Page 284 - HOW TO PROVE IT: A Structured Approach, Second Edition

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270 Mathematical Induction
Base case: n = 0. Suppose R is a partial order on A and A has 0 elements.
Then clearly A = R = ∅. It is easy to check that∅ is a total order on A,so
we are done.
Induction step: Let n be an arbitrary natural number, and suppose that
every partial order on a set with n elements can be extended to a total or-
der. Now suppose that A has n + 1 elements and R is a partial order on A.By
the theorem in the last example, there must be some a ∈ A such that a is an

R-minimal element of A. Let A = A \{a} and let R = R ∩ (A × A ). You

are asked to show in exercise 1 that R is a partial order on A . By inductive

hypothesis, we can let T be a total order on A such that R ⊆ T . Now let

T = T ∪ ({a}× A). You are also asked to show in exercise 1 that T is a total
order on A and R ⊆ T , as required.
The theorem in the last example can be extended to apply to partial orders
on infinite sets. For a step in this direction, see exercise 17 in Section 7.1.
Example 6.2.3. Prove that for all n ≥ 3, if n distinct points on a circle are
connected in consecutive order with straight lines, then the interior angles of
the resulting polygon add up to (n − 2)180 .
◦
Solution
Figure 1 shows an example with n = 4. We won’t give the scratch work
separately for this proof.
Theorem. For all n ≥ 3, if n distinct points on a circle are connected in consec-
utive order with straight lines, then the interior angles of the resulting polygon
◦
add up to (n − 2)180 .
Proof. We use induction on n.
Base case: Suppose n = 3. Then the polygon is a triangle, and it is well
◦
known that the interior angles of a triangle add up to 180 .
Induction step: Let n be an arbitrary natural number, n ≥ 3, and assume the
statement is true for n. Now consider the polygon P formed by connecting
some n + 1 distinct points A 1 , A 2 ,..., A n+1 on a circle. If we skip the last
point A n+1 , then we get a polygon P with only n vertices, and by inductive

◦
hypothesis the interior angles of this polygon add up to (n − 2)180 . But now
as you can see in Figure 2, the sum of the interior angles of P is equal to the sum

of the interior angles of P plus the sum of the interior angles of the triangle
A 1 A n A n+1 . Since the sum of the interior angles of the triangle is 180 , we can
◦
conclude that the sum of the interior angles of P is
◦
◦
◦
(n − 2)180 + 180 = ((n + 1) − 2)180 ,
as required.