Real Analysis: The Real Numbers
Algebra on $\mathbb{R}$
Axiomatic Definition
The real numbers are a set of numbers, denoted as $\mathbb{R}$, that extend the rational numbers to ensure that there are no "gaps" in the number line. For example, there is neither a rational number whose square is $2$ nor one that measures the ratio of the circumference of a circle to its diameter, but there are real numbers that fit these criteria, namely $\sqrt{2}$ and $\pi$.
The real numbers are defined according to a list of axioms, or statements we take as true by definition. In rigorous mathematics, every true statement must follow from some other, more fundamental true statement, but at some point we have to dig our heels in and assume a few very fundamental things to be true. The list of axioms should therefore ideally be small, where each one defines a key aspect of the set. One noteworthy detail is that no axiom should be able to be derived from a combination of the other ones  if this were the case, we would have no reason to assume it as an axiom, as it would (by definition) be implied by the others.
Let $\mathbb{R}$ be the real numbers, which form a set with the following properties:

$\mathbb{R}$ is closed under the addition function $+ : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, which has the following properties

Commutativity of addition: $a + b = b + a$ for all $a, b \in \mathbb{R}$.

Associativity of addition: $a + (b + c) = (a + b) + c$ for all $a, b, c \in \mathbb{R}$.

Additive identity: There exists an element $0 \in \mathbb{R}$ called zero such that $a + 0 = a$ for all $a \in \mathbb{R}$.

Additive inverse: For each $a \in \mathbb{R}$ there exists an element $a \in \mathbb{R}$ such that $a + (a) = 0$.


$\mathbb{R}$ is closed under the multiplication function $\cdot : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, which has the following properties:

Commutativity of multiplication: $a \cdot b = b \cdot a$ for all $a, b \in \mathbb{R}$.

Associativity of multiplication: $a + (b + c) = (a + b) + c$ for all $a, b, c \in \mathbb{R}$.

Multiplicative identity: There exists an element $1 \in \mathbb{R}$ called one such that $a \cdot 1 = a$ for all $a \in \mathbb{R}$.

Multiplicative inverse: For each $a \neq 0 \in \mathbb{R}$ there exists an element $\dfrac{1}{a}$ such that $a \cdot \dfrac{1}{a} = 1$. The multiplicative inverse of a value is also called its reciprocal.


The distributive law holds: $a \cdot (b + c) = a \cdot b + a \cdot c$ for all $a, b, c \in \mathbb{R}$.

The Least Upper Bound property holds on $\mathbb{R}$.

$\mathbb{R}$ contains $\mathbb{Q}$ as a subfield.

There exists an order relation < with the following properties

If $y < z$, then $x + y < x + z$.

If $x > 0$ and $y > 0$, then $xy > 0$.

