Solve number 12-15

Given the sets:
$A = \{1, 2\}$
$B = \{2, 3\}$
$C = \{3, 4\}$

a) Compute $A \cup (B \cap C)$
First, calculate the intersection of $B$ and $C$:

Next, calculate the union of $A$ with the result:

b) Compute $(A \cup B) \cap (A \cup C)$
First, calculate the union of $A$ and $B$:

Next, calculate the union of $A$ and $C$:

Finally, calculate the intersection of these two results:

The answers for Question 12 are:

*a) $\boxed{\{1, 2, 3\}}$

*b) $\boxed{\{1, 2, 3\}}$

Given the sets:
$A = \{1, 2\}$
$B = \{a, b\}$
Find the Cartesian product $A \times B$.
The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in B$.

The answer for Question 13 is:
$\boxed{\{(1, a), (1, b), (2, a), (2, b)\}}$

Given the sets:
$A = \{1, 2, 3\}$
$B = \{a, b, c\}$
A function $f: A \to B$ is defined such that $f(1) = a$, $f(2) = b$, and $f(3) = c$.
Determine if this function is injective.
A function is injective (one-to-one) if every distinct element in the domain maps to a distinct element in the codomain. In other words, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
In this case:
$f(1) = a$
$f(2) = b$
$f(3) = c$
Each element in $A$ (1, 2, 3) maps to a unique element in $B$ (a, b, c). No two distinct elements in $A$ map to the same element in $B$.
Therefore, the function is injective.
The answer for Question 14 is:
$\boxed{\text{Yes, the function is injective.}}$

Given the sets:
$A = \{1, 2\}$
$B = \{2, 3\}$
Show that $A \cup B = B \cup A$.

First, calculate $A \cup B$:

Next, calculate $B \cup A$:

Since both $A \cup B$ and $B \cup A$ result in the set $\{1, 2, 3\}$, we have shown that $A \cup B = B \cup A$.
The answer for Question 15 is:
$\boxed{A \cup B = \{1, 2, 3\} \text{ and } B \cup A = \{1, 2, 3\}, \text{ therefore } A \cup B = B \cup A.}$