Question ID: #786
Let $A=\{-2,-1,0,1,2,3,4\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $2x+y\le2$. Let $l$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations respectively. Then $l+m+n$ is equal to:
- (1) 32
- (2) 34
- (3) 33
- (4) 35
Solution:
First, we list the elements of $R$ satisfying $2x+y \le 2$ for $x, y \in A$.
Case $x = -2$: $2(-2) + y \le 2 \Rightarrow -4 + y \le 2 \Rightarrow y \le 6$. All 7 elements of $A$ work. (7 pairs)
Case $x = -1$: $2(-1) + y \le 2 \Rightarrow -2 + y \le 2 \Rightarrow y \le 4$. All 7 elements work. (7 pairs)
Case $x = 0$: $2(0) + y \le 2 \Rightarrow y \le 2$. Elements: $\{-2, -1, 0, 1, 2\}$. (5 pairs)
Case $x = 1$: $2(1) + y \le 2 \Rightarrow y \le 0$. Elements: $\{-2, -1, 0\}$. (3 pairs)
Case $x = 2$: $2(2) + y \le 2 \Rightarrow y \le -2$. Element: $\{-2\}$. (1 pair)
Case $x = 3$: $2(3) + y \le 2 \Rightarrow y \le -4$. No solution in $A$.
Case $x = 4$: $2(4) + y \le 2 \Rightarrow y \le -6$. No solution in $A$.
Total elements in $R$, $l = 7 + 7 + 5 + 3 + 1 = 23$.
**For Reflexive ($m$):**
We need $(x, x) \in R$ for all $x \in A$.
Pairs required: $(-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3), (4,4)$.
Checking against $2x+y \le 2$:
$(-2,-2): -6 \le 2$ (Present)
$(-1,-1): -3 \le 2$ (Present)
$(0,0): 0 \le 2$ (Present)
$(1,1): 3 \le 2$ (False, Missing)
$(2,2): 6 \le 2$ (False, Missing)
$(3,3): 9 \le 2$ (False, Missing)
$(4,4): 12 \le 2$ (False, Missing)
Missing pairs are $(1,1), (2,2), (3,3), (4,4)$.
Thus, $m = 4$.
**For Symmetric ($n$):**
We need $(y, x) \in R$ whenever $(x, y) \in R$.
Let’s identify asymmetric pairs.
Pairs present where $2x+y \le 2$ but $2y+x > 2$:
1. $(0, 2) \in R$ ($2 \le 2$), but $(2, 0) \notin R$ ($4 > 2$). Need to add $(2, 0)$.
2. $(-1, 3) \in R$ ($1 \le 2$), but $(3, -1) \notin R$ ($5 > 2$). Need to add $(3, -1)$.
3. $(-1, 4) \in R$ ($2 \le 2$), but $(4, -1) \notin R$ ($7 > 2$). Need to add $(4, -1)$.
4. $(-2, 3) \in R$ ($-1 \le 2$), but $(3, -2) \notin R$ ($4 > 2$). Need to add $(3, -2)$.
5. $(-2, 4) \in R$ ($0 \le 2$), but $(4, -2) \notin R$ ($6 > 2$). Need to add $(4, -2)$.
6. $(-1, 2)$ is present $(-2+2 \le 2)$. Reverse $(2, -1)$ check: $2(2)+(-1) = 3 > 2$. Need to add $(2, -1)$.
Total pairs to add for symmetry: $n = 6$.
Finally, $l + m + n = 23 + 4 + 6 = 33$.
Ans. (3)
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