1 How to Read and Do Proofs [1]

Transcription

208581 Nonlinear Systems in Mechanical Engineering
Lesson 4-1
1.2 Terms and Symbols
1 How to Read and Do Proofs [1]
Symbols mostly used in mathematical proofs are as follows:
1.1 Introduction
Symbol
⇒
⇔
∈
⊆
∅
∼
∀
∃
∋
A proof is a convincing argument expressed in the language of
mathematics.
Given two statements A and B, each of which may be either true
or false, a fundamental problem of interest in mathematics is to show that
the following statement, called an implication is true:
If A is true, then B is true.
If A, then B.
A implies B.
A ⇒ B.
:
∧
∨
Statement A is called hypothesis and B is called conclusion.
The truth table of “A implies B” is as follows:
A
B
A⇒ B
True
True
False
False
True
False
True
False
True
False
True
True
Meaning
implies
if and only if
is an element of
subset
empty set
not
for all (for each, for any, for every)
there is (there are, there exists)
such that
such that
and
or
Q.E.D. (which was to be determined)
Some useful terms are given as follows:
Axiom -- A statement whose truth is accepted without a proof.
Corollary -- A proposition whose truth follows almost immediately from a
theorem.
Definition -- An agreement, by all parties concerned, as to the meaning of
a particular term.
Lemma -- A proposition that is used in the proof of a subsequent theorem.
Proposition -- A true statement of interest.
Theorem -- An important proposition.
From the truth table, if we want to prove that A implies B is true. The only
case we need to do is to assume that A is true then our job is to conclude
that B is true.
A statement
A is true if and only if B is true
A⇔ B
is identical to A ⇒ B AND B ⇒ A, therefore its proof involves proving
the implication in both directions.
1
Copyright  2007 by Withit Chatlatanagulchai
Lesson 4-1
1.3 Forward-Backward Method
All other proof methods rely on this method. As a general rule, the
forward-backward method is probably the first method to try on a problem
unless you have reason to use a different approach based on the form of
B.
The following table summarizes this proof method
When to use it
As a first attempt, or when B does not have
a recognizable form.
What to assume
What to conclude
How to do it
A
B
Solution
The backward process starts from asking “how can I conclude that
B is true?”. The answer is a triangle is isosceles when its two sides are
equal, that is, x = y or x − y = 0. So we have
Work forward using the facts given in A and
apply the backward process to B by asking
“How or when can I conclude that the
statement B is true?”
B1: x − y = 0.
The forward process uses the facts given in A as follows:
A1: xy / 2 = z / 4.
♥ Example 1: [1] Prove the following proposition.
Proposition 1: If the right triangle XYZ with the sides of lengths x and y
2
and hypotenuse of length z has an area of z / 4 , then the triangle XYZ is
isosceles.
2
A2: x + y = z .
2
2
2
(
A3: xy / 2 = x + y
2
2
) / 4.
A4: x − 2 xy + y = 0.
2
2
A5: ( x − y ) = 0.
2
A6: x − y = 0.
The condense proof may be given as follows:
Proof of Proposition 1: The hypothesis together with the Pythagorean
theorem yield x + y = 2 xy and hence
2
2
( x − y)
2
= 0. Thus the triangle is
isosceles, as required.
2
Lesson 4-1
1.4 Construction Method
When to use it
When B has the quantifiers “there is, there
are, there exists.”
B is in the form:
There is an “object” with a “certain
property” such that “something happens.”
What to assume
What to conclude
How to do it
A
That there is the desired object.
Construct the object, then show that it has
the certain property and that the something
happens.
Solution
Proposition 2: If m < n are consecutive integers and m is even, then 4
B can be rewritten as “There is an integer p such that
m + n − 1 = 4 p. Therefore we can use construction method where
2
divides m + n − 1.
2
2
2
Object: p
Certain property: an integer
2
2
Something happens: m + n − 1 = 4 p
By using the facts in A, we need to find an object p that has the certain
property and that something happens.
