Why is all the rational numbers in the interval $(a-\delta,a+\delta)$ have denominator greater than $N$? 
Guys, Why is this weird statement true? It seems counterintuitive to me I cannot understand or lack creativity understanding it can you help me explain it? Guys please if possible make it visual.
 A: It all comes from the three lines before the part you highlighted.
You might be thinking that no matter how small we make $\delta,$ there is always a rational number in the interval $(a - \delta, a + \delta).$
If so, you would be right.
In fact there are infinitely many rational numbers there.
But they might all have very large denominators.
Consider the closest integer to $a.$ There is only one.
(Really only one; since  $a$ is irrational, it can't be exactly halfway between two integers.)
Now consider the closest rational number with denominator $2.$
Next, the the closest rational number with denominator $3.$
Then $4.$ Then $5.$
Keep on going like that until you have the closest rational number with denominator $N.$
Now you have $N$ rational numbers and $N$ distances from each of those numbers to $a.$
Pick the number that is closest to $a$. Make $\delta$ less than the distance between that number and $a.$
Now which of those $N$ rational numbers can be in $(a - \delta, a + \delta)$?
None of them: we chose an interval that all of them are outside of.
What about all the other rational numbers with denominators $1,$ $2,$ $3,\ldots, N$?
Also outside $(a - \delta, a + \delta)$, because we made sure we had already looked at the closest rational number with each denominator. All the others are further.
This is what the passage is saying, although I gave a lot of unnecessary detail.
We know that only finitely many rationals with denominators $1,$ $2,$ $3,\ldots, N$
can exist in the interval $[0,1]$ where we found $a.$
One of those numbers has to be the closest one to $a.$ We just need to make
$a - \delta$ and $a + \delta$ closer.
A: This is immediate from the definition of $\delta$.  If $x$ is a rational number with denominator at most $N$, that means $x\in Q_N$, so by definition $|a-x|\geq\delta$ and thus $x\not\in (a-\delta,a+\delta)$.  So, any rational number which is in $(a-\delta,a+\delta)$ must have denominator greater than $N$.
A: If there is a rational number $x$ in this interval with  $q \leq N$ then $x \in Q_N$ and the  definition of $\delta$ shows that $\delta \leq |x-a|$ This contradicts the fatc that $x \in (a-\delta, a+\delta)$. 
