Discrete Math Quiz: Truth Of Quantified Statements

19 questions · exam conditions

0:00

Truth Of Quantified StatementsQuestion 1 of 19

Consider predicates $P(x)$ : " $x < 3$ " and $Q(x)$ : " $x$ is odd" over domain $\{1, 2, 3, 4, 5\}$ . Let $W$ be the statement " $\exists x (P(x) \land Q(x)) \leftrightarrow \forall x (P(x) \rightarrow Q(x))$ ". Which analysis of $W$ is correct?

W

is true because both sides of the biconditional have the same truth value

W

is true because the biconditional statement is tautologically true

W

is false because the left side is false but the right side is true

W

is false because the left side is true but the right side is false

← Back to quizzes

Discrete Math Quiz

Discrete Math Quiz: Truth Of Quantified Statements

Practice Truth Of 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 Truth Of 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.

All questions

Question 1

$W$ is true because both sides of the biconditional have the same truth value
$W$ is true because the biconditional statement is tautologically true
$W$ is false because the left side is false but the right side is true
$W$ is false because the left side is true but the right side is false (correct answer)

Explanation: When evaluating biconditional statements involving quantifiers, you need to determine the truth value of each side independently, then check if they match. Let's analyze each side of

W

: "

\exists x (P(x) \land Q(x)) \leftrightarrow \forall x (P(x) \rightarrow Q(x))

" over domain

\{1, 2, 3, 4, 5\}

. Left side:

\exists x (P(x) \land Q(x))

asks "Does there exist an

x

where

x < 3

AND

x

is odd?"

For $x = 1$ : $P(1)$ is true ( $1 < 3$ ) and $Q(1)$ is true (1 is odd)
So $P(1) \land Q(1)$ is true, making the existential statement true.

Right side:

\forall x (P(x) \rightarrow Q(x))

asks "For all

x

, if

x < 3

, then

x

is odd?"

For $x = 1$ : $1 < 3$ and 1 is odd, so $P(1) \rightarrow Q(1)$ is true
For $x = 2$ : $2 < 3$ but 2 is even, so $P(2) \rightarrow Q(2)$ is false
Since one case fails, the universal statement is false.

The left side is true but the right side is false, so the biconditional

W

is false. Answer D correctly identifies this situation. Answer A wrongly claims both sides have the same truth value. Answer B incorrectly suggests the biconditional is always true regardless of the predicates. Answer C reverses which side is true and which is false. Study tip: Always evaluate quantified statements systematically by checking each domain element, especially when mixing existential and universal quantifiers.

Question 2

Let $R(x, y)$ be the relation " $|x - y| \leq 2$ " over the domain $\{1, 2, 3, 4, 5, 6\}$ . Consider the statement $S$ : " $\forall x \exists y (R(x, y) \land y > x + 1)$ ". Which of the following correctly describes the truth value of $S$ and provides the best reasoning?

$S$ is true because for every $x$ , we can find some $y$ satisfying both conditions within the domain
$S$ is false because when $x = 5$ , no $y$ in the domain satisfies $y > x + 1 = 6$
$S$ is false because when $x = 6$ , no $y$ in the domain satisfies both $|6 - y| \leq 2$ and $y > 7$ (correct answer)
$S$ is true because the relation $R(x, y)$ is reflexive and symmetric over the given domain

Explanation: For

S

to be true, every

x

must have some

y

such that

|x - y| \leq 2

and

y > x + 1

. When

x = 6

, we need

y > 7

and

|6 - y| \leq 2

. The second condition means

4 \leq y \leq 8

, but our domain only goes up to 6. So we need

y \in \{4, 5, 6\}

and

y > 7

, which is impossible. (A) is wrong because we can't always find such

y

. (B) is wrong because when

x = 5

, we could have

y = 6

(since

|5-6| = 1 \leq 2

but

6 \not> 6

), but we need

y > 6

. (D) is irrelevant to the specific statement.

Question 3

Let $L(x, y)$ represent " $x \leq y$ " over the finite domain $\{1, 2, 3\}$ . Consider the statement $N$ : " $\neg(\exists x \forall y L(x, y)) \land \neg(\forall x \exists y \neg L(x, y))$ ". Which of the following correctly describes what $N$ asserts?

