Determine if a relation is reflexive, symmetric, transitive - Discrete Math

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 $S$ ?
Answer: $S$ is not reflexive because $2^2 + 2^2 = 8 \equiv 3 \pmod{5} \not\equiv 0$ , violating the reflexive property 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.
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.
Answer: Reflexive, symmetric, and transitive, making it an equivalence relation partitioning reals into positive and negative 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).
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 $Z$ ?
Answer: $Z$ is reflexive, symmetric, and transitive 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.
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?
Answer: Reflexive, symmetric, and transitive, since sine creates an equivalence relation based on periodic and symmetric properties 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.
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?
Answer: $W$ is reflexive, symmetric, and transitive 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.
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.
Answer: $S$ is an equivalence relation on $\mathbb{Z}$ 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.
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 $Q$ ?
Answer: Reflexive and transitive, but not symmetric because $\max(\{1,2\}) \leq \max(\{2,3\})$ while $\max(\{2,3\}) \not\leq \max(\{1,2\})$ 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.
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?
Answer: Reflexive, symmetric, and transitive, partitioning positive reals into equivalence classes based on powers of $2$ 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.
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?
Answer: Reflexive, symmetric, and transitive, since $x^3 \equiv x \pmod{3}$ and this creates an equivalence relation based on residues modulo $3$ 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.
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?
Answer: Reflexive, symmetric, and transitive (equivalence relation) 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.
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?
Answer: Reflexive and symmetric, but not transitive since $3R1$ , $1R(-1)$ , but $3 \not R (-1)$ 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.
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?
Answer: $R$ is reflexive because $x - x = 0$ is always divisible by $x^2 + 1$ 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.
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.
Answer: $X$ is reflexive, symmetric, and transitive 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.
Define relation $U$ on $\mathbb{R}$ by $xUy$ if and only if $\sin(x) = \sin(y)$ . Analyze the properties of $U$ .
Answer: $U$ is reflexive, symmetric, and transitive 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.
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 $P$ .
Answer: $P$ is reflexive and transitive but not symmetric 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.
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.
Answer: Reflexive, symmetric, and transitive, representing the equivalence of rational numbers when treating pairs as fractions 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 $x,y$ .
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?
Answer: $U$ 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 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 $(m,n)$ in $\mathbb{N} \times \mathbb{N}$ 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.
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?
Answer: $Y$ is reflexive, symmetric, and transitive 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.
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 $T$ ?
Answer: $T$ is reflexive and transitive but not symmetric 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.

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