Chapter 2 Proofs
Ch.2 Proofs
In this section we introduce the notion of a proof and describe methods for constructing proofs. A proof is a valid argument that establishes the truth of a mathematical statement.
In our discussion we move from formal proofs of theorems toward more informal proofs.
The methods of proof discussed in this chapter are important not only because they are used
to prove mathematical theorems, but also for their many applications to computer science.
These applications include verifying that computer programs are correct, establishing that operating systems are secure, making inferences in artificial intelligence, showing that system specifications are consistent, and so on. Consequently, understanding the techniques used in proofs is essential both in mathematics and in computer science.
[References]
1. Discrete Math - 7th Edition , Richard Johnsonbaugh, Pearson
2.1 Mathematical Systems, Direct Proofs, and
Counterexamples
2.2 More Methods of Proof Problem-Solving Corner:
Proving Some Properties of Real Numbers
2.3 Resolution Proofs (Skipped)
2.4 Mathematical Induction Problem-Solving Corner:
Mathematical Induction
2.5 Strong Form of Induction and
the Well-Ordering Property
2.1 Mathematical Systems, Direct Proofs, and Counterexamples
A mathematical system consists of Axioms(propositions that are assumed TRUE, Axioms are TRUE), Definitions and Undefined terms.
Definitions are used to give a precise meaning to a new term.
Theorem is a proposition that has been proved to be true.
Special kinds of theorems are lemmas and corollary.
Lemma is useful in proving another theorem.
Corollary is a theorem that follows easily from another theorem.
An argument that establishes the truth of a theorem is a proof.
Point and line are undefined terms
Two triangles are congruent if their vertices can be paired so that the corresponding
sides and corresponding angles are equal.
Two angles are supplementary if the sum of their measures is
.
Example 1.2 |
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The real numbers furnish another example of a mathematical system. Among the axioms are
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Example 1.5 |
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Example of theorems real numbers are
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Example 1.6 |
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An example of a lemma about real numbers is
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*Direct Proofs
For all ,
,
,
,
.
Using direct proof
is true
is true.
Definition 1.7 |
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If |
Example 1.8 |
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The integer |
Example 1.9 |
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The integer |
Example 1.10 |
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Give a direct proof of the theorem “if |
Proof
Let be arbitrary integers.
By definition, is odd, there exists an integer
such that
.
.
By definition of an odd integer, is odd.
Example 1.11 |
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Given a direct proof of the following statement. For all sets
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Direct Proof
Another way:
Example 1.12 |
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For all real numbers if |
Example 1.13 |
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We illustrate by giving two proofs of the statement
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Proof 1
Show that for all ,
and
Let
or
.
In either case, .
Let
or
.
.
Therefore, .
In either case, .
Proof 2
Letting denote the universal set, we obtain
[Distributive law]
[Complement law]
[Identity law]
Disproving a Universally Quantified Statement
To disprove
we simply need to find one member in the domain of discourse that makes
false. Such a value for
is a counterexample.
Example 1.14 |
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The statement
is false. Counterexample, |
Example 1.15 |
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If the statement
is true, prove it: otherwise, give a counterexample (disprove). |
Proof
2.2 More Methods of Proof
Discuss several more methods of proof:
Proof by contradiction, Proof by contrapositive, Proof by cases, Proofs of equivalence, existence Proofs
* Proof by Contradiction (or Reductio ad Absurdum)
Assuming that the hypothesis is true and the conclusion
is false, and finding a contradiction to Establishes
.
(If and
are true then using
and
(as well as other axioms, theorems) derives a contradiction.)
* A contradiction is a proposition of the form .
A proof by contradiction is an indirect proof
since to establish using proof by contradiction.
We derive and then conclude that
is true.
The propositions
and
Using proof by contradiction
are equivalent.
Using a truth table: The equivalence relation is
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Example 2.1 |
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Using proof by contradiction, prove that
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Proof
Using proof by contradiction. Induce contradictions.
Example 2.2 |
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We will give a proof by contradiction of the following statement:
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Proof
Using proof by contradiction.
and
.
Suppose that the conclusion is false, that is, that is true.
By De Morgan’s laws of logic
.
,
.
At this point, we have derived a contradiction : and
.
We conclude that and
,
either
or
.
Example 2.3 |
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We will prove that |
Proof (Using proof by contradiction.)