There is no smallest element, and every element has some element it does not exceed
There is no element that is $\leq$ all others, and it's not true that every element has something greater
No element is minimal, and no element is maximal in the ordering
There is no element $\leq$ everything, and not every element fails to be $\leq$ something (correct answer)

Explanation:

N

has two conjuncts:

\neg(\exists x \forall y L(x, y))

means "there is no

x

such that

x \leq y

for all

y

", i.e., "no element is

\leq

everything." The second part

\neg(\forall x \exists y \neg L(x, y))

means "it's not the case that for every

x

, there exists

y

such that

x \not\leq y

", which is equivalent to "not every element fails to be

\leq

something." Option (D) correctly captures both parts of this logical structure.

Question 4

Consider the domain $D = \{-2, -1, 0, 1, 2\}$ and predicates $P(x)$ : " $x^2 \leq 1$ " and $Q(x)$ : " $x + 1 > 0$ ". Which of the following compound quantified statements has the same truth value as $\neg(\exists x (P(x) \land \neg Q(x)))$ ?

$\forall x (P(x) \rightarrow Q(x))$ (correct answer)
$\forall x (\neg P(x) \lor Q(x))$
$\exists x (\neg P(x) \land Q(x))$
$\neg(\forall x (\neg P(x) \lor Q(x)))$

Explanation: First, apply De Morgan's law:

\neg(\exists x (P(x) \land \neg Q(x))) \equiv \forall x \neg(P(x) \land \neg Q(x)) \equiv \forall x (\neg P(x) \lor \neg\neg Q(x)) \equiv \forall x (\neg P(x) \lor Q(x))

. Since

\neg P(x) \lor Q(x) \equiv P(x) \rightarrow Q(x)

, the answer is (A). Let's verify:

P(x)

is true for

x \in \{-1, 0, 1\}

and

Q(x)

is true for

x \in \{0, 1, 2\}

. The original statement asks if there exists

x

where

P(x)

is true but

Q(x)

is false. This happens when

x = -1

. So the original is true, meaning its negation is false. Statement (A) says whenever

P(x)

is true,

Q(x)

must be true, which fails for

x = -1

, so (A) is false. This matches.

Question 5

Let $Q(x)$ be " $x$ is even" and $R(x)$ be " $x > 2$ " over the domain $\{1, 2, 3, 4, 5, 6\}$ . Consider the statement $T$ : " $\exists x (Q(x) \land R(x)) \rightarrow \forall y (Q(y) \rightarrow R(y))$ ". What is the truth value of $T$ ?

True, because the antecedent is true and the consequent is true
True, because the antecedent is false, making the implication vacuously true
False, because the antecedent is true but the consequent is false (correct answer)
False, because both the antecedent and consequent are false

Explanation: First, evaluate the antecedent

\exists x (Q(x) \land R(x))

: this asks if there's an even number greater than 2. Yes:

x = 4

and

x = 6

are both even and greater than 2. So the antecedent is true. Next, evaluate the consequent

\forall y (Q(y) \rightarrow R(y))

: this asks if every even number is greater than 2. The even numbers in the domain are

\{2, 4, 6\}

. Since

2

is even but not greater than 2, the statement

Q(2) \rightarrow R(2)

is false (true

\rightarrow

false). Therefore the consequent is false. Since we have true

\rightarrow

false, the overall statement

T

is false.

Question 6

Consider the domain $D = \{0, 1, 2, 3\}$ and let $R(x, y, z)$ be the relation " $x + y = z$ ". Which of the following statements about the quantification $\forall x \exists y \exists z R(x, y, z)$ is correct?

The statement is true because for any $x$ , we can choose $y = 0$ and $z = x$ to satisfy the relation (correct answer)
The statement is false because when $x = 3$ , no values of $y, z \in D$ can satisfy $3 + y = z$
The statement is true because the relation $R$ is reflexive over the domain $D$
The statement is false because we need $z \leq 3$ but $x + y$ might exceed this bound for some choices

Explanation: The statement asks whether for every

x \in D

, there exist

y, z \in D

such that

x + y = z

. For

x = 0

: choose

y = 0, z = 0

giving

0 + 0 = 0

✓. For

x = 1

: choose

y = 0, z = 1

giving

1 + 0 = 1

✓. For

x = 2

: choose

y = 0, z = 2

giving

2 + 0 = 2

✓. For

x = 3

: choose

y = 0, z = 3

giving

3 + 0 = 3

✓. So (A) is correct. (B) is wrong because

y = 0, z = 3

works for

x = 3

. (C) is irrelevant since reflexivity doesn't apply to a three-place relation in this context. (D) identifies a potential issue but misses that we can always choose

y = 0

Question 7

Consider the domain $D = \{a, b, c\}$ where $a = 1, b = 2, c = 4$ . Let $S(x, y)$ mean " $x$ divides $y$ " and consider the statement $U$ : " $\neg(\exists x \exists y (S(x, y) \land S(y, x) \land x \neq y))$ ". Which interpretation of $U$ is correct?

