1. Algebra — Relations and Functions

Symmetric relation

A symmetric relation on a set $A$ is a binary relation $R$ where the presence of $(a, b) \in R$ universally guarantees the presence of $(b, a) \in R$ for all $a, b \in A$.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Definition of Symmetric Relation

Let $A$ be a non-empty set. A relation $R$ on $A$ (i.e., $R \subseteq A \times A$) is said to be symmetric if and only if \forall a, b \in A, \, (a, b) \in R \implies (b, a) \in R.

📐 Geometric Interpretation

On the Cartesian plane $A \times A$, a symmetric relation exhibits complete reflectional symmetry across the principal diagonal line $y = x$. If a point $(x_0, y_0)$ belongs to the relation graph, its mirror point $(y_0, x_0)$ also necessarily lies on the graph.

⚙️ Algebraic & Structural Meaning

Algebraically, symmetry means the inverse relation $R^{-1}$ is identical to $R$ itself ($R = R^{-1}$). In terms of adjacency matrices over a finite set, the boolean matrix representation is a symmetric matrix equal to its transpose ($M_R = M_R^T$).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-empty base set $A$	`Domain set $A$`	Strictly Mandatory / Necessary	If $A$ is empty, the relation is vacuously symmetric, but standard problem-solving assumes non-empty sets where domain and codomain are well-defined.
Universal Quantifier Condition	`For all ordered pairs $(a, b)`	Strictly Mandatory	If even a single pair $(a, b) \in R$ exists such that $(b, a) \notin R$, the relation fails symmetry entirely.
Identity Relation inclusion	`Diagonal pairs $(a, a)`	Optional / Not required	Symmetric relations do not require reflexive elements; $(a, a) \in R$ is neither necessary nor sufficient for symmetry by itself.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Matrix Transpose Test for Finite Sets

$$M_R = M_R^T$$

Rule: Construct the boolean adjacency matrix of the relation. If the matrix is symmetric across its main diagonal, the relation is symmetric instantly without checking pairs one by one.

Application: Use this result in the stated type of problem.

Inverse Relation Identity

$$R = R^{-1}$$

Rule: If you can algebraically or logically show that swapping coordinates yields the exact same condition defining the relation, it is symmetric.

Counting Symmetric Relations

$$2^{\frac{n(n+1)}{2}}$$තික

Rule: For a finite set $A$ containing $n$ elements, the total number of symmetric relations is given by $2^{\frac{n(n+1)}{2}}$, where $\frac{n(n+1)}{2}$ counts the diagonal and upper-triangular positions.

Equality and Congruence Relations

$$xRy \iff x \equiv y \pmod m$$

Rule: Equivalence relations (like congruence modulo $m$, equality, perpendicularity of lines) are automatically symmetric. Never waste time testing $(b, a)$ for them.

Algebraic Inequality Trap Check

$$x < y \implies y < x \text{ (False)}$$

Rule: Strict inequalities ($<$ or $>$) are never symmetric because $a < b$ does not imply $b < a$. Non-strict inequalities ($\le$ or $\ge$) are antisymmetric, not symmetric.

Application: Classic option-elimination trick in state board PGT tests.

Rapid Elimination via Counterexample

$$ \text{Find } (a,b) \in R \text{ such that } (b,a) \notin R $$

Rule: To disprove symmetry, test integers or small values like $(1, 2)$. If $1$ relates to $2$ but $2$ does not relate to $1$, eliminate the option immediately.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Reflexive with Symmetric

Common Pitfall: Candidates assume that if a relation contains all diagonal elements $(a, a)$, it must be symmetric.

Counterexample: R = \{(1,1), (2,2), (1,2)\} \text{ on } \{1, 2\}

Why it fails: Here $(1, 2) \in R$, but $(2, 1) \notin R$. Hence, the relation is reflexive but NOT symmetric.

🛡️ Exam Defense: Always check off-diagonal pairs independently of diagonal elements.

⚠️ COMMON ERROR #02

Assuming Antisymmetric means Not Symmetric

Common Pitfall: Believing that a relation cannot be both symmetric and antisymmetric.

