# Real Analysis: The Real Numbers

## Exponents

### Extending the Pattern

Multiplication on $\mathbb{R}$ is defined as a rather abstract algebraic operation that has several specific properties, but in its simplest form when dealing with just the natural numbers, it is a shorthand for repeated addition. When we write $a \cdot b$, we are saying "add $a$ to $0$ $b$ times." This operation is easy to generalize when $b$ becomes an integer, since we simply change the sign of the result if $b$ is negative, and it becomes only somewhat trickier when we want to add up fractional bits of $a$ when $b$ is a rational number. However, in the end, we see multiplication as a compact way of doing lots of addition. But this leads to the question...is there a compact way of doing lots of multiplication?

The answer is yes. Just as multiplication "does a lot" of addition, the concept of exponentiation "does a lot" of multiplication. We start with the simplest definition involving natural number exponents and then generalize from there.

### Natural Number Exponents

Given $x \in \mathbb{R}$ and $n \in \mathbb{N}^+$, define the notation $x^n$ to denote the product

$$x^n = \underbrace{x \cdot x \cdot \ldots \cdot x}_{n \text{ times}}.$$

For example, $x^4 = x \cdot x \cdot x \cdot x$. Define $x^0 = 1$ if $x \neq 0$; we leave $0^0$ undefined. The notation $x^n$ is read aloud as "$x$ to the $n$" or "$x$ to the $n$th power," and $x$ is called the base and $n$ is called the exponent or power. This terminology is the same for integer, rational, and real exponents.

Exponents of natural number powers have the following two algebraic properties:

1. $x^n \cdot x^m = x^{n + m}$

2. $\left(x^n\right)^m = x^{nm}$

### Integer Exponents

Integers generalize the natural numbers by adding an additive inverse $-x$ for every $x$ such that $x + (-x) = 0.$ To generalize exponentiation to include integers, we just need to add the negative integers, since exponentiation of non-negative integers (i.e. the natural numbers) are already defined. Now, how should we do this? Ideally, the definition should elegantly expand on the algebraic properties. Namely, we'd like to maintain that $x^n \cdot x^m = x^{n+m}$.

Consider the special case of $x^n \cdot x^{-n} = x^{0} = 1.$ For this to hold, it must be that $x^{-n}$ is the multiplicative inverse of $x^{n}$, i.e. $(x^{n})^{-1} = \dfrac{1}{x^n}.$ In fact, this is exactly the definition we will choose. For a positive natural $n$, define $x^{-n}$ as

$$x^{-n} = \left(x^n\right)^{-1} = \dfrac{1}{x^n}.$$

This definition expands the existing algebraic properties for natural numbers to hold for integers as well.

### Rational Exponents

Rational numbers generalize the integers by adding a multiplicative inverse $x^{-1}$ for every $x$ such that $x \cdot x^{-1} = 1.$ To generalize exponentiation to include rational numbers, we need to add all of the fractions between integers, since the integers themselves are already defined. The definition should elegantly expand on the algebraic properties of integer exponents, just as integer exponents elegantly expand those of natural number exponents. The constraint to maintain this time is that $(x^{\frac{a}{b}})^{\frac{b}{a}} = 1.$ However, we need to be careful with our definition, since fractional exponents and negative bases don't always make sense.

To start, given a non-negative real number $x$ and a nonzero integer $n$, define $x^{\frac{1}{n}}$ to be the real number $y$ such that $y^n = x$. This is equivalent to saying $\left(x^{\frac{1}{n}}\right)^n = x.$ We can also use the notation $\sqrt[n]{x}$ instead of $x^{\frac{1}{n}},$ which we typically read aloud as "the $n$th root of $x$." In the special case where $n = 2$, we simply write $\sqrt{x}.$ Additionally, we call $\sqrt{x}$ the square root of $x$ and $\sqrt[3]{x}$ as the cube root of $x.$

From here, we expand the definition to include all rational exponents. Given a positive real number $x$, an integer $a$, and a nonzero integer $b,$ define $x^\frac{a}{b}$ as

$$x^{\frac{a}{b}} = \left(x^{\frac{1}{b}}\right)^a.$$

The properties of rational exponents are more involved to prove, since rational numbers do not have unique representations (e.g. $\frac{1}{2} = \frac{2}{4}$). However, everything works as we'd like it to in the end.

