Discrete Math Quiz: Relation Properties

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Relation PropertiesQuestion 1 of 19

Consider the relation $W$ on the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ where $fWg$ if and only if $f(0) = g(0)$ . Which properties does $W$ have?

W

is reflexive and symmetric but not transitive

W

is reflexive and transitive but not symmetric

W

is symmetric and transitive but not reflexive

W

is reflexive, symmetric, and transitive

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Discrete Math Quiz

Discrete Math Quiz: Relation Properties

Practice Relation Properties 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 Relation Properties, 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

Consider the relation $W$ on the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ where $fWg$ if and only if $f(0) = g(0)$ . Which properties does $W$ have?

$W$ is reflexive and symmetric but not transitive
$W$ is reflexive and transitive but not symmetric
$W$ is symmetric and transitive but not reflexive
$W$ is reflexive, symmetric, and transitive (correct answer)

Explanation: W is reflexive because f(0) = f(0) for any function f, so fWf holds. W is symmetric because if f(0) = g(0), then g(0) = f(0), so if fWg then gWf. W is transitive because if f(0) = g(0) and g(0) = h(0), then f(0) = h(0), so if fWg and gWh then fWh. Therefore W is an equivalence relation on the set of all functions from ℝ to ℝ.

Question 2

Let $X$ be the relation on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a,b)X(c,d)$ if and only if $a + d = b + c$ . Determine which properties $X$ satisfies.

$X$ is reflexive and symmetric but not transitive
$X$ is reflexive and transitive but not symmetric
$X$ is reflexive, symmetric, and transitive (correct answer)
$X$ is symmetric and transitive but not reflexive

Explanation: When you encounter a relation and need to determine its properties, systematically check each property using the definitions. A relation is reflexive if every element relates to itself, symmetric if

aRb

implies

bRa

, and transitive if

aRb

and

bRc

together imply

aRc

. For relation

X

where

(a,b)X(c,d)

if and only if

a + d = b + c

, let's verify each property: Reflexive: Does

(a,b)X(a,b)

for all

(a,b)

? We need

a + b = b + a

, which is always true by commutativity of addition. So

X

is reflexive. Symmetric: If

(a,b)X(c,d)

, does

(c,d)X(a,b)

? If

a + d = b + c

, then

c + b = d + a

(just rearranging the same equation). So

X

is symmetric. Transitive: If

(a,b)X(c,d)

and

(c,d)X(e,f)

, does

(a,b)X(e,f)

? From the first relation:

a + d = b + c

. From the second:

c + f = d + e

. Adding these equations:

a + d + c + f = b + c + d + e

. Canceling

c + d

from both sides gives

a + f = b + e

, which means

(a,b)X(e,f)

. So

X

is transitive. Since

X

satisfies all three properties, answer C is correct. Option A incorrectly claims transitivity fails, B incorrectly claims symmetry fails, and D incorrectly claims reflexivity fails. Study tip: When checking relation properties, work through the algebra carefully and remember that equivalence relations (like this one) always satisfy all three properties.

Question 3

Let $Z$ be a relation on $\mathbb{N}$ where $aZb$ if and only if $a$ and $b$ have the same number of prime factors (counting multiplicities). Which statement is correct about $Z$ ?

$Z$ is reflexive, symmetric, and transitive (correct answer)
$Z$ is reflexive and transitive but not symmetric
$Z$ is symmetric and transitive but not reflexive
$Z$ is reflexive and symmetric but not transitive

Explanation: When analyzing relations, you need to systematically check three key properties: reflexivity (every element relates to itself), symmetry (if

a

relates to

b

, then

b

relates to

a

), and transitivity (if

a

relates to

b

and

b

relates to

c

, then

a

relates to

c

). Let's examine relation

Z

where

aZb

means

a

and

b

have the same number of prime factors with multiplicities. For example,

12 = 2^2 \cdot 3

has 3 prime factors (counting

2

twice), so

12Z8

since

8 = 2^3

also has 3 prime factors. Reflexivity: Every natural number has the same number of prime factors as itself, so

aZa

always holds. Symmetry: If

a

and

b

have the same number of prime factors, then

b

and

a

also have the same number of prime factors. The relationship is perfectly symmetric. Transitivity: If

a

and

b

have

n

prime factors, and

b

and

c

have

n

prime factors, then

a

and

c

both have

n

prime factors, so

aZc

holds. Since

Z

satisfies all three properties, it's an equivalence relation, making choice A correct. Choice B incorrectly claims

Z

isn't symmetric, but we've shown it is. Choice C wrongly states

Z

isn't reflexive, but every number relates to itself. Choice D incorrectly claims

Z

isn't transitive, but having the same count is clearly transitive. Study tip: When checking if a relation is an equivalence relation, systematically verify all three properties with concrete examples. Relations based on "sameness" (like equal counts) are often equivalence relations.

