Let $A = \{1, 2, 3, 4, 5\}$. Define $R = \{(x, y) : x + y \in N;\ x, y \in A\}$.

Question

Accepted Answer

> [!TLDR]
> Since $x,y\in A$ are always natural numbers, their sum $x+y$ is always a natural number too — so $R$ equals the entire set $A	imes A$, the universal relation.

> [!FORMULA]
> $$A=\{1,2,3,4,5\},\qquad R=\{(x,y): x+y\in\mathbb{N},\ x,y\in A\}$$
> A relation that contains **every possible pair** of $A	imes A$ is called the **universal relation**, $R=A	imes A$, with $n(R)=n(A)	imes n(A)$.

1. **Recall the domain.** $A=\{1,2,3,4,5\}$ is a finite subset of the natural numbers $\mathbb{N}=\{1,2,3,\dots\}$.

2. **Analyse the sum $x+y$ for $x,y\in A$.** Since both $x$ and $y$ are positive integers, their sum is also a positive integer:
 $$x,y\in\mathbb{N}\ \Rightarrow\ x+y\in\mathbb{N}\quad	ext{always, for every pair}$$
 For example, the smallest possible sum is $1+1=2\in\mathbb{N}$ and the largest is $5+5=10\in\mathbb{N}$ — every value in between is also a natural number.

3. **Compare with the condition defining $R$.** $R$ requires $x+y\in\mathbb{N}$, which — as shown — is **true for every single pair** $(x,y)\in A	imes A$, with no exceptions.

4. **Conclude.**
 $$R=\{(x,y): x,y\in A\}=A	imes A$$
 This is the **universal relation** on $A$ — it relates every element to every other (including itself).

5. **Count the elements.**
 $$n(R)=n(A	imes A)=n(A)	imes n(A)=5	imes5=25$$

**Self-check:** Testing the extreme pairs $(1,1)$: $1+1=2\in\mathbb{N}$ ✓; and $(5,5)$: $5+5=10\in\mathbb{N}$ ✓ — both, and every pair in between, satisfy the condition, confirming $R=A	imes A$.

> [!ANSWER]
> $R=A	imes A$ — the **universal relation**, containing all $25$ ordered pairs of $A$, since $x+y\in\mathbb{N}$ holds for every $x,y\in A$.

Type	Reflexive?	Symmetric?	Transitive?	Example
Empty relation	No (unless $A = \emptyset$ )	Yes (vacuously)	Yes (vacuously)	$R = \emptyset$ on $A = \{1,2\}$
Universal relation	Yes	Yes	Yes	$R = A \times A$
Identity relation	Yes	Yes	Yes	$R = \{(a,a) \mid a \in A\}$
Equivalence relation	Yes	Yes	Yes	"Same city", "same remainder mod 3"

Type	Definition	Example
Antisymmetric	If $(a,b) \in R$ and $(b,a) \in R$ , then $a = b$	$\leq$ on real numbers
Asymmetric	If $(a,b) \in R$ then $(b,a) \notin R$	$<$ on real numbers
Partial order	Reflexive + antisymmetric + transitive	$\subseteq$ on sets
Total order	Partial order where every pair is comparable	$\leq$ on real numbers

Q.Let $A = \{1, 2, 3, 4, 5\}$ . Define $R = \{(x, y) : x + y \in N;\ x, y \in A\}$ .

Concept understanding — Types of Relations

Types of Relations – From Intuition to Precision

The Intuition: Three Key Properties

The Precise Definitions

The Four Types of Relations You Must Know

Equivalence Relations — The Most Important Type

Other Important Types (for completeness)

Quick Check — Test Yourself

Q.Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5}. Define R={(x,y):x+y∈N; x,y∈A}R = \{(x, y) : x + y \in N;\ x, y \in A\}R={(x,y):x+y∈N; x,y∈A}.

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Q.Let $A = \{1, 2, 3, 4, 5\}$ . Define $R = \{(x, y) : x + y \in N;\ x, y \in A\}$ .

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