There exist distinct elements that divide each other mutually in the domain
No two distinct elements in the domain divide each other mutually (correct answer)
Every element in the domain divides itself but no other element
At least one element in the domain divides all other elements

Explanation: The statement

U

\neg(\exists x \exists y (S(x, y) \land S(y, x) \land x \neq y))

, which means "it is not the case that there exist distinct

x, y

such that

x

divides

y

AND

y

divides

x

". This is equivalent to saying "no two distinct elements divide each other mutually". Let's verify: In our domain,

1|2, 1|4, 2|4

, but

2 \nmid 1, 4 \nmid 1, 4 \nmid 2

. So indeed, no two distinct elements have mutual divisibility. (A) contradicts

U

. (C) is true but doesn't capture what

U

says. (D) is about one element dividing others, not mutual divisibility.

Question 8

Let $T(x, y)$ be the statement " $x^2 + y^2 = 5$ " over the domain of integers from $-3$ to $3$ . Consider the nested quantification $\exists x \forall y T(x, y)$ . Which statement best explains why this quantification is false?

No single value of $x$ makes $T(x, y)$ true for every possible $y$ in the domain (correct answer)
The equation $x^2 + y^2 = 5$ has no integer solutions within the given domain
For each $x$ , there exists at least one $y$ such that $T(x, y)$ is false
The statement $\forall y \exists x T(x, y)$ would be true instead of the given quantification

Explanation: The statement

\exists x \forall y T(x, y)

asks whether there exists some

x

such that for ALL values of

y

in the domain,

x^2 + y^2 = 5

. This would require finding an

x

where the equation holds for

y = -3, -2, -1, 0, 1, 2, 3

simultaneously, which is impossible since

y^2

takes different values. (B) is wrong because there are solutions like

(1, 2), (2, 1), (-1, 2)

, etc. (C) describes why

\forall x \forall y T(x, y)

is false, not our statement. (D) is irrelevant to why the given statement is false.

Question 9

Let $F(x, y)$ be the statement " $x^2 - y^2 = 3$ " over the domain $\{-2, -1, 0, 1, 2\}$ . Consider the quantified statement $V$ : " $\forall x \exists y F(x, y) \lor \exists x \forall y \neg F(x, y)$ ". What is the truth value of $V$ ?

True, because the first disjunct is true
True, because at least one solution to the equation exists in the domain
False, because both disjuncts are false
True, because the second disjunct is true (correct answer)

Explanation: When you encounter complex quantified statements with disjunctions, break them down systematically by evaluating each part separately over the given domain. Let's analyze statement

V

: "

\forall x \exists y F(x, y) \lor \exists x \forall y \neg F(x, y)

" where

F(x, y)

means "

x^2 - y^2 = 3

" and our domain is

\{-2, -1, 0, 1, 2\}

. The first disjunct "

\forall x \exists y F(x, y)

" asks: "For every

x

, does there exist some

y

such that

x^2 - y^2 = 3

?" Testing

x = 0

: we need

0 - y^2 = 3

, so

y^2 = -3

. Since no real number squared gives

-3

, no

y

in our domain works for

x = 0

. The first disjunct is false. The second disjunct "

\exists x \forall y \neg F(x, y)

" asks: "Does there exist some

x

such that for all

y

F(x, y)

is false?" Again using

x = 0

: for every

y

in the domain,

0^2 - y^2 = 3

becomes

-y^2 = 3

, which is impossible. So

\neg F(0, y)

is true for all

y

. The second disjunct is true. Since

V

is a disjunction and the second part is true,

V

is true. Answer A is wrong because the first disjunct is actually false. Answer B misses the logical structure entirely—individual solutions don't determine the truth of universal quantifiers. Answer C incorrectly claims both disjuncts are false when the second is true. Remember: in logic problems, always check quantified statements systematically by testing specific values, especially potential counterexamples like boundary cases.