While this list of properties may seem a bit long, you will quickly discover that many familiar algebraic operations must be derived from them. Even something as seemingly obvious as the fact that if $x + y = x$, then $y = 0$ must be proven from the axioms.
The Field Axioms
The above arithmetic axioms pertaining to addition, multiplication, and distributivity are collectively called the field axioms. A field is any set that fulfills these axioms, and the letter $F$ (or $\mathbb{F}$) is often used instead of $\mathbb{R}$ to denote this generalization. A field that also fulfills the two ordering axioms above is, naturally, an ordered field. A subfield is a field that is a subset of another field and uses the same arithmetic operations.
Other examples of fields include the rational numbers and complex numbers, although the natural numbers and integers are not fields. Fields don't necessarily need to be "numbers," as any set that fulfills the axioms, by definition, a field. (One such more exotic example of a field is the set of all rational functions over some field $F$, i.e. all functions made up of one polynomial divided by another.) However, fields themselves are more closely studied in abstract algebra, along with other algebraic structures like groups and rings. Real analysis is interested in the particular field of the real numbers.
Notation
Some familiar notation regarding parenthesis, addition, and multiplication is used for real analysis just as it is for high school algebra. Namely, parenthesis are only used when necessary, and the multiplication dot is elided entirely. Thus, we write $ab + c$ instead of $(a \cdot b) + c$. Likewise, multiplicative inverses are written as the denominator of fractions, with the terms by which they are multiplied placed in the numerator. Thus we write $\dfrac{b}{a}$ instead of $a^{1}b$. Other shortcuts, such as the fact that $a = 1 \cdot a$, are results we must first prove.
Rational and Irrational Numbers
$\mathbb{R}$ contains $\mathbb{Q}$ as a subfield. This means that all of the rational numbers are also real numbers, and the algebraic properties defined for them are the same as for $\mathbb{Q}$ itself. Note that this does not mean more properties can't be defined for them as part of $\mathbb{R}$. Indeed, $\sqrt{2}$ not a rational number, as proved below, but that does mean it's not a real number. Which leads us to irrational numbers.
The irrational numbers are the real numbers that are not rational numbers. Formally, the irrational numbers are $\mathbb{R}  \mathbb{Q}$. There is unfortunately no standardized symbol for the irrational numbers, but currently the most popular seems to be $\mathbb{P}$. If we wish to use $\mathbb{P}$ to denote the set of irrational numbers, we will explicitly say so.
The Extended Real Numbers
While there is no greatest real number, we can define the symbol $\infty$, called infinity or positive infinity, such that $\infty > x$ for all $x \in \mathbb{R}$. Likewise, we can define $\infty$, accordingly called negative infinity, such that $\infty < x$ for all $x \in \mathbb{R}$. We call the set $\mathbb{R} \cup \{\infty, \infty\}$ the extended real numbers, as it includes both positive and negative infinity, and denote it as $\overline{\mathbb{R}}$.
The extended real numbers are defined to have the following algebraic properties:

