📂Analysis

Proof of the Density of Real Numbers

Theorem

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

Explanation

On the real numbers, no matter what interval we consider, there always exists another real number in between. This means that no matter how small we split it, there is always a point that can be split further. It may seem obvious, but keep in mind that this is not only far from obvious but also an extremely abstract property. For example, even matter and energy dealt with in physics reach a limit when split into smaller and smaller pieces.

Proof

Strategy: The proof is done separately for rational and irrational numbers. If between two real numbers there exists a rational number as well as an irrational number, the proof is complete. The expression without loss of generality is mentioned because the positive numbers appearing in the proof can always be expressed as differences of real numbers, so there is no need to consider numbers less than or equal to $0$. For example, even if the proof starts with two negative numbers $c < d < 0$, as long as equality does not hold, a positive number can be made as in $d - c > 0$.

The essential premises needed are as follows.

Field axioms:

• (A1) Closure under addition: $a+b \in \mathbb{R}$
• (A5) Additive inverse: There exists $(-a)$ satisfying $a + (-a) = (-a) + a = 0$
• (M1) Closure under multiplication: $a\cdot b \in \mathbb{R}$
• (M5) Multiplicative inverse: There exists ${a^{-1}}$ satisfying $a \cdot a^{-1} = a^{-1} \cdot a = 1$
• (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 $n$ satisfying $an>b$.

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• Part 1. Density of the rationals ¹

  Let us show that there always exists $q \in \mathbb{Q}$ satisfying $a<q<b$. First, without loss of generality, considering the positive number $(b-a) > 0$ satisfying $0 < a < b$ and the real number $1 \in \mathbb{R}$, there exists a set of natural numbers $\left\{ n \in \mathbb{N} : (b-a) n > 1 \right\}$ satisfying the inequality of the Archimedean principle, and by the existence of additive inverses, closure, the distributive law, and additivity, we can see that $$ bn-an > 1 \implies an + 1 < bn \implies an < an + 1 < bn $$ Since the difference between $an$ and $bn$ is greater than $1$, there exists at least one integer between them, and letting it be $m$, we have $$ an < m < bn $$ Multiplying each side by the multiplicative inverse $n^{-1}$ of $n$ yields the following. $$ a < {{ m } \over { n }} < b $$ Here, letting $\displaystyle q := {{ m } \over { n }}$, $q$ is none other than a rational number, a ‘ratio of natural numbers’, and we obtain the following inequality. $$ a < q < b $$

• Part 2. Density of the irrationals

  Let us show that there always exists $\xi \in \mathbb{Q^{c}}$ satisfying $a<\xi<b$. Without loss of generality, considering real numbers satisfying $0 < a < b$ and an irrational number $c>0$, by multiplicativity, if $a<b$ then $ac<bc$. Since the real numbers are closed under multiplication, $ac$ and $bc$ are also real numbers, and by the density of the rationals, there exists a rational number $q \ne 0$ satisfying $ac<q<bc$. Multiplying each side of $ac<q<bc$ by the multiplicative inverse $\displaystyle {1 \over c}$ of $c$ gives the following.

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

  Here, letting $\displaystyle \xi := {q \over c}$, $\xi$ is the product of a nonzero rational number and an irrational number, hence irrational, and we obtain the following inequality.

  $$ a<\xi<b $$

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