Let A = {2, 3, 5, 7, 9}. Let R be the relation on A defined by xRy if and only if 2x ≤ 3y. Let l be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then l + m is equal to

A Binary Relation $R$ from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$. If the relation is on a single set $A$, then $R \subseteq A \times A$. In the context of JEE, relations are often defined by algebraic inequalities or equations. For example, in our problem: $$\text{Given } A = \{2, 3, 5, 7, 9\}, \quad R = \{(x, y) \in A \times A : 2x \le 3y\}$$ The total number of possible pairs in $A \times A$ is $n(A) \times n(A) = 5 \times 5 = 25$. Our relation $R$ is a specific subset containing 19 of these 25 possible pairs.

Understanding these types is crucial for solving JEE problems:

1. Reflexive Relation: A relation $R$ on set $A$ is reflexive if every element is related to itself. $\forall a \in A, (a, a) \in R$ In our problem, $2x \le 3x$ is always true for positive $x$, so $(2,2), (3,3), \dots$ are all in $R$. $R$ is reflexive.

2. Symmetric Relation: If one element is related to another, the second must be related back to the first. $(x, y) \in R \implies (y, x) \in R$ Our relation $R$ is not symmetric because $(2, 5) \in R$ (as $4 \le 15$) but $(5, 2) \notin R$ (as $10 \not\le 6$).

3. Transitive Relation: If $aRb$ and $bRc$, then $aRc$. $(a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R$

4. Equivalence Relation: A relation that is Reflexive, Symmetric, and Transitive all at once.

The problem asks for $m$, the “minimum number of elements to make $R$ symmetric.” This is mathematically known as finding the Symmetric Closure of $R$. The symmetric closure $S$ is defined as $S = R \cup R^{-1}$, where $R^{-1}$ is the inverse relation.

Formula for $m$: $$m = n(R \cup R^{-1}) – n(R)$$ Essentially, you look for all pairs $(x, y)$ in $R$ where the flipped version $(y, x)$ is missing. For every such missing pair, you must add it to satisfy the definition of symmetry.

When dealing with inequalities in relations, it helps to rewrite them: $$y \ge \frac{2}{3}x$$ By plugging in each $x$ from the set $\{2, 3, 5, 7, 9\}$, we can find the lower bound for $y$.
• If $x=2$, $y \ge 1.33 \implies y \in \{2, 3, 5, 7, 9\}$
• If $x=9$, $y \ge 6 \implies y \in \{7, 9\}$
This systematic approach ensures no elements are missed when calculating $l$.

1. Confusion between “Minimum added” and “Total in symmetric”: Some students calculate $n(R \cup R^{-1})$ which would be $19 + 6 = 25$ in this case, but they might misinterpret what $m$ stands for. Always read if the question asks for the total count or just the added count.

2. Missing the “Equal to” sign: The inequality $2x \le 3y$ includes equality. For example, if $2x = 3y$, the pair belongs to $R$. If you miss this, your count for $l$ will be wrong.

3. Set boundary errors: Always ensure the values of $x$ and $y$ stay within the given set $A$. Calculations resulting in values outside $A$ must be discarded.

For a finite set $A$ with $n$ elements:
• Total number of relations $= 2^{n^2}$
• Total number of reflexive relations $= 2^{n^2-n}$
• Total number of symmetric relations $= 2^{\frac{n(n+1)}{2}}$

In this question, $n=5$. So there are $2^{25}$ possible relations and $2^{15}$ symmetric relations. This helps visualize how specific our $R$ is.

$n(A \times A) = n^2$ $R \text{ is symmetric } \iff R = R^{-1}$ Elements to add for symmetry $= n(R^{-1} – R)$ Symmetric Closure $= R \cup \{(y,x) : (x,y) \in R\}$