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Q1. Let ๐ด = {0,1,2,3,4,5,6,7} suppose ๐‘… ๐‘Ž๐‘›๐‘‘ ๐‘‡ are two relations on ๐ด such that ๐‘… = {(๐‘ฅ, ๐‘ฆ): ๐‘ฅ + 2๐‘ฆ โ‰ฅ 4}, ๐‘‡ = {(๐‘ฅ, ๐‘ฆ): 2๐‘ฅ + 3๐‘ฆ โˆˆ ๐ด} Write ๐‘…, ๐‘‡, ๐‘Ž๐‘›๐‘‘ ๐‘…ยฐ๐‘‡

Question

Q1. Let ๐ด = {0,1,2,3,4,5,6,7} suppose ๐‘… and ๐‘‡ are two relations on ๐ด such that

๐‘… = {(๐‘ฅ, ๐‘ฆ): ๐‘ฅ + 2๐‘ฆ โ‰ฅ 4},
๐‘‡ = {(๐‘ฅ, ๐‘ฆ): 2๐‘ฅ + 3๐‘ฆ โˆˆ ๐ด}
Write ๐‘…, ๐‘‡, and ๐‘…ยฐ๐‘‡.

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Solution

First, let's find the pairs for ๐‘… and ๐‘‡.

For ๐‘… = {(๐‘ฅ, ๐‘ฆ): ๐‘ฅ + 2๐‘ฆ โ‰ฅ 4}, we substitute the values from ๐ด for ๐‘ฅ and ๐‘ฆ and find the pairs that satisfy the condition.

The pairs for ๐‘… are: {(2,1), (3,1), (4,0), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (4,7), (5,0), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (5,7), (6,0), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (6,7), (7,0), (7,1), (7,2), (7,3), (7,4), (7,5), (7,6), (7,7)}

For ๐‘‡ = {(๐‘ฅ, ๐‘ฆ): 2๐‘ฅ + 3๐‘ฆ โˆˆ ๐ด}, we substitute the values from ๐ด for ๐‘ฅ and ๐‘ฆ and find the pairs that satisfy the condition.

The pairs for ๐‘‡ are: {(0,0), (0,1), (0,2), (1,0), (1,1), (2,0)}

Now, let's find ๐‘…ยฐ๐‘‡, which is the composition of ๐‘… and ๐‘‡. A pair (๐‘Ž,๐‘) is in ๐‘…ยฐ๐‘‡ if there exists a ๐‘ in ๐ด such that (๐‘Ž,๐‘) is in ๐‘… and (๐‘,๐‘) is in ๐‘‡.

The pairs for ๐‘…ยฐ๐‘‡ are: {(4,0), (4,1), (5,0), (5,1), (6,0), (6,1), (7,0), (7,1)}

This problem has been solved

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