Solutions

(i)

$$\begin{aligned} a < b \ \wedge \ c < d & \implies b - a > 0 \ \wedge \ d - c > 0 \\ & \implies (b - a) + (d - c) > 0 && \text{Closure under addition} \\ & \implies b + d > a + c \\ & \implies a + c < b + d && \text{By definition} \end{aligned}$$

(ii)

$$\begin{aligned} a < b & \implies b - a \in P && \text{By definition} \\ & \implies - a + b \in P \\ & \implies - a + b > 0 \\ & \implies - a > - b \end{aligned}$$

(iii)

$$\begin{aligned} a < b & \implies b - a \in P \\ & \implies (b - a) \times c \in P && \text{Closure under multiplication} \\ & \implies (b - a) c > 0 \\ & \implies bc - ac > 0 \\ & \implies bc > ac \\ & \implies ac < bc \end{aligned}$$

(iv) We first notice that:

$$\begin{aligned} c < 0 & \implies c \in -P \\ & \implies -c \in P \end{aligned}$$

Then,

$$\begin{aligned} a < b & \implies b - a \in P \\ & \implies (b - a) \times -c \in P && \text{Closure under multiplication} \\ & \implies (b - a) (-c) > 0 \\ & \implies -bc + ac > 0 \\ & \implies ac > bc \end{aligned}$$

(v)

$$\begin{aligned} a > 1 & \implies a - 1 \in P \\ & \implies (a - 1) \times a \in P && \text{Closure under multiplication} \\ & \implies (a - 1) a > 0 \\ & \implies a^2 - a > 0 \\ & \implies a^2 > a \end{aligned}$$

(vi)

$$\begin{aligned} 0 < a < 1 & \implies a \in P \ \wedge \ 1 - a \in P \\ & \implies (1 - a) a \in P && \text{Closure under multiplication} \\ & \implies (1 - a) a > 0 \\ & \implies a - a^2 > 0 \\ & \implies a > a^2 \\ & \implies a^2 < a \end{aligned}$$

(vii)

$$\begin{aligned} b - a, c \in P \ \wedge \ d - c, b \in P & \implies (b - a) c \in P \ \wedge \ (d - c) b \\ & \implies bc - ac \in P \ \wedge \ bd - bc \in P \\ & \implies bc > ac \ \wedge \ bd > bc \\ & \implies bd > ac && \text{By the transitivity property} \end{aligned}$$

(viii)

$$\begin{aligned} 0 \leq a < b \ \wedge \ 0 \leq a < b & \implies a^2 < b^2 && \text{Using 9}, c = a, b = d \end{aligned}$$

(ix)

$$\begin{aligned} a^2 < b^2 & \implies b^2 - a^2 \in P \\ & \implies (b - a) (b + a) \in P && \text{Square completion} \\ & \implies [\; b - a, b + a \in P \;] \vee [\; - (b - a), -(b + a) \in P \;] \end{aligned}$$

Notice that the second case would lead to a contradiction with the
initial premise that $a, b \geq 0$, so we ignore it. With some
simple algebraic manipulation, we obtain:

$$\begin{aligned} b - a > 0 \ \wedge \ b + a > 0 & \implies b > a \ \wedge \ b > -a \\ & \implies a < b && \text{Ignoring the second inequality} \end{aligned}$$