Let the relation R on the set $M = \{1, 2, 3, \ldots, 16\}$ be given by $R = \{(x, y) : 4y = 5x – 3, x, y \in M\}$. Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to
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Solution Steps
1
Find all pairs in relation R
Given: $R = \{(x, y) : 4y = 5x – 3, x, y \in M\}$ where $M = \{1, 2, 3, \ldots, 16\}$
From $4y = 5x – 3$, we get $y = \frac{5x – 3}{4}$
For $y$ to be an integer, $(5x – 3)$ must be divisible by 4.
We need: $5x – 3 \equiv 0 \pmod{4}$
2
Solve the congruence condition
$5x – 3 \equiv 0 \pmod{4}$
$5x \equiv 3 \pmod{4}$
Since $5 \equiv 1 \pmod{4}$, we have: $x \equiv 3 \pmod{4}$
Therefore, $x \in \{3, 7, 11, 15\}$ from the set $M$.
3
Calculate corresponding y values
For $x = 3$: $y = \frac{5(3) – 3}{4} = \frac{15 – 3}{4} = \frac{12}{4} = 3$ ✓
For $x = 7$: $y = \frac{5(7) – 3}{4} = \frac{35 – 3}{4} = \frac{32}{4} = 8$ ✓
For $x = 11$: $y = \frac{5(11) – 3}{4} = \frac{55 – 3}{4} = \frac{52}{4} = 13$ ✓
For $x = 15$: $y = \frac{5(15) – 3}{4} = \frac{75 – 3}{4} = \frac{72}{4} = 18$ ✗ (not in M)
4
List all elements of R
From calculations above, the relation R contains:
$R = \{(3, 3), (7, 8), (11, 13)\}$
These are the only three ordered pairs that satisfy the given condition and have both elements in M.
5
Understand symmetric relation property
A relation R is symmetric if: whenever $(a, b) \in R$, then $(b, a) \in R$
In other words, for every pair in R, its reverse must also be in R.
We need to check each pair in R and see if its symmetric counterpart exists.
6
Check symmetry for each pair
• $(3, 3) \in R$: Check if $(3, 3) \in R$ → Yes, already present ✓ (reflexive pair)
• $(7, 8) \in R$: Check if $(8, 7) \in R$ → No, need to add $(8, 7)$
• $(11, 13) \in R$: Check if $(13, 11) \in R$ → No, need to add $(13, 11)$
Pairs to add: $(8, 7)$ and $(13, 11)$
7
Calculate minimum elements to add
To make R symmetric, we must add:
1. $(8, 7)$ — symmetric counterpart of $(7, 8)$
2. $(13, 11)$ — symmetric counterpart of $(11, 13)$
Minimum number of elements required = $\boxed{2}$
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Theory Topics
1. Symmetric Relations
A relation R on a set A is called symmetric if for every ordered pair $(a, b)$ in R, the ordered pair $(b, a)$ is also in R. Mathematically, $\forall a, b \in A: (a, b) \in R \Rightarrow (b, a) \in R$. This property creates a two-way relationship between elements. For example, if “is sibling of” is a relation on people, it’s symmetric because if A is sibling of B, then B is sibling of A. To make any relation symmetric, we identify all pairs $(x, y)$ where $(y, x)$ is missing and add these missing pairs. Reflexive pairs like $(a, a)$ are automatically symmetric with themselves. Understanding symmetry is fundamental in set theory, graph theory, and discrete mathematics, appearing frequently in JEE Main problems involving relations and their properties.
2. Finding Elements in a Relation
To find all elements satisfying a relation defined by an equation like $4y = 5x – 3$, we must ensure both coordinates belong to the given set and satisfy the equation. The key technique is solving for one variable in terms of the other: $y = \frac{5x – 3}{4}$. For y to be an integer, the numerator must be divisible by the denominator. This leads to modular arithmetic: $5x – 3 \equiv 0 \pmod{4}$. Solving congruences is essential—here $5 \equiv 1 \pmod{4}$ simplifies to $x \equiv 3 \pmod{4}$, giving $x \in \{3, 7, 11, 15\}$. We then verify each x produces a valid y in the set. This systematic approach prevents missing elements or including invalid ones, crucial for accurate problem-solving in relations, functions, and number theory questions in competitive exams.