### Real Exponents?

If we graph the two functions $f(x) = x^a$ and $g(x) = x^b$ where $x > 0$ and $a, b \in \mathbb{Q}^+,$ we can see that $f$ and $g$ move closer together as $a$ and $b$ get closer together. We haven't defined exponents of real powers yet, but the smoothness of this phenomenon suggests that we would like irrational exponents to behave in a predictable manner. Namley, if $q$ is irrational, $a$ and $b$ are rational, and $a < q < b$, then we'd like the graph of $x^q$ to lie between the graphs of $x^a$ and $x^b$.

However, the definition of the exponential function, which indeed has this property, requires the more sophisticated machinery of either infinite sequences or infinite series, which we haven't gotten to yet. As a result, exponents of real (namely irrational) powers remains a mystery for now.

## Problems

1. Natural Number Exponents - Algebra: Let $n \in \mathbb{N}$. Show the following:

1. $x^n \cdot x^m = x^{n + m}$

2. $\left(x^n\right)^m = x^{nm}$

1. Proof by induction. Base case: $x^n \cdot x^0 = x^n \cdot 1 = x^n = x^{n + 0}$. Inductive Step: Assume $x^nx^m=x^{n+m}$. Then $x^nx^{m+1} = x^nx^mx = x^{n+m}x = x^{n+m+1}$. The result follows by induction.

2. Proof by induction. Base case: $(x^n)^0 = 1 = x^{0} = x^{n \cdot 0}$. Inductive Step: Assume $(x^n)^m = x^{nm}$. Then $(x^n)^{m+1} = (x^n)^mx^n = x^{nm}x^n = x^{nm+n} = x^{n(m+1)}$. The result follows by induction.

2. Show that $x^2 \geq 0$ for all $x \in \mathbb{R}$.

If $x > 0$, then $x^2 = x \cdot x > 0$ by the second ordering axiom. If $x < 0$, then $-x > 0$, and $x^2 = (-1)^2x^2 = ((-1)x)^2 = (-x)^2 > 0$. Finally, if $x = 0$, then $x^2 = 0$.

3. Natural Number Exponents - Ordering I: Show that for all non-negative $x \in \mathbb{R}$ and all $n, m \in \mathbb{N}^+$ where $n < m$ that

1. $x^n > 0.$

2. $0 < x^n < 1$ when $x \in (0, 1)$ and $n > 1.$

3. $0 < x^m < x^n < 1$ when $x \in (0, 1).$

4. $x < x^n$ when $x > 1$ and $n > 1.$

5. $x^n < x^m$ when $x > 1.$

All five proofs will be by induction.

1. Base case: $x > 0$ is given, and $x^0 = 1 > 0$. Inductive step: Assume $x^n > 0$. By definition, $x^{n+1} = x^nx$. Since $x^n > 0$ and $x > 0$, it follows from the ordering axioms on $\mathbb{R}$ that $x^{n}x > 0,$ so therefore $x^{n+1} > 0.$

2. Assume $x \in (0, 1).$ Base case: Since $0 < x < 1,$ it follows that $0 < x \cdot x < x.$ By definition of natural number exponents, $x^2 = x \cdot x.$ Therefore $0 < x^2 < x < 1.$ Inductive step: Assume $0 < x^n < x < 1.$ Multiplying by $x$ gives us $0 < x^n \cdot x < x \cdot x < x < 1.$ By definition of natural number exponents, this simplifies to $0 < x^{n+1} < x < 1.$ The result follows by induction.

3. Assume $x \in (0, 1).$ Base case: Since $0 < x < 1,$ it follows from the above that $0 < x^n < x$ for all $n > 1.$ Since $0 < x < 1,$ it also follows that $x^{n+1} = x^n \cdot x < x^n.$ Inductive step: Assume $x^m < x^n$ when $m > n.$ Since $0 < x < 1$, it follows that $x^{m+1} = x^m \cdot x < x^m.$ It follows by transitivtiy that $x^{m+1} < x^n.$ The result follows by induction.

4. Assume $x > 1$. Base case: Since $x > 1$, it follows that $x \cdot x > x$. By definition of natural number exponents, $x^2 = x \cdot x$. Therefore $x^2 > x.$ Inductive step: Assume $n > 1$ and $x^n > x.$ Since $x > 1$, it follows that $x^{n+1} = x^n \cdot x > x^n.$ It follows by transitivity that $x^{n+1} > x.$ The result follows by induction.

