exists nn such that for every prime pp, either p≤n2p \leq n^2 or pp does not divide n!+1n! + 1". Choice B incorrectly keeps the original quantifier order. Choice C adds an unnecessary conditional structure. Choice D changes both quantifiers incorrectly.

Question 14

Let P(x,y)P(x,y) be a predicate over the domain of positive integers. Consider the statement SS: "There exists a positive integer xx such that for every positive integer y≥xy \geq x, we have P(x,y)P(x,y)." Which of the following is equivalent to ¬S\neg S?

  1. For every positive integer xx, there exists a positive integer y<xy < x such that ¬P(x,y)\neg P(x,y).
  2. There exists a positive integer xx such that for every positive integer y≥xy \geq x, we have ¬P(x,y)\neg P(x,y).
  3. For every positive integer xx, there exists a positive integer y≥xy \geq x such that ¬P(x,y)\neg P(x,y). (correct answer)
  4. There exists a positive integer yy such that for every positive integer x≤yx \leq y, we have ¬P(x,y)\neg P(x,y).
Explanation: When you encounter questions about negating complex logical statements involving quantifiers, the key is systematically applying De Morgan's laws to flip each quantifier and negate the final predicate. The original statement SS can be written symbolically as: ∃x∀y≥x P(x,y)\exists x \forall y \geq x \, P(x,y). To find ¬S\neg S, you must negate this step by step. The negation of "there exists" (∃\exists) becomes "for all" (∀\forall), and the negation of "for all" (∀\forall) becomes "there exists" (∃\exists). Finally, you negate the predicate itself. Starting with ¬(∃x∀y≥x P(x,y))\neg(\exists x \forall y \geq x \, P(x,y)), this becomes ∀x¬(∀y≥x P(x,y))\forall x \neg(\forall y \geq x \, P(x,y)), which further becomes ∀x∃y≥x ¬P(x,y)\forall x \exists y \geq x \, \neg P(x,y). In words: "For every positive integer xx, there exists a positive integer y≥xy \geq x such that ¬P(x,y)\neg P(x,y)." This matches option C exactly. Option A is wrong because it changes the constraint from y≥xy \geq x to y<xy < x, which fundamentally alters the meaning. Option B incorrectly keeps the original quantifier structure (∃x∀y\exists x \forall y) instead of flipping them. Option D scrambles the roles of xx and yy and uses x≤yx \leq y instead of y≥xy \geq x. Remember this pattern: when negating nested quantifiers, flip each quantifier in sequence from outside to inside, preserve all constraints exactly as written, and negate only the final predicate. Practice writing statements symbolically first—it makes the negation process much clearer.

Question 15

Consider the nested quantified statement: "For every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that for all xx, if 0<∣x−a∣<δ0 < |x - a| < \delta, then ∣f(x)−L∣<ϵ|f(x) - L| < \epsilon." Which of the following correctly negates this statement?