Question 4

Let $T$ be a relation on the set of all non-empty subsets of $\{1, 2, 3\}$ where $ATB$ if and only if $\max(A) \leq \max(B)$ . Which statement correctly describes $T$ ?

$T$ is reflexive and transitive but not symmetric (correct answer)
$T$ is symmetric and transitive but not reflexive
$T$ is reflexive and symmetric but not transitive
$T$ is reflexive, symmetric, and transitive

Explanation: The non-empty subsets are: {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}. T is reflexive since max(A) ≤ max(A) for any set A. T is not symmetric: consider A = {1} and B = {2}. We have max({1}) = 1 ≤ 2 = max({2}), so ATB. However, max({2}) = 2 ≰ 1 = max({1}), so BTA does not hold. T is transitive: if ATB and BTC, then max(A) ≤ max(B) and max(B) ≤ max(C), which implies max(A) ≤ max(C), so ATC. Therefore T is reflexive and transitive but not symmetric.

Question 5

Let $S = \{1, 2, 3, 4\}$ and define relation $R$ on $S$ by $aRb$ if and only if $a^2 \equiv b^2 \pmod{3}$ . Which combination of properties does $R$ possess?

Reflexive and symmetric, but not transitive
Reflexive, symmetric, and transitive (equivalence relation) (correct answer)
Symmetric and transitive, but not reflexive
Reflexive and transitive, but not symmetric

Explanation: First, compute squares mod 3:

1^2 \equiv 1 \pmod{3}

2^2 \equiv 1 \pmod{3}

3^2 \equiv 0 \pmod{3}

4^2 \equiv 1 \pmod{3}

. So

R = \{(1,1), (1,2), (1,4), (2,1), (2,2), (2,4), (3,3), (4,1), (4,2), (4,4)\}

. Reflexive: All pairs

(a,a)

are present since

a^2 \equiv a^2 \pmod{3}

always. Symmetric: If

a^2 \equiv b^2 \pmod{3}

, then

b^2 \equiv a^2 \pmod{3}

. Transitive: If

a^2 \equiv b^2 \pmod{3}

and

b^2 \equiv c^2 \pmod{3}

, then

a^2 \equiv c^2 \pmod{3}

. Choice A is wrong because transitivity holds. Choice C is wrong because reflexivity holds. Choice D is wrong because symmetry holds.

Question 6

Consider the relation $P$ on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a,b)P(c,d)$ if and only if $ad = bc$ . Determine which properties $P$ satisfies.

Symmetric and transitive but not reflexive, because pairs with zero components can't satisfy the multiplicative condition $ab = ba$
Reflexive and symmetric but not transitive, since the cross-multiplication condition doesn't chain properly through three pairs
Reflexive, symmetric, and transitive, representing the equivalence of rational numbers when treating pairs as fractions (correct answer)
Reflexive and transitive but not symmetric, since $ad = bc$ doesn't necessarily imply $cb = da$ due to order of operations

Explanation: When you encounter a relation on ordered pairs with a cross-multiplication condition like

ad = bc

, you should immediately think about rational number equivalence. This relation essentially treats pairs

(a,b)

as fractions

\frac{a}{b}

and asks when two "fractions" are equal. Let's verify each property systematically. For reflexivity, we need

(a,b)P(a,b)

, which means

a \cdot b = b \cdot a

. This is always true by commutativity of multiplication, even when components are zero (since

0 \cdot anything = 0

). For symmetry, if

(a,b)P(c,d)

, then

ad = bc

. We need

(c,d)P(a,b)

, which requires

cb = da

. Since multiplication is commutative,

cb = bc = ad = da

, so symmetry holds. For transitivity, if

(a,b)P(c,d)

and

(c,d)P(e,f)

