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1. Let set A = {1, 2} and C be {3, 4} then A X B (Cartesian product of set A and B) is?a) {1, 2, 3, 4}b) {(1, 3),(2, 4)}c) {(1, 3), (2, 4), (1, 4), (2, 3)}d) {(3, 1), (4, 1)}

Question

  1. Let set A = {1, 2} and C be {3, 4} then A X B (Cartesian product of set A and B) is?a) {1, 2, 3, 4}b) {(1, 3),(2, 4)}c) {(1, 3), (2, 4), (1, 4), (2, 3)}d) {(3, 1), (4, 1)}
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Solution 1

Para resolver el problema, sigamos los siguientes pasos:

  1. Identifiquemos los conjuntos dados:

    • Conjunto A = {1, 2}
    • Conjunto C = {3, 4}
  2. Definamos el producto cartesiano A X C. El producto cartesiano de dos conjuntos A y C es el conjunto de todos los pares ordenados (a, c) donde 'a' pertenece a A y 'c' pertenece a C.

  3. Generemos todos los pares ordenados posibles:

    • (1, 3)
    • (1, 4)
    • (2, 3)
    • (2, 4)
  4. Listemos el conjunto resultante:

    • A X C = {(1, 3), (1, 4), (2, 3), (2, 4)}

Por lo tanto, la respuesta correcta es:

c) {(1, 3), (2, 4), (1, 4), (2, 3)}

This problem has been solved

Solution 2

There seems to be a mistake in the question. You mentioned set A and set C but you're asking for the Cartesian product of set A and B. Assuming you meant the Cartesian product of set A and C, the answer would be:

c) {(1, 3), (2, 4), (1, 4), (2, 3)}

The Cartesian product of two sets is a set of all ordered pairs (a, b) where a is in set A and b is in set C. So, for set A = {1, 2} and set C = {3, 4}, the Cartesian product A x C would be {(1, 3), (2, 4), (1, 4), (2, 3)}.

This problem has been solved

Solution 3

You seem to have a small mistake in your question. You mentioned sets A and C but then asked for the Cartesian product of A and B. Assuming you meant to find the Cartesian product of A and C, the answer would be:

c) {(1, 3), (2, 4), (1, 4), (2, 3)}

The Cartesian product of two sets A and C, denoted by A X C, is the set of all ordered pairs (a, c) where a is in A and c is in C. So, for A = {1, 2} and C = {3, 4}, the Cartesian product A X C = {(1, 3), (2, 4), (1, 4), (2, 3)}.

This problem has been solved

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