Pregunta 1: dado A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, encuentra (A × B) ∩ (B × C).
Solución:
Dado:
A = {1, 2, 3}, B = {3, 4} y C = {4, 5, 6}
Encontremos: (A × B) ∩ (B × C)
(A × B) = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(B × C) = {3, 4} × {4, 5, 6}
= {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
Por lo tanto,
(A × B) ∩ (B × C) = {(3, 4)}
Pregunta 2: Si A = {2, 3}, B = {4, 5}, C = {5, 6} encuentra A × (B ∪ C), (A × B) ∪ (A × C).
Solución:
Dado:
A = {2, 3}, B = {4, 5} y C = {5, 6}
Encontremos: A x (B ∪ C) y (A x B) ∪ (A x C)
(B ∪ C) = {4, 5, 6}
A × (B ∪ C) = {2, 3} × {4, 5, 6}
= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) = {2, 3} × {4, 5}
= {(2, 4), (2, 5), (3, 4), (3, 5)}
(A × C) = {2, 3} × {5, 6}
= {(2, 5), (2, 6), (3, 5), (3, 6)}
Por lo tanto,
(A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Pregunta 3: Si A = {1, 2, 3}, B = {4}, C = {5}, verifica que:
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) A × (B – C) = (A × B) – (A × C)
Solución:
Dado:
A = {1, 2, 3}, B = {4} y C = {5}
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
Supongamos LHS: (B ∪ C)
(B ∪ C) = {4, 5}
A × (B ∪ C) = {1, 2, 3} × {4, 5}
= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Ahora, RHS
(A × B) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(A × C) = {1, 2, 3} × {5}
= {(1, 5), (2, 5), (3, 5)}
(A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}
Por lo tanto,
LHS = RHS
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Supongamos LHS: (B ∩ C)
(B ∩ C) = ∅ (Sin elemento común)
A × (B ∩ C) = {1, 2, 3} × ∅
= ∅
Ahora, RHS
(A × B) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(A × C) = {1, 2, 3} × {5}
= {(1, 5), (2, 5), (3, 5)}
(A × B) ∩ (A × C) = ∅
Por lo tanto,
LHS = RHS
(iii) A × (B − C) = (A × B) − (A × C)
Supongamos LHS: (B − C)
(B − C) = ∅
UN × (B – C) = {1, 2, 3} × ∅
= ∅
Ahora, RHS
(A × B) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(A × C) = {1, 2, 3} × {5}
= {(1, 5), (2, 5), (3, 5)}
(A × B) − (A × C) = ∅
Por lo tanto,
LHS = RHS
Pregunta 4: Sean A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} y D = {5, 6, 7, 8}. Comprueba eso:
(i) A × C ⊂ B × D
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Solución:
Dado:
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} y D = {5, 6, 7, 8}
(i) A x C ⊂ B x D
Consideremos LHS A x C
A × C = {1, 2} × {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}
Ahora, RHS
B × P = {1, 2, 3, 4} × {5, 6, 7, 8}
= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Ya que todos los elementos de A × C están en B × D.
