📂Analysis

Proof of the Density of Real Numbers

Theorem

For two real numbers $a<b$, there exists a $r \in \mathbb{R}$ that satisfies $a<r<b$.

Explanation

In the real number space, no matter what interval you consider, there is always another real number in between. No matter how much you split it, there is a point that can be further divided. Although it seems obvious, keep in mind that this is not only non-obvious but also highly abstract. As an example, even the matter and energy dealt with in physics have their limits when split into smaller and smaller pieces.

Proof

Strategy: The proof is divided for rational and irrational numbers respectively. If there exist both a rational number and an irrational number between two real numbers, then the proof is complete. The phrase “without loss of generality” is mentioned because the positive numbers that appear in the proof can always be represented by the difference between real numbers, so there is no need to specifically consider numbers below $0$. For example, even if the proof starts from two negative numbers $c < d < 0$, as long as inequality holds, a positive number can be formed as in $d - c > 0$.

The necessary foundational premises are as follows:

Field Axioms:

• (A1) Closure under addition: $a+b \in \mathbb{R}$
• (A5) Additive inverse: There exists a $(-a)$ that satisfies $a + (-a) = (-a) + a = 0$
• (M1) Closure under multiplication: $a\cdot b \in \mathbb{R}$
• (M5) Multiplicative inverse: There exists a ${a^{-1}}$ that satisfies $a \cdot a^{-1} = a^{-1} \cdot a = 1$
• (D) Distributive law: $a \cdot (b + c) = a \cdot b + a \cdot c$

Order Axioms:

• Additivity: If $a<b$ and $c\in \mathbb{R}$, then $a+ c< b + c$
• Multiplicativity: If $a<b$ and $c>0$, then $ac< bc$, or if $c<0$, then $ac> bc$

Archimedean Principle: For a positive number $a$ and a real number $b$, there exists a natural number $n$ that satisfies $an>b$.

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• Part 1. Density of Rational Numbers ¹

  Let’s show that there always exists a $q \in \mathbb{Q}$ that satisfies $a<q<b$. Without loss of generality, considering a positive number $(b-a) > 0$ and a real number $1 \in \mathbb{R}$ that satisfy $0 < a < b$, there exists a set of natural numbers $\left\{ n \in \mathbb{N} : (b-a) n > 1 \right\}$ that satisfies the inequality of Archimedes’ principle, and by the existence of an additive inverse, closure, and distributive law, and additivity $$bn-an > 1 \implies an + 1 < bn \implies an < an + 1 < bn$$ is known. $an$ and $bn$ have a difference greater than $1$, so there is at least one integer between them, let’s denote it as $m$, then $$an < m < bn$$ If each side is multiplied by the multiplicative inverse of $n$ $n^{-1}$, we get the following: $$a < {{ m } \over { n }} < b$$ Here, if we set $\displaystyle q := {{ m } \over { n }}$, then $q$ is a ‘ratio of natural numbers’, a rational number, and we obtain the following inequality: $$a < q < b$$

• Part 2. Density of Irrational Numbers

  Let’s show that there always exists a $\xi \in \mathbb{Q^{c}}$ that satisfies $a<\xi<b$. Without loss of generality, considering real numbers and irrational numbers $c>0$ that satisfy $0 < a < b$, if $a<b$ then $ac<bc$ by multiplicativity. Since real numbers are closed under multiplication, $ac$ and $bc$ are also real numbers, and by the density of rational numbers, there exists a rational number $q \ne 0$ that satisfies $ac<q<bc$. If each side of $ac<q<bc$ is multiplied by the multiplicative inverse of $c$ $\displaystyle {1 \over c}$, it’s as follows:

  $$a<{q \over c}<b$$

  Here, if we set $\displaystyle \xi := {q \over c}$, then $\xi$ is the product of a non-$0$ rational number and an irrational number, thus an irrational number, and we obtain the following inequality:

  $$a<\xi<b$$

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