Between any two distinct real numbers there is a rational number

[jrsweet]

Homework Statement

Let x and y be real numbers with x<y and write an inequality involving a rational
number p/q capturing what we need to prove. Multiply everything in your inequality by q,
then explain why this means you want q to be large enough so that q(y-x)>1 . Explain
how you can rewrite this inequality and use the Archimedean property to find such a q.

The Attempt at a Solution

So, this is a question on a worksheet our teacher gave us to go along with the theorem in the book. Here is what I did so far:

x < p/q < y
Then multiply both sides by q as the question states:
qx < p < yq
0 < p-qx < yq-qx
0 < p-qx < q(y-x)

I am having trouble seeing why q(y-x)>1. It is obviously great than zero as the inequality states, but can someone help me see why it has to be >1??

 

[ystael]

What you are trying to do is exhibit a rational number p/q between x and y, that is, to choose p and q so that x &lt; p/q &lt; y. The argument you're being led down is a strategy for choosing q, then p, so that they satisfy the conditions you want.

So, what you need to do is:
1. prove that you can choose q so that q(y - x) &gt; 1, i.e., prove that such a q exists; then
2. figure out why that means that you can choose p so that 0 &lt; p - qx &lt; q(y - x), i.e., give a proof that if you have q satisfying part 1 then such a p exists.

 

[jrsweet]

So, we chose q>1/(y-x)?

So then,
0 < p-(x/(y-x)) < 1
0 < p < (y/(y-x))

I'm still lost on how we choose p then.

 

[ystael]

You need to give one sentence (really, one phrase) of proof for why one can choose q &gt; 1/(y - x).

Now you need p to satisfy 0 &lt; p - qx &lt; q(y - x). Rewrite this as qx &lt; p &lt; qy. You need to find an integer p satisfying this inequality. Why does your choice of q guarantee that there is at least one such p? (This is the only condition you need p to satisfy!)