* Proof by Contrapositive
We have proved
.
is true
is true.
and
are equivalent.
This special case of proof by contradiction is proof by contrapositive (대우).
Example 2.4 |
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Use proof by contrapositive (대우) to show that
|
Proof
Letting be an arbitrary real number.
The contrapositive of the given statement is
is not irrational
is not irrational
or equivalently
is rational
is rational.
Suppose that is rational.
Then for some integers
and
.
Now .
Since is the quotient of integers,
is rational.
The proof is complete.
* Proof by Cases (Exhaustive proof)
Proof by cases is used when the original hypothesis naturally divides itself into various cases. (and Prove each case separatedly)
For example, the hypothesis “ is a real number” can be divided into cases:
(a) is a nonnegative real number
(b) is a negative real number.
Suppose that the task is to prove and that
is equivalent to
(
are the cases).
Instead of proving
,
we prove
.
is true.
and
are equivalent.
* Exhaustive proof
Sometimes the number of cases to prove is finite and not tool large so we can check them all one by one.
Example 2.5 |
Exhaustive proof |
Show that there are no solutions in positive integers |
Proof
Example 2.6 |
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consists of all nonnegative integers. Show that |
Proof
The proposition is “If
, then
.
”Because , the conclusion of the conditional statement “If
, then
”is true.
Hence, this conditional statement, which is , is true.
This is an example of a trivial proof.
* Proof of Equivalence
Some theorems are of the form
.
Such theorems are proved by using the equivalence
that is, to prove “
,” prove “
” and “
.”
Example 2.7 |
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Prove that |
Proof
Prove that is odd
is even.
If is odd then
for some integer
.
.
is even.
Prove that is even
is odd.
If is even then
for some integer
.
.
Therefore, is odd.
Example 2.8 |
|
Prove that for all real numbers
|
Proof
If , then
.
If , then
.
That is,
.
To show
.
If , then
.
If , then
. Since
.
.
In either case, we have proved that .
Example 2.9 |
|
Let (a) |
Proof
* Existence Proofs
A proof of
(2.4)
is an existence proof.
One way to prove (2.4) is to exhibit one member in the domain of discourse that makes
true.
Example 2.10 |
|
Let Prove that there exists a real numbers |
Proof
It suffices to find one real number satisfying
.
The real number
,
halfway between and
, surely satisfies
.
Example 2.11 |
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Prove that there exists a prime |
Proof [Constructive proof]
For the prime ,
is composite:
.
http://en.wikipedia.org/wiki/Mersenne_prime
Example 2.12 |
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Let be the average of the real numbers Prove that there exists |
Proof
We use proof by contradiction and assume the negation of the conclusion
.
By the generalized De Morgan’s laws for logic, this latter statement is equivalent to
or
.
Thus we assume
.
Adding these inequality yields
,
which contradicts the hypothesis
.
Therefore, there exists such that
.
2.4 Mathematical Induction
The Mathematical Induction is a proof technique to prove a statement is true for every positive integer
. (by showing 2 steps)
A sequence of blocks numbered ,
,
sits on an (infinitely) long table and that some blocks are marked with an “
.” Suppose that
(2.4.1) The first is marked.
(2.4.2) For all , if block
is marked, then block
is also marked.
We claim that (2.4.1) and (2.4.2) imply that every block is marked.
[Principle of Mathematical Induction]
Suppose that we have a propositional function whose domain of discourse is the set of positive integers.
Suppose that
1. Basis Step is true.
2. Inductive Step If is true, then
is true, for all
.
Then is true for every positive integer
. (Induction means Mathematical Induction.)
Definition 4.1 |
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For nonnegavive integer
If |
Special case, .
Example 4.3 |
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Use induction to show that (4.9) n ! >= 2^{n-1} for all |
Verify that the statements
,
,
, are true, where
.
Basic Step
[Show : is true.]
The Basic Step is to prove that the propositional function is true for the smallest value
in the domain of discourse.
Inductive Step
[Assume : for all , if
is true], then [Show :
is true].
Example 4.4 |
Geometric Sum |
Use induction to show that if (4.12) |
Basis Step
For ,
is true.
Inductive Step
[Assume : If , then
is true for
.]
[Show true for .]
Now
is true.
By Principle of Mathematical Induction,
is true for all
.
Example 4.5 |
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Use induction to show that |
If a set has
elements, the power set of
,
has
elements.