Question 10

Let $P(x)$ be the statement " $x^2 + 3x - 4 > 0$ " and let the domain be the set of integers from $-5$ to $5$ inclusive. Which of the following statements about the quantified expressions is true?

$\exists x P(x)$ is true and $\forall x P(x)$ is true
$\exists x P(x)$ is true and $\forall x P(x)$ is false (correct answer)
$\exists x P(x)$ is false and $\forall x P(x)$ is true
$\exists x P(x)$ is false and $\forall x P(x)$ is false

Explanation: First, solve

x^2 + 3x - 4 > 0

. Factoring:

(x+4)(x-1) > 0

. This is true when both factors have the same sign, which occurs when

x < -4

x > 1

. In the domain

\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}

P(x)

is true for

x = -5, 2, 3, 4, 5

. Since at least one value makes

P(x)

true,

\exists x P(x)

is true. Since not all values make

P(x)

true (e.g.,

P(-3)

is false),

\forall x P(x)

is false.

Question 11

Let $P(n)$ be the statement " $n$ is prime" over the domain $\{10, 11, 12, 13, 14, 15\}$ . Consider the statement $S$ : " $\neg(\forall n P(n)) \land \neg(\forall n \neg P(n))$ ". Which of the following best describes what $S$ asserts about the domain?

All numbers in the domain are composite, and no numbers in the domain are prime
Not all numbers in the domain are prime, and not all numbers are composite (correct answer)
Some numbers in the domain are prime, and some numbers in the domain are composite
Either all numbers are prime or all numbers are composite, but not both

Explanation:

S

states

\neg(\forall n P(n)) \land \neg(\forall n \neg P(n))

. The first part

\neg(\forall n P(n))

means "not all

n

are prime," which is equivalent to "at least one

n

is not prime." The second part

\neg(\forall n \neg P(n))

means "not all

n

are non-prime," which is equivalent to "at least one

n

is prime." Together, this means the domain contains both prime and non-prime numbers. (A) contradicts the statement. (C) is close but uses "some" which could be interpreted as stronger than "at least one." (D) contradicts

S

entirely. (B) correctly captures that it's not the case that all are prime, and not the case that all are composite.

Question 12

Let $M(x, y)$ be " $\max(x, y) = x + 1$ " over domain $\{0, 1, 2, 3\}$ . Consider the statement $Z$ : " $\exists x \forall y M(x, y) \lor \forall x \exists y M(x, y)$ ". Which reasoning correctly determines the truth value of $Z$ ?

$Z$ is true because the first disjunct holds with $x = 3$
$Z$ is true because the second disjunct holds since each $x$ has a suitable $y$
$Z$ is false because neither disjunct can be satisfied in the given domain (correct answer)
$Z$ is true because $\max(x, y) = x + 1$ when $y = x + 1$ for valid values

Explanation: First disjunct:

\exists x \forall y M(x, y)

asks if there's an

x

such that for ALL

y

\max(x, y) = x + 1

. For this to work, we'd need

\max(x, 0) = \max(x, 1) = \max(x, 2) = \max(x, 3) = x + 1

. This is impossible since

\max(x, y)

varies with

y

. Second disjunct:

\forall x \exists y M(x, y)

asks if for every

x

, there's some

y

with

\max(x, y) = x + 1

. For

x = 3

: need

\max(3, y) = 4

, but

4 \notin \{0,1,2,3\}

, so

\max(3, y) \leq 3

always. Thus no

y

works for

x = 3

. Both disjuncts are false, so

Z

is false.

Question 13

Consider the predicate $S(x, y): |x - y| < 2$ where the domain for both variables is $\{0, 1, 2, 3, 4\}$ . Evaluate the truth of the nested quantified statement $\forall x \exists y \, (y \neq x \wedge S(x, y))$ .

True, because for every x, there is always a different y within distance 2 (correct answer)
False, because when x = 0, the only y satisfying |x - y| < 2 is y = 1
False, because when x = 4, no different y satisfies |4 - y| < 2 in the domain
True, because the domain is large enough that boundary cases still work

Explanation: We need to verify that for each x in {0,1,2,3,4}, there exists a y ≠ x such that |x - y| < 2. For x = 0: |0 - y| < 2 means -2 < -y < 2, so -2 < y < 2. In our domain, y ∈ {1} works (since y ≠ 0). For x = 1: |1 - y| < 2 means -2 < 1-y < 2, so -1 < y < 3. Values y ∈ {0, 2} work. For x = 2: |2 - y| < 2 means -2 < 2-y < 2, so 0 < y < 4. Values y ∈ {1, 3} work. For x = 3: |3 - y| < 2 means -2 < 3-y < 2, so 1 < y < 5. Values y ∈ {2, 4} work. For x = 4: |4 - y| < 2 means -2 < 4-y < 2, so 2 < y < 6. Only y = 3 works in our domain. Since every x has at least one valid y ≠ x, the statement is true.

