Integer upper bound to a subset of real numbers?

Question

Exploring the properties of real numbers from "Mathematical Thinking" by John P. D'Angelo, I am asked to prove that

For every positive real number $r$ there exists an integer $k$ such that $5 \lt |k|r$

I would say that it is true by the Completeness Axiom, I think that this means "There exists an integer $k$ that is an upper bound to the set of real numbers of the form $\frac 5r$ where $r$ is a positive real number, which I would express as:

$$(\exists\,k \in \Bbb Z)\,(\forall \,r\in \Bbb R^+) \,:\, k=\,\sup \bigg\lbrace \frac 5r \bigg |\; r \in \Bbb R^+\bigg \rbrace$$

I'm not sure whether my reasoning is correct or how I would go about proving it though.

Answer 1

"I think that this means "There exists an integer k that is an upper bound to the set of real numbers of the form 5/r where r is a positive real number""

So that's obviously not true. As $r \rightarrow 0$ then $5/r \rightarrow \infty$ and is not bounded above.

Consider this: Let $A_r = \{|k|*r|k \in \mathbb Z\}$. This is clearly not empty. Are any of the $|k|r > 5$? If not then $5$ is an upper bound of $A_r$ and $A_r$ is bounded above. Is that possible?

If it were, let $a = \sup A_r$. $r > 0$ so $a - r < a$ so $a-r$ is not an upper bound. So there is a $|k|r$ so that $a - r < |k|r$ so $a < |k|r + r = (|k| + 1)r$ as $|k|+1 = ||k|+1|$ [why in the world is this problem making a deal about absolute values... that's just confusing and irelevent], we can conclude $(|k| + 1)r \in A_r$.

But that is a contradiction! We have an element of $A_r$ that is bigger than it upper limit! So $A_r$ is unbounded. So it isn't bounded by $5$ (or anything else).

So it is not that case that all $|k|r \le 5$. So some $|k|r > 5$.

---

This is the archimedian property that can be stated as.

For any $r > 0$ and any $y$. There is some positive integer $k$ so that $kr > y$.

As direct consequences we can say, for any $y$ there is an integer $k$ so that $k > y$. Also that there is in integer $m$ so that $m \le y < m+1$. And for any $x < y$ there is a rational $q$ so that $x < q < y$.

Handy stuff to know.

Answer 2

You can prove this from the completeness axiom, but it's a bit tricky. What you want to do is take the infimum of the set of reciprocals of counterexamples. That is, take the set $$A=\{R\in\mathbb{R}^+:|k|/R\leq 5\text{ for all }k\in\mathbb{Z}\}.$$ This set is bounded below in $\mathbb{R}$, so if it is nonempty then it has an infimum $x$. But if $R\in A$, then $R-1\in A$ as well (I'll leave proving that to you). Since $x=\inf A$, there must be an element of $A$ that is less than $x+1$, and then subtracting $1$ gives another element of $A$ which is less than $x$. This is a contradiction, which implies $A$ must be empty.