5. Assume $x > 1.$ Base case: Since $x > 1$, it follows from the above that $x^n > x$ for all $n > 1.$ Since $1 < x,$ it also follows that $x^{n+1} = x^n \cdot x > x^n.$ Inductive step: Assume $x^m > x^n$ where $m > n.$ Since $x > 1$, it follows that $x^{m+1} = x^m \cdot x > x^m.$ It follows by transitivity that $x^{m+1} > x^n.$ The result follows by induction.

4. Natural Number Exponents - Ordering II: Consider $a, b \in \mathbb{R}$ where $0 < a < b$ and $n \in \mathbb{N}^+$. Show that $a^n < b^n.$

Proof by induction.

Base case: It is given that $a^1 = a < b = b^1$.

Inductive step: Assume $a^n < b^n$. Since $a > 0$, we can multiply both sides by $a$ to get $a^{n+1} < ab^n$. Next, because $a < b$, it follows that $1 < \dfrac{b}{a}$. Therefore $ab^n < \dfrac{b}{a}(ab^n) = b^{n+1}$. Thus by transitivity we see that $a^{n+1} < b^{n+1}.$ The result follows by induction.

5. Let $x \in \mathbb{R}^+$ and $n \in \mathbb{N}^+.$ Show that $x^n = 1$ if and only if $x = 1$.

Assume $x = 1$. Proof by induction. Base case: $x^1 = x = 1.$ Inductive step: Assume $x^{n} = 1.$ Then $x^{n+1} = x^nx = 1 \cdot 1 = 1.$ The result follows by induction.

Conversely, assume $x^n = 1$. By trichotomy, either $x < 1$, $x = 1$, or $x > 1.$ If $x < 1,$ then $x^n < 1,$ and if $x > 1,$ then $x^n > 1.$ Therefore $x = 1.$

6. Integer Exponents - Algebra: Let $x \in \mathbb{R} - \{0\}$ and $n, m \in \mathbb{Z}.$ Show the following:

1. Equivalent definition: $x^{-n} = (x^{-1})^n.$

2. $x^nx^m = x^{n+m}$

3. $\left(x^n\right)^m = x^{nm}$

1. By commutativity, it follows that $x^n \cdot (x^{-1})^n = (x \cdot x^{-1})^n.$ The righthand side then simplifies to $1^n = 1.$ It then follows by the definition of multiplicative inverse that $(x^{-1})^n = (x^n)^{-1},$ and thus $(x^{-1})^n = x^{-n}.$

2. The result for when $n$ and $m$ are both non-negative is already proved. If $n$ is nonnegative and $m$ is negative, then

$x^nx^m = x^n(x^{-m})^{-1} \\ x^nx^m = x^n(x^{-m + n - n})^{-1} \\ x^nx^m = x^n(x^nx^{-m - n})^{-1} \\ x^nx^m = x^n(x^{n})^{-1}(x^{-m - n})^{-1} \\ x^nx^m = (x^{-m - n})^{-1} \\ x^nx^m = x^{m + n}$

The case when $n$ is negative and $m$ is nonnegative is proved identically. If both $n$ and $m$ are negative, then $x^nx^m = (x^{-n})^{-1}(x^{-m})^{-1} = (x^{-n}x^{-m})^{-1} = (x^{-n-m})^{-1} = x^{n + m}$.

3. The result for when $n$ and $m$ are both nonnegative is already proved. If $n$ is positive and $m$ is negative, then $(x^n)^m = \dfrac{1}{(x^n)^{-m}} = \dfrac{1}{x^{-nm}} = x^{nm}$. The case for when $m$ is positive and $n$ is negative is proved identically. If both $n$ and $m$ are negative, then $(x^n)^m = \dfrac{1}{\frac{1}{(x^{-n})^{-m}}} = \dfrac{1}{\frac{1}{x^{nm}}} = x^{nm}$.