  1. There exists ϵ>0\epsilon > 0 and δ>0\delta > 0 such that for all xx, if 0<∣x−a∣<δ0 < |x - a| < \delta, then ∣f(x)−L∣≥ϵ|f(x) - L| \geq \epsilon.
  2. For all ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that there exists xx with 0<∣x−a∣<δ0 < |x - a| < \delta and ∣f(x)−L∣≥ϵ|f(x) - L| \geq \epsilon.
  3. There exists ϵ>0\epsilon > 0 such that for all δ>0\delta > 0, there exists xx with 0<∣x−a∣<δ0 < |x - a| < \delta and ∣f(x)−L∣≥ϵ|f(x) - L| \geq \epsilon. (correct answer)
  4. For all ϵ>0\epsilon > 0 and all δ>0\delta > 0, there exists xx such that 0<∣x−a∣<δ0 < |x - a| < \delta and ∣f(x)−L∣≥ϵ|f(x) - L| \geq \epsilon.
Explanation: When you encounter nested quantifiers, the key to finding the correct negation is systematically moving the negation symbol through each quantifier and logical connective, flipping ∀ to ∃, ∃ to ∀, and negating the final statement. The original statement has the logical structure: ∀ε > 0, ∃δ > 0, ∀x, (P(x) → Q(x)), where P(x) is "0 < |x - a| < δ" and Q(x) is "|f(x) - L| < ε". To negate this, you flip each quantifier as you move through: ∃ε > 0 such that ∀δ > 0, ∃x where ¬(P(x) → Q(x)). Since ¬(P → Q) is equivalent to (P ∧ ¬Q), the final negation becomes: "There exists ε > 0 such that for all δ > 0, there exists x with 0 < |x - a| < δ and |f(x) - L| ≥ ε." This matches choice C. Choice A incorrectly keeps "there exists δ > 0" instead of flipping it to "for all δ > 0," and it maintains the universal quantifier over x. Choice B fails to negate the first quantifier, keeping "for all ε > 0" instead of "there exists ε > 0." Choice D incorrectly negates both the ε and δ quantifiers to universal quantifiers, when only the ε should become existential. Remember: when negating nested quantifiers, work from left to right, flip each quantifier (∀ becomes ∃, ∃ becomes ∀), and carefully negate the final conditional statement using the equivalence ¬(P → Q) ≡ (P ∧ ¬Q).

Question 16

Let SS be the statement: "There exists a real number xx such that for all real numbers y>0y > 0, we have x2+y2<xyx^2 + y^2 < xy." The negation of SS is equivalent to which of the following?

  1. For all real numbers xx, there exists a real number y>0y > 0 such that x2+y2≥xyx^2 + y^2 \geq xy. (correct answer)
  2. There exists a real number xx such that for all real numbers y>0y > 0, we have x2+y2≥xyx^2 + y^2 \geq xy.
  3. For all real numbers xx and all real numbers y>0y > 0, we have x2+y2≥xyx^2 + y^2 \geq xy.
  4. There exists a real number y>0y > 0 such that for all real numbers xx, we have x2+y2≥xyx^2 + y^2 \geq xy.
Explanation: The original statement has the form ∃x∀y(P(y)→Q(x,y))\exists x \forall y (P(y) \rightarrow Q(x,y)) where P(y)P(y) is "y>0y > 0" and Q(x,y)Q(x,y) is "x2+y2<xyx^2 + y^2 < xy". Since the domain restriction y>0y > 0 is built into the quantification, we can treat this as ∃x∀y>0Q(x,y)\exists x \forall y_{>0} Q(x,y). The negation flips both quantifiers and negates the predicate: ∀x∃y>0¬Q(x,y)\forall x \exists y_{>0} \neg Q(x,y), which gives us choice A. Choice B keeps the wrong quantifier order, choice C changes both quantifiers to universal, and choice D swaps the variables' roles entirely.

Question 17

Consider the statement: "For every integer nn, if nn is prime, then there exists a positive integer kk such that n=2k−1n = 2^k - 1." What is the correct negation of this statement?

  1. There exists an integer nn such that nn is prime and for all positive integers kk, n≠2k−1n \neq 2^k - 1. (correct answer)
  2. For every integer nn, if nn is not prime, then for all positive integers kk, n≠2k−1n \neq 2^k - 1.
  3. There exists an integer nn such that nn is not prime or there exists a positive integer kk such that n≠2k−1n \neq 2^k - 1.
  4. For every integer nn, nn is not prime and for all positive integers kk, n≠2k−1n \neq 2^k - 1.
Explanation: The original statement has the form ∀n(P(n)→∃kQ(n,k))\forall n (P(n) \rightarrow \exists k Q(n,k)). Its negation is ∃n¬(P(n)→∃kQ(n,k))\exists n \neg(P(n) \rightarrow \exists k Q(n,k)), which is equivalent to ∃n(P(n)∧¬(∃kQ(n,k)))\exists n (P(n) \wedge \neg(\exists k Q(n,k))), or ∃n(P(n)∧∀k¬Q(n,k))\exists n (P(n) \wedge \forall k \neg Q(n,k)). This translates to: there exists an integer nn such that nn is prime and for all positive integers kk, n≠2k−1n \neq 2^k - 1. Choice B incorrectly negates the antecedent instead of the entire implication. Choice C uses 'or' instead of 'and' when connecting the negated components. Choice D incorrectly applies universal quantification instead of existential.

