Prove that $\sup \{\frac{a}{b}; a \in \mathbb{N}, b \in \mathbb{N}, a < b\}=1$ Consider a subset of rational numbers $S = \{\frac{a}{b}; a \in \mathbb{N}, b \in \mathbb{N}, a < b\}$. I want to  prove that $\sup S = 1$. By the definition of supremum, for $\epsilon > 0$, it suffices to show that there exists $\frac{a}{b} \in S$ such that $1 - \epsilon < \frac{a}{b}$. 
I tried to prove it using archmedian property ($\forall x \in \mathbb{R}, \exists n \in \mathbb{N}$ such that $n \geq x$) by setting $a = 1$ or $b$ being a multiple of $a$ for the purpose of deriving the value of remaining variable by fixing one variable. However, none of them worked. I feel like I ran out of trick. How should one prove it?

Let $s=sup(S)$
Suppose $s<1$
Between two real numbers $x<y$ there's always $q\in\mathbb{Q}$ (i.e. $q \in (x, y)$)
Let $q=\frac{q_1}{q_2}$ be a rational number in $(max(s, 0), 1)$
By definition $s<q$
then choose $a=q_1$ and $b=q_2$ $\Rightarrow s<q=\frac{a}{b}$
But $q\in S$.
So $s \geq 1$
But $S$ is bounded by $1$, then $s = 1$