7. Integer Exponents - Ordering: Let $x \in \mathbb{R}^+,$ and let $n, m \in \mathbb{N}^+$ where $n < m.$ Show the following:

1. $x^{-n} > 0.$

2. $1 < x^{-n}$ when $x \in (0, 1).$

3. $1 < x^{-n} < x^{-m}$ when $x \in (0, 1).$

4. $0 < x^{-n} < x$ when $x > 1.$

5. $0 < x^{-m} < x^{-n} < 1$ when $x > 1.$

Hint: Do not use induction. Make use of the ordering properties for multiplicative inverses.

All proofs will make use of the fact that $(x^n)^{-1} = x^{-n}$.

1. Let $x > 0.$ Then $x^n > 0$ for all $n \in \mathbb{N}^+.$ By trichotomy, either $x^{-n} < 0,$ $x^{-n} = 0,$ or $x^n > 0.$ It cannot be that $x^{-n} = 0$, as otherwise $x^n \cdot 0 = 1.$ It also cannot be the case that $x^{-n} < 0$, as otherwise $1 = x^n \cdot x^{-n} < 0.$ Therefore $x^{-n} > 0.$

2. Assume $x \in (0, 1).$ Then $0 < x^n < 1$ for all $n \in \mathbb{N}^+.$ Therefore $x^{-n} > 1.$

3. Assume $x \in (0, 1).$ Then $0 < x^m < x^n < 1$ for $m > n.$ Therefore $1 < x^{-n} < x^{-m}.$

4. Assume $x > 1.$ Then $x < x^n$ and $1 < x^n$ by transitivity. Therefore $0 < x^{-n} < 1$, and by transitivity again $0 < x^{-n} < x.$

5. Assume $x > 1.$ Then $1 < x^n < x^m.$ Therefore $0 < x^{-m} < x^{-n} < 1.$

8. Rational Exponents I: For a positive real number $x$ and positive natural number $n$, define $x^{\frac{1}{n}}$ to be the positive real number $y$ such that $y^n = x$. The expression $x^{\frac{1}{n}}$ can also be written as $\sqrt[n]{x}.$

Prove that there exists exactly one real number $y$ such that $y = x^{\frac{1}{n}}.$

Hint: Use the identity $a^n - b^n = (b - a)(b^{n-1} + b^{n-2}a + \ldots + a^{n-1})$ and consider what happens when $a < b$.

First we show that one $y$ exists, then show that there is only one.

Let $T = \{ t > 0 : t^n < x \}$. Consider the number $t = \dfrac{x}{1+x}$. Note that $0 \leq t^n < 1$. It follows that $t^n \leq t < x$. Therefore $T$ is not empty. Consider the number $b = x + 1$. Then $x < b \leq b^n$, and so $b$ is an upper bound of $T$. Thus it follows that $T$ has a least upper bound, call it $y$.

Next we show that $y^n = x$ by eliminating the possibilities that $y^n < x$ and $y^n > x$.

Assume $y^n < x$. We would like to show that there exists some $t^n$ between $y^n$ and $x$, as this will lead to a contradiction, specifically with the fact that $y = \text{sup}(T)$.

If $y^n < x$, then $x - y^n > 0$. Note the identity $b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \ldots + a^{n-1})$. If $0 < a < b$, we get the inequality $0 < b^n - a^n < (b-a)(b^{n-1} + b^{n-2}b + \ldots + b^{n-1})) = (b-a)(nb^{-1})$. Dividing through by $(nb^{-1})$ on both sides gives $0 < \dfrac{b^n - a^n}{(nb^{-1})} < b-a$. Now pick a $t$ such that $0 < t < 1$ and $t < \dfrac{b^n - a^n}{(nb^{-1})}$. Set $a = y$ and $b = y + h$. Then $(y + h)^n - y^n < ((y+h)-y)n(y + h)^{n-1} = hn(y+h)^{n-1} < hn(y+1)^{n-1} < x - y^n$. Therefore $(y+h)^n < x$, and $y + h \in A$. But this is a contradiction, since $y < y + h$ and $y = \text{sup}(T)$. Therefore $y \nless \sqrt[n]{x}$.

Conversely, assume $y^n > x$. Let $k = \dfrac{y^n - x}{ny^{n-1}}$. Then $k = \dfrac{y}{n} - \dfrac{x}{ny^{n-1}} < y$. If $t \geq y - k$, then $y^n - t^n \leq y^n - (y-k)^n < kny^{n-1} = y^n - x$. Therefore $t^n > x$, so $t \notin T$. Therefore $y-k$ is an upper bound of $T$. But $y-k < y$, and $y = \text{sup}(T)$, which is a contradiction. Therefore $y \ngtr \sqrt[n]{x}$. Therefore $y = \sqrt[n]{x}$.

