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78 The Nuts and Bolts of Proof, Third Edition
Proof: Use mathematical induction on the number of sets.
Let Ai and A2 be two convex sets. If y4i fi ^2 is either empty or
contains one element, then it is convex.
Let us assume that v4i n ^2 has at least two distinct elements, x and y.
Then x and y are elements of both Ai and A2. Because Ai and A2 are
convex, the Une segment joining x and y is contained in both sets Ai
and A2. Therefore, it is contained in their intersection, Ai fi ^2-
Assume that if ^1, ^2, • • •, ^n are convex sets, then ^1 n A2 H ... fl ^„
is a convex set.
Prove that if Ai,A2,..., An, An+i are convex sets, then
^1 n y42 n ... n ^„ n An+i is a convex set.
We can use the associative property of intersection (See Exercise 4)
to write:
^1 n ^2 n... n ^„ n An+i = (AinA2n...nAn)n An+i.
The set ^1 n ^2 n ... n v4„ is convex by inductive hypothesis.
So ^1 n ^2 n ... n yl„ n An+i is convex because it is the intersection
of two convex sets.
EQUALITY OF NUMBERS
There are at least three common ways to prove that two numbers, call
them a and b, are equal. We can do so by showing that:
1. The two inequahties a<b and b<a hold.
2. The equahty a-b = 0 holds.
3. The equality a/b = 1 holds (in this case we need to be sure that b ^ 0).
It is preferable to use the second and third ways when we can set up
algebraic expressions involving a and b. The first way is more useful when we
have to compare numbers through the examination of their definitions and
properties.
EXAMPLE 1. If a and b are two positive integers, then their least common
multiple is equal to the quotient between ab and the greatest common
divisor of a and b; that is,
Icm (a, b)
GCD(a,b)'
Proof: Let d = GCD{a, b) and L = lcm(a, b). We want to prove that:
ab