Counterexample: R = \{(1,1), (2,2)\} \text{ on } \{1, 2\}

Why it fails: This relation is both symmetric and antisymmetric (since $(a,b) \in R \implies a=b$).

🛡️ Exam Defense: Check definitions strictly: symmetry and antisymmetry are mutually exclusive only for distinct elements, but can overlap on diagonal pairs.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Symmetric Relation	$(a, b) \in R \implies (b, a) \in R$	$R = R^{-1}$	Reflectional symmetry across $y = x$	Swap variables and check if condition holds
Reflexive Relation	$(a, a) \in R$ for all $a \in A$	Contains full diagonal	All points on line $y = x$ are present	Look for self-pairing for every element
Transitive Relation	$(a, b) \in R \land (b, c) \in R \implies (a, c) \in R$	$R \circ R \subseteq R$	Chain closure property	Look for chain links $a \to b \to c$
Antisymmetric Relation	$(a, b) \in R \land (b, a) \in R \implies a = b$	$R \cap R^{-1} \subseteq \Delta$	No mutual cross-links between distinct elements	Inequalities like $\le$ or divisibility relation

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Let $A = \{1, 2, 3\}$. How many symmetric relations can be defined on set $A$?

(A) 64(B) 32(C) 512(D) 16

📚 Method A: Formal Textbook Derivation

Step 1: Identify cardinality $n = 3$. Step 2: Apply the formula for the number of symmetric relations $2^{\frac{n(n+1)}{2}}$. Step 3: Compute $2^{\frac{3(4)}{2}} = 2^6 = 64$.

⚡ Method B: 25-Second Exam Shortcut

Direct formula recall: $n=3 \implies 2^{3 \times 4 / 2} = 64$. Mark option (A) in 5 seconds.

💡 Tip: Memorize values for $n=1, 2, 3, 4$ ($2, 8, 64, 1024$) to save calculation time.

CANONICAL PROBLEM 02 Dual-Method Breakdown

On the set of integers $\mathbb{Z}$, let relation $R$ be defined by $xRy \iff x + y$ is an even integer. Then $R$ is:

(A) Reflexive only(B) Symmetric only(C) Equivalence relation(D) Transitive only

📚 Method A: Formal Textbook Derivation

Step 1: Reflexivity: $x + x = 2x$ (even), so reflexive. Step 2: Symmetry: If $x + y$ is even, $y + x$ is also even, so symmetric. Step 3: Transitivity: If $x+y$ and $y+z$ are even, their sum $(x+y)+(y+z) = x+z + 2y$ is even, implying $x+z$ is even, hence transitive.

⚡ Method B: 25-Second Exam Shortcut

Recognize modular arithmetic properties: congruence modulo 2 or parity relations are always equivalence relations.

💡 Tip: Any relation defined by sum/difference being divisible by a fixed integer $m$ is always an equivalence relation.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $R$ is a symmetric relation on a set $A$, then $R^{-1}$ is:

A $R^{-1} \neq R$

B $R^{-1} = R$

C $R^{-1} = \emptyset$

D $R^{-1} = A \times A$

Derivation & Analysis:

By definition, symmetry means $(a,b) \in R \iff (b,a) \in R$, which is precisely the definition of the inverse relation being equal to $R$.

MCQ 2: The number of symmetric relations on a set containing $4$ elements is:

A $64$

B $256$

C $512$

D $1024$

Derivation & Analysis:

Using formula $2^{\frac{n(n+1)}{2}}$ with $n=4$: $2^{\frac{4(5)}{2}} = 2^{10} = 1024$.

MCQ 3: Let $R = \{(1,2), (2,1), (1,1)\} \n on $A = \{1, 2}$. The relation $R$ is:

A Symmetric only

B Equivalence relation

C Symmetric and Reflexive

D Not symmetric

Derivation & Analysis:

It contains $(1,1)$ but lacks $(2,2)$, so it is NOT reflexive. However, for every $(1,2)$ there is $(2,1)$, and for $(1,1)$ its reverse is $(1,1)$, making it symmetric. It also includes $(1,1)$, so it has reflexive elements for ${1}$ but fails full set reflexivity. Wait, $(2,2)$ is missing, so it is NOT reflexive on $A=\{1,2\}$. Let's check options: it is symmetric.

