# Proving double inequalities using well-ordering property

I was given a proposition: $∀x ∈ R$, if $0 < x < 1$, then $∃ n ∈ N$, $\frac1n < x \le \frac {1}{n-1}$. I understand that the well-ordering property states that for a set $S ∈ Z$, if $S$ is a non-zero set, and $x ∈ S$, then there is an element $a ∈ S$ such that $a \le x$. However, I don't get how that property can be used to prove a proposition such as this one. Can someone please explain how to use the well-ordering property?

• Where did the "m" come from? Did you mean"1/(n-1}"? Mar 10 '17 at 19:36
• yes sorry, I meant to say n Mar 10 '17 at 19:37

My guess is that this is what you want to prove:

For $n \in \mathbb{N}^+$, let $R(n) =\{x | \frac1{n+1} < x \le \frac1{n}\}$.

Then, $0 < x \le 1 \implies \exists n \in \mathbb{N}^+$ with $x \in R(n)$.

Here is my proof (as usual, off the top of my head with editing as I go).

Since $0 < x \le 1$, $1 \le \frac1{x}$. Let $y = \frac1{x}$.

By the axiom of Archimedes. there is an integer $m$ such that $m \gt y$.

Let $G(y) =\{m | m \gt y\}$.

We have shown that $G(y)$ is non-empty.

Also, all integers $j \le 1$ are not in $G(y)$.

By the well-ordering principle, $G(y)$ has a smallest member. Call this $k$.

For this, we must have $k \gt y$ and $k-1 \le y$.

Therefore $\frac1{k} < \frac1{y} \le \frac1{k-1}$ or $\frac1{k} < x \le \frac1{k-1}$, so that $x \in R(k)$.

And we are done.

The definition of the real number system $\mathbb R$ implies that there is no $x\in \mathbb R$ such that $0<x\leq q$ for every positive rational $q.$ (It is possible to extend $\mathbb R$ to a larger number-field in which such $x$ do exist.)

One of the consequences is that if $0<x\in \mathbb R$ then there exists $n\in \mathbb N$ such that $n> 1/x.$

Otherwise for any $q\in \mathbb Q^+$ we have $q=m/n$ for some $m,n\in \mathbb N,$ and $n\leq 1/x,$ so $q=m/n\geq 1/n\geq x$. But this cannot be true for every $q\in \mathbb Q$.

So for $0<x<1$ let $n_0$ be the least $n\in \mathbb N$ such that $n> 1/x.$ Then $n_0>1$ because $1/x>1.$ We have $$(\bullet )\quad n_0> 1/x\geq n_0-1$$ by def'n of $n_0.$..... And $n_0-1\in \mathbb N$ because $n_0>1.$ So $n_0-1>0$, so $(\bullet)$ is equivalent to $$1/n_0<x\leq 1/(n_0-1).$$