1. Mathematical Induction

2. Motivation Mathematics uses 2 kinds of arguments:
deductive
inductive
P(n): … + n = n(n + 1)/2.
Observe that P(1), P(2), P(3), & P(4). Conjecture: nN, P(n).
Mathematical induction is a finite proof pattern for proving nN, P(n).

3. The Principle of Mathematical Induction
Let P(n) be a proposition function: n N, P(n) is either true or false.
To prove nN P(n), it suffices to prove:
P(1) is true.
n ( P(n)  P(n + 1) ).
We are applying modus ponens, ad infinitum.
It is a framework for constructing a finite proof that n P(n).

4. Proving P(3) Given P(1)  n  1 (P(n)  P(n + 1)). Proof:
1. P(1). [premise 1]
2. P(1)  P(2). [U.S. of premise 2]
3. P(2). [step 1, 2, & modus ponens]
4. P(2)  P(3). [[U.S. of premise 2]
5. P(3). [step 2, 3, & modus ponens]
Construct a finite proof for P(1,999,765).

5. Mathematical Induction as the Domino PrincipleIf
the 1st domino falls over
and
the nth domino falls over implies that the (n + 1)st domino falls over
then
domino n falls over for all n  N.

6. Mathematical Induction as the Domino Principle1 2 3 n
n +1

7. A rule of inference A tautological implication is a rule of inference.
The Principle of Mathematical Induction is a rule of inference:
( P(1)  n (P(n)  P(n + 1)) )  n P(n)

8. The 3-Step Method Typically, the proof of step 2 is direct.
Thus, the framework has 3 steps:
Prove P(1) [called the basis]
Assume P(n) [called the induction hypothesis]
Prove P(n + 1) [called the inductive step]
The last 2 steps are for arbitrary n  N.

9. Induction as a Creative Process
Mathematical induction is similar to, but not identical to, scientific induction.
In both cases, a “theory” is created.
Look at specific cases; perceive a pattern.
Hypothesizing a pattern, a theory, is a creative process (only people who are bad at mathematics say otherwise - sour grapes).
With mathematical induction, a “theory” can be proved.

10. Scientific theories cannot be proved. They can be disproved.
Mathematical models of scientific theories are amenable to mathematical proof.
Like axioms, the relationship between:
the mathematical model
the physical reality
cannot be proven correct.

11. Example
1 = 1
3 = 1 + 2
6 =
10 =
What is general formula, if any, for … + n?
F(n): n.
F(n + 1) = F(n) + (n + 1).

12. A Geometric Interpretation1: 2: 3:
Put these blocks, which represent numbers, together to form sums:
1 + 2 = =

13. n n
Area is n2/2 + n/2 = n(n + 1)/2

14. Mathematical Induction Proof
Assume F(n) = n(n + 1)/2
Show F(n + 1) = (n + 1)(n + 2)/2.
F(n + 1) = n + (n + 1)
= F(n) + n + 1
= n(n + 1)/2 + n + 1 [Induction hyp.]
= n(n + 1)/2 + (n + 1)(2/2)
= (n + 1)(n + 2)/2.

15. In proving the formula by mathematical induction, we focused on the:
similarities
differences
for successive values of n.
Very roughly speaking:
Isolate difference for successive values of n.
Find a pattern in this sequence of differences.

16. Example: 13 + 23 + . . . + n3 = ? Let F(n) = 13 + 23 + . . . + n3.
What is a formula for F(n)?
1 = 13
9 =
36 =
100 =
Do you see a pattern?

17. Try to prove that n F(n) = [n(n + 1)/2]2
2. Assume F(n) = [n(n + 1)/2]  I.H.
3. Prove F(n + 1) = [(n + 1)(n + 2)/2]2 .
F(n + 1) = n3 + (n+1)3  Defn. of F(n + 1)
= F(n) + (n+1)  Enable use of I.H.
= [n(n + 1)/2]2 + (n+1)  Use I. H.
= (n+1)2[(n/2)2 + (n+1)]
= (n+1)2[n 2 /4 + (4/4)(n+1)]
= (n+1)2[(n 2 + 4n + 4)/4] = [(n+1) (n+2)/2]2.

