Solutions

(a) We use mathematical induction to prove the assertion. Assuming that
the predicate holds for $n$, we will prove it for $n + 1$.

$$\begin{aligned} 0 \leq x < y & \implies x^{n} < y^{n}, n = 1, 2, 3, \dots \\ & \implies x^{n + 1} = x^{n} x < y^{n} x < y^{n} y = y^{n + 1} \end{aligned}$$

Of course, we also have to prove the base case, however trivial it
is:

$$\begin{aligned} n = 1 & \implies 0 \leq x < y \implies x^{1} < y^{1} \\ & \implies x < y \end{aligned}$$

(b) The previous exercise already proved this one for the case where
$x, y \geq 0$. We must now preoccupy ourselves with the other cases.
If $\bm{x < y < 0}$:

$$\begin{aligned} x (x^{n - 1}) < y (x^{n - 1}) < y (y^{n - 1}) < 0 \\ & \implies x^{n} < y^{n} \end{aligned}$$

(c) Suppose $\bm{x, y < 0}$. Then, we can set
$- x = a > 0, - y = b > 0$.

$$\begin{aligned} x^{n} = y^{n} & \implies { (- a) }^{n} = { (- b) }^{n} \\ & \implies { (-1) }^{n} a^{n} = { (-1) }^{n} b^{n} \\ & \implies a^{n} = b^{n} \end{aligned}$$

Thus, solving for $x, y < 0$ ends up being solving for $x, y > 0$.
But what about for other number combinations?

Let us consider when one of the variables is equal to 0. Without
loss of generality, let $x = 0, y \in \mathbb{R}$,

$$\begin{aligned} x^{n} = y^{n} & \implies 0^{n} = y^{n} \\ & \implies 0 = y^{n} \\ & \implies y_{1} = 0 \vee y_{2} = 0, \dots, \vee y_{n - 1} = 0 \vee y_{n} = 0 \\ & \implies y = 0 && \text{Since } y_{1} = y_{2} = \dots = y_{n} \end{aligned}$$

Now then, if the reader is frowning his eyes at the proof above,
then good! Notice that we've used the proposition that if the
product of numbers is equal to 0, then one of them has to be equal
to. If you're like me, and you prefer mathematical notation over
literal notation, then here's what the previous proposition is in
that form:

$$\begin{aligned} a \times b \times \times \dots \times m \times n = 0 \implies a = 0 \vee b = 0 \vee \dots \vee m = 0 \vee n = 0 \end{aligned}$$

How do we prove it? If we prove that
$xy = 0 \implies x = 0 \vee y = 0$, then we will have proven the
proposition above.

$$\begin{aligned} xy = 0 & \implies x^{-1} x y = x^{-1} 0 \vee x y y^{-1} = 0 y^{-1} \\ & \implies y = 0 \vee x = 0 && a \times 0 = 0 \end{aligned}$$

Of course $x = 0, y = 0$ also is a possibility but it is left as an
exercise.

We now know for certain that we only need to consider the case where
$x, y > 0$. The second part to the question requires of us two
concepts that are introduced later in the book; the one is proofs by
contradiction and the other is mathematical induction.

For those who are not faint of heart, let us proceed. Suppose
$\bm{0 < x < y}$ and let $\bm{n}$ be an odd number. We then
have:

$$\begin{aligned} x^{n + 2} = x^{n} * x^{2} < x^{n} * y^{2} < y^{n} * y^{2} = y^{n + 2} \end{aligned}$$

The base case is obviously true, because of our initial premise[^1]:

$$\begin{aligned} 0 < x^{1} < y^{1} \implies 0 < x < y \end{aligned}$$

We thus have proven that:

$$\begin{aligned} 0 < x < y & \implies x^{n} < y^{n} \\ & \implies x^{n} \neq y^{n} && \text{(P10)} \end{aligned}$$

The other case $0 < y < x$ is also taken care of by having proven
the above (why?[^2]).

By elimination, $x$ must then be equal to $y$.

(d) Hint: Difference of squares is our very good friend for this proof.

$$\begin{aligned} x^{n} = y^{n} & \implies x^{n} - y^{n} = 0 \\ & \implies x^{n/2} - y^{n/2} = 0 \ \vee \ x^{n/2} + y^{n/2} = 0 \end{aligned}$$

The second equality is impossible (why?). We therefore have to prove
that the sum of numbers with even powers, individually, is positive.
Suppose that $x \in \mathbb{R}$ and $n$ is an even number.

$$\begin{aligned} x^{n} & = \underbrace{(x^{2}) (x^{2}) \dots (x^{2})}_{n/2 \text{ terms}} > 0 \end{aligned}$$

We know that $x^{2} > 0$ and, using the property of closure under
mutliplication, we have proven the proposition above. With that, we
know that only the first equation is possible:

$$\begin{aligned} x^{n/2} - y^{y/2} = 0 & \implies (x^{n/2} - y^{n/2})(x^{n/2} + y^{n/2}) = 0 && \text{D.o.s} \\ & \implies (x - y)(x + y) \dots (x^{n/4} + y^{n/4})(x^{n/2} + y^{n/2}) = 0 \\ & \implies x - y = 0 \ \vee \ x + y = 0 \ \vee \ \dots \ \vee \ x^{n/2} + y^{n/2} = 0 \end{aligned}$$

NB. The equality $x^{n/4} + y^{n/4} = 0$ was dropped because of
a lack of space.

Again, we have terms in the product above that are necessarily
positive. We can therefore ignore them and concentrate on the
leftmost terms:

$$\begin{aligned} x - y = 0 \ \vee \ & \implies x + y = 0 \\ & \implies x = y \vee x = - y \end{aligned}$$