Download Class #2 - Mathematical Induction and Binomial Theorem

Class #2 - Mathematical Induction and Binomial
Theorem
1. To prove a statement that is true for each positive integer n, the proof technique,
Mathematical Induction, is devised.
2. Example: Show
Pn
i=1 (2i
− 1) = n2 for every positive integer n.
3. Mathematical Induction
(a) What is the S(n)?
(b) Use properties of Natural numbers:
A list that starts with 1 and (*) holds:
((*) given any number in the list (call it n), its successor, n + 1, is also in the list),
then the list includes all natural numbers.
4. Example: Define a functon recursively for all positive integers n by f (1) = 1, f (2) = 5,
and for n ≥ 2, f (n + 1) = f (n) + 2f (n − 1). Show that f (n) = 2n + (−1)n .
What is the starting number?
What is the difference between this example and the previous one?
5. Mathematical Induction (2nd principle)
S(1) is true, and (**) holds:
((**)) given S(n) is true and S(k) are true for all k = 1, 2, · · · , n − 1, then S(n + 1) is
also true)
then S(n) is true for all natural numbers n.
6. Justification of the 2nd principle:
is in the list.
Let n = 1. By (**),
Let n = 2. Then
is in the list.
and
are in the set. So by (**),
the set. etc.
1
is in
7. Classwork The Fibonacci sequence F1 , F2 , F3 , . . . is defined by F1 = 1, F2 = 1,
Fn = Fn−1 + Fn−2 for n ≥ 3. Prove that for every natural number n,
F1 + · · · + Fn = Fn+2 − 1.
8. Binomial Theorem: (a + b)n = (a + b)(a + b) · · · (a + b)
Out of
possible a, can choose k many a for k =
the term ak bn−k with
. This leads to
many ways.
(a + b)n =
.
9. Def: An integer
is called a prime number, or simply a prime, if its only
factors are
. An integer
that is not a prime is called
.
10. Divisibility: 9|36
• 9 divides 36.
• 9 is a
of 36.
• 9 is a
of 36.
• 36 is a multiple of 9.
• 36 is
by 9.
Def: Let a, d ∈ Z. We say that d divides a if there exists q ∈ Z such that
Notation: d|a.
If d does not divide a, then we write d6 | a.
11. True or False: Mark your answer by True or False. Justify your answer:
For all a, b, c ∈ Z, ac|bc if and only if a|b. (a and c both nonzero.)
12. Classwork True or False: Mark your answer by True or False. Justify your answer:
Let a, b, c, d ∈ Z. If a|b and c|d, then ac|bd. (a and c both nonzero.)
2
.