We use the following steps:
S1: Let n = m + 1.
S2: Then m + n − 1 = m + ( m + 1) − 1 = 2m ( m + 1) .
2
2
2
2
S3: Because m is even, there is an interger k such that m = 2k .
S4: Letting p = k ( m + 1) , it follows that
m 2 + n 2 − 1 = 2m ( m + 1) = 4k ( m + 1) = 4 p,
and so 4 divides m + n − 1. This completes the proof.
2
3
2
Lesson 4-1
1.5 Choose Method
When to use it
When B has the quantifiers “for all, for
each, for every, for any.”
B is in the form:
What to assume
What to conclude
How to do it
For every “object” with a “certain property,”
“something happens.”
A, and choose a general object with the
certain property.
That the something happens.
Work forward from A and the fact that the
general object has the certain property to
conclude that something happens.
Solution
To show that S = T is to show that S is a subset of T and T is
a subset of S . In this proof, we will only prove that S is a subset of T .
The other direction can be done similarly.
S is a subset of T can be rewritten as
Be sure that the general object can be
replaced with any object to satisfy the “for
all” quantifier.
“For all elements x ∈ S , x ∈ T . ”
Proposition 3: If S and T are the two sets defined by
Therefore we can use choose method where
Object: x
Certain property: x ∈ S
Something happens: x ∈ T
S = {real numbers x : x 2 − 3 x + 2 ≤ 0}
T = {real numbers x :1 ≤ x ≤ 2} ,
By using the facts in A, we need to find a general object x that has the
certain property and that something happens and the general object can
be extended to all objects in the set.
We use the following steps:
then S = T .
S1: x − 3 x + 2 ≤ 0.
S2: ( x − 2 )( x − 1) ≤ 0.
2
S3: x ≤ 2 and x ≥ 1.
S4: Therefore for all elements x ∈ S , x ∈ T .
4
Lesson 4-1
1.6 Specialization Method
When A has the quantifiers “for all, for
each, for every, for any.”
When to use it
A is in the form:
What to assume
What to conclude
How to do it
A
B
Identify, in the for-all statement, the
object, the certain property, and the
something that happens.
Look for one particular object to apply
specialization to. (Usually it is the same
object as when the choose method is used
in the backward process.)
Verify that this object has the certain
property and that something happens for
this particular object.
Solution
From the backward process, we have
♥ Example 4: [1] Use the following definition
∗
∗
∗
∗
B1: v ≤ w and w ≤ v .
Definition 1: A real number u is a least upper bound for a set S of real
numbers if and only if (1) u is an upper bound for S and (2) for every
upper bound v for S , u ≤ v.
Using the forward process, we first rewrite A from the Definition 1
∗
A1: “For every upper bound u for T , v ≤ u. ”
∗
∗
A3: “For every upper bound u for T , w ≤ u. ”
to prove the following proposition.
∗
Proposition 4: If v
∗
∗
A2: Using specialization method, we have v ≤ w .
∗
A4: Using specialization method, we have w ≤ v .
∗
and w are least upper bounds for a set T , then
∗
∗
∗
∗
A5: Therefore v ≤ w and w ≤ v .
v∗ = w∗ .
The proof is then completed.
5
Lesson 4-1
1.7 Nested Quantifiers
When a statement contains more than one quantifier, we say that
the statement has nested quantifiers. When processing such statements,
always work from left to right.
Solution
♥ Example 5: [1] Use the following definition
B is identical to “For every real number y , there is a real number
x such that mx + b = y. ” Since B contains nested quantifiers, we process
Definition 2: A function f of one variable is onto if and only if for every
from left to right to have
real number y , there is a real number x such that f ( x ) = y.
Object: y
Certain property: real number
Something happens: there is a real number x such that mx + b = y.
to prove the following proposition.
Proposition 5: If m and b are real numbers with m ≠ 0, then the function
Object: x
Something happens: mx + b = y.
f ( x ) = mx + b is onto.
Using the forward-backward method, we have
B1: For every real number y , there is a real number x such that
mx + b = y.
