Let $A = \{1, 2, 3, 4, 5\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $4x \leq 5y$. Let $m$ be the number of elements in $R$ and $n$ be the minimum number of elements from $A \times A$ that are required to be added to $R$ to make it a symmetric relation. Then $m + n$ is equal to:
To solve this problem, we need to analyze the relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5\} \) by the condition \( xRy \) if and only if \( 4x \leq 5y \).
First, consider and list all pairs \((x, y)\) satisfying the condition \( 4x \leq 5y \):
\( x \)
Possible \( y \)
1
\( (1, 1), (1, 2), (1, 3), (1, 4), (1, 5) \)
2
\( (2, 2), (2, 3), (2, 4), (2, 5) \)
3
\( (3, 3), (3, 4), (3, 5) \)
4
\( (4, 4), (4, 5) \)
5
\( (5, 5) \)
Counting all these pairs, we have \( m = 15 \) elements in \( R \).
Next, we need to make \( R \) symmetric. A relation is symmetric if, whenever \((x, y) \in R\), then \((y, x) \in R\) as well. Our task is to make this relation symmetric by adding the minimum number of pairs.
Let's analyze:
For \( 1 \): All pairs \((1, y)\) are already symmetric because no extra \((y, 1)\) is needed as they exist already or are the same pair.
For \( 2 \): The pairs \((2, 3)\), \((2, 4)\), \((2, 5)\) imply we need to add \((3, 2)\), \((4, 2)\), \((5, 2)\).
For \( 3 \): Pairs needing symmetry are \((4, 3)\), \((5, 3)\).
For \( 4 \): Pair needing symmetry is \((5, 4)\).
Summing the extra pairs needed: \((3, 2), (4, 2), (5, 2), (4, 3), (5, 3), (5, 4)\), we must add \( n = 10 \) pairs to ensure all relations are symmetric.
Adding these pairs to the existing 15 elements in \( R \), we have \( m + n = 15 + 10 = 25 \).
The final answer is 25, which means:
The relation has 15 original pairs satisfying \( 4x \leq 5y \).
10 additional pairs are needed to make it symmetric.
