Show that (A∪B)∩C =(A∩C)∪(B∩C) Proof (i) show that (A∪B)∩C⊂(A∩C) (B∩C) take any x∈(A∪B)∩C obvious x∈(A∪B)∩C ⇔x∈(A∪B) ∧ x∈C ⇔(x∈A ∨ x∈B) ∧ x∈C (distributif) ⇔x∈A ∧ x∈C ∨ x∈B ∧ x∈C ⇔x∈(A∩C) ∨ x∈ (B∩C) ⇔x∈(A∩C) ∪ (B∩C) so...
we get for all x∈(A∪B) ∩C then x∈(A∩C)∪ (B∩C) it means (A∪B)∩C =(A∩C)∪(B∩C)
ii. Show that A ⊂ B if and only if A ∩ B = A Proof: 1) Show that A ⊂ B ⇒ A ∩ B = A We have A ⊂ B It means ∀ x ∈ A,x ∈ B Show that A ⊂ A ∩ B Take any x ∈ A Obvious x ∈ A ∧ x ∈ A ⇔ x ∈ A ∧ x ∈ B (because ∀ x ∈ A,x ∈ B ) ⇔ x ∈ A ∩ B We get for all x ∈ A then x ∈ A ∩ B It means A ⊂ A ∩ B....(1) 2) Show that A ∩ B ⊂ A Take any x ∈ A ∩ B Obvious x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B ⇔ x ∈ A (simplifikasi) We get for all x ∈ A ∩ B,x ∈ A It means A ∩ B ⊂ A....(2) From (1) and (2) we conclude that A ∩ B = A So if A ⊂ B then A ∩ B = A
PBE 1. A ⊄ B 2. Tidak benar bahwa A ⊂ B 3. Tidak benar bahwa ∀ x ∈ A,x ∈ B 4. Tidak setiap x ∈ A berlaku x ∈ B 5. Ada x ∈ A yang bukan x ∈ B 6. Terdapat x ∈ A,x ∈ B 7. ∃ x ∈ A, x ∉ B 8. x ∈ A ∧ x ∉ B
Take any x ∈ A ∩ B Show that A ∩ B ⊂ A Andaikan A ∩ B ⊄ A maka ∃ x ∈ A ∩ B, akan tetapi x ⊄ A maka x ∈ A ∩ B ∧ x ∈ A komplemen ⇔ ( x ∈ A ∧ x ∈ B) ∧ x ∈ A komplemen (asosiatif) ⇔ x ∈ A ∧ x ∈ A komplemen ∧ x ∈ B ⇔ x ∈ ( A ∩ A komplemen) ∧ x ∈ B ⇔ x ∈ ∅ ∧ x ∈ B ⇔ x ∈ ( ∅ ∩ B ) ⇔ x ∈ ∅ Maka terjadi kontradiksi oleh sebab terdapat x ∈ A ∩ B dan x ∉ A berlaku x ∉ ∅ Jadi pengandaian ditolak Jadi yang benar adalah A ∩ B ⊂ A
1.)Let A , B set and x is an element . While : a) x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B b) x ∉ A ∩ B ⇔ x ∉ A ∧ x ∉ B
Answer: a) x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B b) PBE : 1.) x ∉ A ∩ B 2.) Tidak benar bahwa x ∈ A ∩ B 3.) Tidak benar bahwa x ∈ A ∧ x ∈ B 4.) x ∉ A ∧ x ∉ B
2.) Show that : a) A ∩ A = A b) A ∩ B = B ∩ A c) (A ∩ B) ∩ C = A ∩ (B ∩ C )
Answer: a) Proof i) Show that A ∩ A ⊂A Take any x ∈ A ∩ A Obvious x ∈ A ∩ A ⇔ x ∈ A ∧ x ∈ A ⇔ x ∈ A (idempoten) ....(i) So A ∩ A ⊂A ⇔ x ∈ A ii)Show that A ⊂A ∩ A Take any x ∈ A ∩ A Obvious x ∈ A ∩ A ⇔ x ∈ A ∧ x ∈ A ⇔ x ∈ A (idempoten) ....(ii) So A ∩ A ⊂A ⇔ x ∈ A From that (i) and (ii) we conclude that A ∩ A = A b) Proof i) Show that A ∩ B ⊂ B ∩ A Take any x ∈ A ∩ B Obvious x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B ⇔ x ∈ B ∩ A So A ∩ B ⊂ B ∩ A (komutatif )...(i) ii) Show that B ∩ A ⊂A ∩ B Take any x ∈ B ∩ A Obvious x ∈ B ∩ A ⇔ x ∈ B ∧ x ∈ A ⇔ x ∈ A ∩ B So B ∩ A ⊂ A ∩ B (komutatif )...(ii) From that (i) and (ii) we conclude that A ∩ B = B ∩ A c) Proof i) Show that (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C ) Take any x ∈ (A ∩ B) ∩ C Obvious x ∈ (A ∩ B) ∩ C ⇔ (x ∈ A ∧ x ∈ B) x ∈ C ⇔ x ∈ A ∧ ( x ∈ B x ∈ C) ⇔ x ∈ A ∩ (B ∩ C ) So (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C ) (asosiatif)...(i) ii) i) Show tha) A ∩ (B ∩ C )⊂ (A ∩ B) ∩ C Take any x ∈ A ∩ (B ∩ C ) Obvious x ∈ A ∩ (B ∩ C ) ⇔ x ∈ A ∧ ( x ∈ B x ∈ C) ⇔ (x ∈ A ∧ x ∈ B ) x ∈ C ⇔ x ∈ (A ∩ B ) ∩ C So A ∩ (B ∩ C ) ⊂ (A ∩ B) ∩ C (asosiatif)...(ii) From that (i) and (ii) we conclude that (A ∩ B) ∩ C = A ∩ (B ∩ C )