Answer
Claim: For all rational numbers r and s, if r < s, then there is another rational number between r and s.
Proof: Suppose that r and s are both particular but arbitrarily chosen rational numbers.
If r < s, then r + r < r + s = 2r < r + s = r < (r + s)/2 (by T19 of Appendix A).
If r < s, then r + s < s + s = r + s < 2s = (r + s)/2 < s (by T19 of Appendix A).
Since r < (r + s)/2 and (r + s)/2 < s, then r < (r + s)/2 < s (by the Transitive Law T18 of Appendix A).
Let m = (r + s) and n = 2, then (r + s)/2 = m/n.
By substitution, r < m/n < s.
By definition of a rational number, it follows that m/n is a rational that is between r and s.
Therefore, there is a rational number between r and s.
Work Step by Step
Method of Direct Proof
1. Express the statement to be proved in the form “∀x ∈ D, if P(x) then Q(x).”
(This step is often done mentally.)
2. Start the proof by supposing x is a particular but arbitrarily chosen element of D
for which the hypothesis P(x) is true. (This step is often abbreviated “Suppose
x ∈ D and P(x).”)
3. Show that the conclusion Q(x) is true by using definitions, previously established
results, and the rules for logical inference.