Question 14

Let $V(m, n)$ represent the statement " $\gcd(m, n) = 1$ " where the domain is $\{2, 3, 4, 5, 6\}$ . Which of the following statements about the truth values is correct?

$\exists m \forall n \, V(m, n)$ is true because m = 3 is coprime to all other values
$\forall m \exists n \, V(m, n)$ is false because m = 4 shares factors with multiple values
$\forall m \exists n \, (m \neq n \wedge V(m, n))$ is true for this domain (correct answer)
$\exists m \forall n \, (m \neq n \rightarrow V(m, n))$ is true because some m works for all different n

Explanation: Let's analyze each option systematically. For option A: We need one m that is coprime to ALL n in the domain. Let's check m = 3: gcd(3,2) = 1, gcd(3,3) = 3, gcd(3,4) = 1, gcd(3,5) = 1, gcd(3,6) = 3. Since gcd(3,3) ≠ 1 and gcd(3,6) ≠ 1, this is false. For option B: We need every m to have some n where gcd(m,n) = 1. For m = 4: gcd(4,3) = 1, gcd(4,5) = 1, so this works. Similarly, every m in the domain has at least one n where they're coprime, so this statement is actually true. For option C: For each m, we need some different n where gcd(m,n) = 1. m = 2: gcd(2,3) = 1, gcd(2,5) = 1. m = 3: gcd(3,2) = 1, gcd(3,4) = 1, gcd(3,5) = 1. m = 4: gcd(4,3) = 1, gcd(4,5) = 1. m = 5: gcd(5,2) = 1, gcd(5,3) = 1, gcd(5,4) = 1, gcd(5,6) = 1. m = 6: gcd(6,5) = 1. Every m has at least one different n that is coprime to it, so this is true.

Question 15

Let $Z(a, b)$ be the predicate " $a^2 - b^2$ is divisible by 4" where the domain for both variables is $\{1, 2, 3, 4, 5\}$ . Which statement correctly describes the truth value of $\exists a \forall b \, Z(a, b)$ ?

True, because when a = 2, we have 4 - b² ≡ 0 (mod 4) for all b
False, because a² - b² = (a-b)(a+b) cannot be divisible by 4 for all b
True, because when a = 4, we have 16 - b² ≡ 0 (mod 4) for all b
False, because no value of a makes a² - b² divisible by 4 for all values of b (correct answer)

Explanation: We need to find if there exists an 'a' such that a² - b² is divisible by 4 for ALL values of b in {1,2,3,4,5}. Note that a² - b² ≡ 0 (mod 4) means a² ≡ b² (mod 4). First, let's find the possible values of squares mod 4: 1² ≡ 1, 2² ≡ 0, 3² ≡ 1, 4² ≡ 0, 5² ≡ 1 (mod 4). So squares can only be ≡ 0 or 1 (mod 4). For any fixed a, we need a² ≡ b² (mod 4) for all b. But since b² takes both values 0 and 1 (mod 4) as b varies, and a² can only be one specific value (either 0 or 1), there's no single a that works for all b. For example, if a = 2 (so a² ≡ 0 (mod 4)), then when b = 1, we get a² - b² ≡ 0 - 1 ≡ 3 (mod 4), which is not divisible by 4.

Question 16

Let $T(a, b, c)$ be the statement " $a + b + c$ is divisible by 3" where $a, b, c \in \{0, 1, 2\}$ . Which of the following is the correct truth value of $\exists a \forall b \exists c \, T(a, b, c)$ ?

