Answer:
[tex]A\cap(B\cup C)=\{3\}[/tex]
[tex](A \cap B)\cup (A \cap C)=\{3\}[/tex]
Step-by-step explanation:
Given sets:
[tex]A = \{x \mid 3 \leq x < 4, x \in \mathbb{R}\}[/tex]
[tex]B = \{x \mid x < 5, x \in \mathbb{N}\}[/tex]
[tex]C = \{-5, -3, -1, 0, 1, 3\}[/tex]
Set A consists of all real numbers greater than or equal to 3 and less than 4: [3, 4).
Since set B consists of natural numbers less than 5, then:
[tex]B = \{1, 2, 3, 4\}[/tex]
The union of sets B and C consists of all the elements that are present in either of the two sets, or in both sets. Therefore:
[tex]B\cup C=\{-5, -3, -1, 0, 1,2,3,4\}[/tex]
The intersection of set A and set B∪C is the set of elements that are common to both sets. The only common element to both sets is the number 3, so:
[tex]A\cap(B\cup C)=\{3\}[/tex]
The intersection of set A and set B is the set of elements that are common to both sets. Therefore:
[tex]A \cap B=\{3\}[/tex]
Similarly, the intersection of set A and set C is the set of elements that are common to both sets. Therefore:
[tex]A \cap C=\{3\}[/tex]
The union of sets A∩B and A∩C consists of all elements that belong to set A∩B, set A∩C, or both sets A∩B and A∩C. As the only element in both sets is 3, then:
[tex](A \cap B)\cup (A \cap C)=\{3\}[/tex]
Since we have found that A∩(B∪C) = {3} and (A∩B)∪(A∩C) = {3}, we have proved that A∩(B∪C) = (A∩B)∪(A∩C).
