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Sagot :
Let's solve the questions step-by-step based on the given sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex].
### a) List the members of the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex].
Set [tex]\( E \)[/tex]:
[tex]\( E \)[/tex] is defined as the set of all even numbers less than 10. The even numbers less than 10 are:
[tex]\[ 0, 2, 4, 6, 8 \][/tex]
Thus, the members of set [tex]\( E \)[/tex] are:
[tex]\[ E = \{0, 2, 4, 6, 8\} \][/tex]
Set [tex]\( F \)[/tex]:
[tex]\( F \)[/tex] is defined as the set of all factors of 10. The factors of 10 are:
[tex]\[ 1, 2, 5, 10 \][/tex]
Thus, the members of set [tex]\( F \)[/tex] are:
[tex]\[ F = \{1, 2, 5, 10\} \][/tex]
### b) Are the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] equal or equivalent? Give reason.
Equality:
Two sets are equal if they contain the same elements. Comparing the members of sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ E = \{0, 2, 4, 6, 8\} \][/tex]
[tex]\[ F = \{1, 2, 5, 10\} \][/tex]
Since [tex]\( E \)[/tex] contains 0, 4, 6, and 8 which are not in [tex]\( F \)[/tex] and [tex]\( F \)[/tex] contains 1, 5, and 10 which are not in [tex]\( E \)[/tex], the sets do not have the same elements.
Therefore, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are not equal.
Equivalence:
Two sets are equivalent if they have the same number of elements. The number of elements in each set is:
[tex]\[ |E| = 5 \][/tex]
[tex]\[ |F| = 4 \][/tex]
Since [tex]\( E \)[/tex] has 5 elements and [tex]\( F \)[/tex] has 4 elements, they do not have the same number of elements.
Therefore, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are not equivalent.
### c) Are the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] disjoint or overlapping? Give reason.
Disjoint sets:
Two sets are disjoint if they have no elements in common. Let's check for common elements between [tex]\( E \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ E = \{0, 2, 4, 6, 8\} \][/tex]
[tex]\[ F = \{1, 2, 5, 10\} \][/tex]
The common element between [tex]\( E \)[/tex] and [tex]\( F \)[/tex] is:
[tex]\[ \{2\} \][/tex]
Since there is a common element, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are not disjoint but overlapping.
Therefore, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are overlapping.
### d) Show the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] in a Venn-diagram.
Since this is a textual solution and we cannot actually draw a Venn-diagram, we will describe it.
1. Draw two overlapping circles.
2. Label the left circle as [tex]\( E \)[/tex] and the right circle as [tex]\( F \)[/tex].
3. Place the common element [tex]\( \{2\} \)[/tex] in the overlapping region.
4. Place the remaining elements of [tex]\( E \)[/tex] (\{0, 4, 6, 8\}) in the non-overlapping part of the left circle.
5. Place the remaining elements of [tex]\( F \)[/tex] (\{1, 5, 10\}) in the non-overlapping part of the right circle.
The Venn-diagram should visually represent the intersection at [tex]\( \{2\} \)[/tex], and the other elements distributed in their respective sets.
This detailed solution addresses each part of the question comprehensively using the information provided.
### a) List the members of the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex].
Set [tex]\( E \)[/tex]:
[tex]\( E \)[/tex] is defined as the set of all even numbers less than 10. The even numbers less than 10 are:
[tex]\[ 0, 2, 4, 6, 8 \][/tex]
Thus, the members of set [tex]\( E \)[/tex] are:
[tex]\[ E = \{0, 2, 4, 6, 8\} \][/tex]
Set [tex]\( F \)[/tex]:
[tex]\( F \)[/tex] is defined as the set of all factors of 10. The factors of 10 are:
[tex]\[ 1, 2, 5, 10 \][/tex]
Thus, the members of set [tex]\( F \)[/tex] are:
[tex]\[ F = \{1, 2, 5, 10\} \][/tex]
### b) Are the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] equal or equivalent? Give reason.
Equality:
Two sets are equal if they contain the same elements. Comparing the members of sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ E = \{0, 2, 4, 6, 8\} \][/tex]
[tex]\[ F = \{1, 2, 5, 10\} \][/tex]
Since [tex]\( E \)[/tex] contains 0, 4, 6, and 8 which are not in [tex]\( F \)[/tex] and [tex]\( F \)[/tex] contains 1, 5, and 10 which are not in [tex]\( E \)[/tex], the sets do not have the same elements.
Therefore, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are not equal.
Equivalence:
Two sets are equivalent if they have the same number of elements. The number of elements in each set is:
[tex]\[ |E| = 5 \][/tex]
[tex]\[ |F| = 4 \][/tex]
Since [tex]\( E \)[/tex] has 5 elements and [tex]\( F \)[/tex] has 4 elements, they do not have the same number of elements.
Therefore, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are not equivalent.
### c) Are the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] disjoint or overlapping? Give reason.
Disjoint sets:
Two sets are disjoint if they have no elements in common. Let's check for common elements between [tex]\( E \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ E = \{0, 2, 4, 6, 8\} \][/tex]
[tex]\[ F = \{1, 2, 5, 10\} \][/tex]
The common element between [tex]\( E \)[/tex] and [tex]\( F \)[/tex] is:
[tex]\[ \{2\} \][/tex]
Since there is a common element, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are not disjoint but overlapping.
Therefore, the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] are overlapping.
### d) Show the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] in a Venn-diagram.
Since this is a textual solution and we cannot actually draw a Venn-diagram, we will describe it.
1. Draw two overlapping circles.
2. Label the left circle as [tex]\( E \)[/tex] and the right circle as [tex]\( F \)[/tex].
3. Place the common element [tex]\( \{2\} \)[/tex] in the overlapping region.
4. Place the remaining elements of [tex]\( E \)[/tex] (\{0, 4, 6, 8\}) in the non-overlapping part of the left circle.
5. Place the remaining elements of [tex]\( F \)[/tex] (\{1, 5, 10\}) in the non-overlapping part of the right circle.
The Venn-diagram should visually represent the intersection at [tex]\( \{2\} \)[/tex], and the other elements distributed in their respective sets.
This detailed solution addresses each part of the question comprehensively using the information provided.
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