Download Lecture 10 - 188 200 Discrete Mathematics and Linear Algebra

Lecture 10
188 200
Discrete Mathematics and Linear Algebra
Pattarawit Polpinit
Department of Computer Engineering
Khon Kaen Uiversity
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Overview
In the last two lectures, we learned rules of inference which enables
us to logically show (prove) that a mathematical statement can be
concluded from given premises.
This is often called formal proof.
In this lecture, we will learn several informal proof methods. We
will be examining a lot of examples that come from variety of
mathematical topics to give you an idea of how to do each proof
method.
References
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Chapter 4 : 4.1-4.7
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Methods of Proof
In this lecture, we will be discussing the following methods of
proof:
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Existence Proofs
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Counterexamples
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Direct Proofs
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Proof by Contraposition (Indirect Proofs)
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Proof by Contradiction
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Vacuous Proofs
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Trivial Proofs
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Proof by Cases
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Proof of Equivalences
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Uniqueness Proofs
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Existential Proofs
Given a statement : ∃x P(x), how do we show it is true?
We only have to show that P(x) is true for at least one x.
There are two types of existential proofs:
1. Constructive proofs — find a specific value of x for which
P(x) is true.
2. Non-constructive proofs — show that such x exists, but do
not actually find it.
— Assume that it does not exist, and show a contradiction.
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Example : Existential Proofs
Constructive existence proof examples
Example: Show that a square that is the sum of two other squares
exists.
Proof : 32 + 42 = 52 .
Q.E.D.
Example: Show that a cube exists that is the sum of three other
cubes.
Proof : 33 + 43 + 53 = 63
Q.E.D.
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Example 2 : Existential Proofs
Non-constructive existence proof examples
Prove that either 2 · 10500 + 15 or 2 · 10500 + 16 is not a
perfect square.
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A perfect square is a square of an integer, e.g. 16, 25, 100 etc.
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In other words: show that a non-perfect square exists in the
set {2 · 10500 + 15, 2 · 10500 + 16}.
Proof : The only two perfect squares that differ by 1 is 0 and 1.
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Thus, any other numbers that differ by 1 cannot both be
perfect squares
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Thus, a non-perfect square must exist in any set that contains
two numbers that differ by 1
Q.E.D.
Note that we didn’t specify which one it was!
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Counterexample
Given a universal quantified statement, ∀x, P(x), we find a single
example that is not true.
This is to disprove a statement, i.e., to show that a statement is
false.
Example : Every positive integer is the square of another integer.
Proof: The square root of 5 is approximately 2.23 which is not an
integer. Hence the statement is false.
Q.E.D.
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Counterexamples II
You cannot prove a universal statement by example.
Example : Prove that all numbers are even.
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Counterexample: 1 is not even.
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Proof by example: 2, 4, 6, 8, . . . are all even ←−
invalid.
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Direct Proof
Proof of a statement :
∀x ∈ D, P(x) → Q(x)
Method of Direct Proof
1. Express the statement to be proved in the form
∀x ∈ D, P(x) → Q(x) (often done mentally).
2. Start the proof by supposing x is any member of D for which
P(x) is true.
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P(x) is called “hypothesis”.
3. Show that the conclusion Q(x) follows from definitions,
previously proved results and the rules of inference.
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Direct Proof: Example
Example If n is odd integer, n2 is odd.
Let’s first look at the definition of odd and even number.
Definition
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An integer n is even iff n = 2k for some integer k.
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An integer n is odd iff n = 2k + 1 for some integer k.
Let P(x) denote x is an odd integer
Q(x) denote x 2 is odd integer
Step 1:
Express the statement as a conditional form:
∀n ∈ D, P(n) → Q(n) where D is the set of all odd integers.
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Direct Proof : Example 1 continues
Proof:
Step 2: We assume P(n) is true for any n ∈ D, i.e. every n is
odd integer.
Step 3: We show Q(n) is true for any n ∈ D, i.e. n2 is odd
integer.
I By definition, n = 2k + 1 where k is some integer.
I Therefore
n2 = (2k + 1)2
= 4k 2 + 4k + 1
= 2(2k 2 + 2k) + 1
= 2(k 0 ) + 1
where k 0 is an integer equal to 2k 2 + 2k which shows that n2
is odd number
Q.E.D.
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Direct Proof : Example 2
Example: Prove the following statement using direct proof:
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If x is even integer and y is odd integer then x-y is odd
integer.
Proof:
Step 1: Let P(x) be “x is even integer”
Q(x) be “x is odd integer”
∀x∀y (P(x) ∧ Q(y )) → Q(x − y )
Step 2: Assume that P(x) ∧ Q(x) is true.
That is assume x is even integer and y is odd integer.
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Proof: (cont.)
Step 3: Show Q(x − y ) is true.
By definition of odd and even number we have
where k ∈ Z
x = 2k
0
y = 2k + 1
where
k0
∈Z
(1)
(2)
Careful not to use the same constant (k 6= k 0 ). Hence
x − y = 2k − 2k 0 − 1
from (1) and (2) where k, k 0 ∈ Z
= 2(k − k 0 − 1) + 1
= 2m + 1
m = k − k0 − 1
∴ x − y is odd integer.
Q.E .D.
Note:
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A sum and difference of integers is integer.
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A product of integers is integer.
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Direct Proof : Example 3
Example: For all integers x, y and z if x | y and x | z then
x | (y + z).
Definition: x | y denotes “x divides y ” or “y is divisible by x”
y is divisible by x (x | y ) iff y = k · x for some integer k.
x | y ↔ ∃k ∈ Z such that y = k · x where Z is the set of integers
Properties:
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∀x, y , z ∈ Z such that if x | y and y | z then x | z.
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Any integer greater than 1 is divisible by a prime number
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For all integers x and y with y 6= 0, x | y
integer.
↔
y
x
is not an
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Direct Proof : Example 3, (2)
Proof: (of Example 3)
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∀x, y , z ∈ Z, if x | y and x | z then x | (y + z)
Suppose
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x, y and z are integer
x|y and x|z. By the definition of divisible, this implies that
where k ∈ Z
y = kx
0
where k ∈ Z
z =k x
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(1)
0
(2)
Show x|(y + z) is true. From (1) and (2),
y + z = kx + k 0 x
(3)
0
= x(k + k )
= xm
(4)
where m = k + k
0
(5)
This implies by the definition of divisible that x|(y + z)
Q.E.D.
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