Question 18

A mathematics professor states: "Every student who studies discrete mathematics for at least 10 hours per week will pass every exam in the course." Which statement correctly negates this claim?

  1. There exists a student who studies discrete mathematics for at least 10 hours per week but fails at least one exam in the course. (correct answer)
  2. Every student who studies discrete mathematics for less than 10 hours per week will fail at least one exam in the course.
  3. There exists a student who studies discrete mathematics for less than 10 hours per week and fails at least one exam in the course.
  4. No student who studies discrete mathematics for at least 10 hours per week will pass every exam in the course.
Explanation: The original statement can be written as ∀s(S(s)→∀eP(s,e))\forall s (S(s) \rightarrow \forall e P(s,e)) where S(s)S(s) means "student ss studies at least 10 hours per week" and P(s,e)P(s,e) means "student ss passes exam ee". The negation is ∃s¬(S(s)→∀eP(s,e))=∃s(S(s)∧¬(∀eP(s,e)))=∃s(S(s)∧∃e¬P(s,e))\exists s \neg(S(s) \rightarrow \forall e P(s,e)) = \exists s (S(s) \wedge \neg(\forall e P(s,e))) = \exists s (S(s) \wedge \exists e \neg P(s,e)). This means there exists a student who studies at least 10 hours per week but fails at least one exam. Choice B incorrectly negates the antecedent. Choice C changes the study time condition incorrectly. Choice D uses universal quantification instead of existential.

Question 19

Consider the statement: "∃x∈N∀y∈N∃z∈N(xy=z2→gcd⁡(x,y)=1)\exists x \in \mathbb{N} \forall y \in \mathbb{N} \exists z \in \mathbb{N} (xy = z^2 \rightarrow \gcd(x,y) = 1)". What is the correct negation of this statement?

