Proof of continuity of Thomae Function at irrationals.

In Thomae’s Function:

t(x)={0if x is irrational1nif x=mn where gcd

I can prove the discontinuity at rational b by taking a sequence of irrationals x_n which converge to b.

But while going through an argument for continuity at irrationals.
I found this in a book.

On the other hand if b is an irrational number and \epsilon > 0
then there is a natural number n_0 such that 1/n_0 < \epsilon.
There are only finite number of rationals with denominator less than
n_0 in the interval (b-1,b+1).
Hence we can find a \delta > 0
such that \delta neighbourhood of b contains no rational with
denominator less than n_0.

I understand the rest of the proof. But I am unable to prove the emphasized text. Although I find it intuitive.

Answer

Let m=n_0-1, so we want to consider rationals with denominators 1,\cdots,m in the interval (b-1,b+1). Since consecutive rationals with denominator k differ by 1/k and the interval (b-1,b+1) has length 2, there are at most 2k rationals with denominator k in (b-1,b+1).

Therefore there are at most 2\cdot1+2\cdot2+\cdots+2m rationals in (b-1,b+1) with denominator less than n_0, so we can choose a \delta with 0<\delta<|b-r|, where r is the rational with denominator less than n_0 in (b-1,b+1) which is closest to b.

Attribution
Source : Link , Question Author : Alvis , Answer Author : user84413

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