$x + \infty = \infty$

$x  \infty = \infty$

$\dfrac{x}{\infty} = \dfrac{x}{\infty} = 0$

$x\infty = \infty$ if $x > 0$ and $x\infty = \infty$ if $x < 0$

$x(\infty) = \infty$ if $x > 0$ and $x(\infty) = \infty$ if $x < 0$

$0\infty$ is undefined.
To distinguish the strictly real numbers from the two infinities, the former are called finite.
Construction of $\mathbb{R}$
What is a construction? It's a process by which a more complicated set is assembled out of more primitive sets that are known to exist according to some prior and more fundamental set of axioms. How do we know which objects exist? For one, the axioms of set theory define a few very primitive sets into existence, such as the empty set, out of which the natural numbers can be constructed. In turn, there is a construction of the integers from the natural numbers, and a construction of the rational numbers from the integers, and finally a construction of the real numbers from the rational numbers. In fact, there are a number of ways of performing this last construction. Many of them are fairly obscure, but two are well known. One construction uses the concept of a Dedekind cut, and the other uses the concept of a Cauchy sequence. The one thing common to all of these constructions is that they form a set that fulfills the axioms listed above.
You might be asking  which construction is the "real" one? Which one do we "use" when doing real analysis? The answer is actually none of them. We just use the axioms as listed. To wit, the entire mark of a successful construction of $\mathbb{R}$ is that it perfectly fulfills all of the axioms. Once we have constructed some set that does that, it makes no difference whatsoever which one may be lurking in the background, as all of the interesting properties of $\mathbb{R}$ derive from the axioms and not any details particular to the construction. Indeed, we may skip all this construction and simply define the sets $\mathbb{N}$ through $\mathbb{R}$ to exist in the first place if we wanted. So then why construct them at all? One reason is that it is philosophically satisfying to be able to build up more sophisticated mathematical objects out of simpler ones, i.e. from the "ground up." We like to be confident that we can break down even the loftiest theorems of integrals, probability, and topology into statements concerning the fundamental axioms of set theory and logic. Formally constructing $\mathbb{R}$ from $\mathbb{Q}$ lets us sleep easy at night.
In contrast, sets that are possible to define but not construct may be less interesting  if we can't actually come up with an example of such an object, what exactly are we studying in the first place? Additionally, the more extensive the set of axioms we require in order to construct something, the harder it is to map it onto physical phenomena in the real world. In fact, some stricter mathematicians object to the existence of the real numbers themselves, despite the constructions, because the constructions rely on certain axioms pertaining to the existence of infinite sets that the mathematicians don't think reflect physical reality very well. For example, computers use finite bytes to do calculations on either integers or rational numbers within a certain amount of precision, and all of our scientific and engineering tools use finite precision, so who needs infinity? Why make a mathematical construction that clearly has a hard time mapping onto reality? Alternatively, $\sqrt{2}$ is clearly a useful representation of the length of the hypotenuse of an isosceles right triangle, and every finite approximation we make of it tends towards that number. Ultimately the answer to these philosophical chestnuts is that we like the real numbers around here, so we're going to study them.
