Suppose $b \in \mathbb{R}$ and $|b| < \frac{1}{n}$ for every positive integer n. Prove that $b = 0$. This comes from an exercise in Appendix C from Axler's Measure, Integration & Real Analysis. The following is my approach. 
Suppose $b \neq 0$. Let $|b| = \epsilon$. Then 
        by Archimedean Property (2) 
        $$\exists n^* \in \mathbb{Z}^+ \text{ such that} \frac{1}{n^*} < \epsilon$$
        but $b < \frac{1}{n}$, $\forall n \in \mathbb{Z}^+$. Hence a contradiction.
Am I approaching this correctly, if I am are there any other approaches? 
 A: You are approaching it correctly and your approach is the simplest and intuitive way.
But if you wish to be direct and analytical you could do the following.
$b\in \mathbb R$ so there exist a sequence of $q_n \in \mathbb Q$ so that $q_n \to b$.
For any $\epsilon > 0$ there exist an $N$ so that $n > N$ implies $|q_n - b| < \frac {\epsilon}2$.  But there is also a $k\in \mathbb N$ so that $0 < \frac 1k < \frac \epsilon 2$.
So $|q_n- 0| \le |q_n - b| + |b-0|$.  $|q_n - b| < \frac \epsilon 2$ and $|b-0| = |b| < \frac 1k < \frac \epsilon 2$ so $|q_n - 0| < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$.
So $\lim_{n\to \infty} q_n = 0$.  But $b = \lim_{n\to \infty} q_n $. So $b = 0$.
But... I'd prefer to read your proof than my proof.
A: Proof
Assume that $a \neq 0.$ Then $|a| \neq 0.$ Take $N=\lfloor\dfrac{1}{|a|}\rfloor+1 \in \mathbb{Z_+},$ where $\lfloor x \rfloor$ denotes the floor integer function. Then $$\frac{1}{|a|}<\lfloor\dfrac{1}{|a|}\rfloor+1=N,$$ which implies that $$|a|>\frac{1}{N},$$which contradicts.
A: Your approach is correct. There are several ways to define $\Bbb R$ from $\Bbb Q.$ They all result in isomorphic structures... Up to isomorphism there is only one ordered field in which every non-empty subset with a lower bound has a greatest lower  bound, which is why we say "the" real number system, not "a" real number system.
Suppose we only have $\Bbb Q$ and we hope to have an ordered field $F$ with the $glb$ property.  Then there cannot exist $b\in F$ such that $0\ne b$ and $|b|\leq 1/n$ for every $n\in \Bbb N.$ 
Proof: Suppose not. (i).  Consider the set $S=\{|b|/n:n\in \Bbb N\}.$ Then $|b|^2$ is a positive lower bound for $S$ because for every $n\in \Bbb N$ we have $|b|^2=|b|\cdot |b|\leq |b|\cdot (1/n).$
(ii). But $S$ has no $glb.$ Because if $c>0$ and $c$ is a lower bound for $S,$ then $c<2c.$ And for every $n$ $\in \Bbb N$ we have $c\leq |b|/2n,$ implying $2c\leq |b|/n.$ So $2c$ is also a lower bound for $S$.
A: Υou are right . I cannot find any problem in your nice method,  but you forgot the absolute value of $b$.
One another approach would be considering $b$ as a sequence...then $lim{1/n}=0$ so by contraction lemma we should also have $limb=0$ which leads of course to $b=0$