, then

ad = bc

and

cf = de

. We need

af = be

. From the equations:

ad \cdot cf = bc \cdot de

, so

acdf = bcde

. If

c \neq 0

, we can divide both sides by

cd

to get

af = be

. When

c = 0

, both original equations force specific constraints that still preserve transitivity. Choice A incorrectly claims the relation isn't reflexive due to zero components, but

ab = ba

always holds. Choice B wrongly states transitivity fails, but the cross-multiplication does chain properly. Choice D incorrectly claims symmetry fails, missing that

ad = bc

is equivalent to

cb = da

by commutativity. Study tip: Relations involving cross-multiplication typically model rational number equivalence. Always check if zero cases break your reasoning, but remember that

0 \cdot x = 0 \cdot y

for any

x,y

Question 7

Define relation $Q$ on the set of all non-empty subsets of $\{1, 2, 3\}$ by $AQB$ if and only if $\max(A) \leq \max(B)$ . Which statement correctly identifies the properties of $Q$ ?

Reflexive and transitive, but not symmetric because $\max(\{1,2\}) \leq \max(\{2,3\})$ while $\max(\{2,3\}) \not\leq \max(\{1,2\})$ (correct answer)
Reflexive and symmetric, but not transitive since maximum operations don't preserve ordering through chains of three sets
Symmetric and transitive, but not reflexive since some subsets don't relate to themselves under maximum comparison
Only transitive, lacking both reflexivity and symmetry due to the strict ordering imposed by maximum elements

Explanation: The domain is

\{\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}

. Reflexive:

\max(A) \leq \max(A)

always holds. Transitive: If

\max(A) \leq \max(B)

and

\max(B) \leq \max(C)

, then

\max(A) \leq \max(C)

by transitivity of

\leq

. Not symmetric:

\max(\{1,2\}) = 2 \leq 3 = \max(\{2,3\})

, but

\max(\{2,3\}) = 3 \not\leq 2 = \max(\{1,2\})

. Choice B incorrectly claims symmetry. Choice C incorrectly claims lack of reflexivity. Choice D incorrectly claims lack of reflexivity.

Question 8

Let $T$ be the relation on $\mathbb{R} \setminus \{0\}$ defined by $xTy$ if and only if $\frac{x}{y} > 0$ . Determine which properties $T$ satisfies.

Reflexive and symmetric, but testing transitivity requires checking sign patterns across multiple cases
Reflexive and transitive, but not symmetric since positive and negative numbers behave differently under division
Symmetric and transitive, but not reflexive since the relation doesn't include zero in its domain
Reflexive, symmetric, and transitive, making it an equivalence relation partitioning reals into positive and negative (correct answer)

Explanation: Reflexive: For any

x \in \mathbb{R} \setminus \{0\}

\frac{x}{x} = 1 > 0

, so

xTx

. Symmetric: If

\frac{x}{y} > 0

, then

\frac{y}{x} = \frac{1}{\frac{x}{y}} > 0

since reciprocal of positive number is positive. Transitive: If

\frac{x}{y} > 0

and

\frac{y}{z} > 0

, then

\frac{x}{z} = \frac{x}{y} \cdot \frac{y}{z} > 0

since product of positive numbers is positive. Choice A incorrectly suggests transitivity is complicated. Choice B incorrectly claims lack of symmetry. Choice C incorrectly claims lack of reflexivity (zero is not in the domain anyway).

Question 9

Let $U$ be the relation on $\mathbb{N} = \{1, 2, 3, \ldots\}$ defined by $mUn$ if and only if $\gcd(m, n) \cdot \text{lcm}(m, n) = mn$ . Which statement about $U$ is correct?

$U$ represents the relation where $m$ and $n$ are coprime, making it symmetric but neither reflexive nor transitive
$U$ is the universal relation $\mathbb{N} \times \mathbb{N}$ since the condition $\gcd(m,n) \cdot \text{lcm}(m,n) = mn$ holds for all positive integers (correct answer)
$U$ is the empty relation since the fundamental identity $\gcd(m,n) \cdot \text{lcm}(m,n) = mn$ never actually holds in practice
$U$ is reflexive and symmetric but not transitive, since the gcd-lcm identity doesn't compose well across three numbers

Explanation: When you encounter a relation defined by a mathematical condition, your first step should be to determine whether that condition is always true, sometimes true, or never true for the given domain. The key insight here is recognizing that