3. Types of Relations
Relations on a set can have various properties: (1) Reflexive: $(a, a) \in R$ for all $a \in A$. (2) Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$. (3) Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$. (4) Antisymmetric: $(a, b), (b, a) \in R \Rightarrow a = b$. (5) Equivalence relation: reflexive, symmetric, and transitive. Understanding these properties helps classify relations and predict their behavior. In this problem, we transform a non-symmetric relation into a symmetric one by adding minimal pairs. The strategy involves listing all pairs, identifying missing symmetric counterparts, and counting additions needed. This concept extends to equivalence classes, partial orders, and graph representations, making it fundamental for higher mathematics and computer science applications.
4. Modular Arithmetic in Relations
Modular arithmetic is powerful for determining which elements satisfy divisibility conditions in relations. When an equation like $y = \frac{5x – 3}{4}$ requires integer solutions, we use congruences. The condition $(5x – 3) \equiv 0 \pmod{4}$ means $5x – 3$ leaves remainder 0 when divided by 4. Simplifying: $5x \equiv 3 \pmod{4}$. Since $5 = 4 + 1$, we have $5 \equiv 1 \pmod{4}$, thus $x \equiv 3 \pmod{4}$. This gives arithmetic sequence: $x = 3, 7, 11, 15, \ldots$ within our set. Modular arithmetic is essential for number theory problems, cryptography, and discrete structures. JEE frequently tests this in relations, sequences, and divisibility problems. Mastery involves understanding equivalence classes modulo n, solving linear congruences, and applying Chinese Remainder Theorem for complex systems.
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FAQs
Q
What is a symmetric relation?
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A relation R on set A is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ also. Every pair must have its reverse pair present in the relation.
Q
How to find pairs satisfying $4y = 5x – 3$?
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Solve $y = \frac{5x-3}{4}$. For integer y, need $5x – 3 \equiv 0 \pmod{4}$. This gives $x \equiv 3 \pmod{4}$, so $x \in \{3, 7, 11, 15\}$ from M. Calculate y for each x and verify $y \in M$.
Q
Why is $(3,3)$ already symmetric?
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For $(3, 3)$, the reverse pair is also $(3, 3)$. Since it equals itself, it’s automatically symmetric and requires no additional pair to be added.
Q
What pairs exist in relation R?
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$R = \{(3, 3), (7, 8), (11, 13)\}$. For $x = 15$, we get $y = 18$ which is not in M = {1, 2, …, 16}, so $(15, 18)$ is not included.
Q
Which pairs need to be added?
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Need to add $(8, 7)$ and $(13, 11)$. These are symmetric counterparts of $(7, 8)$ and $(11, 13)$ respectively. Total = 2 pairs.
Q
Why not $(15, 18)$ in R?
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While $4(18) = 5(15) – 3$ is true (72 = 72), the pair $(15, 18)$ cannot be in R because $18 \notin M = \{1, 2, 3, \ldots, 16\}$. Both coordinates must be in M.
Q
How solve $x \equiv 3 \pmod{4}$?
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This means x leaves remainder 3 when divided by 4. So $x = 4k + 3$ for integer k ≥ 0. In M: $x = 3, 7, 11, 15, \ldots$ until exceeding 16.
Q
Can R be reflexive with these additions?
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No. Reflexive requires $(a, a) \in R$ for ALL $a \in M$. Currently only $(3, 3) \in R$. Would need to add $(1,1), (2,2), (4,4), \ldots, (16,16)$ — 15 more pairs for reflexivity.
Q
What if we add only $(8,7)$?
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Then R would still not be symmetric because $(11, 13) \in R$ but $(13, 11) \notin R$. We must add BOTH $(8, 7)$ and $(13, 11)$ for complete symmetry. Minimum = 2.
Q
Similar JEE problems on relations?
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JEE often asks about making relations reflexive, symmetric, transitive, or equivalence. Common questions: minimum elements to add/remove, finding equivalence classes, determining relation types, and working with modular arithmetic constraints on sets.
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