Q1.
Try to explain why the average of two rational numbers a and b, which equals (a + b)/2, is always a rational number between a and b.
Answer
Two things have to be shown: that the average is rational, and that it lies between a and b.
1. It is rational. ℚ is closed under addition, and closed under division except by zero.
a = p/q and b = r/s with q, s ≠ 0
a + b = (ps + qr)/(qs)
(a + b)/2 = (ps + qr)/(2qs)
a + b = (ps + qr)/(qs)
(a + b)/2 = (ps + qr)/(2qs)
The numerator ps + qr and the denominator 2qs are integers, and 2qs ≠ 0. So (a + b)/2 has the form of a rational number.
2. It lies between. Suppose a < b. Then
(a + b)/2 − a = (a + b − 2a)/2 = (b − a)/2 > 0
b − (a + b)/2 = (2b − a − b)/2 = (b − a)/2 > 0
b − (a + b)/2 = (2b − a − b)/2 = (b − a)/2 > 0
Both gaps are positive and, in fact, equal — so a < (a + b)/2 < b, with the average exactly halfway.
Why it happens: this single fact is what makes ℚ dense. The average of a and b can be averaged with a again, and again, without end — so between any two rational numbers, however close, there are infinitely many more. Between 1 and 3/2 the chapter finds 5/4; between 1 and 5/4 you would find 9/8, then 17/16, and so on for ever.
Tip: the argument needs a ≠ b. If a = b the “average” is a itself and there is nothing strictly between them.