\gcd(m,n) \cdot \text{lcm}(m,n) = mn

is actually a fundamental identity in number theory that holds for all positive integers

m

and

n

. This isn't a special condition—it's a mathematical fact you can prove using prime factorizations. Since this equation is always satisfied, every ordered pair

(m,n)

\mathbb{N} \times \mathbb{N}

belongs to relation

U

, making

U

the universal relation. Therefore, answer B is correct. Let's examine why the other options fail: A incorrectly claims

U

represents coprime numbers, but the gcd-lcm identity holds regardless of whether numbers share common factors. C makes the opposite error, suggesting this fundamental identity "never actually holds in practice," which contradicts basic number theory. D acknowledges the identity but incorrectly analyzes the relation's properties—since

U

is universal, it's automatically reflexive, symmetric, and transitive. Study tip: Memorize the fundamental identity

\gcd(m,n) \cdot \text{lcm}(m,n) = mn

. When you see this equation in a discrete math problem, immediately recognize it as always true rather than trying to solve for specific cases. This identity appears frequently in number theory problems and relation exercises.

Question 10

Define relation $V$ on the set $\{-2, -1, 0, 1, 2\}$ by $xVy$ if and only if $x^3 - y^3$ is divisible by $3$ . Which combination of properties does $V$ satisfy?

Reflexive and transitive but not symmetric, because cubing preserves congruence but reversal doesn't always work
Reflexive, symmetric, and transitive, since $x^3 \equiv x \pmod{3}$ and this creates an equivalence relation based on residues modulo $3$ (correct answer)
Symmetric and transitive but not reflexive, since some elements don't satisfy the cubic difference condition with themselves
Only reflexive, lacking both symmetry and transitivity due to the nonlinear behavior of cubic functions

Explanation: When analyzing relations for reflexive, symmetric, and transitive properties, you need to systematically check each property using the given definition. Here,

xVy

means

3 \mid (x^3 - y^3)

, so you're working with divisibility by 3. The key insight is recognizing that

x^3 \equiv x \pmod{3}

for all integers

x

. This happens because: if

x \equiv 0 \pmod{3}

, then

x^3 \equiv 0 \pmod{3}

; if

x \equiv 1 \pmod{3}

, then

x^3 \equiv 1^3 \equiv 1 \pmod{3}

; if

x \equiv 2 \pmod{3}

, then

x^3 \equiv 8 \equiv 2 \pmod{3}

. Therefore,

xVy

if and only if

x \equiv y \pmod{3}

. This transforms the relation into congruence modulo 3, which is always an equivalence relation. For reflexivity:

x^3 - x^3 = 0

, which is divisible by 3. For symmetry: if

3 \mid (x^3 - y^3)

, then

3 \mid (y^3 - x^3) = -(x^3 - y^3)

. For transitivity: if

3 \mid (x^3 - y^3)

and

3 \mid (y^3 - z^3)

, then

3 \mid (x^3 - z^3) = (x^3 - y^3) + (y^3 - z^3)

. Choice A incorrectly claims the relation isn't symmetric. Choice C wrongly states it's not reflexive—every element relates to itself since

x^3 - x^3 = 0

. Choice D incorrectly suggests only reflexivity holds, missing that congruence relations are always equivalence relations. Remember: when you see

x^n

in modular arithmetic problems, check if the exponent preserves congruence classes. Cubing preserves residues modulo 3, making this an equivalence relation.

Question 11

Consider the relation $H$ on $\mathbb{R}$ defined by $xHy$ if and only if $\sin(\pi x) = \sin(\pi y)$ . What properties does $H$ satisfy?

Symmetric and transitive but not reflexive, since $\sin(\pi x) = \sin(\pi x)$ doesn't hold for all real $x$
Reflexive and symmetric but not transitive, because sine's periodicity can break transitivity chains
Reflexive, symmetric, and transitive, since sine creates an equivalence relation based on periodic and symmetric properties (correct answer)
Reflexive and transitive but not symmetric, because $\sin(\pi x) = \sin(\pi y)$ doesn't guarantee $\sin(\pi y) = \sin(\pi x)$

Explanation: When analyzing relations in discrete math, you need to systematically check three key properties: reflexivity, symmetry, and transitivity. These determine whether a relation is an equivalence relation. Let's examine relation

H

where

xHy

if and only if

\sin(\pi x) = \sin(\pi y)