18. Translating the starting point
Suppose we know P(n) is false for 1  n  9 but think P(n) is true for n > 9.
Define Q(n) = P(n + 9).
Use mathematical induction to show that n N, Q(n).
Thus, we can start the induction at any integer, not just 1.

19. Example: Stamps
Suppose the US Post Office prints only 5 & 9 cent stamps.
Prove n > 34, you can make postage for n cents, using only 5 & 9 cent stamps.
S(n): you can make postage for n cents
1. Basis: For n = 35, use seven 5 cent stamps.
2. I.H.: S(n).

20. 3. Prove S(n + 1). Case 1: Zero 9-cent stamps are used for S(n):
only 5 cent stamps are used for S(n).
Then at least 7 such stamps were used.
Replace those with four 9 cent stamps.
Case 2: At least one 9 cent stamp is used for S(n).
Replace it with two 5 cent stamps.

21. Can we start from a negative number?
Can we use this technique for something that is true only for positive odd integers?
How general is this?
(Think of composing functions)

22. 1 = 12
1 + 3 = 22
= 32
What is the generalization?
What is the proposition function?

23. Proof by induction Basis: P(0): 2(0) + 1 = (0 +1)2
Let P(n) be … + (2n + 1) = (n + 1)2
Induction: Show P(n)  P(n + 1).
Assume … + (2n + 1) = (n + 1)2
Show … + (2n + 1) + (2(n+1) + 1) = (n + 2)2
… + (2n + 1) + (2(n+1) + 1) = (n + 1)2 + (2(n+1) + 1) = n2 + 2n n = (n + 2) 2

24. Generalizing the Basis
To prove nN, P(n), if suffices to show:
P(1)  P(2).
n ( ( P(n)  P(n + 1) )  P(n + 2) ).
That is, if:
We can push over the first 2 dominos AND
Pushing over any 2 adjacent dominos implies pushing over the domino following those.
then we can push over all the dominos.

25. The Fibonacci Formula Let F(0) = 0, F(1) = 1, F(n) = F(n-1) + F(n-2).
Prove
F(n) = 5-1/2 ([(1 + 51/2)/2]n - [(1 - 51/2)/2]n ).
Basis: F(0) = 5-1/2 ([(1 + 51/2)/2]0 - [(1 - 51/2)/2]0 )
= 0.
F(1) = 5-1/2 ([(1 + 51/2)/2]1 - [(1 - 51/2)/2]1 )
= (5-1/2/2) (1 + 51/ /2 )
= 1.

26. Assume Show F(n) = 5-1/2 ([(1 + 51/2)/2]n - [(1 - 51/2)/2]n ) and
1.Let a = (1 + 51/2)/2 and b = (1 - 51/2)/2.
2. Then, by the induction hypothesis,
F(n) = 5-1/2 (an - bn) and F(n+1) = 5-1/2 (an+1 - bn+1) .

27. 3. F(n+2) = F(n+1) + F(n), by definition.
4. F(n+2) = 5-1/2 (an+1 - bn+1) + 5-1/2 (an - bn)
= 5-1/2 (an + an+1 - bn - bn+1)
= 5-1/2 ( an(a+1) - bn (b+1) )
5. But, a + 1 = a2 and b + 1 = b2.
6. F(n+2) = 5-1/2 (an+2 - bn+2) .

28. Generalizing this ... If P(1)  P(2)  . . . P(k) and
n ( (P(n+1)  P(n+2)  P(n+k))  P(n+k+1) )
then, n P(n).

29. Strong Mathematical InductionIf
P(1)  P(2)  P(k) and
for n  k,
if P(i) is true, for 1  i  n then P(n+1)
then, n P(n).

30. Example: Fundamental Theorem of Arithmetic
Prove that all natural numbers  2 can be represented as a product of primes.
Basis: 2: 2 is a prime.
Assume that 1, 2, . . ., n can be represented as a product of primes.

31. Show that n+1can be represented as a product of primes.
Case n + 1 is a prime: It can be represented as a product of 1 prime, itself.
Case n + 1 is composite: n + 1 = ab, for some a,b < n + 1.
By induction hypothesis, a = p1p pk & b = q1q ql, where the pis & qis are primes.
Then, n + 1 = p1p pkq1q ql.