Problems
Cancellation Law (Addition): Show that if $x + y = x + z$, then $y = z$.
$ x + y = x + z \\ x + (x + y) = x + (x + z) \\ (x + x) + y = (x + x) + z \\ 0 + y = 0 + z \\ y = z $
Zero is Unique: Show that if $x + y = x$, then $y = 0$.
$ x + y = x \\ (x) + (x + y) = x + x \\ (x + x) + y = x + x \\ 0 + y = 0 \\ y = 0 \\ $
Additive Inverses are Unique: Show that if $x + y = 0$, then $y = x$.
$ x + y = 0 \\ x + (x + y) = x + 0 \\ (x + x) + y = x \\ 0 + y = x \\ y = x \\ $
Additive Inverse of Additive Inverse is the Original Value: Show that $(x) = x$.
$x + (x) = 0 \\ x + ((x)) = 0 \\ x + (x) = x + ((x)) \\ x = (x)$
Cancellation Law (Multiplication): Show that if $x \neq 0$, and $x \cdot y = x \cdot z$, then $y = z$.
If $x \neq 0$, then $\dfrac{1}{x}$ exists. Multiplying both sides of the equation gives the desired result:
$ xy = xz \\ \left(\dfrac{1}{x}\right)(xy) = \left(\dfrac{1}{x}\right)(xz) \\ \left(\dfrac{1}{x} \cdot x \right) \cdot y = \left(\dfrac{1}{x} \cdot x \right) \cdot z \\ 1y = 1z \\ y = z \\$
Multiplicative Identity is Unique: Show that if $x \neq 0$ and $xy = x$, then $y = 1$.
If $x \neq 0$, then $\dfrac{1}{x}$ exists. Multiplying both sides by $\dfrac{1}{x}$ gives the desired result:
$ xy = x \\ \left( \dfrac{1}{x} \right) (xy) = \left( \dfrac{1}{x} \right) x \\ \left( \dfrac{1}{x} \cdot x \right) \cdot y = \left( \dfrac{1}{x} \right) x\\ 1y = 1 \\ y = 1 \\ $
Reciprocals Are Unique: Show that if $x \neq 0$ and $xy = 1$, then $y = \dfrac{1}{x}$.
$ xy = 1 \\ \left( \dfrac{1}{x} \right)(xy) = \dfrac{1}{x}(1) \\ \left( \dfrac{1}{x} \cdot x \right) \cdot y = \dfrac{1}{x} \\ 1y = \dfrac{1}{x} \\ y = \dfrac{1}{x} \\ $
Show that if $x \neq 0$, then $\dfrac{1}{\frac{1}{x}} = x$.
Since $x$ is nonzero, $\dfrac{1}{x}$ exists. Likewise, since $\dfrac{1}{x}$ is itself nonzero, $\dfrac{1}{\frac{1}{x}}$ exists. Arithmetic and the cancellation law produce the desired result:
$ x \cdot \dfrac{1}{x} = 1 \\ \dfrac{1}{\frac{1}{x}} \cdot \dfrac{1}{x} = 1 \\ x \cdot \dfrac{1}{x} = \dfrac{1}{\frac{1}{x}} \cdot \dfrac{1}{x} \\ x = \dfrac{1}{\frac{1}{x}} \\ $
Prove that $0x = 0$ for all $x \in \mathbb{R}$.
$ 0x = (0 + 0) \cdot x \\ 0x = 0x + 0x \\ 0 = 0x \\ $
Show that if $x \neq 0$ and $y \neq 0$, then $xy \neq 0$.
Proof by contradiction. Assume $xy = 0$. By the multiplicative inverse axiom, we have
$ 1 = \left(\dfrac{1}{x}\right)x \\ 1 = \left(\dfrac{1}{y}\right)y \\ 1 \cdot 1 = \left(\dfrac{1}{x}\right)x \cdot \left(\dfrac{1}{y}\right)y \\ 1 = \left(\dfrac{1}{x}\right)\left(\dfrac{1}{y}\right)xy \\ 1 = \left(\dfrac{1}{x}\right)\left(\dfrac{1}{y}\right)0 \\ 1 = 0 \\ $
Clearly $1$ is not equal to $0$. Thus $xy \neq 0$.
Show that $x = 1 \cdot x$.
$ 1x + x = 1x + 1x \\ 1x + x = (1 + 1)x \\ 1x + x = 0 \\ (1x + x)  x = x \\ 1x + (x  x) = x \\ 1x = x \\ $
Show that $(x)y = (xy) = x(y)$.
$(x)y + xy = (x +x)y \\ (x)y + xy = 0y \\ (x)y + xy = 0 \\ (x)y = (xy)$
The proof that $(xy) = x(y)$ is the same:
$x(y) + xy = x(y+y) \\ x(y) + xy = x0 \\ x(y) + xy = 0 \\ x(y) = (xy)$
Show that $x^{1}y^{1} = (xy)^{1}$.
$ x^{1}y^{1} = 1 \cdot x^{1}y^{1} \\ x^{1}y^{1} = (xy)(xy)^{1}x^{1}y^{1} \\ x^{1}y^{1} = (x^{1}x)(y^{1}y)(xy)^{1} \\ x^{1}y^{1} = 1\cdot1\cdot(xy)^{1} \\ x^{1}y^{1} = (xy)^{1} \\ $