Por lo tanto,
Podemos decir A × C ⊂ B × D
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Supongamos LHS A × (B ∩ C)
(B ∩ C) = ∅
A × (B ∩ C) = {1, 2} × ∅
= ∅
Ahora, RHS
(A × B) = {1, 2} × {1, 2, 3, 4}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
(A × C) = {1, 2} × {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}
Por lo tanto, no hay un elemento común entre A × B y A × C
(A × B) ∩ (A × C) = ∅
Por lo tanto,
A × (B ∩ C) = (A × B) ∩ (A × C)
Pregunta 5: Si A = {1, 2, 3}, B = {3, 4} y C = {4, 5, 6}, encuentra
(i) A × (B ∩ C)
(ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C)
(iv) (A × B) ∪ (A × C)
Solución:
Dado:
A = {1, 2, 3}, B = {3, 4} y C = {4, 5, 6}
(i) A × (B ∩ C)
(B ∩ C) = {4}
A × (B ∩ C) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(ii) (A × B) ∩ (A × C)
(A × B) = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(A × C) = {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}
(iii) A × (B ∪ C)
(B ∪ C) = {3, 4, 5, 6}
A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
(iv) (A × B) ∪ (A × C)
(A × B) = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(A × C) = {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), ( 2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
Pregunta 6: Demuestre que:
(i) (A ∪ B) × C = (A × C) = (A × C) ∪ (B × C)
(ii) (A ∩ B) × C = (A × C) ∩ (B × C)
Solución:
(i) (A ∪ B) × C = (A × C) = (A × C) ∪ (B × C)
Sea (x, y) un elemento arbitrario de (A ∪ B) × C
(x, y) ∈ (A ∪ B) C
Como (x, y) son elementos del producto cartesiano de (A ∪ B) × C
x ∈ (A ∪ B) y y ∈ C
(x ∈ A o x ∈ B) y y ∈ C
(x ∈ A y y ∈ C) o (x ∈ Band y ∈ C)
(x, y) ∈ A × C o (x, y) ∈ B × C
(x, y) ∈ (A × C) ∪ (B × C) … (1)
Sea (x, y) un elemento arbitrario de (A × C) ∪ (B × C).
(x, y) ∈ (A × C) ∪ (B × C)
(x, y) ∈ (A × C) o (x, y) ∈ (B × C)
(x ∈ A y y ∈ C) o (x ∈ B y y ∈ C)
(x ∈ A o x ∈ B) y y ∈ C
x ∈ (A ∪ B) y y ∈ C
(x, y) ∈ (A ∪ B) × C … (2)
De 1 y 2, obtenemos: (A ∪ B) × C = (A × C) ∪ (B × C)
(ii) (A ∩ B) × C = (A × C) ∩ (B × C)
Sea (x, y) un elemento arbitrario de (A ∩ B) × C.
(x, y) ∈ (A ∩ B) × C
Dado que (x, y) son elementos del producto cartesiano de (A ∩ B) × C
x ∈ (A ∩ B) y y ∈ C
(x ∈ A y x ∈ B) y y ∈ C
(x ∈ A y y ∈ C) y (x ∈ Band y ∈ C)
(x, y) ∈ A × C y (x, y) ∈ B × C
(x, y) ∈ (A × C) ∩ (B × C) … (1)
Sea (x, y) un elemento arbitrario de (A × C) ∩ (B × C).
(x, y) ∈ (A × C) ∩ (B × C)
(x, y) ∈ (A × C) y (x, y) ∈ (B × C)
(x ∈A y y ∈ C) y (x ∈ Band y ∈ C)
(x ∈A y x ∈ B) y y ∈ C
x ∈ (A ∩ B) y y ∈ C
(x, y) ∈ (A ∩ B) × C … (2)
De 1 y 2, obtenemos: (A ∩ B) × C = (A × C) ∩ (B × C)
Pregunta 7: Si A × B ⊆ C × D y A ∩ B ∈ ∅, Demuestre que A ⊆ C y B ⊆ D.
Solución:
Dado:
A × B ⊆ C x D y A ∩ B ∈ ∅
A × B ⊆ C x D denota que A × B es un subconjunto de C × D, es decir, cada elemento A × B está en C × D.
Y A ∩ B ∈ ∅ denota A y B no tienen ningún elemento común entre ellos.
A × B = {(a, b): a ∈ A y b ∈ B}
Por lo tanto,
Podemos decir (a, b) ⊆ C × D [Ya que, A × B ⊆ C x D está dado]
a ∈ C y b ∈ D
un ∈ UN = un ∈ C
UN ⊆ C
Y
segundo ∈ segundo = segundo ∈ re
segundo ⊆ re
Por lo tanto probado.
Publicación traducida automáticamente
Artículo escrito por sudhasinghsudha90 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA