Relations and Functions - Cartesian Product of Sets-advanced

Find the values of $x$ and $y$ if $(\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})$.

By the definition of equality of ordered pairs, we equate the corresponding components:

1. $\frac{x}{3} + 1 = \frac{5}{3}$ Subtracting $1$ from both sides: $\frac{x}{3} = \frac{5}{3} - 1$ $\frac{x}{3} = \frac{2}{3}$ $x = 2$

2. $y - \frac{2}{3} = \frac{1}{3}$ Adding $\frac{2}{3}$ to both sides: $y = \frac{1}{3} + \frac{2}{3}$ $y = \frac{3}{3}$ $y = 1$

Therefore, $x = 2$ and $y = 1$.

Two ordered pairs are equal if their first elements are equal and their second elements are equal. We solve the resulting linear equations for the variables.

If $A = \{1, 2\}$ and $B = \{3, 4\}$, find $A \times B$ and show that $A \times B \neq B \times A$.

$A \times B = \{ (1, 3), (1, 4), (2, 3), (2, 4) \}$ Now, find $B \times A$: $B \times A = \{ (3, 1), (3, 2), (4, 1), (4, 2) \}$ Comparing the sets, we see that $(1, 3) \in A \times B$ but $(1, 3) \notin B \times A$. Since the ordered pairs are different, $A \times B \neq B \times A$.

The order of elements in an ordered pair matters. $(1, 3)$ is not the same as $(3, 1)$.

The Cartesian product $A \times A$ has $9$ elements, among which are found $(-1, 0)$ and $(0, 1)$. Find the set $A$ and the remaining elements of $A \times A$.

We know $n(A \times A) = n(A) \cdot n(A) = 9$. This implies $n(A) = \sqrt{9} = 3$. From the given elements $(-1, 0)$ and $(0, 1)$, the components belong to set $A$. Therefore, the elements of $A$ are $\{-1, 0, 1\}$. $A = \{-1, 0, 1\}$ Now, $A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}$. The remaining elements are: $\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$.

Since the product is $A \times A$, both elements of any ordered pair must belong to set $A$. By identifying all unique first and second components from the given pairs, we reconstruct set $A$.

1. Solve for $A$: $x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0 \implies A = \{2, 3\}$.
2. Given $B = \{2, 4\}$.
3. Calculate $A - B$: Elements in $A$ not in $B$. $A - B = \{3\}$.
4. Calculate $B - A$: Elements in $B$ not in $A$. $B - A = \{4\}$.
5. Find the product: $\{3\} \times \{4\} = \{(3, 4)\}$.

First, the sets are defined by solving the quadratic equation. Then, set difference is applied to find unique elements. Finally, the Cartesian product of the resulting single-element sets yields one ordered pair.

1. Set $A$ is the set of all first elements in the ordered pairs: $A = \{a, b, a, b, a, b\} = \{a, b\}$.
2. Set $B$ is the set of all second elements in the ordered pairs: $B = {1, 3, 3, 1, 2, 2} = {1, 2, 3}.
3. $n(A) = 2$, $n(B) = 3$. Total elements = $2 \times 3 = 6$.

By definition of the Cartesian product, the domain (Set A) consists of all first components and the codomain (Set B) consists of all second components of the pairs.