# Simple Rudin

1. Jan 29, 2009

### completenoob

Hello,

I have decided to study analysis on my own and am starting with principles of mathematical analysis by rudin.
I am having trouble understanding pg. 9 on the density of Q in R, part b.

It states:
If $x \in R, y \in R$ and $x<y$ the there exists a $p \in Q$ such that $x < p < y$
Proof:
Since $x<y$, we have $y-x>0$ and the Archemedian Property furnishes a positive integer n such that:
$n(y-x)>1$
Applying the AP again, to obtain positive integers m1 & m2 such that $m1>nx$, $m2>-nx$
Then: $-m2<nx<m1$
Hence there is an integer m such that $m-1 \le nx<m$........

Can someone explain to me the last line? Where does this less then or equal come from?

2. Jan 30, 2009

### completenoob

3. Jan 30, 2009

### JinM

4. Jan 30, 2009

### completenoob

Thank you JinM, that helped a lot. However, on your last post in that thread, It does not make sense to me that there exists a m-1 when m is the minimum. Wouldn't it make sense to say that m-1 is the minimum?

5. Jan 31, 2009

### completenoob

MMmm....For being my first analysis book Principles of Mathematical Analysis by Rudin might be a little too hard. Should I study Apostol's Principles of Mathematical Analysis first? Then go back to Rudin's text?

6. Jan 31, 2009

### uman

Probably depends how much you already know.

7. Jan 31, 2009

### completenoob

Calc 1-3. No analysis experience.

8. Jan 31, 2009

### completenoob

Yah i need a text for self study. I have no experience with proofs.

9. Jan 31, 2009