1. Agenda Proofs (Konsep Pembuktian) Direct Proofs & Counterexamples
Indirect Proofs
Proof by Contraposition
Proof by Contradiction
Exhaustive & Resolution Proof
Mathematical Induction
Latihan Soal

2. Proof Pembuktian tentang kebenaran suatu mathematical statement.
Formal proof: - utilize rule of inference
- all steps were supplied
- long and hard to follow
- suitable for computer
Informal proof: - steps maybe skip
- utilize assumption
- suitable for human

3. Proof Example: Suppose you did commit the crime.
Then at the time of the crime, you would have had to be at the scene of the crime.
In fact, you were in a meeting with 10 people at that time, as they will testify.
This contradicts the assumption that you committed the crime.
Hence the assumption is false.

4. Mathematical Systems Axiom: statement we assume to be true.
Definition: statement that are used to create new concepts in terms of existing ones
Theorem: statement that can be shown to be true.
Lemma: theorem that is usually not too interesting in its own right, but useful in proving another theorem
Corollary: a less important theorem (follows easily from another theorem.
Use proof to demonstrate that a theorem is true.

5. Methods of Proving Theorem
p  q
Proof:
Demonstrate that q is true if p is true
Methods:
Direct Proofs
Indirect Proofs:
Proof by Contraposition
Proof by Contradiction
Exhaustive Proof

6. Direct Proof & Counterexample

7. Direct Proof p  q Direct Proof: Assume that p is true
Use axiom, definition, rule of inference etc.
Show that q must also be true

8. Give a direct proof of the theorem:
Example
Give a direct proof of the theorem:
“IF n is an odd integer, THEN n2 is odd”
Definition:
The integer n is even if there exists an integer k such that n = 2k, and n is odd if there exists an integer k such that n = 2k + 1.
Solution:
Assume that “n is odd” is true
Use axiom, definition, rule of inference etc
n = 2k + 1 (from definition)
n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1
Conclusion: it is proved that n2 is an odd integer

9. Example Give a direct proof of the theorem:
“IF m is an odd integer and n is even integer, THEN m + n is odd”
Definition:
The integer n is even if there exists an integer k1such that n = 2k1, and m is odd if there exists an integer k2 such that n = 2k2 + 1.
Solution:
Assume that “m is odd and n is even” is true
Use axiom, definition, rule of inference etc
m = 2k2 + 1 and n = 2k1 (from definition)
m + n = 2 (k1 + k2 ) + 1
Conclusion: it is proved that m + n is an odd integer

10. Give a direct proof that:
Example
Give a direct proof that:
“IF m and n are both perfect squares, THEN nm is also a perfect square”
Definition:
An integer a is a perfect square if there is an integer b such that a = b2
Solution:
Assume that “m and n are both perfect square” is true
Use axiom, definition, rule of inference etc
There are integers s and t such that m = s2 and n = t2.
mn = s2t2, mn = (st)2
Conclusion: it is proved that mn is a perfect square, because it is the square of st, which is an integer.

11. Counterexample
To disprove a universally quantified statement:
x P(x)
We simply need to find one member x in the domain of discourse that makes P(x) false
The value of x is called a counterexample

12. Example
Show that the statement
n  Z+ (2n + 1 is prime)
Is false
A counterexample is n=3, since 23+1 = 9, which is not prime

13. Existence Proof
A proof of
x P(x)
Is called an existence proof. Show that there exists one number a in the domain of discourse that makes P(a) true.

14. Example

15. Indirect Proof: - Contraposition - Contradiction

16. Proof by Contraposition
Indirect Proof: proof that do not start with the hypothesis and end with the conclusion.
Contraposition:
p  q is logically equivalent to ~q  ~p.
Proof by Contraposition:
Assume that ~q is true
Use axiom, definition, rule of inference etc.
Show that ~p must also be true

17. 3n + 2 is even, and therefore not odd = ~p
Example
Prove that:
“IF n is an integer and 3n + 2 is odd, THEN n is odd”
Solution:
Assume that “~q = n is even” is true
Use axiom, definition, rule of inference etc
n = 2k (from definition)
3n + 2 = 6k + 2 = 2 ( 3k + 1)
3n + 2 is even, and therefore not odd = ~p
Conclusion: it is proved by contraposition

18. Example ab > √n . √n = n So ab ≠ n Prove that:
“IF n = ab, where a and b are positive integers,
THEN a ≤ √n or b ≤ √n ”
Solution:
¬q is a > √n and b > √n
¬p is n ≠ ab
Assume that ¬q is true
Use axiom, definition, rule of inference etc
ab > √n . √n = n
So ab ≠ n
Conclusion: it is proved by contraposition

19. Proof by Contradiction
is a statement form that is always false regardless of the truth values of the individual statements substituted for its statement variables.
q Ù ~q
Proof by Contradiction:
To prove that a statement k is true:
Suppose the statement to be proved is false. Means its negation ~k is true.
Show that this supposition leads logically to a contradiction, so the assumption is false.
Conclude that the statement to be proved is true.

