A Rational Prelude

Let’s start our discussion on real analysis with the set of rational numbers, denoted $\mathbb Q$ . We can construct them using integers and even natural numbers, but roughly speaking, the rational numbers are fractions of the form $m/n$ , whenever $n$ is a nonnegative integer.

Definition 1. Define the rational number equality (technically: equivalence relation) via

$\displaystyle \frac ab = \frac cd \quad \iff \quad ad = bc,$

where $b,d$ are nonzero.

Intuitively (which can be formalised in a rather mundane manner), setting $b = 1$ yields an integer $a/1 \equiv a$ , and if $a > 0$ , we obtain a natural number. For real analysis, we will adopt the convention that $\mathbb N = \mathbb N^+$ so that $0 \notin \mathbb N$ . Let’s first convince ourselves that without loss of generality, we can assume the denominators $b,d$ are positive.

Theorem 1. $a/(-b) = (-a)/b$ .

Proof. By the double-negative property of the integers,

$ab = (-a)(-b) \Rightarrow a/(-b) = (-a)/b.$

The rationals are, in a sense, defined using addition, multiplication, and ordering, as follows.

Definition 2. Furthermore, define addition and multiplication of rational numbers as follows:

$\displaystyle \begin{aligned} \frac ab + \frac cd := \frac{ad + bc}{bd},\quad \frac ab \cdot \frac cd := \frac{ac}{bd}. \end{aligned}$

We abbreviate $ab = a\cdot b$ . Furthermore define ordering by

$\displaystyle \frac ab \leq \frac cd \quad \iff \quad ad \leq bc.$

The rational numbers form what is known as a field, which basically means that we can perform the usual actions of addition, subtraction, multiplication, and division.

Theorem 2. The rational numbers $\mathbb Q$ form an abelian group under addition in the following sense:

For any $x,y \in \mathbb Q$ , $x+y \in \mathbb Q$ .
For any $x,y \in \mathbb Q$ , $x+y = y + x$ .
For any $x,y,z \in \mathbb Q$ , $x+(y+z)=(x+y)+z$ .
There exists $0 \in \mathbb Q$ such that for any $x \in \mathbb Q$ , $x+0 = 0+x = x$ .
For any $x \in \mathbb Q$ , there exists a unique $-x$ such that $x+(-x) = (-x)+x= 0$ .

Furthermore, $\mathbb Q$ form an abelian group under multiplication in a similar sense:

For any $x,y \in \mathbb Q$ , $xy \in \mathbb Q$ .
For any $x,y \in \mathbb Q$ , $xy = yx$ .
For any $x,y,z \in \mathbb Q$ , $x(yz)=(xy)z$ .
There exists $1 \in \mathbb Q$ such that for any $x \in \mathbb Q$ , $x\cdot 1 = 1\cdot x = x$ .
For any $x \in \mathbb Q$ with $x \neq 0$ , there exists a unique $1/x$ such that $x \cdot (1/x) = (1/x) \cdot x= 1$ .

Finally, addition and multiplication are connected through distributivity:

$\displaystyle \begin{aligned}x(y+z) &= xy + xz, \\ (x+y)z &= xz + yz.\end{aligned}$

Combining all properties, we call $\mathbb Q$ a field.

Proof. Exercise.

There are several implications of these algebraic properties of $\mathbb Q$ , which we may explore in some exercises. Interestingly, we have not actually used any of the order properties $\leq$ of $\mathbb Q$ , other than a quick sanity check that we can assume the denominators are positive. The idea is that our definition of $\leq$ ought to match our definitions of addition and multiplication, so that $\mathbb Q$ is not just (totally) ordered, but an ordered field.

Theorem 3. The rational numbers form an ordered field in the following sense:

For $a, b, c \in \mathbb Q$ , $a \leq b \Rightarrow a+c \leq b+c$ .
For $a, b \in \mathbb Q$ , $a \geq 0 \wedge b\geq 0 \Rightarrow ab \geq 0$ .

Here, we denote $r \geq s$ to mean $s \leq r$ . Furthermore,

$r < s \quad \iff \quad r\leq s \wedge r \neq s,$

and $r > s \iff s < r$ . We say that $r$ is positive if $r > 0$ and negative if $r < 0$ .

