Page 21 - E-Modul Himpunan

A ∪ (B ∪ C) = {p, q, r, s} ∪ {q, r, s, t}

A ∪ (B ∪ C) = {p, q, r, s, t}

Dengan demikian, dapat ditunjukkan bahwa (A ∪ B) ∪ C = A ∪ (B ∪ C).

E. Sifat Distributif

Sifat distributif pada operasi himpunan hanya berlaku pada operasi irisan

dan gabungan, yaitu A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) dan A ∪ (B ∩ C)

= (A ∪ B) ∩ (A ∪ C).

Contoh:

Diketahui himpunan A = {1, 2, 3, 4, ..., 10}, B = {2, 4, 6, 8, 10} dan C =

{1, 3, 5, 7, 9}. Tunjukkan bahwa A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

Penyelesaian:

Langkah pertama, tentukan hasil dari A ∩ (B ∪ C).

A ∩ (B ∪ C) = {1, 2, 3, 4, ..., 10} ∩ ({2, 4, 6, 8, 10} ∪ {1, 3, 5, 7, 9})

A ∩ (B ∪ C) = {1, 2, 3, 4, ..., 10} ∩ {1, 2, 3, 4, ..., 10}

A ∩ (B ∪ C) = {1, 2, 3, 4, ..., 10}

Langkah kedua tentukan hasil dari (A ∩ B) ∪ (A ∩ C).

(A ∩ B) = {1, 2, 3, 4, ..., 10} ∩ {2, 4, 6, 8, 10}

(A ∩ B) = {2, 4, 6, 8, 10}

(A ∩ C) = {1, 2, 3, 4, ..., 10} ∩ {1, 3, 5, 7, 9}

(A ∩ C) = {1, 3, 5, 7, 9}

(A ∩ B) ∪ (A ∩ C) = {2, 4, 6, 8, 10} ∪ {1, 3, 5, 7, 9}
(A ∩ B) ∪ (A ∩ C) = {1, 2, 3, 4, ..., 10}

Dengan membandingkan hasil akhir langkah pertama dan kedua, dapat

ditunjukkan bahwa A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ B).

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