A1: Choose a real number y (Choose method.)
B2: There is a real number x such that mx + b = y.
A2: From m ≠ 0, construct the real number x = ( y − b ) / m (Construction
method.)
B3: mx + b = y.
A3: mx + b = m ( y − b ) / m  + b = ( y − b ) + b = y.
Since A3 = B3, the forward-backward method is finished and the proof is
completed.
A condensed proof may look like
Proof of Proposition 5: To show that f is onto, let y be a real number.
Because, by hypothesis, m ≠ 0, let x = ( y − b ) / m. It is easy to see that
f ( x ) = mx + b = y, and so the proof is complete.
6
Lesson 4-1
Proposition 6: If a, b, and c are real numbers with a < 0, then there is a
real number y such that for every real number x , ax + bx + c ≤ y.
2
Solution
B contains nested quantifiers, we process from left to right to have
Object: y
2
Something happens: for every real number x , ax + bx + c ≤ y.
Object x
2
Something happens: ax + bx + c ≤ y.
Using the forward-backward method, we have
4ac − b 2
(Construction method.)
4a
2
B1: For every real number x, ax + bx + c ≤ y.
A2: Choose a real number x (Choose method.)
2
B2: ax + bx + c ≤ y.
A1: Construct y =
A3: It follows that
2
b  4ac − b 2

ax + bx + c = a  x +  +
.
2a 
4a

A4: Because a < 0, we have
4ac − b 2
ax 2 + bx + c ≤
= y.
4a
2
Since A4 = B2, the forward-backward method is finished and the proof is
completed.
7
Lesson 4-1
2
Proposition 7: If r is a real number such that r = 2, then r is irrational.
1.8 Contradiction Method
Consider the following truth table
A
B
True
True
False
False
True
False
True
False
NOT B
False
True
False
True
NOT A
False
False
True
True
A⇒ B
True
False
True
True
NOT B ⇒ NOT A
True
False
True
True
We can see that the truth of A AND NOT B are opposite to that of A ⇒ B .
This leads to a proof method called “contradiction.”
The proof begins by assuming that A and NOT B are true and
then tries to conclude some contradiction.
Proofs done by the contradiction method are shorter and easier
than those done by the forward-backward method because you do not
have to create the desired object (as in the construction method.)
However, a disadvantage of this method is that you do not know
exactly what the contradiction is going to be. Another disadvantage is that
this method produces no meaningful result since the outcome is some
contradiction not a constructed object.
When to use it
Solution
The proof contains the following steps.
2
S1: r = 2.
S2: r is a rational number.
S3: From the definition of a rational number, there are integers p and q
with q ≠ 0 such that r = p / q. p and q must have no common divisor.
S4: r = p / q .
2
2
2
S5: 2 = p / q .
2
When NOT B gives useful information.
2
S6: 2q = p .
2
2
When the statement B is one of the two
possible alternatives.
S7: p is even.
S8: p is even.
S9: p = 2k , for some integer k .
What to assume
When the statement B contains the key
word “no” or “not.”
A and NOT B
What to conclude
How to do it
Some contradiction
Work forward from A and NOT B to
2
S10: 2q = ( 2k ) = 4k .
2
2
2
S11: q = 2k .
2
2
2
S12: q is even.
S13: q is even.
S14: Since both p and q are even. They have 2 as their common divisor
and we have reached a contradiction with S3.
reach a contradiction.
8
Lesson 4-1
Proposition 8: Assume that a and b are integers with a ≠ 0. If a does
1.9 Contrapositive Method
Consider the following truth table
A
B
True
True
False
False
True
False
True
False
NOT B
False
True
False
True
NOT A
False
False
True
True
not divide b, then ax + bx + b − a has no positive integer root.
2
A⇒ B
True
False
True
True
NOT B ⇒ NOT A
True
False
True
True
We can see that NOT B ⇒ NOT A has the same truth as A ⇒ B . This
leads to a proof method called “contrapositive.”