MCQ 4: Which of the following relations on $\mathbb{R}$ is symmetric?

A $R = \{(x,y) : x \le y\}$

B $R = \{(x,y) : x - y \in \mathbb{Q}\}$

C $R = \{(x,y) : x > 2y\}$

D $R = \{(x,y) : x = 2y\}$

Derivation & Analysis:

If $x - y$ is rational, then $y - x = -(x - y)$ is also rational. Thus $R$ is symmetric. Other options fail symmetry.

MCQ 5: If $R_1$ and $R_2$ are two symmetric relations on set $A$, then $R_1 \cap R_2$ is:

A Always symmetric

B Never symmetric

C Transitive only

D Reflexive only

Derivation & Analysis:

If $(a,b) \in R_1 \cap R_2$, then $(a,b) \in R_1$ and $(a,b) \in R_2$. By symmetry, $(b,a) \in R_1$ and $(b,a) \in R_2$, so $(b,a) \in R_1 \cap R_2$.

MCQ 6: The empty relation $\Phi$ on a non-empty set $A$ is:

A Not symmetric

B Symmetric vacuously

C Reflexive

D Equivalence

Derivation & Analysis:

The condition $(a,b) \in \Phi \implies (b,a) \in \Phi$ is vacuously true because the hypothesis $(a,b) \in \Phi$ is never satisfied.

MCQ 7: Universal relation $U = A \times A$ on a non-empty set $A$ is:

A Symmetric only

B Reflexive only

C An equivalence relation

D Not symmetric

Derivation & Analysis:

The universal relation is always reflexive, symmetric, and transitive, hence an equivalence relation.

MCQ 8: Let $R$ be a relation on $\mathbb{Z}$ defined by $xRy \iff xy > 0$. Then $R$ is:

A Symmetric and Reflexive

B Symmetric but not Reflexive

C Transitive only

D None of these

Derivation & Analysis:

If $xy > 0$, then $yx > 0$, so symmetric. If $x=0$, $0 \times 0 = 0 \ngtr 0$, so not reflexive. Thus symmetric but not reflexive.

MCQ 9: If a relation is both symmetric and antisymmetric on a set $A$, then $R$ must be a subset of:

A The diagonal relation $\Delta$

B $A \times A$

C An empty set

D The inverse relation

Derivation & Analysis:

If $(a,b) \in R$, then by symmetry $(b,a) \in R$. By antisymmetry, this implies $a = b$. Thus, $R$ can only contain pairs of the form $(a,a)$, meaning $R \subseteq \Delta$.

Q10

MCQ 10: Let $A = \{1, 2, 3\}$. The maximum number of elements in a symmetric relation on $A$ which is not reflexive is:

A 9

B 7

C 6

D 8

Derivation & Analysis:

Total symmetric relations is 64. Reflexive symmetric relations have all 3 diagonal elements fixed, leaving 3 pairs above diagonal, giving $2^3 = 8$ reflexive symmetric relations. Thus non-reflexive symmetric relations = $64 - 8 = 56$. Wait, looking at maximum elements: a symmetric relation on 3 elements can have at most $3 + 2(3) = 9$ elements. If not reflexive, it must omit at least one diagonal element, so maximum elements = $9 - 1 = 8$.

Common questions

Yes. The definition of a symmetric relation is an implication: if $(a,b) \in R$, then $(b,a) \in R$. Since the antecedent $(a,b) \in \Phi$ is false for the empty relation, the conditional statement is vacuously true.

Yes. Any subset of the identity/diagonal relation (such as the empty relation or any subset of ${(1,1), (2,2)}$) is simultaneously symmetric and antisymmetric.

Instantly swap variables $x$ and $y$ in the given defining equation or condition. If the resulting expression is logically equivalent to the original, the relation is symmetric.