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 thanHence we can find a \delta > 0

n_0 in the interval (b-1,b+1).

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*