Prove that the rational numbers are dense in the reals. I.e., if with , then there exists an such that . It follows that there are then infinitely many such.

*Proof.*Since , we know . Therefore, there exists an such that

We also know (I.3.12, Exercise #4) that there exists such that . Putting these together we have,

Since we have . Hence, letting we have found such that

This then guarantees infinitely many such rationals since we can just replace by (and note that ) and apply the theorem again to find such that . Repeating this process we obtain infinitely many such rationals

I know that the proof establishes that n is an element of the positive integers, but I think it is insufficient to say that (m+1)/n is a rational number because m and n are elements of just the integers.

I think it’s important that n is explicitly an element of the positive integers because then n is non-zero, otherwise, without the context of the full proof, we have that if m is an integer and n = 0, (m+1)/n is rational.

Theorem I.30. tells that n is a postive integer i.e., n>0. so by trichotomy law n is not equal to 0. Hence, division by zero error will not occur.