The contrapositive method assumes that NOT B is true then tries
Solution
The proof uses the forward-backward method and contains the
following steps.
to conclude that NOT A is true.
The disadvantage of the contrapositive method compared to the
contradiction method is that you work forward from only one statement
( NOT B ) instead of two. However, the advantage is that you know what
A1: ( NOT B ) x > 0 is an integer with ax + bx + b − a = 0.
2
you are looking for ( NOT A .) Thus, you can apply the forward-backward
B1: ( NOT A ) a divides b.
method.
When to use it
A2:
x=
When NOT A and NOT B give useful
information.
What to conclude
How to do it
Work forward from NOT B and backward
−b ± ( b − 2a )
,
2a
b
.
a
B2: There is an integer c such that b = ca.
A3: Since x > 0, x = 1 −
When the statement A and B contain the
key word “no” or “not.”
NOT B
NOT A
2a
=
b
x = −1 and x = 1 − .
a
When the statement A and B are one of
the two possible alternatives.
What to assume
−b ± b 2 − 4 a ( b − a )
A4:
b = (1 − x ) a,
c = 1 − x.
from NOT A.
Since A4 = B2, the forward-backward method is finished and the
proof is completed.
9
Lesson 4-1
c) For every real number x between -1 and 1, there is a real number y
1.10 NOT of Statements
between -1 and 1 such that x + y ≤ 1.
2
Both the contrapositive and contradiction methods require that you
be able to write the NOT of a statement. The following list summarizes the
rules for taking the NOT of statements that have a special form.
[
]
2. NOT [ A AND B ] becomes ( NOT A ) OR ( NOT B )  .
3. NOT [ A OR B ] becomes ( NOT A ) AND ( NOT B )  .
4. NOT [ there is an object with a certain property such that something
happens ] becomes “For all objects with the certain property, the
2
1. NOT NOT A becomes A.
something does not happen.”
5. NOT for all objects with a certain property, something happens
[
Solution
There is a real number x between -1 and 1 such that, for all real
numbers y between -1 and 1, x + y > 1.
2
2
d) There is a real number x between -1 and 1 such that, for all real
numbers y between -1 and 1, x + y ≤ 1.
2
2
]
becomes “There is an object with the certain property such that the
something does not happen.”
Solution
For all real numbers x between -1 and 1, there is a real number y
between -1 and 1 such that x + y > 1.
2
2
♥ Example 9: [1] Find the NOT of the following statements.
a) For every real number x ≥ 2, x + x − 6 ≥ 0.
2
Solution
2
There is a real number x ≥ 2 such that x + x − 6 < 0.
b) There is a real number x ≥ 2 such that x + x − 6 ≥ 0.
2
Solution
2
For all real numbers x ≥ 2, x + x − 6 < 0.
10
Lesson 4-1
1.11 Uniqueness Methods
These methods are used when you want to show that there is only
one object with a certain property such that something happens. There are
two methods: direct and indirect.
The following table summarizes the direct uniqueness method.
When to use it
What to assume
What to conclude
How to do it
When B has the word “unique” in it.
There are two such objects, and A.
That the two objects are equal.
Work forward using A and the properties
of the objects. Also work backward to show
the objects are equal.
Proposition 9: If a, b, c, d , e, and f are real numbers such that
ad − bc ≠ 0, then there are unique real numbers x and y such that
ax + by = e and cx + dy = f .
Solution
We need to prove both “there are” and “unique”. We can use the
construction method to prove the “there are” part and therefore is omitted
here. In this problem, we will only prove uniqueness.
Using the direct uniqueness method, we assume that ( x1 , y1 ) and
( x2 , y2 ) are two objects. Using the forward-backward method, we have
A1: ax1 + by1 = e and cx1 + dy1 = f .
A2: ax2 + by2 = e and cx2 + dy2 = f .
B1: ( x1 , y1 ) = ( x2 , y2 ) .
B2: x1 − x2 = 0 and y1 − y2 = 0.