Theorem 4.6 |
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If (4.13) for all |
Proof
The proof is by induction on .
Basis Step ()
If ,
is the empty set.
The only subset of the empty set is the empty set itself; this,
.
Thus, is true for
.
Inductive Step
[Assume : holds for
.]
[Show true for .]
Let be a set with
elements. Choose
.
Exactly half of the subsets of contain
, and exactly half of the subset of
do not contain
.
Each subset of
that contain
can be paired uniquely with the subset obtained by removing
from
.
Thus exactly half of the subsets of contain
, and exactly half subsets of
do not contain
.
is the set obtained from
by removing
,
has
elements.
By the inductive assumption, | P(Y) | = 2^n .
But the subsets of are precisely the subsets of
that do not contain
.
From the argument in the preceding paragraph, we conclude that
.
Therefore
.
(2.4.13) hold for and the inductive step is complete.
By the Principle of Mathematical Induction,
(2.4.13) holds for all .
Example 2.4.7 A Tiling Problem
Loop invariant [Factorial Computation]
Statement is true After each iteration of the loop is true
A loop invariant is true after the loop finishes.
For example, a loop invariant for a while loop
while(condition)
// loop body
Example 4.8 |
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We use a loop invariant to prove that when the pseudocode which ( fact = fact* } |
terminates, is equal to
.
We prove that is an invariant for the while loop.
Before the while loop begins executing, and
, so
.
Assume that .
If is true,
becomes
and
becomes
*
*
We have proved the Inductive Step.
is an invariant for the while loop.
The while loop terminates when .
is an invariant, at this point,
.
2.5 Strong Form of Induction and the Well-Ordering Property
Strong Form of Mathematical Induction |
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To prove 1. Basis step: Show 2. Inductive step: For all |
[Another way to state (Strong From of Mathematical Induction)]
1. Basis step: Show is true.
2. Inductive step: Assume is true for all
, 1 <= k < n (the inductive hypothesis), Show
is true. Then
is true for every integer
.
Example 5.1 |
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Show that postage of four cents or more can be achieved by using only 2-cent, 5-cent stamps. |
is the floor of
.
is the greatest integer less than or equal to
.
Example 5.2 |
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Suppose that the sequence
We want to prove a statement for all n >= 1 involving
|
Example 5.3 |
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Define the sequence
We want to prove a statement for all |
The Inductive Step will assume the truth of the statement involving .
What are the Basis Steps?
In this example, and
.
In the Inductive Step,
taking
.
Because , it follows then
.
If , then
.
Thus we must add to the Basis Step
.
If , then
and so
.
Therefore if and
, inequality
(
) is satisfied.
Thus the Basis Step are and
.
* Well-Ordering Property
The Well-Ordering Property for nonnegative integers states that every nonempty set of nonnegative integers has a least element.
This property is equivalent to the two forms of induction.
We use the Well-Ordering Property to prove something familiar from long division:
divided by
is dividend,
is divisor,
is quotient,
is remainder,
Example 5.5 |
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We divide The quotient
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Theorem 5.6 |
Quotient-Remainder Theorem |
If
then |
Proof
Let
.
Show that is nonempty.
If , then
so .
Suppose that .
Since is a positive integer,
.
Thus
.
In this case, .
Therefore is nonempty.
Since is nonempty set of nonnegative integers,
by Well-Ordering Property,
has a smallest element, which we denote
.
We let denote the specific of
for which
. Then
.
Since ,
. [Show that
.] (BWOC) Suppose that
. Then
.
Thus . Also
.
But is the smallest integer in
.
This contradiction shows that .
Shown that if and
,
, there exist integers
and
satisfying
(2) [uniqueness] We turn now to the uniqueness of and
. Suppose that
and
.
We must show that and
.
Subtracting the previous equations, we obtain
,
which can be rewritten
The preceding equation shows that divides
.
However, because and
,
.
But the only integer strictly between and
divisible by
is
.
Therefore,
.
Thus,
:
hence
.
The proof is complete.
HW Exercises ( in ..., p. 86, p.93, p.102, p.113 , ... )
Solve problems # 1+ 5k or Exercises 1,16,19
Quiz 1
Copyright by SGLee, SKKU sglee@skku.edu
Copyright @ 2018 SKKU Matrix Lab. All rights reserved.
Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee
*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).