20. Example
Prove that:
“IF 3n + 2 is odd, then n is odd”
Proof by Contradiction:
p = 3n + 2 is odd, q = n is odd.
Assume that ~(p  q) is true OR
(p Ù ~q) is true. It means that both p AND ~q must be true
To show contradiction, show that q is also true, OR Show that ~p is also true.
~q = n is not odd = n is even
n = 2k
3n + 2 = 6k + 2 = 2 (3k + 1) = 2t is even
~p is true
Thus, p and ~p are both true  contradiction

21. Example
Prove that:
“IF n2 is even, then n is even”
Proof by Contradiction:
p = n2 is even, q = n is even.
Assume that ~(p  q) is true OR (p Ù ~q) is true.
It means that both p AND ~q must be true
To show contradiction, show that q is also true, OR Show that ~p is also true.
~q = n is odd
n = 2k + 1
n2 = 2 (2k2 + 2k) + 1 = (2 t + 1) is odd
~p is true
p and ~p are true  contradiction

22. More Methods of Proof: - Exhaustive - Resolution

23. Exhaustive Proof
Some theorems can be proved by examining a relatively small number of examples.
Prove that (n+1)3  3n if n is a positive integer with n ≤ 4.
Solution:
n = 1; (1+1)3  31
n = 2; (2+1)3  32
n = 3; (3+1)3  33
n = 4; (4+1)3  34

24. Proof Strategies: When to use Exhaustive Proof
Analyze what the hypothesis (premises) and conclusion mean.
If it is a conditional statement:
Try a Direct Proof
Try a Proof by Contraposition
Try a Proof by Contradiction
If it has a relatively small number of domain:
Try an Exhaustive Proof

25. If (p  q) and (~ p  r) are both true, then (q  r) is true
Resolution
Used to automate the task of reasoning and proving theorems, proposed by JA. Robinson in 1965 that depends on a single rule:
If (p  q) and (~ p  r) are both true,
then (q  r) is true
((p  q) Ù (~ p  r))  (q  r)
(p  q)
(~ p  r)
 (q  r)

26. Proof by Resolution
In a proof by resolution, the hypothesis and the conclusion are written as clauses.
A clause consists of terms separated by or’s.
A direct proof by resolution proceeds by repeatedly applying resolution to pairs of statement to derive new statements until the conclusion is derived.
If a hypothesis is not a clause, it must be replaced by an equivalent expression that is a clause.

27. Example Jasmine is skiing or it is not snowing.
It is snowing or Bart is playing hockey.
Conclusion:
Jasmine is skiing or Bart is playing hockey.
p: it is snowing
q: Jasmine is skiing
r: Bart is playing hockey
(~p  q)
(p  r)
 (q  r)

28. Example (a  b) (~a c) (~c  d) (b  d) From 1 and 2: b  c  4
From 3 and 4: b  d

29. Mathematical Induction

30. Proof by Mathematical Induction
How to prove n P(n)?
Utilize Mathematical Induction
Basic Step:
Show that the statement is true for the positive integer 1.
Inductive Step:
Show that if the statement is true for a positive integer (k) then it must also be true for the next positive integer (k+1).

31. Proof by Mathematical Induction
Basic Step:
P(1) is true
Inductive Step:
Assume that P(k) is true for an arbitrary integer k.
P(k)  P(k+1) is true for all positive integer k.
Notation:
[P(1)  k (P(k)  P(k+1))]  n P(n)
It is not a tool for discovering formula or theorems.

32. P(1) is true, because 1 = 1(1+1)/2
Example 1
… + n = n(n+1)/2
Basic step:
P(1) is true, because 1 = 1(1+1)/2
Inductive step:
Assume that P (k) is true for an arbitrary integer k.
… + k = k(k+1)/2
Show that P(k+1) is true.

33. Example 1 Show that P(k+1) is true
… + k + (k+1) = (k+1)[(k+1)+1]/2
= (k+1)(k+2)/2
Proof:
Add element (k+1) to P(k)
… + k + (k+1) = k(k+1) / 2 + (k+1)
= (k(k+1) + 2(k+1))/2
= (k2 + 3k + 2)/2
= (k+1) (k+2)/2

34. Assume that P (k) is true for an arbitrary integer k.
Example 2
… + (2n -1) = n2
Basic step:
P(1) is true, because 1 = 12
Inductive step:
Assume that P (k) is true for an arbitrary integer k.
… + (2k – 1) = k2
Show that P(k+1) is true.

35. Example 2 Show that P(k+1) is true
… + (2k-1) + (2k+1) = (k+1)2
Proof:
Add element (k+1) to P(k)
… + (2k-1) + (2k+1) = k2 + (2k+1)
= k2 + 2k+1
= (k+1)2

36. Assume that P (k) is true for an arbitrary integer k.
Example 3
… + (3n -2) = n2
Basic step:
P(1) is true, because 1 = 12
Inductive step:
Assume that P (k) is true for an arbitrary integer k.
… + (3k – 2) = k2
Show that P(k+1) is true.

