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Consequences of Completeness (Part 2)

MATH 4630 · Chapter 1: The Real Numbers

In the previous handout, we established the Archimedean Property and the Nested Interval Property as consequences of the Axiom of Completeness. In this handout, we focus on two further consequences that highlight approximation and existence in $\mathbb{R}$.

3. The Density of $\mathbb{Q}$ in $\mathbb{R}$

Theorem: Density of $\mathbb{Q}$

For any two real numbers $a $$a < r < b.$$

No matter how close two real numbers are, we can always find a rational number strictly between them. This says that $\mathbb{Q}$ has no gaps inside $\mathbb{R}$.

Proof Sketch

Step 1: Choose a large enough $n$

Because $a.

By the Archimedean Property, there exists $n\in\mathbb{N}$ such that

$$\frac{1}{n} < b-a.$$

Step 2: Find an integer $m$ between $na$ and $nb$

Multiply the inequality $\frac{1}{n} $$1 < n(b-a).$$

Expanding: $na < na + 1 < nb$ (since $1 < n(b-a)$).

Define the set:

$$S = \{\,k\in\mathbb{Z} : k > na\,\}.$$

Verification

  1. Explain why $S$ is nonempty.
  2. Is $S$ bounded below? $\square$ Yes $\quad$ $\square$ No. Justify your choice.

From the previous handout, the set $S$ has a greatest lower bound (infimum) in $\mathbb{R}$. Let $m$ be the minimum (smallest integer) of $S$.

Step 3: Define the rational number

Define

$$r = \frac{m}{n}.$$

Verification

Explain why $r$ is a rational number:

Step 4: Show $a < r < b$

From the definition of $m$, we have $na < m$. Dividing by $n$, conclude that

$$a < \frac{m}{n}=r.$$

Verification

Use the minimality of $m$ to show that $m \le na+1$.

Hint: What can you say about $m-1$? Is it in $S$?

Divide by $n$ and use Step 1 to conclude

$$r < b.$$

Therefore, $a < r < b$. $\square$

4. Existence of $n^{\text{th}}$ Roots

This is the moment we've been waiting for! We can now prove that $\sqrt{2}$ exists—the fifth item on our "Wish List".

Theorem: Existence of $\sqrt{2}$

There exists a real number $\alpha \in \mathbb{R}$ such that $\alpha^2=2$.

We denote this number by $\alpha=\sqrt{2}$ (or $2^{1/2}$).

Proof Sketch

Define the set

$$S = \{s \in \mathbb{R} : s^2 < 2\}.$$

Verification

Explain why the set $S$ is nonempty and bounded above:

Therefore by Axiom 3, the set $S$ has a in $\mathbb{R}$, call it $x \in \mathbb{R}$.

Verification

Explain why $x>1$:

Since the real numbers satisfy the Trichotomy Law, exactly one of the following must hold:

$$x^2 < 2, \quad x^2 = 2, \quad x^2 > 2.$$

The strategy is to show that $x^2=2$ by ruling out the other two possibilities.

Case 1: Assume $x^2 < 2$

Our goal is to find a number slightly larger than $x$ that still belongs to $S$.

Contradiction

Explain why finding such a number would contradict the fact that $x = \sup S$:

Consider:

$$\left(x + \frac{1}{n}\right)^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2}.$$

Show

Show that: $\left(x + \frac{1}{n}\right)^2 \le x^2 + \frac{1}{n}(2x+1)$.

We want: $\left(x + \frac{1}{n}\right)^2 < 2$. This will happen provided

$$\frac{1}{n}(2x+1) < 2 - x^2.$$

By the Archimedean Property, choose $n \in \mathbb{N}$ such that $\frac{1}{n}$ is sufficiently small. Then $x + \frac{1}{n} \in S$, which contradicts that $x = \sup S$.

Therefore, we cannot have $x^2 < 2$.

Case 2: Assume $x^2 > 2$

This time, we try to find a number slightly smaller than $x$ that is still an upper bound for $S$.

Contradiction

Explain why this would contradict the definition of supremum:

By the Archimedean Property, choose $m \in \mathbb{N}$ such that $\left(x - \frac{1}{m}\right)^2 > 2$. Then $x - \frac{1}{m}$ is an upper bound for $S$ smaller than $x$, contradicting $x = \sup S$.

Therefore, we cannot have $x^2 > 2$.

Conclusion

Since both $x^2 < 2$ and $x^2 > 2$ lead to contradictions, we conclude

$$x^2 = 2. \,\square$$

Corollary

There exists an irrational number. (For example, $\sqrt{2}$ is irrational.)

Activity

Modify the argument above to show that there exists a positive real number $y$ such that $y^2=3$.

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