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Find the inverse relation R-1 in each of the following cases:
(i) R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
(ii) R= {(x, y) : x, y ∈ N; x + 2y = 8}
(iii) R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3

1 Answer

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(i) Given as

R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}

Therefore, R‑1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}

(ii) Given as

R= {(x, y): x, y ∈ N; x + 2y = 8}

Since, x + 2y = 8

x = 8 – 2y

As y ∈ N, Putting the values of y = 1, 2, 3,…… till x ∈ N

If, y = 1, x = 8 – 2(1) = 8 – 2 = 6

If, y = 2, x = 8 – 2(2) = 8 – 4 = 4

If, y = 3, x = 8 – 2(3) = 8 – 6 = 2

If, y = 4, x = 8 – 2(4) = 8 – 8 = 0

Then, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

∴ R = {(2, 3), (4, 2), (6, 1)}

R‑1 = {(3, 2), (2, 4), (1, 6)}

(iii) Given as

R is the relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3

Since,

x = {11, 12, 13} and y = (8, 10, 12}

y = x – 3

If, x = 11, y = 11 – 3 = 8 ∈ (8, 10, 12}

If, x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}

If, x = 13, y = 13 – 3 = 10 ∈ (8, 10, 12}

∴ R = {(11, 8), (13, 10)}

Thus, R‑1 = {(8, 11), (10, 13)}

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