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Sagot :
Certainly! We will need to determine the Cartesian product [tex]\(A \times B\)[/tex] where [tex]\(A = \{1, 2, 3\}\)[/tex] and [tex]\(B = \{4, 5, 6\}\)[/tex], and then analyze the various conditions for relations from [tex]\(A\)[/tex] to [tex]\(B\)[/tex].
### Cartesian Product [tex]\(A \times B\)[/tex]:
The Cartesian product [tex]\(A \times B\)[/tex] consists of all possible ordered pairs [tex]\((a, b)\)[/tex] where [tex]\(a \in A\)[/tex] and [tex]\(b \in B\)[/tex]. Let's enumerate these pairs:
[tex]\[ A \times B = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \][/tex]
### Relation determined by each condition:
(a) For [tex]\(x < y\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(a < b\)[/tex]:
- [tex]\((1, 4)\)[/tex]: [tex]\(1 < 4\)[/tex]
- [tex]\((1, 5)\)[/tex]: [tex]\(1 < 5\)[/tex]
- [tex]\((1, 6)\)[/tex]: [tex]\(1 < 6\)[/tex]
- [tex]\((2, 4)\)[/tex]: [tex]\(2 < 4\)[/tex]
- [tex]\((2, 5)\)[/tex]: [tex]\(2 < 5\)[/tex]
- [tex]\((2, 6)\)[/tex]: [tex]\(2 < 6\)[/tex]
- [tex]\((3, 4)\)[/tex]: [tex]\(3 < 4\)[/tex]
- [tex]\((3, 5)\)[/tex]: [tex]\(3 < 5\)[/tex]
- [tex]\((3, 6)\)[/tex]: [tex]\(3 < 6\)[/tex]
Thus, the relation is:
[tex]\[\{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}\][/tex]
(b) For [tex]\(x + y = 5\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(a + b = 5\)[/tex]:
- [tex]\((1, 4)\)[/tex]: [tex]\(1 + 4 = 5\)[/tex]
Thus, the relation is:
[tex]\[\{(1, 4)\}\][/tex]
(c) For [tex]\(x > y\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(a > b\)[/tex]:
There are no such pairs where the element from [tex]\(A\)[/tex] is greater than the element from [tex]\(B\)[/tex] because all elements in [tex]\(A\)[/tex] are less than the elements in [tex]\(B\)[/tex].
Thus, the relation is:
[tex]\[\{\}\][/tex]
(d) For [tex]\(y = 2x\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(b = 2a\)[/tex]:
- [tex]\((2, 4)\)[/tex]: [tex]\(4 = 2 \cdot 2\)[/tex]
- [tex]\((3, 6)\)[/tex]: [tex]\(6 = 2 \cdot 3\)[/tex]
Thus, the relation is:
[tex]\[\{(2, 4), (3, 6)\}\][/tex]
### Results Summarized:
- [tex]\(A \times B\)[/tex]:
[tex]\[ \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \][/tex]
- Relation by condition:
- (a) [tex]\(x < y\)[/tex]:
[tex]\[ \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \][/tex]
- (b) [tex]\(x + y = 5\)[/tex]:
[tex]\[ \{(1, 4)\} \][/tex]
- (c) [tex]\(x > y\)[/tex]:
[tex]\[ \{\} \][/tex]
- (d) [tex]\(y = 2x\)[/tex]:
[tex]\[ \{(2, 4), (3, 6)\} \][/tex]
### Cartesian Product [tex]\(A \times B\)[/tex]:
The Cartesian product [tex]\(A \times B\)[/tex] consists of all possible ordered pairs [tex]\((a, b)\)[/tex] where [tex]\(a \in A\)[/tex] and [tex]\(b \in B\)[/tex]. Let's enumerate these pairs:
[tex]\[ A \times B = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \][/tex]
### Relation determined by each condition:
(a) For [tex]\(x < y\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(a < b\)[/tex]:
- [tex]\((1, 4)\)[/tex]: [tex]\(1 < 4\)[/tex]
- [tex]\((1, 5)\)[/tex]: [tex]\(1 < 5\)[/tex]
- [tex]\((1, 6)\)[/tex]: [tex]\(1 < 6\)[/tex]
- [tex]\((2, 4)\)[/tex]: [tex]\(2 < 4\)[/tex]
- [tex]\((2, 5)\)[/tex]: [tex]\(2 < 5\)[/tex]
- [tex]\((2, 6)\)[/tex]: [tex]\(2 < 6\)[/tex]
- [tex]\((3, 4)\)[/tex]: [tex]\(3 < 4\)[/tex]
- [tex]\((3, 5)\)[/tex]: [tex]\(3 < 5\)[/tex]
- [tex]\((3, 6)\)[/tex]: [tex]\(3 < 6\)[/tex]
Thus, the relation is:
[tex]\[\{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}\][/tex]
(b) For [tex]\(x + y = 5\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(a + b = 5\)[/tex]:
- [tex]\((1, 4)\)[/tex]: [tex]\(1 + 4 = 5\)[/tex]
Thus, the relation is:
[tex]\[\{(1, 4)\}\][/tex]
(c) For [tex]\(x > y\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(a > b\)[/tex]:
There are no such pairs where the element from [tex]\(A\)[/tex] is greater than the element from [tex]\(B\)[/tex] because all elements in [tex]\(A\)[/tex] are less than the elements in [tex]\(B\)[/tex].
Thus, the relation is:
[tex]\[\{\}\][/tex]
(d) For [tex]\(y = 2x\)[/tex]:
We need to find all pairs [tex]\((a, b) \in A \times B\)[/tex] such that [tex]\(b = 2a\)[/tex]:
- [tex]\((2, 4)\)[/tex]: [tex]\(4 = 2 \cdot 2\)[/tex]
- [tex]\((3, 6)\)[/tex]: [tex]\(6 = 2 \cdot 3\)[/tex]
Thus, the relation is:
[tex]\[\{(2, 4), (3, 6)\}\][/tex]
### Results Summarized:
- [tex]\(A \times B\)[/tex]:
[tex]\[ \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \][/tex]
- Relation by condition:
- (a) [tex]\(x < y\)[/tex]:
[tex]\[ \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \][/tex]
- (b) [tex]\(x + y = 5\)[/tex]:
[tex]\[ \{(1, 4)\} \][/tex]
- (c) [tex]\(x > y\)[/tex]:
[tex]\[ \{\} \][/tex]
- (d) [tex]\(y = 2x\)[/tex]:
[tex]\[ \{(2, 4), (3, 6)\} \][/tex]
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