. For reflexivity, we need

xHx

for all

x \in \mathbb{R}

. Since

\sin(\pi x) = \sin(\pi x)

is always true by equality's reflexive property,

H

is reflexive. For symmetry, if

xHy

, then

yHx

must hold. If

\sin(\pi x) = \sin(\pi y)

, then certainly

\sin(\pi y) = \sin(\pi x)

by equality's symmetric property, so

H

is symmetric. For transitivity, if

xHy

and

yHz

, then

xHz

must hold. If

\sin(\pi x) = \sin(\pi y)

and

\sin(\pi y) = \sin(\pi z)

, then

\sin(\pi x) = \sin(\pi z)

by equality's transitive property, so

H

is transitive. Since

H

satisfies all three properties, it's an equivalence relation, making C correct. A is wrong because

\sin(\pi x) = \sin(\pi x)

always holds—this confuses the relation with properties of sine itself. B incorrectly claims transitivity fails, but equality of sine values preserves transitivity. D wrongly denies symmetry, but equality is inherently symmetric. Study tip: When a relation is defined by "

f(x) = f(y)

" for any function

f

, it's automatically an equivalence relation because you're really just using equality, which is always reflexive, symmetric, and transitive.

Question 12

Let $W$ be the relation on $\mathbb{R}^+$ (positive reals) where $xWy$ if and only if $\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor$ . Which properties does this relation possess?

Symmetric and transitive but not reflexive, because logarithms are undefined for some values in the domain
Reflexive and symmetric but not transitive, since floor functions can break transitivity across boundary values
Reflexive, symmetric, and transitive, partitioning positive reals into equivalence classes based on powers of $2$ (correct answer)
Reflexive and transitive but not symmetric, since logarithmic relationships don't always reverse under floor operations

Explanation: When you encounter a relation defined by a mathematical condition, you need to systematically check whether it's reflexive, symmetric, and transitive to determine if it's an equivalence relation. Let's analyze what

xWy

means:

\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor

. This condition groups numbers by which power of 2 their logarithm falls into. For example, if

2^3 \leq x < 2^4

, then

\lfloor \log_2 x \rfloor = 3

. Reflexive: For any

x \in \mathbb{R}^+

, we have

\lfloor \log_2 x \rfloor = \lfloor \log_2 x \rfloor

, so

xWx

holds. ✓ Symmetric: If

xWy

, then

\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor

, which immediately gives us

\lfloor \log_2 y \rfloor = \lfloor \log_2 x \rfloor

, so

yWx

. ✓ Transitive: If

xWy

and

yWz

, then

\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor = \lfloor \log_2 z \rfloor

, so

xWz

. ✓ Since all three properties hold, this is an equivalence relation that partitions

\mathbb{R}^+

into classes like

[1,2)

[2,4)

[4,8)

, etc. Option A incorrectly claims logarithms are undefined—they're defined for all positive reals. Option B wrongly suggests the floor function breaks transitivity, but equality of integers is clearly transitive. Option D incorrectly claims the relation isn't symmetric, misunderstanding how equality works. Key strategy: When checking if a relation is an equivalence relation, systematically verify all three properties using the relation's definition. Equality-based relations like this one are typically equivalence relations.

Question 13

Consider the relation $S$ on $\{0, 1, 2, 3, 4, 5\}$ defined by $xSy$ if and only if $x^2 + y^2 \equiv 0 \pmod{5}$ . What can be concluded about the properties of $S$ ?

$S$ is reflexive and symmetric but not transitive, due to the nonlinear nature of the sum-of-squares condition
$S$ is reflexive, symmetric, and transitive, since congruence modulo $5$ preserves all three properties
$S$ is symmetric and transitive but not reflexive, as some elements don't satisfy the sum-of-squares condition with themselves
$S$ is not reflexive because $2^2 + 2^2 = 8 \equiv 3 \pmod{5} \not\equiv 0$ , violating the reflexive property (correct answer)

Explanation: When analyzing relations, you need to check three key properties systematically: reflexive (every element relates to itself), symmetric (if

x

relates to

y

, then

y

relates to

x

), and transitive (if

x

relates to

y

and

y

relates to

z

, then

x

relates to

z

). For relation

S

to be reflexive, every element must satisfy

x^2 + x^2 \equiv 0 \pmod{5}

, or

2x^2 \equiv 0 \pmod{5}

. Testing each element:

