Let $ A = \{-2, -1, 0, 1, 2, 3\} $. Let $ R $ be a relation on $ A $ defined by $ (x, y) \in R $ if and only if $ |x| \le |y| $. Let $ m $ be the number of reflexive elements in $ R $ 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
Let $ A = \{-2, -1, 0, 1, 2, 3\} $. Let $ R $ be a relation on $ A $ defined by $ (x, y) \in R $ if and only if $ |x| \le |y| $. Let $ m $ be the number of reflexive elements in $ R $ 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
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The Correct Option is A
Approach Solution - 1
The problem provides a set \(A = \{-2, -1, 0, 1, 2, 3\}\) and a relation R on A defined by \(xRy\) if and only if \(y = \max\{x, 1\}\). We need to find the number of elements in R (\(l\)), the minimum number of elements to add to make R reflexive (\(m\)), and the minimum number of elements to add to make R symmetric (\(n\)). Finally, we need to compute the sum \(l + m + n\).
Concept Used:
A relation R on a set A is a subset of the Cartesian product \(A \times A\).
A relation R on a set A is reflexive if for every element \(a \in A\), the ordered pair \((a, a)\) is in R.
A relation R on a set A is symmetric if for every ordered pair \((a, b)\) in R, the pair \((b, a)\) is also in R. That is, if \(aRb\), then \(bRa\).
Step-by-Step Solution:
Step 1: Determine the elements of the relation R and find its size, \(l\).
The relation is defined by \(y = \max\{x, 1\}\) for \(x, y \in A\). We test each value of \(x\) from set A:
- For \(x = -2\), \(y = \max\{-2, 1\} = 1\). Since \(1 \in A\), the pair \((-2, 1)\) is in R.
- For \(x = -1\), \(y = \max\{-1, 1\} = 1\). Since \(1 \in A\), the pair \((-1, 1)\) is in R.
- For \(x = 0\), \(y = \max\{0, 1\} = 1\). Since \(1 \in A\), the pair \((0, 1)\) is in R.
- For \(x = 1\), \(y = \max\{1, 1\} = 1\). Since \(1 \in A\), the pair \((1, 1)\) is in R.
- For \(x = 2\), \(y = \max\{2, 1\} = 2\). Since \(2 \in A\), the pair \((2, 2)\) is in R.
- For \(x = 3\), \(y = \max\{3, 1\} = 3\). Since \(3 \in A\), the pair \((3, 3)\) is in R.
So, the relation R is the set of these ordered pairs:
\[ R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\} \]The number of elements in R is \(l\). By counting the pairs, we get:
\[ l = 6 \]Step 2: Determine the number of elements (\(m\)) to be added to make R reflexive.
For R to be reflexive on A, it must contain all pairs \((a, a)\) for every \(a \in A\). The required set of pairs for reflexivity is:
\[ R_{\text{reflexive\_pairs}} = \{(-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)\} \]Comparing this with the elements of R, we see that R already contains \((1, 1)\), \((2, 2)\), and \((3, 3)\).
The pairs that are missing from R are:
\[ \{(-2, -2), (-1, -1), (0, 0)\} \]Therefore, we need to add these 3 elements to R to make it reflexive. The minimum number of elements to be added is:
\[ m = 3 \]Step 3: Determine the number of elements (\(n\)) to be added to make R symmetric.
For R to be symmetric, if \((a, b) \in R\), then \((b, a)\) must also be in R. We check each pair in R:
- For \((-2, 1) \in R\), we need \((1, -2)\). This is not in R.
- For \((-1, 1) \in R\), we need \((1, -1)\). This is not in R.
- For \((0, 1) \in R\), we need \((1, 0)\). This is not in R.
- For \((1, 1) \in R\), we need \((1, 1)\), which is already in R.
- For \((2, 2) \in R\), we need \((2, 2)\), which is already in R.
- For \((3, 3) \in R\), we need \((3, 3)\), which is already in R.
The pairs that need to be added to make R symmetric are:
\[ \{(1, -2), (1, -1), (1, 0)\} \]Therefore, the minimum number of elements to be added is:
\[ n = 3 \]Step 4: Calculate the final sum \(l + m + n\).
We have found the values \(l=6\), \(m=3\), and \(n=3\). We now compute their sum.
\[ l + m + n = 6 + 3 + 3 = 12 \]The value of \(l + m + n\) is 12.
Approach Solution -2
Let the set \( A = \{-2, -1, 0, 1, 2, 3\} \)
Let the relation \( R = \{(-2,1), (-1,1), (0,1), (1,1), (2,2), (3,3)\} \)
\[\begin{align*} \lambda &= 6 \\m &= 3 \\n &= 3 \\\lambda + m + n &= 6 + 3 + 3 = 12 \end{align*}\]
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