Given the sets: mu = Z, P = \2k : k Z Q = : -7 < x < 7 R = : x > 0\. Find P R, P' Q, and n(Q R).

Here are the solutions to the problems.

:
Given the sets:
Universal set $\mu = \{x : x \in \mathbb{Z}\}$ (all integers)
$P = \{..., -6, -4, -2, 0, 2, 4, 6, ...\}$ (all even integers)
$Q = \{x : -7 < x < 7\}$
$R = \{x : x > 0\}$

First, let's list the elements of $Q$ and $R$ as integers:
$Q = \{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$
$R = \{1, 2, 3, 4, 5, 6, 7, ...\}$ (all positive integers)

a) $P \cap R$

Step 1: Find the intersection of set $P$ (even integers) and set $R$ (positive integers). This means finding the elements that are both even and positive.

$$P \cap R = \{x : x \text{ is an even integer and } x > 0\}$$ $$P \cap R = \{2, 4, 6, 8, ...\}$$

The intersection is $\boxed{\{2, 4, 6, 8, ...\}}$.

b) $P' \cap Q$

Step 1: Determine $P'$, the complement of $P$. Since $P$ is the set of all even integers, $P'$ is the set of all odd integers.

$$P' = \{..., -5, -3, -1, 1, 3, 5, ...\}$$

Step 2: Find the intersection of $P'$ (odd integers) and $Q = \{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$. This means finding the elements that are both odd and within the range of $Q$.

$$P' \cap Q = \{-5, -3, -1, 1, 3, 5\}$$

The intersection is $\boxed{\{-5, -3, -1, 1, 3, 5\}}$.

c) $n(Q \cap R)$

Step 1: Find the intersection of set $Q$ and set $R$.

$$Q \cap R = \{x : -7 < x < 7 \text{ and } x > 0, x \in \mathbb{Z}\}$$ $$Q \cap R = \{1, 2, 3, 4, 5, 6\}$$

Step 2: Find the number of elements in the set $Q \cap R$.

$$n(Q \cap R) = 6$$

The number of elements is $\boxed{6}$.

:
A binary operation $*$ is defined on the set of real numbers, $\mathbb{R}$, by $p * q = \frac{3p + 2q}{2p + 3q}$.

a) Evaluate $(\sqrt{3}) * (\sqrt{2})$.

Step 1: Substitute $p = \sqrt{3}$ and $q = \sqrt{2}$ into the given formula.

$$(\sqrt{3}) * (\sqrt{2}) = \frac{3\sqrt{3} + 2\sqrt{2}}{2\sqrt{3} + 3\sqrt{2}}$$

Step 2: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{3} - 3\sqrt{2}$.

$$(\sqrt{3}) * (\sqrt{2}) = \frac{(3\sqrt{3} + 2\sqrt{2})(2\sqrt{3} - 3\sqrt{2})}{(2\sqrt{3} + 3\sqrt{2})(2\sqrt{3} - 3\sqrt{2})}$$

Step 3: Expand the numerator and the denominator.
Numerator:

$$(3\sqrt{3} + 2\sqrt{2})(2\sqrt{3} - 3\sqrt{2}) = (3\sqrt{3})(2\sqrt{3}) - (3\sqrt{3})(3\sqrt{2}) + (2\sqrt{2})(2\sqrt{3}) - (2\sqrt{2})(3\sqrt{2})$$ $$= (6 \times 3) - 9\sqrt{6} + 4\sqrt{6} - (6 \times 2)$$ $$= 18 - 9\sqrt{6} + 4\sqrt{6} - 12$$ $$= 6 - 5\sqrt{6}$$

Denominator:

$$(2\sqrt{3} + 3\sqrt{2})(2\sqrt{3} - 3\sqrt{2}) = (2\sqrt{3})^2 - (3\sqrt{2})^2$$ $$= (4 \times 3) - (9 \times 2)$$ $$= 12 - 18$$ $$= -6$$

Step 4: Combine the simplified numerator and denominator.

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