Ross Analysis Question 4.10 (using archimedean property)

Question

Ross defines the Archimedean Property as follows:

If $a>0$ and $b>0$ , then for some integer $n$, we have $na>b$.

Here is problem 4.10:

Prove that if $a>0$ , then there exists $n$ (element of natural numbers) such that $\frac{1}{n}< a < n$ .

Solution: We consider two cases:

Case 1: $a>1$ , then since $a>0 $we can choose a positive integer $n$ such that $a < n$. From $a< n$ we deduce that $\frac{1}{a}>\frac{1}{n}$ and because $a>1$ we have $a>\frac{1}{a}$, hence $a>\frac{1}{n}$. Therefore $\frac{1}{n}< a < n$.

Case 2: $0 < a < 1$. Then since $a>0$ we can choose a positive integer $n$ such that $\frac{1}{n}< a$. From $\frac{1}{n}< a$ we deduce that $n>\frac{1}{a}$ and because $0< a < 1$ we have $\frac{1}{a}>a$ , hence $n >a$ .Therefore $\frac{1}{n}< a < n$.

I have seen a shorter proof online, I am only asking if my solution is correct.

Answer 1

Seems perfectly fine to me. However, you are missing the case $a=1$; of course this is trivial as for $a=1$ then any $n \in \mathbb{N}$ strictly greater than $1$ satisfies $\frac{1}{n} < a < n$, but one should include this in the proof.

Answer 2

It seems to me you are assuming what you are trying to prove.

How do you know that there exists an $n$ so that $a < n$? And how do you know there exists and $n$ so that $\frac 1n < a$?

I think you are supposed to use the archmediean principle. Let $\beta = a$ and let $\alpha = 1$ so there exists an $n$ so that $n\alpha > \beta$. i.e. $n\cdot 1 > a$.

Proving there is an $n$ so that $\frac 1n < a$ requires a twist in proving that there is an $n > \frac 1a$ but that's the same thing. By AP there is an $n$ so that $n\cdot 1 > \frac 1a$.