Show that if $x > 0$ and $y > 0$, then $x + y > 0$.
Let $x \gt 0$. By the first ordering property, $x + y \gt y$. Since $y \gt 0$, $x + y \gt 0$ by transitivity of order relations.
Show that if $x > 0$, then $x < 0$. Likewise, show that if $x < 0$, then $x > 0$.
If $x > 0$, then by the first ordering axiom $x + (x) > 0 + (x)$, so $x < 0$. Likewise, if $x < 0$, then $x + (x) < 0 + (x)$, so $x > 0$.
Show that if $x >0$ and $y < 0$, then $xy < 0$. Likewise, show that $xy > 0$ when $x < 0$ instead.
If $x > 0$ and $y < 0$, then $0 < y$. By the second ordering axiom, $0 < x(y) = (xy)$. Adding $xy$ to both sides shows that $xy < 0$.
If instead $x < 0$, then $0 < x$. By the second ordering axiom, $0 < (x)(y) = (1)x(1)y = xy$.
Show that if $z > 0$, then $x < y$ if and only if $zx < zy$.
Assume $x < y$. By trichotomy, either $zx < zy$, $zx = zy$, or $zx > zy$. If $zx = zy$, then $x = y$, which is a contradiction. If instead $zx > zy$, then $zx  zy = z(xy) > 0$. But $x  y < 0$, and therefore by the prior proof $z(xy) < 0$, which is a contradiction. Therefore $zx < zy$.
Conversely, assume $zx < zy$. By trichotomy, either $x < y$, $x = y$, or $x > y$. If $x = y$, then $zx = zy$, which is a contradiction. If instead $x > y$, then $y  x < 0$. Likewise, $zy  zx = z(yx) > 0$. But this is a contradiction, as by the prior proof $z(yx) < 0$. Therefore $x < y$.
Show that $0 < a < 1$ and $b > 0$ if and only if $0 < ab < b$.
Assume $0 < a < 1$ and $b > 0$. Then $ab > 0$ follows directly from the second ordering axiom. To show $ab < b$, note that $a < 1$, so $0 < 1  a$. Therefore by the second ordering axiom, $0 < (1a)b = b  ab$. Adding $ab$ to both sides reveals $ab < b$.
Conversely, assume $0 < ab < b$. Then $b > 0$ by transitivity. Likewise, since $ab < b$, then $0 < b  ab = (1a)b$. Note that $b^{1} > 0$, so by the second ordering axiom, $0 < (1a)b \cdot b^{1} = 1a$. Therefore $a < 1$. If $a = 0$, then $ab = 0$, which is a contradiction. If instead $a < 0$, then $ab < 0$, which is also a contradiction. Thus $a > 0$ by trichotomy. Therefore $0 < a < 1$.
Define the absolute value of a number $ \cdot  : \mathbb{R} \rightarrow \mathbb{R}$ as follows:
$$x = \left\{ \begin{array}{ll} x & \text{if } x \geq 0 \\ x & \text{if } x < 0 \end{array}\right.$$
Show that $x \geq x$ for all $x \in \mathbb{R}$.
If $x \geq 0$, then $x = x$. If $x < 0$, then $x < 0 < x = x$. Therefore $x \geq x$.
Triangle Inequality: Show that $x + y \leq x + y$.
Either $x + y = x + y$ or $x + y = (x + y)$. Because $x \leq x$, we see that $x + y \leq x + y$. By the same logic, we see that $x + y \leq x + y$. Therefore $x + y \leq x + y$. Likewise, $(x+y) = x  y \leq x  y \leq x + y = x + y$. Therefore $x + y \leq x + y$.
Show that $xy = xy$.
Consider the first case when $xy = xy$. Then either $x$ and $y$ are both positive or both negative. If the former, $xy = xy$. If the former, $xy = (x)(y) = xy$. Consider the second case when $xy = xy$. Then one of $x$ and $y$ is negative and the other is positive. If $x<0$ and $y>0$, then x=x and $y=y$, so $xy = xy$. If instead $x>0$ and $y<0$, then $x=x$ and $y=y$, so $xy = xy$. Thus in both cases $xy = xy$.
Summation Notation: Let $\{x_1, \ldots, x_n\}$ be a finite set of real numbers. Their sum is denoted with the $\Sigma$ symbol:
$$\sum\limits_{i=1}^{n} x_i = x_1 + \ldots + x_n$$
The index variable $i$ can also be used algebraically as well:
$$\sum\limits_{i=1}^{n} i = 1 + 2 + \ldots + n$$
Compute the following sums:

