Is this proof correct? Show $\mathbb{Q}$ is dense in $\mathbb{R}$

Question

I like proof by contradictions in showing that $\mathbb{Q}$ is dense in $\mathbb{R}$. But I can't understand this one>

https://math.dartmouth.edu/archive/m54x12/public_html/m54densitynote.pdf

Suppose we fix $b = 2, a = 1$

Then $b - a = 2 - 1 = 1$

Let $N > 1/(b-a)$, where $N \in \mathbb{N}$, then $N = 2$ Then $A = \{1/2, 2/2, 3/2, 4/2, \ldots\}$

I don't understand the statements after the "Assume Otherwise".

Take $m_1$ the greatest integer such that $m_1/N < a$. Since $a = 1, N = 2$, then $m_1 = 1$

Then the proof states "$\frac{m_1+1}{n} > b$", but clearly $(m_1 + 1) / 2 = 2/2 = 1 < b$. So this "$\frac{m_1+1}{n} > b$" line is not verified!

Is there a problem in the proof or is there a problem in my example?

Answer 1

The problem is, the proof is proceeding by contradiction. That is why it says "Then ${m_1+1\over N}>b$" -- because it's assumed there's no number of the form $m/N$ between $a$ and $b$. It then proves that this leads to a contradiction, just as you proved that in fact $(m_1+1)/N$ in your situation was actually $<b$ and not $>b$ as assumed. The contradiction is what establishes the theorem.

This is a case where perhaps a proof by contradiction was not necessary.