Solve the following short questions: a) Prove (A - B) (A B) = A using set identities. b) Given X = , b\ and Y = , v list four binary relations from X to Y that are not functions.

$$A \cap U = A$$

Thus, $(A - B) \cup (A \cap B) = A$.
$\boxed{\text{Proven}}$

b)
The set $X = \{a, b\}$ and $Y = \{u, v\}$. The Cartesian product $X \times Y = \{(a,u), (a,v), (b,u), (b,v)\}$.
A binary relation is a subset of $X \times Y$. A relation is not a function if either:

1. An element in $X$ is not mapped to any element in $Y$.
2. An element in $X$ is mapped to more than one element in $Y$.

Here are four binary relations from $X$ to $Y$ that are not functions:

1. $R_1 = \{(a,u)\}$ (Element $b$ is not mapped).
2. $R_2 = \{(a,u), (a,v)\}$ (Element $a$ is mapped to two elements).
3. $R_3 = \{(a,u), (a,v), (b,u)\}$ (Element $a$ is mapped to two elements).
4. $R_4 = \{(b,u)\}$ (Element $a$ is not mapped).

$\boxed{\text{Four relations: } R_1 = \{(a,u)\}, R_2 = \{(a,u), (a,v)\}, R_3 = \{(a,u), (a,v), (b,u)\}, R_4 = \{(b,u)\}}$

c)
To prove that if $f$ and $g$ are one-to-one functions, then their composition $g \circ f$ is one-to-one, we must show that if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then $x_1 = x_2$.

Applying this, we have:

$$f(x_1) = f(x_2)$$

$$x_1 = x_2$$

Since we started with $(g \circ f)(x_1) = (g \circ f)(x_2)$ and concluded $x_1 = x_2$, the composition $g \circ f$ is one-to-one.
$\boxed{\text{Proven}}$

d)
The sequence is defined by $b_j = 1 - 2^j$ for all integers $j \ge 0$.

$$b_3 = 1 - 2^3 = 1 - 8 = -7$$

The first four terms of the sequence are $0, -1, -3, -7$.
$\boxed{0, -1, -3, -7}$

Q3. Solve the following Long Questions.

a)
To show that $f(x) = 3x + 2$ and $g(x) = \frac{x-2}{3}$ are inverse functions, we need to prove that $f(g(x)) = x$ and $g(f(x)) = x$.

$$f(g(x)) = f\left(\frac{x-2}{3}\right)$$

Substitute $\frac{x-2}{3}$ into $f(x) = 3x + 2$:

$$f\left(\frac{x-2}{3}\right) = 3\left(\frac{x-2}{3}\right) + 2$$ $$= (x-2) + 2$$ $$= x$$

$$g(f(x)) = g(3x+2)$$

Substitute $3x+2$ into $g(x) = \frac{x-2}{3}$:

$$g(3x+2) = \frac{(3x+2)-2}{3}$$ $$= \frac{3x}{3}$$ $$= x$$

Since $f(g(x)) = x$ and $g(f(x)) = x$, the functions $f$ and $g$ are inverses of each other.
$\boxed{\text{Proven}}$

b)
Given two distinct rational numbers $r$ and $s$ with $r < s$. We need to prove that there is a rational number $x$ such that $r < x < s$.

Since $r$ and $s$ are rational numbers, they can be expressed as $r = \frac{a}{b}$ and $s = \frac{c}{d}$, where $a, b, c, d$ are integers and $b, d \ne 0$.

$$x = \frac{\frac{a}{b} + \frac{c}{d}}{2} = \frac{\frac{ad+bc}{bd}}{2} = \frac{ad+bc}{2bd}$$

Since $a, b, c, d$ are integers, $ad+bc$ is an integer, and $2bd$ is a non-zero integer (because $b \ne 0$ and $d \ne 0$). Therefore, $x$ is a rational number.

Since $r < s$, we can add $r$ to both sides of the inequality:

$$r+r < s+r$$ $$2r < r+s$$

Divide by 2:

$$r < \frac{r+s}{2}$$

So, $r < x$.

Similarly, since $r < s$, we can add $s$ to both sides of the inequality:

$$r+s < s+s$$ $$r+s < 2s$$

Divide by 2:

$$\frac{r+s}{2} < s$$

So, $x < s$.

Combining these results, we have $r < x < s$.
Thus, for any two distinct rational numbers $r$ and $s$ with $r < s$, there exists a rational number $x = \frac{r+s}{2}$ such that $r < x < s$.
$\boxed{\text{Proven}}$

c)
We need to prove by mathematical induction that for any integer $n \ge 1$, $x^n - y^n$ is divisible by $x-y$, where $x$ and $y$ are any two integers with $x \ne y$.

Base Case (n=1):
For $n=1$, we need to show that $x^1 - y^1$ is divisible by $x-y$.
$x^1 - y^1 = x-y$.
Since $x-y$ is clearly divisible by $x-y$ (with a quotient of 1), the base case holds.

Inductive Hypothesis:
Assume that for some integer $k \ge 1$, $x^k - y^k$ is divisible by $x-y$.
This means $x^k - y^k = M(x-y)$ for some integer $M$.

Inductive Step:
We need to prove that $x^{k+1} - y^{k+1}$ is divisible by $x-y$.
Consider the expression $x^{k+1} - y^{k+1}$:

$$x^{k+1} - y^{k+1} = x \cdot x^k - y \cdot y^k$$

We can rewrite this by adding and subtracting $x \cdot y^k$:

$$x^{k+1} - y^{k+1} = x \cdot x^k - x \cdot y^k + x \cdot y^k - y \cdot y^k$$

Factor out $x$ from the first two terms and $y^k$ from the last two terms:

$$x^{k+1} - y^{k+1} = x(x^k - y^k) + y^k(x - y)$$

Now, apply the inductive hypothesis, substituting $x^k - y^k = M(x-y)$:

$$x^{k+1} - y^{k+1} = x(M(x-y)) + y^k(x - y)$$

Factor out $(x-y)$:

$$x^{k+1} - y^{k+1} = (xM + y^k)(x - y)$$

Since $x, y, M, k$ are integers, $xM + y^k$ is also an integer.
Therefore, $x^{k+1} - y^{k+1}$ is divisible by $x-y$.

Conclusion:
By the principle of mathematical induction, $x^n - y^n$ is divisible by $x-y$ for all integers $n \ge 1$.
$\boxed{\text{Proven}}$