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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 ${a=x}$.

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3 Answers

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).

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Given the context I provided, I don't see how, by definition, the rationals satisfy the archimedean property. –  Isaac Kleinman Jun 7 '13 at 18:54
    
I mean use how we define the rationals, and then given the definition of the Archimedean Property, your proof is simply a verification that the rationals satisfy the property: write out what being Archimedean means & what being rational means & verify the latter satisfies the former. –  amWhy Jun 7 '13 at 18:57
    
@amWhy: You are an answering machine! :-) +1 –  Amzoti Jun 8 '13 at 1:09
    
@amWhy: :-)..... –  B. S. Jun 9 '13 at 4:42
    
@amWhy: I'm trying to prove that there is no implication from the archimedean property that there are real numbers which are not rational, are we on the same page? –  Isaac Kleinman Jun 9 '13 at 19:35
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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.

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Given the context I provided, I don't see how, by definition, the rationals satisfy the archimedean property. –  Isaac Kleinman Jun 7 '13 at 19:18
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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 $n=p+1$.

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You are proving "For all rational a,x,y, if a≤x<a+y/n for all positive integers n, then a=x." I'm trying to show that there is no implication from the archimedean property that there are real numbers which are not rational. –  Isaac Kleinman Jun 9 '13 at 19:31
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