Proof. We will prove the first property for illustration. Interestingly, this proof will highlight the core heuristic of reverse-engineering a proof that real analysis is replete of.

Let $a = a_1/a_2, b = b_1/b_2, c=c_1/c_2$ , where $a_2,b_2,c_2 > 0$ . Suppose $a \leq b$ . By definition,

$\displaystyle \frac{a_1}{a_2} \leq \frac{b_1}{b_2} \iff a_1 b_2 \leq a_2 b_1.$

We aim to prove that $a+c \leq b+c$ . This translates to

$\displaystyle \frac{a_1 c_2 + a_2 c_1}{a_2c_2} \leq \frac{b_1c_2 + b_2c_1}{b_2c_2},$

which is in turn equivalent to

$(a_1 c_2 + a_2 c_1)b_2c_2 \leq (b_1c_2 + b_2c_1)a_2c_2.$

Since $c_2 > 0$ , by the ordering properties on $\mathbb Z$ , this is equivalent to canceling the $c_2$ to obtain

$(a_1 c_2 + a_2 c_1)b_2 \leq (b_1c_2 + b_2c_1)a_2.$

By distributivity in $\mathbb Z$ , this is equivalent to

$a_1 b_2 c_2 + a_2 b_2 c_1 \leq a_2b_1c_2 + a_2b_2c_1.$

Some algebra shows us that we can remove that $a_2b_2c_1$ on both sides and cancel the positive $c_2$ to obtain the inequality

$a_1 b_2 \leq a_2 b_1,$

which is what we assumed in the first place! While this does suffice as a proof, if we want to be completely convinced, let’s write the proof out in its proper sequence:

$\begin{aligned} a \leq b \quad &\Rightarrow \quad \frac{a_1}{a_2} \leq \frac{b_1}{b_2} \\ &\Rightarrow \quad a_1 b_2 \leq a_2 b_1 \\ &\Rightarrow \quad a_1 b_2c_2 \leq a_2 b_1c_2 \\ &\Rightarrow \quad a_1 b_2 c_2 + a_2 b_2 c_1 \leq a_2b_1c_2 + a_2b_2c_1 \\ &\Rightarrow \quad (a_1 c_2 + a_2 c_1)b_2 \leq (b_1c_2 + b_2c_1)a_2 \\ &\Rightarrow \quad (a_1 c_2 + a_2 c_1)b_2c_2 \leq (b_1c_2 + b_2c_1)a_2c_2 \\ &\Rightarrow \quad \frac{a_1 c_2 + a_2 c_1}{a_2c_2} \leq \frac{b_1c_2 + b_2c_1}{b_2c_2}\\ &\Rightarrow \quad \frac{a_1}{a_2} + \frac{c_1}{c_2} \leq \frac{b_1}{b_2} + \frac{c_1}{c_2} \\ &\Rightarrow \quad a + c \leq b + c. \end{aligned}$

Either presentations are valid proofs. In fact, we could replace each $\Rightarrow$ with $\Leftrightarrow$ , though this will unfortunately be one of the few rare moments where such a move is legitimate. Yet, what is interesting is this challenge—that we needed to plan the proof in reverse order to actually writing the proof. This is the biggest paradigm shift you’ll experience in learning real analysis.

We will state the various order properties that we need for our discussions when we get there. For now, let’s first ask a simple question: Is there any rational number $r$ with the property that $r^2 = 2$ ? Can you prove your answer?

—Joel Kindiak, 17 Dec 24, 0035H

Responses

subratv

January 30, 2025 at 6:14 pm

I hope there are more follow up posts to this. Take a look at my blog: https://mathmusings17.wordpress.com/

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1. joelkindiak
  
  January 30, 2025 at 6:44 pm
  
  Thanks Subrat; subscribed to your blog so that we can inspire one another~
  
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2. joelkindiak
  
  January 30, 2025 at 9:22 pm
  
  Just updated my website to reflect my math blog instead of my personal blog (though feel free to read those too haha); kindiakmath.com
  
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A Rational Prelude

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