Why is all the rational numbers in the interval $(a-\delta,a+\delta)$ have denominator greater than $N$?

Question

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.

Answer 1

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.

Answer 2

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 $N$.

Answer 3

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 $x \in (a-\delta, a+\delta)$.