Prove: If a and b are rational numbers such that a<b, then there is a rational number c such that a<c<b.
HINT:
Proof from problem below.
Prove: If a and b are real numbers such that a<b, then a<(a+b)/2 and (a+b)/2<b.
Proof
Suppose that a and b are frequently but arbitrarily chosen real numbers such that a<b. Show that a<(a+b)/2 and (a+b)/2<b. So there exists a real positive number k such that b=a+k. Thus (a+b)/2=(a+a+k)/2=(2a+k)/2=a+k/2, which is greater than a. Since b=a+k, then a+k/2 is less than b.