[Math] Suppose $a$ is a positive real number. Prove that there exists a natural number $n$ such that $0 < 1/n < a$.

Question

Suppose $a$ is a positive real number. Prove that there exists a natural number $n$ such that $0 < 1/n < a$.

Could you guys help with this? I'm awful at writing proofs, I know why this is true but I don't know how to write the proof in a way that others could follow. Any help would be greatly appreciated. Thanks.

Answer 1

To derive this we need to use the completeness of $\mathbb{R}$:

Every nonempty set of real numbers that is bounded above has a supremum in $\mathbb{R}$.

There are two approaches to this, the axiomatic approach (where it is an axiom) and the constructive approach (where it is a theorem). More information can be found in analysis texts.

Now we prove the Archimedean property of $\mathbb{R}$:

$\mathbb{N}$ is not bounded above in $x\in\mathbb{R}$. That is, for each $x\in\mathbb{R}$ there exists $n\in\mathbb{N}$ such that $n>x$.

For $x<0$ it is trivial. Suppose $x\geq0$. Let $A=\{n\in\mathbb{N}\ ;n\leq x\}$. By completeness we have $\sup A\in\mathbb{R}$. Obviously, there is $a\in A$ such that $\sup A-1/2<a$ (Why?), then let $n=a+1$ and $n>x$ and the theorem is proved.

Finally, we turn to the question:

For any real number $a>0$, there exists a natural number $n$ such that $0<1/n<a$.

We proceed using proof by contradiction. Let $0<1/n<a$ for all $n\in\mathbb{N}^{\times}$, then $n\leq 1/a$ for all $n\in\mathbb{N}^{\times}$. Thus $\mathbb{N}$ is bounded in $x\in\mathbb{R}$, contradicting the Archimedean property.