Page 370 - HOW TO PROVE IT: A Structured Approach, Second Edition

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356 Appendix 1: Solutions to Selected Exercises
We now know that T is a partial order. To see that it is total, suppose
x ∈ A and y ∈ A.If x = a then (x, y) ∈{a}× A ⊆ T . Similarly, if
y = a then (y, x) ∈ T . Now suppose x = a and y = a. Then x ∈ A
and y ∈ A , so since T is a total order, either (x, y) ∈ T ⊆ T or

(y, x) ∈ T ⊆ T .

Finally, to see that R ⊆ T , suppose that (x, y) ∈ R.If x = a then
(x, y) ∈{a}× A ⊆ T . Now suppose x = a.If y = a then the fact that
(x, y) ∈ R would contradict the R-minimality of a. Therefore y = a.

But then (x, y) ∈ R ∩ (A × A ) = R ⊆ T ⊆ T .

4. (a) We will prove the statement: For every n ≥ 1, for every B ⊆ A,if B has
n elements then there is some x ∈ B such that ∀y ∈ B((x, y) ∈ R ◦ R).
We proceed by induction on n.
Base case: Suppose n = 1. If B ⊆ A and B has one element, then for
some x ∈ B, B ={x}. Since R is reflexive, (x, x) ∈ R, and therefore
(x, x) ∈ R ◦ R. But x is the only element in B,so ∀y ∈ B((x, y) ∈
R ◦ R), as required.
Induction step: Suppose that n ≥ 1 and for every B ⊆ A,if B has
n elements then ∃x ∈ B∀y ∈ B((x, y) ∈ R ◦ R). Now suppose that
B ⊆ A and B has n + 1 elements. Choose some b ∈ B, and let

B = B \{b}. Then B ⊆ A and B has n elements, so by inductive
hypothesis there is some x ∈ B such that ∀y ∈ B ((x, y) ∈ R ◦ R).

We now consider two cases.
Case 1: (x, b) ∈ R ◦ R. Then ∀y ∈ B((x, y) ∈ R ◦ R), so we are
done.
Case 2: (x, b) /∈ R ◦ R. In this case, we will prove that ∀y ∈
B((b, y) ∈ R ◦ R). To do this, let y ∈ B be arbitrary. If y = b, then
since R is reflexive, (b, b) ∈ R, and therefore (b, y) = (b, b) ∈ R ◦ R.
Now suppose y = b. Then y ∈ B , so by the choice of x we know

that (x, y) ∈ R ◦ R. This means that for some z ∈ A, (x, z) ∈ R and
(z, y) ∈ R.Wehave(x, z) ∈ R,soif(z, b) ∈ R then (x, b) ∈ R ◦ R,
contrary to the assumption for this case. Therefore (z, b) /∈ R,sobythe
hypothesis on R, (b, z) ∈ R. But then since (b, z) ∈ R and (z, y) ∈ R,
we have (b, y) ∈ R ◦ R, as required.
(b) Hint: Let A = B = the set of contestants and let R ={(x, y) ∈ A ×
A | x beats y}∪ i A . Now apply part (a).
8. (a) Let m = (a + b)/2, the arithmetic mean of a and b, and let d =
(a − b)/2. Then it is easy to check that m + d = a and m − d = b,so

√

√ a + b
2 2 2
ab = (m + d)(m − d) = m − d ≤ m = m = .
2