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Step-by-Step Solution
Step 1: Understand the Problem
We have two binary relations on the set of natural numbers $N$ defined as follows:
$R_{1} = \{(x, y) \in N \times N : 2x + y = 10\}$
$R_{2} = \{(x, y) \in N \times N : x + 2y = 10\}$
We need to determine the range of each relation and then see which of the given statements is correct. The correct statement provided is that the range of $R_{2}$ is $\{1, 2, 3, 4\}$.
Step 2: Find All Possible Ordered Pairs in $R_{1}$
For $R_{1}$, we require $2x + y = 10$. Since $x$ and $y$ are natural numbers (typically $N = \{1,2,3,\ldots\}$), we list all solutions:
$x=1 \implies 2(1) + y = 10 \implies y = 8 \implies (1, 8)$
$x=2 \implies 2(2) + y = 10 \implies y = 6 \implies (2, 6)$
$x=3 \implies 2(3) + y = 10 \implies y = 4 \implies (3, 4)$
$x=4 \implies 2(4) + y = 10 \implies y = 2 \implies (4, 2)$
No larger $x$ can be used because $2x$ would exceed 10 if $x \geq 5$. So,
$R_{1} = \{(1,8), (2,6), (3,4), (4,2)\}.$
Step 3: Determine the Range of $R_{1}$
The range (set of all second components) of $R_{1}$ is $\{8, 6, 4, 2\} = \{2, 4, 6, 8\}.$
Step 4: Check Symmetry and Transitivity for $R_{1}$
Symmetry: A relation $R$ is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$. For instance, $(1,8) \in R_{1}$, but $(8,1) \notin R_{1}$. So $R_{1}$ is not symmetric.
Transitivity: A relation $R$ is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. For $R_{1}$, we have $(3,4)$ and $(4,2)$ in $R_{1}$, but $(3,2) \notin R_{1}$. So $R_{1}$ is not transitive.
Step 5: Find All Possible Ordered Pairs in $R_{2}$
For $R_{2}$, we require $x + 2y = 10$. Again, we look for solutions in natural numbers:
$y=1 \implies x + 2(1) = 10 \implies x=8 \implies (8, 1)$
$y=2 \implies x + 2(2) = 10 \implies x=6 \implies (6, 2)$
$y=3 \implies x + 2(3) = 10 \implies x=4 \implies (4, 3)$
$y=4 \implies x + 2(4) = 10 \implies x=2 \implies (2, 4)$
No larger $y$ can be used because $2y$ would exceed 10 if $y \geq 5$. Hence,
$R_{2} = \{(8,1),(6,2),(4,3),(2,4)\}.$
Step 6: Determine the Range of $R_{2}$
The range (second components) of $R_{2}$ is $\{1, 2, 3, 4\}.$
Step 7: Check Symmetry and Transitivity for $R_{2}$
Symmetry: Similarly, $(8,1) \in R_{2}$ does not imply $(1,8) \in R_{2}$. So $R_{2}$ is not symmetric.
Transitivity: We can also check pairs to see if transitivity holds. We find that $R_{2}$ does not satisfy transitivity either (e.g., $(8,1)$ and $(1,\dots)$ is not in $R_{2}$, so we cannot form a transitive chain leading to a matching pair).
Step 8: Conclude the Correct Statement
The correct statement given is that the range of $R_{2}$ is $\{1, 2, 3, 4\}$. From our computation, this is indeed true. The other options about ranges and symmetry/transitivity do not hold.
Final Answer
The statement "Range of $R_{2}$ is $\{1, 2, 3, 4\}$" is correct.
