Key Concepts and Formulas

• Symmetric Relation: A relation $R$ on a set $X$ is symmetric if for every $(x, y) \in R$, it is also true that $(y, x) \in R$.
• Minimum Additions for Symmetry: To make a relation symmetric, for every pair $(x, y) \in R$ where $(y, x) \notin R$, we must add $(y, x)$ to the relation. The minimum number of elements to be added is the count of such missing inverse pairs. Pairs of the form $(x, x)$ are always symmetric.

Step-by-Step Solution

Step 1: Understanding the Problem and the Set

We are given a set $X = \{1, 2, 3, \ldots, 20\}$. Two relations, $R_1$ and $R_2$, are defined on $X$:

1. $R_1 = \{(x, y) : 2x - 3y = 2\}$
2. $R_2 = \{(x, y) : -5x + 4y = 0\}$

We need to find $M$, the minimum number of elements to add to $R_1$ to make it symmetric, and $N$, the minimum number of elements to add to $R_2$ to make it symmetric. Our final goal is to compute $M+N$.

Step 2: Analyzing Relation $R_1$ and Calculating $M$

The relation $R_1$ is defined by $2x - 3y = 2$. We need to find pairs $(x, y)$ where $x, y \in \{1, 2, \ldots, 20\}$. Rearranging the equation for $x$: $2x = 3y + 2$. Since $2x$ is even, $3y + 2$ must be even. This implies $3y$ must be even, which means $y$ must be an even number. We test even values of $y$ from 2 to 20:

• If $y=2$: $2x = 3(2) + 2 = 8 \implies x = 4$. Pair: $(4, 2)$. Both $4, 2 \in X$.
• If $y=4$: $2x = 3(4) + 2 = 14 \implies x = 7$. Pair: $(7, 4)$. Both $7, 4 \in X$.
• If $y=6$: $2x = 3(6) + 2 = 20 \implies x = 10$. Pair: $(10, 6)$. Both $10, 6 \in X$.
• If $y=8$: $2x = 3(8) + 2 = 26 \implies x = 13$. Pair: $(13, 8)$. Both $13, 8 \in X$.
• If $y=10$: $2x = 3(10) + 2 = 32 \implies x = 16$. Pair: $(16, 10)$. Both $16, 10 \in X$.
• If $y=12$: $2x = 3(12) + 2 = 38 \implies x = 19$. Pair: $(19, 12)$. Both $19, 12 \in X$.
• If $y=14$: $2x = 3(14) + 2 = 44 \implies x = 22$. $22 \notin X$, so we stop.

The elements of $R_1$ are $\{(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)\}$. For $R_1$ to be symmetric, for each $(x, y) \in R_1$, we must have $(y, x) \in R_1$. Let's check the inverse pairs:

• For $(4, 2)$, is $(2, 4) \in R_1$? $2(2) - 3(4) = 4 - 12 = -8 \neq 2$. No.
• For $(7, 4)$, is $(4, 7) \in R_1$? $2(4) - 3(7) = 8 - 21 = -13 \neq 2$. No.
• For $(10, 6)$, is $(6, 10) \in R_1$? $2(6) - 3(10) = 12 - 30 = -18 \neq 2$. No.
• For $(13, 8)$, is $(8, 13) \in R_1$? $2(8) - 3(13) = 16 - 39 = -23 \neq 2$. No.
• For $(16, 10)$, is $(10, 16) \in R_1$? $2(10) - 3(16) = 20 - 48 = -28 \neq 2$. No.
• For $(19, 12)$, is $(12, 19) \in R_1$? $2(12) - 3(19) = 24 - 57 = -33 \neq 2$. No.

None of the inverse pairs are present in $R_1$. All elements in these inverse pairs are within $X$. Since there are 6 pairs in $R_1$, and none of their inverses are present, we need to add 6 elements to make $R_1$ symmetric. Thus, $M = 6$.

Step 3: Analyzing Relation $R_2$ and Calculating $N$

The relation $R_2$ is defined by $-5x + 4y = 0$. We need to find pairs $(x, y)$ where $x, y \in \{1, 2, \ldots, 20\}$. Rearranging the equation: $4y = 5x$. This implies $5x$ must be a multiple of 4. Since 5 and 4 are coprime, $x$ must be a multiple of 4. Also, $y = \frac{5x}{4}$. Since $y \le 20$, we have $\frac{5x}{4} \le 20 \implies 5x \le 80 \implies x \le 16$. We test multiples of 4 for $x$ from 4 to 16:

• If $x=4$: $4y = 5(4) = 20 \implies y = 5$. Pair: $(4, 5)$. Both $4, 5 \in X$.
• If $x=8$: $4y = 5(8) = 40 \implies y = 10$. Pair: $(8, 10)$. Both $8, 10 \in X$.
• If $x=12$: $4y = 5(12) = 60 \implies y = 15$. Pair: $(12, 15)$. Both $12, 15 \in X$.
• If $x=16$: $4y = 5(16) = 80 \implies y = 20$. Pair: $(16, 20)$. Both $16, 20 \in X$.
• If $x=20$: $4y = 5(20) = 100 \implies y = 25$. $25 \notin X$, so we stop.

The elements of $R_2$ are $\{(4, 5), (8, 10), (12, 15), (16, 20)\}$. For $R_2$ to be symmetric, for each $(x, y) \in R_2$, we must have $(y, x) \in R_2$. Let's check the inverse pairs:

• For $(4, 5)$, is $(5, 4) \in R_2$? $-5(5) + 4(4) = -25 + 16 = -9 \neq 0$. No.
• For $(8, 10)$, is $(10, 8) \in R_2$? $-5(10) + 4(8) = -50 + 32 = -18 \neq 0$. No.
• For $(12, 15)$, is $(15, 12) \in R_2$? $-5(15) + 4(12) = -75 + 48 = -27 \neq 0$. No.
• For $(16, 20)$, is $(20, 16) \in R_2$? $-5(20) + 4(16) = -100 + 64 = -36 \neq 0$. No.

None of the inverse pairs are present in $R_2$. All elements in these inverse pairs are within $X$. Since there are 4 pairs in $R_2$, and none of their inverses are present, we need to add 4 elements to make $R_2$ symmetric. Thus, $N = 4$.

Step 4: Calculating $M+N$

We have found $M = 6$ and $N = 4$. The required sum is $M+N = 6 + 4 = 10$.

Common Mistakes & Tips

• Forgetting the Set Constraint: Always ensure that both elements of a pair $(x, y)$ and its inverse $(y, x)$ belong to the given set $X$.
• Ignoring $(x,x)$ Pairs: Pairs of the form $(x,x)$ are inherently symmetric and do not require any additions. In this problem, neither $R_1$ nor $R_2$ contained any such pairs.
• Algebraic Errors: Double-check the rearrangement of equations and the arithmetic when finding pairs and their inverses.

Summary

We first identified all the pairs belonging to relations $R_1$ and $R_2$ by satisfying their respective equations and the set constraint $X=\{1, 2, \ldots, 20\}$. For each relation, we then checked if the inverse of every existing pair was also present in the relation. The number of missing inverse pairs for $R_1$ gives $M$, and for $R_2$ gives $N$. We found $R_1$ to have 6 pairs and $R_2$ to have 4 pairs, none of whose inverses were present in their respective relations. Therefore, $M=6$ and $N=4$, leading to $M+N=10$.

The final answer is $\boxed{10}$.