WTW 285: Discrete Structures
The principle of strong mathematical induction
In WTW 115 you learned (in Epp: Sections 5.2 and 5.3) a principle of mathematical
induction that is sometimes called the principle of weak mathematical induction or the
first principle of mathematical induction. In WTW 285 you will also use a principle of
mathematical induction that is sometimes called the principle of strong mathematical
induction or the second principle of mathematical induction (see Epp: Section 5.4 for
a detailed explanation and examples). The principle of strong mathematical induction
can be summarised as follows:
Principle of strong mathematical induction
Consider a statement P that is formulated in terms of integers. Let a and b be
integers with a 6 b. Suppose that:
1. the statement P is true for all integers n with a 6 n 6 b (basis step), and
2. for each integer k with k > b, if the statement P is true for all integers n with
a 6 n 6 k (induction hypothesis) then P is also true for the integer k+1 (induction
step).
Then the statement P is true for every integer n with n > a (conclusion).
The principle of strong mathematical induction can be used to prove properties about
recurrence relations that have more than one initial value. This will be demonstrated
with the following example.
Example Use the principle of strong mathematical induction to prove that the function
fn = 5n − 1 (with n > 0) solves the recurrence relation
sn = 6sn−1 − 5sn−2 , s0 = 0, s1 = 4 .
Solution It must be proved that fn = sn for each integer n with n > 0.
Explanation of the structure of the proof:
• The basis step of the induction proof will involve showing that fn = sn for each of
the initial values of sn , i.e. that f0 = s0 and f1 = s1 .
• For the induction hypothesis, an integer k with k > 1 will be fixed, and it will be
assumed that fn = sn for all integers n in the range 0 6 n 6 k. The reason for
fixing k to be at least 1, is that the claim being proved (namely that fn = sn when
n > 0) has already been confirmed for n = 0, 1 in the basis step. As an example,
had it instead been shown in the basis step that f0 = s0 , f1 = s1 and f2 = s2 , then
k would have been fixed to be at least 2.
• For the induction step, the induction hypothesis will be used to show that fk+1 =
sk+1 .
• The proof will end by invoking the principle of strong mathematical induction to
conclude that fn = sn when n > 0.
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Basis step: f0 = 50 − 1 = 0 = s0 and f1 = 51 − 1 = 4 = s1 hence fn = sn when
n = 0 and n = 1.
Induction hypothesis: Let k be a fixed but arbitrary integer with k > 1 and
assume that fn = sn (hence sn = 5n − 1) for each integer n with 0 6 n 6 k.
Induction step:
sk+1 =
=
=
=
=
=
6sk − 5sk−1 (definition of sn )
6(5k − 1) − 5(5k−1 − 1) (induction hypothesis)
6 · 5k − 6 − 5k + 5
5 · 5k − 1
5k+1 − 1
fk+1 (definition of fn )
hence fn = sn when n = k + 1.
Conclusion: By the principle of strong mathematical induction, fn = sn for each
integer n with n > 0.
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