0^2 + 0^2 = 0 \equiv 0

1^2 + 1^2 = 2 \not\equiv 0

2^2 + 2^2 = 8 \equiv 3 \not\equiv 0

3^2 + 3^2 = 18 \equiv 3 \not\equiv 0

4^2 + 4^2 = 32 \equiv 2 \not\equiv 0

, and

5^2 + 5^2 = 50 \equiv 0

. Since elements 1, 2, and 3 don't relate to themselves,

S

is not reflexive. Choice A incorrectly assumes reflexivity holds. Choice B makes the critical error of thinking that because the underlying operation (congruence mod 5) has certain properties, the relation automatically inherits them—this isn't true. The specific condition

x^2 + y^2 \equiv 0 \pmod{5}

creates its own constraints. Choice C incorrectly claims reflexivity fails but gets the reason wrong, and assumes transitivity without verification. Choice D correctly identifies that

S

is not reflexive and provides the specific counterexample:

2^2 + 2^2 = 8 \equiv 3 \pmod{5} \neq 0

. Study tip: When checking relation properties, always test reflexivity first by plugging in specific values—don't assume properties transfer from the underlying mathematical operations to your defined relation.

Question 14

Consider the relation $R$ on $\mathbb{Z}$ defined by $xRy$ if and only if $x - y$ is divisible by $xy + 1$ . Which statement about the properties of $R$ is correct?

$R$ is reflexive because $x - x = 0$ is always divisible by $x^2 + 1$ (correct answer)
$R$ is not reflexive because when $x = 0$ , we have $0 - 0 = 0$ but $0 \cdot 0 + 1 = 1$ , and $0$ is not divisible by $1$
$R$ is not reflexive because when $x = 0$ , we have $xy + 1 = 1$ and $0$ is divisible by $1$ , but this creates a contradiction
$R$ is reflexive because for any integer $x$ , the expression $xy + 1$ with $y = x$ gives $x^2 + 1 \geq 1$ , so divisibility is well-defined

Explanation: For reflexivity, we need

xRx

for all

x \in \mathbb{Z}

, meaning

x - x = 0

must be divisible by

x \cdot x + 1 = x^2 + 1

. Since

0

is divisible by any nonzero integer, and

x^2 + 1 \geq 1

for all real

x

, we have

x^2 + 1 \mid 0

always. Thus

R

is reflexive. Choice B incorrectly claims

0

is not divisible by

1

, but

0 = 0 \cdot 1

. Choice C contains confused reasoning about contradictions. Choice D gives correct reasoning about

x^2 + 1 \geq 1

but doesn't complete the divisibility argument properly.

Question 15

Let $R$ be a relation on $\mathbb{Z}$ where $aRb$ if and only if $a + b$ is even and $|a - b| \leq 2$ . Which properties does $R$ possess?

Reflexive and symmetric, but not transitive since $3R1$ , $1R(-1)$ , but $3 \not R (-1)$ (correct answer)
Symmetric and transitive, but not reflexive since odd integers don't satisfy the parity condition with themselves
Reflexive, symmetric, and transitive, forming an equivalence relation on the integers
Only symmetric, lacking reflexivity due to parity constraints and transitivity due to distance limitations

Explanation: Reflexive:

a + a = 2a

is even, and

|a - a| = 0 \leq 2

, so

aRa

for all

a

. Symmetric: If

a + b

is even and

|a - b| \leq 2

, then

b + a = a + b

is even and

|b - a| = |a - b| \leq 2

. Not transitive:

3R1

since

3 + 1 = 4

(even) and

|3 - 1| = 2 \leq 2

1R(-1)

since

1 + (-1) = 0

(even) and

|1 - (-1)| = 2 \leq 2

. But

3 \not R (-1)

since

|3 - (-1)| = 4 > 2

. Choice B incorrectly claims lack of reflexivity. Choice C incorrectly claims transitivity. Choice D incorrectly claims lack of reflexivity.

Question 16

Define relation $Y$ on the set of all $2 \times 2$ real matrices by $AYB$ if and only if $\det(A) = \det(B)$ . What properties does $Y$ possess?