$\sum\limits_{i=1}^4 i$

$\sum\limits_{i=1}^6 3i$

$\sum\limits_{i=1}^5 2^i$

$\sum\limits_{i=1}^4 i = 1 + 2 + 3 + 4 = 10$

$\sum\limits_{i=1}^6 3i = 3(1) + 3(2) + 3(3) + 4(3) = 30$

$\sum\limits_{i=1}^5 2^i = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 = 2 + 4 + 8 + 16 + 32 = 62$

Show that $\sum\limits_{i=1}^n x_i \leq \sum\limits_{i=1}^n x_i$.
Proof by induction. Base case: Let $n = 1$. Then $\sum\limits_{i=1}^1 x_i = x_i \leq x_i = \sum\limits_{i=1}^1x_i$. Inductive step: Assume $\sum\limits_{i=1}^n x_i \leq \sum\limits_{i=1}^n x_i$. Then
$ \sum\limits_{i=1}^{n+1} x_i = x_{n+1} + \sum\limits_{i=1}^n x_i \\ \sum\limits_{i=1}^{n+1} x_i \leq x_{n+1} + \sum\limits_{i=1}^n x_i \\ \sum\limits_{i=1}^{n+1} x_i \leq x_{n+1} + \sum\limits_{i=1}^n x_i \\ \sum\limits_{i=1}^{n+1} x_i \leq \sum\limits_{i=1}^{n+1} x_i$
Natural Number Exponents: We use the notation $x^n$ to denote the product $\underbrace{x \cdot x \cdot \ldots \cdot x}_{n \text{ times}}$. For example, $x^4 = x \cdot x \cdot x \cdot x$. We define $x^0$ to be 1 if $x \neq 0$; $0^0$ is undefined.
Show the following:

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

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

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.

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.

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 $x^2 = (x)^2 = ((1)x)^2 = (1)^2x^2 = x^2 > 0$. Finally, if $x = 0$, then $x^2 = 0$.
Integer Exponents: For a positive natural number $n$, define for a real number $x \in \mathbb{R}$ the notation $x^{n} = (x^n)^{1} = \dfrac{1}{x^n}$.
Show the following:

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

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

The result for when $n$ and $m$ are both nonnegative 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 $x^nx^m = (x^{n})^{1}(x^{m})^{1} = (x^{n}x^{m})^{1} = (x^{nm})^{1} = x^{n + m}$.

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

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 $\sqrt[n]{x}$ and when $n=2$ as simply $\sqrt{x}$.
Prove that there exists exactly one real number $y$ such that $y = x^{\frac{1}{n}}$.
Hint: Use the identity $a^n  b^n = (b  a)(b^{n1} + b^{n1}a + \ldots + a^{n1})$ 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 $b^n  a^n = (ba)(b^{n1} + b^{n1}a + \ldots + a^{n1})$. If $0 < a < b$, we get the inequality $0 < b^n  a^n < (ba)(b^{n1} + b^{n1}b + \ldots + b^{n1})) = (ba)(nb^{1})$. Dividing through by $(nb^{1})$ on both sides gives $0 < \dfrac{b^n  a^n}{(nb^{1})} < ba$. 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 $(y + h)^n  y^n < ((y+h)y)n(y + h)^{n1} = hn(y+h)^{n1} < hn(y+1)^{n1} < x  y^n$. 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 $k = \dfrac{y^n  x}{ny^{n1}}$. Then $k = \dfrac{y}{n}  \dfrac{x}{ny^{n1}} < y$. If $t \geq y  k$, then $y^n  t^n \leq y^n  (yk)^n < kny^{n1} = y^n  x$. Therefore $t^n > x$, so $t \notin T$. Therefore $yk$ is an upper bound of $T$. But $yk < y$, 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 $y' < y$.
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 $(xy)^{\frac{1}{n}} = ab = x^{\frac{1}{n}}y^{\frac{1}{n}}$.
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 $(x^{\frac{1}{n}})^n = (x^n)^{\frac{1}{n}}$.
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}}$.
Rational Exponents V: For $x, y, \in \mathbb{R}$, does $x^n = y^n$ imply that $x = y$ if $x$ and $y$ are not necessarily nonnegative? Prove or give a counterexample.
No, the implication does not hold. Counterexample: $(1)^2 = 1$, and $1^2 = 1$, but $1 \neq 1$.
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 $\sqrt{2} \notin \mathbb{Q}$.
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 $\sqrt{2} < 2$.
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 $\sqrt{x} < \sqrt{y}$.
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}$
Let $x$ be a nonzero rational number and $y$ be an irrational number. Show the following:

$x + y$ is irrational.