To prove uniqueness, consider some other $y'$ such that $y'^n = x$ and $y' \neq y$. If $y < y'$, then $x = y^n < y'^n$, so $y' \neq \sqrt[n]{x}$ after all, and the same logic holds if $y' < y$.

9. Rational Exponents II: Show that $(xy)^{\frac{1}{n}} = x^{\frac{1}{n}}y^{\frac{1}{n}}$ for positive $n \in \mathbb{N}$ and positive $x, y \in \mathbb{R}$.

Let $a = x^{\frac{1}{n}}$ and $b = y^{\frac{1}{n}}$. Then $xy = a^nb^n = (ab)^n$ by commutativity. Since there is only one $n$th root of any positive real number, it follows that $(xy)^{\frac{1}{n}} = ab = x^{\frac{1}{n}}y^{\frac{1}{n}}$.

10. Rational Exponents III: Show that $(x^{\frac{1}{n}})^n = (x^n)^{\frac{1}{n}}$ for positive $n \in \mathbb{N}$ and positive $x \in \mathbb{R}$.

By definition, $(x^{\frac{1}{n}})^n = x$. Likewise, $(x^n)^{\frac{1}{n}} = y$, where $y^n = x^n$. But this implies $y = x$. Therefore $(x^{\frac{1}{n}})^n = (x^n)^{\frac{1}{n}}$.

11. Rational Exponents IV: Define for $a, b \in \mathbb{N}$ where $b > 0$ and for positive $x \in \mathbb{R}$ that $x^{\frac{a}{b}} = \left(x^{\frac{1}{b}}\right)^a$. Show that for $c, d, \in \mathbb{N}$ where $\dfrac{a}{b} = \dfrac{c}{d}$ that $x^{\frac{a}{b}} = x^{\frac{c}{d}}$.

First we show that $x^{\frac{ac}{bc}} = x^{\frac{a}{b}}$:

By definition, $x^{\frac{ac}{bc}} = \left(x^{ac}\right)^{\frac{1}{bc}}$. Let $y = \left(x^{ac}\right)^{\frac{1}{bc}}$. Then $y^{bc} = x^{ac}$. Therefore $(y^b)^c = (x^a)^c$, so $y^b = x^a$, and thus $y = (x^a)^{\frac{1}{b}} = x^{\frac{a}{b}}$. Therefore $x^{\frac{ac}{bc}} = x^{\frac{a}{b}}$.

Now we show $x^{\frac{a}{b}} = x^{\frac{c}{d}}$ when $\dfrac{a}{b} = \dfrac{c}{d}$. Note that $ad = bc$. Therefore $x^{\frac{a}{b}} = x^{\frac{ad}{bd}} = x^{\frac{bc}{bd}} = x^{\frac{c}{d}}$.

12. Rational Exponents V: Show the following:

1. $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}} = x^{\frac{1}{nm}}.$

2. $\left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = x^{\frac{ac}{bd}}.$

3. $x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{a}{b} + \frac{c}{d}}.$

1. To show that $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}} = x^{\frac{1}{nm}},$ first let $a = \left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}$ such that $a^m = x^{\frac{1}{n}}.$ Next, let $b = x^\frac{1}{n}$ such that $b^n = x.$ It follows that $\left(a^m\right)^n = x = a^{nm}.$ Therefore $x^{\frac{1}{nm}} = a = \left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}.$

2. $\left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(x^{\frac{1}{b}}\right)^a\right)^{\frac{c}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(\left(x^{\frac{1}{b}}\right)^a\right)^c\right)^{\frac{1}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(x^{\frac{1}{b}}\right)^{ac}\right)^{\frac{1}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(x^{ac}\right)^{\frac{1}{b}}\right)^{\frac{1}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(x^{ac}\right)^{\frac{1}{bd}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = x^{\frac{ac}{bd}} \\$

3. $x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{ad}{bd}}x^{\frac{bc}{bd}} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = \left(x^{\frac{1}{bd}}\right)^{ad}\left(x^{\frac{1}{bd}}\right)^{bc} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = \left(x^{\frac{1}{bd}}\right)^{ad+bc} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{ad+bc}{bd}} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{a}{b} + \frac{c}{d}} \\$

13. For $x, y, \in \mathbb{R}$, does $x^n = y^n$ imply that $x = y$ if $x$ and $y$ are not necessarily non-negative? Prove or give a counterexample.

