I have a question about the proof for Theorem 1.20b in Rudin's Principles of Mathematical Analysis.
Theorem 1.20 states:
(a) If $x\in \mathbb{R}$, $y\in \mathbb{R}$, and $x>0$, then there is a positive integer $n$ such that $nx>y$.
(b) If $x\in \mathbb{R}$, $y\in \mathbb{R}$, and $x<y$, then there exists a $p \in \mathbb{Q}$ such that $x<p<y$.
The proof of (b) in the textbook is given below:
Since $x<y$, we have $y-x > 0$, and (a) furnishes a positive integer $n$ such that $n(y-x)>1$. Apply (a) again, to obtain positive integers $m_1$ and $m_2$ such that $m_1 > nx$ and $m_2 > -nx$. Then
$-m_2 < nx < m_1$.
Hence there is an integer $m$ (with $-m_2 \leq m \leq m_1$) such that
$m-1 \leq nx < m$.
If we combine these inequalities, we obtain
$nx < m \leq 1+nx < ny$.
Since $n>0$, it follows that $x< \frac{m}{n} < y$. This proves (b), with $p=\frac{m}{n}$.
I was wondering if this much shorter approach to proving (b) would also be correct:
Since $x<y$, we have $y-x > 0$, and (a) furnishes a positive integer $n$ such that $n(y-x)>1$. Rearranging, we obtain $y> \frac{1+nx}{n}$. Also, notice that $x = \frac{nx}{n} < \frac{1+nx}{n}$. So we choose $p=\frac{1+nx}{n}$, hence proving (b).
(Forgive me if I jumped to any illogical assumptions in my proof)
