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The relation $R$ is defined on the set $A = \{1, 2, 3, 4, 5\}$ such that $(x, y) \in R$ if $|x^2 - y^2| \le 8$. We need to find all pairs $(x, y)$ that satisfy this condition.
We will iterate through all possible pairs $(x, y)$ where $x, y \in A$ and check if the condition $|x^2 - y^2| \le 8$ is satisfied.
For $x = 1$: - $y = 1$: $|1^2 - 1^2| = 0 \le 8$ - $y = 2$: $|1^2 - 2^2| = |1 - 4| = 3 \le 8$ - $y = 3$: $|1^2 - 3^2| = |1 - 9| = 8 \le 8$ - $y = 4$: $|1^2 - 4^2| = |1 - 16| = 15 > 8$ - $y = 5$: $|1^2 - 5^2| = |1 - 25| = 24 > 8$ For $x = 2$: - $y = 1$: $|2^2 - 1^2| = |4 - 1| = 3 \le 8$ - $y = 2$: $|2^2 - 2^2| = 0 \le 8$ - $y = 3$: $|2^2 - 3^2| = |4 - 9| = 5 \le 8$ - $y = 4$: $|2^2 - 4^2| = |4 - 16| = 12 > 8$ - $y = 5$: $|2^2 - 5^2| = |4 - 25| = 21 > 8$ For $x = 3$: - $y = 1$: $|3^2 - 1^2| = |9 - 1| = 8 \le 8$ - $y = 2$: $|3^2 - 2^2| = |9 - 4| = 5 \le 8$ - $y = 3$: $|3^2 - 3^2| = 0 \le 8$ - $y = 4$: $|3^2 - 4^2| = |9 - 16| = 7 \le 8$ - $y = 5$: $|3^2 - 5^2| = |9 - 25| = 16 > 8$ For $x = 4$: - $y = 1$: $|4^2 - 1^2| = |16 - 1| = 15 > 8$ - $y = 2$: $|4^2 - 2^2| = |16 - 4| = 12 > 8$ - $y = 3$: $|4^2 - 3^2| = |16 - 9| = 7 \le 8$ - $y = 4$: $|4^2 - 4^2| = 0 \le 8$ - $y = 5$: $|4^2 - 5^2| = |16 - 25| = 9 > 8$ For $x = 5$: - $y = 1$: $|5^2 - 1^2| = |25 - 1| = 24 > 8$ - $y = 2$: $|5^2 - 2^2| = |25 - 4| = 21 > 8$ - $y = 3$: $|5^2 - 3^2| = |25 - 9| = 16 > 8$ - $y = 4$: $|5^2 - 4^2| = |25 - 16| = 9 > 8$ - $y = 5$: $|5^2 - 5^2| = 0 \le 8$
Based on the above calculations, the relation $R$ is: $R = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)\}$
Final Answer: R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)}
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