No, the implication does not hold. Counterexample: $(-1)^2 = 1$, and $1^2 = 1$, but $-1 \neq 1$.

14. Theorem: Show that $\sqrt{2}$ is irrational.

Proof by contradiction. Assume that there is some rational number $\dfrac{a}{b} \in \mathbb{Q}$ such that $\left(\dfrac{a}{b}\right)^2 = 2$. Without loss of generality, assume that $\dfrac{a}{b}$ is in simplest form, i.e. that $a$ is not divisible by $b$. Then $\dfrac{a^2}{b^2} = 2$, so $a^2 = 2b^2$.  This means that $a^2$ is divisible by 2. Since $a^2$ is a square, $a$ must be divisible by $2$. Let $ac = a$. Then $a^2 = 4c^2 = 2b^2$. Dividing both sides through by $2$ gives us $2c = b^2$. The same logic shows that $b$ must also be divisible by $2$. However, this is a contradiction, as it means that $a$ and $b$ have a common factor. Therefore $\sqrt{2} \notin \mathbb{Q}$.

15. Show that $\sqrt{2} < 2$.

By trichotomy, either $\sqrt{2} < 2$, $\sqrt{2} = 2$, or $\sqrt{2} > 2$. If $\sqrt{2} = 2$, then $\sqrt{2}^2 = 2^2$, but clearly $2 \neq 4$. If instead $\sqrt{2} > 2$. Then $\sqrt{2}^2 > 2\sqrt{2}$, so $2 > 2\sqrt{2}$. But this implies $0 < \sqrt{2} < 1$, which is a contradiction, as we assumed $\sqrt{2} > 2 > 1$. Therefore $\sqrt{2} < 2$.

16. Show that $\sqrt{x} < \sqrt{y}$ if and only if $x < y$ for all positive $x, y \in \mathbb{R}$.

Assume $\sqrt{x} < \sqrt{y}$. By trichotomy, either $x < y$, $x = y$, or $x > y$. If $y = x$, then $\sqrt{x} = \sqrt{y}$, since roots are unique, but this is a contradiction. If instead $y < x$, then $\sqrt{y}\sqrt{y} < \sqrt{x}\sqrt{x}$. But $\sqrt{x}\sqrt{x} < \sqrt{x}\sqrt{y}$, so by transitivity, $\sqrt{y}\sqrt{y} < \sqrt{x}\sqrt{y}$. But this implies $\sqrt{y} < \sqrt{x}$, which is a contradiction. Therefore $x < y$.

Conversely, let $x < y$. By trichotomy, either $\sqrt{x} < \sqrt{y}$, $\sqrt{x} = \sqrt{y}$, or $\sqrt{x} > \sqrt{y}$. If $\sqrt{x} = \sqrt{y}$, then $x = y$, which is a contradiction. If instead $\sqrt{x} > \sqrt{y}$, then $\sqrt{x}\sqrt{y} > \sqrt{y}\sqrt{y}$. But by our assumption $\sqrt{x}\sqrt{x} < \sqrt{y}\sqrt{y}$, so by transitivity $\sqrt{x}\sqrt{y} > \sqrt{x}\sqrt{x}$, so $\sqrt{y} > \sqrt{x}$, which is a contradiction. Therefore $\sqrt{x} < \sqrt{y}$.

17. Show that $\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$.

Hint: Consider $(\sqrt{x + y})^2 - (\sqrt{x} + \sqrt{y})^2$.

$(\sqrt{x + y})^2 - (\sqrt{x} + \sqrt{y})^2 = (x + y) - (x + 2\sqrt{x}\sqrt{y} + y) \\ (\sqrt{x + y})^2 - (\sqrt{x} + \sqrt{y})^2 = -2\sqrt{x}\sqrt{y} \\ (\sqrt{x + y})^2 = (\sqrt{x} + \sqrt{y})^2 - 2\sqrt{x}\sqrt{y} \\ (\sqrt{x + y})^2 \leq (\sqrt{x} + \sqrt{y})^2 \\ \sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$