2

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 ${a=x}$.

amWhy
  • 209,954
Isaac Kleinman
  • 870

3 Answers3

4

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).

amWhy
  • 209,954
  • Given the context I provided, I don't see how, by definition, the rationals satisfy the archimedean property. – Isaac Kleinman Jun 07 '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 07 '13 at 18:57
  • @amWhy: You are an answering machine! :-) +1 – Amzoti Jun 08 '13 at 01:09
  • @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 09 '13 at 19:35
  • That wasn't the question you posted! Sorry if I misunderstood. Perhaps post a separate question to that effect, keeping this thread, but starting a new post. I'd like to see you elaborate on what you are hoping to establish! ;-) – amWhy Jun 09 '13 at 19:40
  • I thought that was the meaning of my question. :( My question was based on an exercise in Apostol's book. Now I'm not certain my understanding of the terminology is correct.

    I did not understand "is satisfied by the rational numbers" to mean "is true of the rational numbers"; I understood it to mean "no reals beyond the rationals are needed for the property hold".

    Please correct me if I'm wrong.

     – Isaac Kleinman Jun 09 '13 at 19:43
  • Actually, it should be understood as follows: "is satisfied by the rational numbers" should be taken to mean "is true for all rational numbers." If the question were "show that the property is satisfied only by the rationals", then you're take would be correct. Feel free to accept the answer ;-). We can discuss this more in comments, or in chat, or you can post a different question, if you need to. – amWhy Jun 09 '13 at 19:49
  • I'd love to chat whenever you're up to it: http://chat.stackexchange.com/rooms/9178/satisfiability – Isaac Kleinman Jun 10 '13 at 00:52
1

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.

Hagen von Eitzen
  • 374,180
0

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 $n=p+1$.

Adriano
  • 41,576
  • 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 09 '13 at 19:31