Prove Satisfiability of Property by Set What is a proof strategy for proving that some property is satisfied by a particular set of numbers. 
For example, what would be an approach for proving that the archimedean property is satisfied by the rational numbers?
In the context I'm coming from (Apostol's Calculus book), the archimedean  property is presented as follows: 

For all real ${a,x,y}$, if ${a \le x < a+y/n}$ for all positive
  integers $n$, then ${a=x}$.

 A: You need only provide a direct proof, showing that the rationals satisfy, by definition, the Archimedean Property.  
Many, many proofs you'll encounter "fall-out" directly from definitions: what does it mean for $x, y$ to be rational numbers? (Definition). What does it mean to say that a particular set of numbers satisfies the Archimedean Property? (Definition). 
A: In this case a very direct strategy by just wrinting down what Archimedean means and what a rational number is should unavoidably lead to the goal.
A: We will instead prove its contrapositive:

For all rational numbers $a,x,y$, if $a\ne x$, then $x<a$ or there exists a positive integer $n$ such that $x \ge a+y/n$.

Choose any $a,x,y\in \mathbb{Q}$ such that $a \ne x$. Now consider $x$. Observe that if $x<a$, then we are done. Thus since $a \ne x$, we may assume that $x>a$. Hence, since $x-a>0$, notice that:
$$ x \ge a+y/n \iff x-a \ge y/n \iff n(x-a) \ge y \iff n \ge \dfrac{y}{x-a}$$
Since $a,x,y\in \mathbb{Q}$ and since $\mathbb{Q}$ is closed under subtraction and (nonzero) division, we know that $\dfrac{y}{x-a} \in \mathbb{Q}$. Hence, by the definition of a rational number, it suffices to prove that for any integers $p,q$ (where $q > 0$), there exists a positive integer $n$ such that $n \ge \dfrac{p}{q}$. There are two cases to consider.
Case 1: If $p<0$, then $\dfrac{p}{q} < \dfrac{0}{q} =0\le 1$, so we can choose the positive integer $n=1$.
Case 2: Otherwise, if $p \ge 0$, then $\dfrac{p}{q} \le \dfrac{p}{1}=p < p+1$, so we can choose the positive integer $n=p+1$.