37. Example 3 Show that P(k+1) is true
… + (3k-2) + (3k+1) = (k+1)2
Proof:
Add element (k+1) to P(k)
… + (3k-2) + (3k+1) = k2 + (3k+1)
= k2 + 3k+1
k2 + 3k + 1 ≠ (k+1)2

38. Example 4: starting from other
… + 2n = 2n+1 -1
Valid for all nonnegative integers (0, 1, 2, …)
n = 0, 1, 2,…
Basic step:
P(0) is true, because 1 = – 1
Inductive step:
Assume that P (k) is true for an arbitrary integer k.
… + 2k = 2k+1 -1
Show that P(k+1) is true.

39. Example 4: starting from other
Show that P(k+1) is true
… + 2k + 2k+1 = 2(k+1)+1 -1
= 2k+2 -1
Proof:
Add element (k+1) to P(k)
… + 2k + 2k+1 = 2k k+1
= 2 . 2k+1 -1

40. Assume that P (k) is true for an arbitrary integer k.
Example 5
Show that:
… +n2 = n(n+1)(2n+1)/6
Basic step:
P(1) is true, because 1 = 1(1+1)(2.1+1)/6
Inductive step:
Assume that P (k) is true for an arbitrary integer k.
… + k2 = k(k+1)(2k+1)/6
Show that P(k+1) is true.

41. Add element (k+1) to P(k)
Example 5
Show that P(k+1) is true
… +(k+1) 2 = (k+1)(k+1+1)(2(k+1)+1)/6
= (k+1)(k+2)(2k+3)/6
Proof:
Add element (k+1) to P(k)
… k2 + (k+1)2 = k(k+1)(2k+1)/6 + (k+1)2
= (k+1)[k(2k+1)/6 + (k+1)]
= (k+1)[(k(2k+1)+ (6k+6))/6]
= (k+1)[(2k2+k+6k+6)/6]
= (k+1)[(2k2+7k+6)/6]
= (k+1)[(k+2)(2k+3)/6]

42. Assume that P (k) is true for an arbitrary integer k. k < 2k
Example 6
Proving Inequality:
n < 2n
Basic step:
P(1) is true, because 1 < 21
Inductive step:
Assume that P (k) is true for an arbitrary integer k.
k < 2k
Show that P(k+1) is true.

43. Example 6 (k+1) < 2k+1 k+1 < 2k + 1  1  2k k+1 < 2k + 2k
Show that P(k+1) is true
(k+1) < 2k+1
Proof:
Add 1 to P(k)
k+1 < 2k + 1
 1  2k
k+1 < 2k + 2k
k+1 < 2. 2k
k+1 < 2k+1

44. Strong Form of Induction & The Well-Ordering Property
Strong Form of Induction and the Well-Ordering Property

45. Inductive Step Inductive Step (section 2.4):
Assume that statement n is true
Then prove that statement n+1 is also true
The Strong Form of Mathematical Induction: Inductive Step (section 2.5):
To prove the truth of statement n, assume the truth of all of the preceding statements (statement n0 to n-1).

46. Strong Form of Mathematical Induction
Suppose that we have a propositional function S(n) whose domain of discourse is the set of integers greater than or equal to n0. Suppose that:
- S(n0) is true;
- For all n > n0, if S(k) is true for all k,
n0 ≤ k < n, then S(n) is true.
Then S(n) is true for every integer n ≥ n0

47. Example
Show that postage of four cents or more can be achieved by using only 2-cent and 5-cent stamps.
Basis Step:
n0 = 4  S(4) is true, because 4 = 2 x 2 cents stamp
n0 = 5  S(5) is true, because 5 = 1 x 5 cents stamp

48. Example
Inductive step:
Assume that n ≥ 6 and that postage of k cents or more can be achieved by using only 2-cent and 5-cent stamps for 4 ≤ k ≤ n
Show that S(n) is true.
Proof:
For any k ≥ n0 , we can add 2 cents to make k-cents postage.
Then, for k = n, we can add 2 cents to make n-cents postage.

49. Well-Ordering Property
The well-ordering property for nonnegative integers: every nonempty set of nonnegative integers has a least (smallest) element.
Let B be a set of integers. An integer m is called a least element of B if m is an element of B, and for every x in B, mx.
Example:
3 is a least element of the set {4,3,5,11}.
Let A be the set of all positive odd integers. Then 1 is a least element of A.
Let U be the set of all odd integers. Then U has no least element.

50. Quotient-Remainder Theorem
If d and n are integers, d > 0, there exist integer q (quotient) and r (remainder) satisfying
n = dq + r 0 ≤ r ≤ d
Furthermore, q and r are unique; that is, if
n = dq1 + r1 0 ≤ r1 ≤ d1
And
n = dq2 + r2 0 ≤ r2 ≤ d2
Then q1 = q2 and r1 = r2

51. Example Divide n = 74 by d = 13 We obtain q = 5 and r = 9
Where: 74 = 13 x and 0 ≤ 9 ≤ 13
n = dq + r and 0 ≤ r ≤ d
If q = 3 then r = 35, which is too large
If q = 6 then r = -4, which is negative