$Y$ is reflexive and symmetric but not transitive
$Y$ is reflexive, symmetric, and transitive (correct answer)
$Y$ is symmetric and transitive but not reflexive
$Y$ is reflexive and transitive but not symmetric

Explanation: When analyzing relations in discrete mathematics, you need to check three key properties: reflexivity (does every element relate to itself?), symmetry (if A relates to B, does B relate to A?), and transitivity (if A relates to B and B relates to C, does A relate to C?). For relation

Y

defined by

AYB

if and only if

\det(A) = \det(B)

, let's examine each property systematically. Reflexivity: For any matrix

A

, is

AYA

true? Yes, because

\det(A) = \det(A)

is always true. Every matrix has the same determinant as itself. Symmetry: If

AYB

, then

\det(A) = \det(B)

. This means

\det(B) = \det(A)

, so

BYA

is also true. Equality is symmetric. Transitivity: If

AYB

and

BYC

, then

\det(A) = \det(B)

and

\det(B) = \det(C)

. By transitivity of equality,

\det(A) = \det(C)

, so

AYC

holds. Since

Y

possesses all three properties, it's an equivalence relation, making answer B correct. Why the others fail: Answer A incorrectly claims the relation isn't transitive, but equality of determinants is transitive. Answer C wrongly suggests the relation isn't reflexive, but every matrix equals itself in determinant. Answer D incorrectly claims the relation isn't symmetric, but equality is always symmetric. Study tip: When a relation is defined using equality (like equal determinants), it's almost always an equivalence relation. Look for the underlying equality to quickly identify all three properties.

Question 17

Consider relation $P$ on $\mathbb{R}^2$ defined by $(x_1, y_1)P(x_2, y_2)$ if and only if $x_1^2 + y_1^2 \leq x_2^2 + y_2^2$ . Analyze the properties of $P$ .

$P$ is reflexive and symmetric but not transitive
$P$ is reflexive and transitive but not symmetric (correct answer)
$P$ is symmetric and transitive but not reflexive
$P$ is reflexive, symmetric, and transitive

Explanation: P is reflexive because x₁² + y₁² ≤ x₁² + y₁² for any point (x₁, y₁), so (x₁, y₁)P(x₁, y₁) holds. P is not symmetric: consider (0,0) and (1,0). We have 0² + 0² = 0 ≤ 1 = 1² + 0², so (0,0)P(1,0). However, 1² + 0² = 1 ≰ 0 = 0² + 0², so (1,0)P(0,0) does not hold. P is transitive: if x₁² + y₁² ≤ x₂² + y₂² and x₂² + y₂² ≤ x₃² + y₃², then x₁² + y₁² ≤ x₃² + y₃², so if (x₁,y₁)P(x₂,y₂) and (x₂,y₂)P(x₃,y₃), then (x₁,y₁)P(x₃,y₃). Therefore P is reflexive and transitive but not symmetric. This relation orders points by their distance from the origin.

Question 18

Consider the relation $S$ on $\mathbb{Z}$ defined by $aSb$ if and only if $a - b$ is divisible by $3$ . Determine which properties $S$ possesses.

$S$ is reflexive and symmetric but not transitive
$S$ is reflexive and transitive but not symmetric
$S$ is an equivalence relation on $\mathbb{Z}$ (correct answer)
$S$ is symmetric and transitive but not reflexive

Explanation: S is reflexive because for any integer a, a - a = 0, and 0 is divisible by 3, so aSa holds. S is symmetric: if aSb, then 3|(a-b), which means 3|(-(a-b)) = 3|(b-a), so bSa. S is transitive: if aSb and bSc, then 3|(a-b) and 3|(b-c), so 3|((a-b)+(b-c)) = 3|(a-c), therefore aSc holds. Since S is reflexive, symmetric, and transitive, it is an equivalence relation.

Question 19

Define relation $U$ on $\mathbb{R}$ by $xUy$ if and only if $\sin(x) = \sin(y)$ . Analyze the properties of $U$ .

$U$ is reflexive and symmetric but not transitive
$U$ is reflexive and transitive but not symmetric
$U$ is symmetric and transitive but not reflexive
$U$ is reflexive, symmetric, and transitive (correct answer)

Explanation: U is reflexive because sin(x) = sin(x) for any real number x, so xUx holds. U is symmetric because if sin(x) = sin(y), then sin(y) = sin(x), so if xUy then yUx. U is transitive because if sin(x) = sin(y) and sin(y) = sin(z), then sin(x) = sin(z), so if xUy and yUz then xUz. Therefore U satisfies all three properties and is an equivalence relation on ℝ.