$xy$ is irrational.

Proof by contradiction: Assume $x + y$ is rational. Then there exist some integers $a$ and $b$ such that $x + y = \dfrac{a}{b}$. Then $y = \dfrac{a}{b}  x$. But the righthand side is a rational number, as $\mathbb{Q}$ is closed under addition. Therefore $x + y$ is irrational after all.

Proof by contradiction: Assume $xy$ is rational. Then there exist some integers $a$ and $b$ such that $xy = \dfrac{a}{b}$. Then $y = \dfrac{a}{b}x^{1}$. But the righthand side is a rational number, as $\mathbb{Q}$ is closed under multiplication. Therefore $xy$ is irrational after all.

Archimedean Property: If $x, y \in \mathbb{R}$, and $x > 0$, then there exists a natural number $n \in \mathbb{N}$ such that $xn > y$.
Proof by contradiction. Assume that there does not exist a natural number $n \in \mathbb{N}$ such that $xn > y$. Then consider the set $A = \{ nx : n \in \mathbb{N} \}$. It follows that $y$ is an upper bound of $A$. Since $A$ has an upper bound, it must have a least upper bound, call it $\alpha$. Since $x > 0$, it follows that $\alpha  x < \alpha$, and $\alpha  x$ is not an upper bound of $A$. Therefore $\alpha  x < a$ for some $a \in A$. Then $a = mx$ for some $ m \in \mathbb{N}$, so $\alpha  x < mx$. Adding $x$ to both sides produces $\alpha < (m+1)x$. But this is a contradiction, since $(m+1)x \in A$ and $\alpha$ is the least upper bound of $A$. Therefore the Archimedean property must hold after all.
$\mathbb{Q}$ is dense in $\mathbb{R}$: Show that for every two distinct real numbers $a < b$, there is a rational number $q$ such that $a < q < b$.
Consider two real numbers $a$ and $b$ such that $a < b$. Then $b  a > 0$. By the Archimedean property, there is some $n \in \mathbb{N}$ such that $(ba)n > 1$. Likewise, there is some $p \in \mathbb{N}$ such that $p > na$, and there is some $q \in \mathbb{N}$ such that $q > na$. It follows that $q < na < p$. Note then that there is some $k \in \mathbb{N}$ such that $q \leq k \leq p$ and $k  1 \leq na < k$. By transitivity, we see that $na < k \leq 1 + na < nb$. Dividing through by $n$ shows that $a < \dfrac{k}{n} < b$.
Show that between every two distinct real numbers there is an irrational number.
Let $a, b \in \mathbb{R}$, and without loss of generality, assume that $a < b$.
Assume both $a$ and $b$ are rational. Note that $0 < \sqrt{2} < 2$. Therefore $0 < \dfrac{\sqrt{2}}{2} < 1$. Therefore $0 < \dfrac{(ba)\sqrt{2}}{2} < ba$, so $a < a + \dfrac{(ba)\sqrt{2}}{2} < b$. Because $\sqrt{2}$ is irrational, it follows that $a + \dfrac{(ba)\sqrt{2}}{2}$ is irrational.
Assume instead that either $a$ or $b$ is irrational. Then $\dfrac{a + b}{2}$ is irrational and $a < \dfrac{a + b}{2} < b$.
Finally, assume that $a$ and $b$ are both irrational. If $a + b$ is irrational, then $\dfrac{a + b}{2}$ is also irrational, and $a < \dfrac{a + b}{2} < b$. If instead $a + b$ is rational, then so is $\dfrac{a + b}{2}$. Then by the prior case, $\dfrac{a + \frac{a + b}{2}}{2} = \dfrac{3a + b}{4}$ is irrational and $a < \dfrac{3a + b}{4} < \dfrac{a + b}{2} < b$.