Combinatorics Through Guided Discovery, 2004a

B.1. The Principle of Mathematical Induction 147
to the assumption that there was an n that made the statement false. In fact, we
will see that we can bypass entirely the use of proof by contradiction. We used it
to help you discover the central ideas of the technique of proof by mathematical
induction.
The central core of mathematical induction is the proof that the truth of a
statement about the integer n for n = k − 1 implies the truth of the statement for
n = k. For example, once we know that a set of size 0 has 2 0 subsets, if we have
proved our implication, we can then conclude that a set of size 1 has 2 1 subsets,
from which we can conclude that a set of size 2 has 2 2 subsets, from which we can
conclude that a set of size 3 has 2 3 subsets, and so on up to a set of size n having 2 n
subsets for any nonnegative integer n we choose. In other words, although it was
the idea of proof by contradiction that led us to think about such an implication,
we can now do without the contradiction at all. What we need to prove a statement
about n by this method is a place to start, that is a value b of n for which we
know the statement to be true, and then a proof that the truth of our statement for
n = k − 1 implies the truth of the statement for n = k whenever k > b.
B.1.2
Mathematical induction
The principle of mathematical induction states that
In order to prove a statement about an integer n,ifwecan
1. Prove the statement when n = b, for some fixed integer b
2. Show that the truth of the statement for n = k − 1 implies the truth
of the statement for n = k whenever k > b,
then we can conclude the statement is true for all integers n ≥ b.
As an example, let us return to Problem 360. The statement we wish to prove is the
statement that “A set of size n has 2 n subsets.”
Our statement is true when n =0, because a set of size 0 is the empty set
and the empty set has 1=2 0 subsets. (This step of our proof is called a
base step.) Now suppose that k > 0 and every set with k − 1 elements
has 2 k−1 subsets. Suppose S = {a 1 , a 2 ,...a k } is a set with k elements.
We partition the subsets of S into two blocks. Block B 1 consists of the
subsets that do not contain a n and block B 2 consists of the subsets that
do contain a n . Each set in B 1 is a subset of {a 1 , a 2 ,...a k−1 }, and each
subset of {a 1 , a 2 ,...a k−1 } is in B 1 . Thus B 1 is the set of all subsets of
{a 1 , a 2 ,...a k−1 }. Therefore by our assumption in the first sentence of
this paragraph, the size of B 1 is 2 k−1 . Consider the function from B 2
to B 1 which takes a subset of S including a n and removes a n from it.
This function is defined on B 2 , because every set in B 2 contains a n . This
function is onto, because if T is a set in B 1 , then T ∪{a k } is a set in B 2
which the function sends to T. This function is one-to-one because if V
and W are two different sets in B 2 , then removing a k from them gives
two different sets in B 1 . Thus we have a bijection between B 1 and B 2 ,
so B 1 and B 2 have the same size. Therefore by the sum principle the
size of B 1 ∪ B 2 is 2 k−1 +2 k−1 =2 k . Therefore S has 2 k subsets. This