32 2. Background
The format of the proof above is typical of inductive proofs of
summation formulas: use the inductive hypothesis to simplify a por-
tion of the next value’s sum.
For the next example we need to know a convex polygon is one
where all the corners point out. If you connect two corners of a convex
polygon with a straight line segment, the segment will lie entirely
within the polygon, cutting it into two smaller convex polygons.
As you get more comfortable with induction, you can write it in a
more natural way, without segmenting off the base case and inductive
step portions of the argument. We’ll do that here. Notice the base
case is not 0 or 1 for this proof.
Example 2.5.3. For n 2, the sum of angle measures of the interior
angles of a convex polygon of n vertices is (n 2) · 180◦.
Proof. We work by induction. For n = 3, the polygon in question is
a triangle, and it has interior angles which sum to
180◦
=
(3−2)·180◦.
Assume the theorem holds for some n 3 and consider a convex
polygon with n + 1 vertices. Let one of the vertices be named x, and
pick a vertex y such that along the perimeter from x in one direction
there is a single vertex between x and y, and in the opposite direction,
(n +1) 3 = n 2 vertices. Join x and y by a new edge, dividing the
original polygon into two polygons. The new polygons’ interior angles
together sum to the sum of the original polygon’s interior angles. One
of the new polygons has 3 vertices and the other has n vertices (x, y,
and the n 2 vertices between them). The triangle has interior angle
sum
180◦,
and by the inductive hypothesis the n-gon has interior
angle sum (n 2) ·
180◦.
The n + 1-gon therefore has interior angle
sum
180◦
+ (n
2)180◦
= (n + 1 2) ·
180◦,
as desired.
Notice also in this example that we used the base case as part of
the inductive step, since one of the two polygons was a triangle. This
is not uncommon.
Exercise 2.5.4. Prove the following statements by induction.
(i) For every positive integer n,
1 + 4 + 7 + . . . + (3n 2) =
1
2
n(3n 1).
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