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 $$x < p/q < y$$. 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 $$q(y - x) > 1$$, i.e., prove that such a $$q$$ exists; then
2. figure out why that means that you can choose $$p$$ so that $$0 < p - qx < q(y - x)$$, 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 $$q > 1/(y - x)$$.

Now you need $$p$$ to satisfy $$0 < p - qx < q(y - x)$$. Rewrite this as $$qx < p < qy$$. 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!)