Mashed Potato Rationals

Definition 1. For sets $K, L \subseteq \mathbb R$ , we say that $K$ is dense in $L$ if for any $x,y \in L$ with $x < y$ , there exists $z \in K$ such that $x < z < y$ . For this definition, we do not assume $K \subseteq L$ .

Problem 1 (Density Theorem). Prove that $\mathbb Q$ is dense in $\mathbb R$ .

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Solution. Fix $x,y \in \mathbb R$ such that $x < y$ . Then $y-x > 0$ . By the Archimedean property, find $n \in \mathbb N$ such that

$\displaystyle 0 < \frac 1n < y-x \quad \Rightarrow \quad nx < nx + 1 < ny.$

Intuitively, there should be one integer between $nx$ and $nx + 1$ . To formulate this rigorously, the definition of the floor function $\lfloor \cdot \rfloor : \mathbb R \to \mathbb Z$ yields the estimate

$\lfloor nx \rfloor \leq nx < \lfloor nx \rfloor + 1.$

Hence,

$\displaystyle nx < \lfloor nx \rfloor +1 \leq nx + 1 < ny \quad \Rightarrow \quad x < \frac{\lfloor nx \rfloor+1}{n} < y.$

Set $z := (\lfloor nx \rfloor+1)/n \in \mathbb Q$ to obtain the desired result.

Problem 2. Prove that $\mathbb Q(\sqrt 2) := \{r\sqrt 2 : r \in \mathbb Q\}$ is dense in $\mathbb R$ .

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Solution. Fix $x,y \in \mathbb R$ such that $x < y$ . Dividing by $\sqrt 2$ yields $x/\sqrt 2 < y/\sqrt 2$ . By Problem 1, find a nonzero rational number $r$ such that

$\displaystyle \frac x{\sqrt 2} < r < \frac y{\sqrt 2}\quad \Rightarrow \quad x < r\sqrt 2 < y.$

Define $z := r\sqrt 2 \in \mathbb R \backslash \mathbb Q$ , as required.

Problem 3. Suppose $K \subseteq L$ and $K$ is dense in $M$ . Then $L$ is dense in $M$ .

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Solution. Fix $x,y \in M$ such that $x < y$ . Since $K$ is dense in $M$ , there exists $z \in K \subseteq L$ such that $x < z < y$ , as required.

Problem 4. Deduce that $\mathbb R \backslash \mathbb Q$ is dense in $\mathbb R$ .

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Solution. We claim that $\mathbb Q(\sqrt 2) \subseteq \mathbb R \backslash \mathbb Q$ , since $\sqrt{2} = (r\sqrt 2)/r$ (so that $r \sqrt 2 \in \mathbb Q \Rightarrow \sqrt 2 \in \mathbb Q$ , a contradiction).

By Problems 2 and 3, $\mathbb R \backslash \mathbb Q$ is dense in $\mathbb R$ .

Problem 5. Suppose $K$ is dense in $L$ and $L$ is dense in $M$ . Then $K$ is dense in $M$ .

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Solution. Fix $x,y \in M$ such that $x < y$ . Since $L$ is dense in $M$ , there exists $u \in L \subseteq M$ such that $x < u < y$ . Apply the density of $L$ in $M$ again to obtain $v \in L$ such that $u < v < y$ .

Since $K$ is dense in $L$ , there exists $z \in K$ such that $u < z < v$ . Therefore,

$x < u < z < v < y \quad \Rightarrow \quad x < z < y,$

as required.

Problem 6. Prove that for continuous functions $f, g : \mathbb R \to \mathbb R$ and $K$ dense in $L$ , if $f|_{K} = g|_{K}$ , then $f|_L = g|_L$ .

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Solution. Fix $x \in L$ . For each $n \in \mathbb N$ , since $K$ is dense in $L$ , find $x_n \in K$ such that $|x_n - x| < 1/n$ . Taking $n \to \infty$ , since $| \cdot |$ is continuous,

$\displaystyle \left| \lim_{n \to \infty} x_n - x \right| = \lim_{n \to \infty} |x_n - x| \leq \lim_{n \to \infty} \frac 1n = 0.$

Therefore, $x_n \to x$ , so that by the continuity of $f, g$ ,

$\displaystyle f(x) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} g(x_n) = g(x).$

Since $x \in L$ is arbitrary, $f|_L = g|_L$ .

Corollary 1. By Problem 1, $\mathbb Q$ is dense in $\mathbb R$ . Hence, by Problem 6, for any $a > 0$ , the continuous exponential $\exp_a$ with base $a > 0$ , defined by $\exp_a(x) := a^x$ , is unique.

—Joel Kindiak, 19 Jan 25, 1633H

KindiakMath

Mashed Potato Rationals

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Mashed Potato Rationals

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