  1. ∃x∈N∀y∈N∀z∈N(xy=z2∧gcd⁡(x,y)>1)\exists x \in \mathbb{N} \forall y \in \mathbb{N} \forall z \in \mathbb{N} (xy = z^2 \wedge \gcd(x,y) > 1)
  2. ∃x∈N∃y∈N∀z∈N(xy≠z2∨gcd⁡(x,y)>1)\exists x \in \mathbb{N} \exists y \in \mathbb{N} \forall z \in \mathbb{N} (xy \neq z^2 \vee \gcd(x,y) > 1)
  3. ∀x∈N∃y∈N∃z∈N(xy≠z2∧gcd⁡(x,y)>1)\forall x \in \mathbb{N} \exists y \in \mathbb{N} \exists z \in \mathbb{N} (xy \neq z^2 \wedge \gcd(x,y) > 1)
  4. ∀x∈N∃y∈N∀z∈N(xy=z2∧gcd⁡(x,y)>1)\forall x \in \mathbb{N} \exists y \in \mathbb{N} \forall z \in \mathbb{N} (xy = z^2 \wedge \gcd(x,y) > 1) (correct answer)
Explanation: When negating complex logical statements with multiple quantifiers and implications, you must systematically apply negation rules at each level. The original statement has the structure ∃x∀y∃z(P→Q)\exists x \forall y \exists z (P \rightarrow Q), where PP is "xy=z2xy = z^2" and QQ is "gcd⁡(x,y)=1\gcd(x,y) = 1". To negate this properly, first flip each quantifier: ∃\exists becomes ∀\forall and vice versa. So ∃x∀y∃z\exists x \forall y \exists z becomes ∀x∃y∀z\forall x \exists y \forall z. Next, negate the inner statement (xy=z2→gcd⁡(x,y)=1)(xy = z^2 \rightarrow \gcd(x,y) = 1). Since ¬(P→Q)≡P∧¬Q\neg(P \rightarrow Q) \equiv P \wedge \neg Q, this becomes xy=z2∧gcd⁡(x,y)≠1xy = z^2 \wedge \gcd(x,y) \neq 1. In this context, gcd⁡(x,y)≠1\gcd(x,y) \neq 1 means gcd⁡(x,y)>1\gcd(x,y) > 1 since gcd values are positive integers. The complete negation is ∀x∈N∃y∈N∀z∈N(xy=z2∧gcd⁡(x,y)>1)\forall x \in \mathbb{N} \exists y \in \mathbb{N} \forall z \in \mathbb{N} (xy = z^2 \wedge \gcd(x,y) > 1), which is answer D. Option A fails to flip the first quantifier (keeps ∃x\exists x). Option B incorrectly flips ∀y\forall y to ∃y\exists y but keeps ∃z\exists z, and uses disjunction instead of conjunction when negating the implication. Option C flips quantifiers correctly but incorrectly negates the implication as xy≠z2∧gcd⁡(x,y)>1xy \neq z^2 \wedge \gcd(x,y) > 1 instead of xy=z2∧gcd⁡(x,y)>1xy = z^2 \wedge \gcd(x,y) > 1. Remember: when negating P→QP \rightarrow Q, you get P∧¬QP \wedge \neg Q, not ¬P∧¬Q\neg P \wedge \neg Q.

Question 20

Consider the nested quantified statement: "∀x∈R∃y∈R∀z∈R(x+y>z2→z=0)\forall x \in \mathbb{R} \exists y \in \mathbb{R} \forall z \in \mathbb{R} (x + y > z^2 \rightarrow z = 0)". What is the correct negation?

  1. ∃x∈R∀y∈R∃z∈R(x+y>z2∧z≠0)\exists x \in \mathbb{R} \forall y \in \mathbb{R} \exists z \in \mathbb{R} (x + y > z^2 \wedge z \neq 0) (correct answer)
  2. ∃x∈R∃y∈R∀z∈R(x+y≤z2∨z≠0)\exists x \in \mathbb{R} \exists y \in \mathbb{R} \forall z \in \mathbb{R} (x + y \leq z^2 \vee z \neq 0)
  3. ∀x∈R∃y∈R∃z∈R(x+y>z2∧z≠0)\forall x \in \mathbb{R} \exists y \in \mathbb{R} \exists z \in \mathbb{R} (x + y > z^2 \wedge z \neq 0)
  4. ∃x∈R∀y∈R∀z∈R(x+y≤z2∧z≠0)\exists x \in \mathbb{R} \forall y \in \mathbb{R} \forall z \in \mathbb{R} (x + y \leq z^2 \wedge z \neq 0)
Explanation: To negate ∀x∃y∀z(P→Q)\forall x \exists y \forall z (P \rightarrow Q), we get ∃x∀y∃z¬(P→Q)\exists x \forall y \exists z \neg(P \rightarrow Q), which is ∃x∀y∃z(P∧¬Q)\exists x \forall y \exists z (P \wedge \neg Q). Here, PP is x+y>z2x + y > z^2 and QQ is z=0z = 0, so ¬Q\neg Q is z≠0z \neq 0. The quantifiers flip: ∀x\forall x becomes ∃x\exists x, ∃y\exists y becomes ∀y\forall y, and ∀z\forall z becomes ∃z\exists z. Choice B incorrectly keeps ∃y\exists y as ∃y\exists y and negates the antecedent. Choice C fails to negate the first quantifier. Choice D incorrectly negates the antecedent and changes the final quantifier incorrectly.