True, because a = 0 works: for any b, we can choose c ≡ -b (mod 3) (correct answer)
False, because no single value of a works for all possible values of b
True, because a = 1 works: for any b, we can choose c ≡ 2-b (mod 3)
False, because when a = 2 and b = 1, no value of c makes the sum divisible by 3

Explanation: We need to find if there exists an 'a' such that for every 'b', there exists a 'c' making a + b + c ≡ 0 (mod 3). Let's try a = 0: For any b ∈ {0,1,2}, we need 0 + b + c ≡ 0 (mod 3), so c ≡ -b ≡ 3-b (mod 3). When b = 0: c ≡ 0 (mod 3), so c = 0 works. When b = 1: c ≡ 2 (mod 3), so c = 2 works. When b = 2: c ≡ 1 (mod 3), so c = 1 works. Since for a = 0, we can always find an appropriate c for any b, the statement is true. We could also verify that a = 1 and a = 2 work similarly, but we only need one value of a to make the existential quantifier true.

Question 17

Consider the three-variable predicate $M(x, y, z): x \cdot y = z$ where all variables range over the domain $\{0, 1, 2\}$ . Evaluate the truth of the statement $\forall x \forall z \exists y \, M(x, y, z)$ .

True, because for any x and z, we can solve xy = z for y when x ≠ 0
False, because when x = 0 and z = 1, no y satisfies 0 · y = 1 (correct answer)
False, because when x = 2 and z = 1, no y in the domain satisfies 2y = 1
True, because the domain is small enough that all equations xy = z are solvable

Explanation: We need to check if for every pair (x,z), there exists a y in {0,1,2} such that xy = z. Let's systematically check all cases: For x = 0: We need 0·y = z, so z must be 0. When z = 0, any y works. When z = 1 or z = 2, no y works since 0·y = 0 always. For x = 1: We need 1·y = z, so y = z. For any z ∈ {0,1,2}, we can choose y = z. For x = 2: We need 2·y = z. When z = 0, y = 0 works. When z = 1, we need y = 1/2, but 1/2 ∉ {0,1,2}. When z = 2, y = 1 works. Since the case x = 0, z = 1 has no solution (we need 0·y = 1, which is impossible), the statement is false.

Question 18

Consider the predicate $W(x, y): x + y \text{ is prime}$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$ . Determine the truth value of $\forall x \exists y \, (x < y \wedge W(x, y))$ .

True, because for each x, there is always a larger y making x + y prime
False, because when x = 6, there is no y > 6 in the domain
False, because when x = 4, no y > 4 in the domain makes x + y prime (correct answer)
True, because the constraint x < y provides enough flexibility for prime sums

Explanation: We need to check if for every x, there exists a y > x such that x + y is prime. Let's check each x: For x = 1: y ∈ {2,3,4,5,6}. Sums: 3,4,5,6,7. Primes: 3,5,7. So y = 2,4,6 work. For x = 2: y ∈ {3,4,5,6}. Sums: 5,6,7,8. Primes: 5,7. So y = 3,5 work. For x = 3: y ∈ {4,5,6}. Sums: 7,8,9. Primes: 7. So y = 4 works. For x = 4: y ∈ {5,6}. Sums: 9,10. 9 = 3² and 10 = 2×5, so neither is prime. No y works. For x = 5: y ∈ {6}. Sum: 11, which is prime. So y = 6 works. For x = 6: No y > 6 in domain. Since x = 4 has no suitable y, the statement is false.

Question 19

Consider the predicate $U(x, y): x^2 \equiv y \pmod{5}$ where both $x$ and $y$ are from the set $\{0, 1, 2, 3, 4\}$ . Determine which quantified statement is FALSE.

$\forall y \exists x \, U(x, y)$ (correct answer)
$\exists x \exists y \, U(x, y)$
$\exists y \forall x \, U(x, y)$
$\forall x \exists y \, U(x, y)$

Explanation: First, let's find the possible values of x² (mod 5): 0² ≡ 0, 1² ≡ 1, 2² ≡ 4, 3² ≡ 4, 4² ≡ 1 (mod 5). So x² can only be congruent to 0, 1, or 4 modulo 5. For option A (∀y ∃x U(x,y)): This claims that for every y, there exists an x such that x² ≡ y (mod 5). But when y = 2 or y = 3, no x satisfies x² ≡ y (mod 5) since x² can only be 0, 1, or 4 (mod 5). Therefore, this statement is false. For option B: This claims some x and y satisfy the relation, which is true (e.g., x = 0, y = 0). For option C: This claims there exists a y such that x² ≡ y (mod 5) for all x. This is false since x² takes on different values (0, 1, 4) for different x values, so no single y works for all x. For option D: This claims that for every x, there exists some y such that x² ≡ y (mod 5), which is true since we can choose y = x² mod 5.