A3: a ( x1 − x2 ) + b ( y1 − y2 ) = 0, and c ( x1 − x2 ) + d ( y1 − y2 ) = 0.
A4: ( ad − bc )( x1 − x2 ) = 0.
A5: Because ad − bc ≠ 0, x1 − x2 = 0.
A6: Similar derivation leads to y1 − y2 = 0.
From B2, A5, and A6, the proof is completed.
11
Lesson 4-1
The following table summarizes the indirect uniqueness method.
When to use it
What to assume
What to conclude
How to do it
There are two different objects, and A.
Some contradiction.
Work forward from A using the properties
of the two objects and the fact that they
are different.
Proposition 10: If r is a positive real number, then there is a unique real
3
number x such that x = r.
Solution
We need to prove both “there is” and “unique”. We can use the
construction method to prove the “there is” part and therefore is omitted
here. In this problem, we will only prove uniqueness.
A1: x and y are two different real numbers such that x = r and y = r.
3
3
A2: x = y , x − y = 0.
3
3
3
(
3
A3: ( x − y ) x + xy + y
2
2
) = 0.
A4: Because x ≠ y , ( x − y ) ≠ 0, therefore x + xy + y = 0.
2
2
− y ± −3 y 2
A5: x =
.
2
2
A6: Because x is real, the only possibility is −3 y = 0 , y = 0.
A7: r = 0 , which leads to contradiction.
12
Lesson 4-1
1.12 Induction Method
Induction is a method for proving that each of the statements in an
infinite list is true.
Use induction when the statement you try to prove has the form,
“For every integer n ≥ n0 , P ( n ) , ” where P ( n ) is some statement that
Proposition 11: For every integer n ≥ 1,
n
∑k =
k =1
n ( n + 1)
.
2
depends on n. The key words to look for are “integer” and “ n ≥ n0 .”
The method can be summarized as follows.
When to use it
When a statement P ( n ) is true for each
integer n beginning with n0 .
What to assume
P ( n ) is true for n.
What to conclude
That P ( n + 1) is true; also show that
P ( n0 ) is true.
How to do it
First show that P ( n0 ) is true by replacing
n in P ( n ) by n0 .
Write P ( n + 1) and relate P ( n + 1) to
Solution
P ( n ) then use the fact that P ( n ) is true
Assume that P ( n ) :
to show that P ( n + 1) is true.
n
∑k =
k =1
Then, P (1) :
Note that induction does not help you to discover the correct form
of the statement P ( n ) . Rather, induction only verifies that a given
1
∑k =
k =1
statement P ( n ) is true for all integers n greater than or equal to some
Since, P ( n + 1) :
initial one.
One possible variation on induction is that P ( n ) , P ( n + 1) can be
n +1
1(1 + 1)
= 1 is true.
2
∑k =
k =1
n ( n + 1)
is true.
2
( n + 1) ( n + 1) + 1 ( n + 1)( n + 2 )
=
.
2
2
Using the fact that P ( n ) is true, we can write
replaced with P ( n − 1) , P ( n ) or with P ( j ) , P ( n ) , where j < n.
P ( n + 1) :
13
n ( n + 1)
( n + 1)( n + 2 ) .
 n 
k
=
+ ( n + 1) =
∑
 ∑ k  + ( n + 1) =
2
2
k =1
 k =1 
n +1
Lesson 4-1
1.13 Elimination Method
1.14 Cases Method
When to use it
When B has the form “ C OR D ”, such
as, “ A implies C OR D. ”
What to assume
The keyword to look for is “either/or.”
A and NOT C (or A and NOT D )
What to conclude
How to do it
D (or C )
Work forward from A and NOT C , and
When to use it
What to assume
What to conclude
How to do it
backward from D. (or work forward from
A and NOT D, and backward from C.)
When A has the form “ C OR D ”, such
as, “ C OR D implies B .”
Case 1: C
Case 2: D
Both cases must imply B .
First prove that C implies B;
then prove that D implies B.
Proposition 13: If a is a negative real number, then y = −b / ( 2a ) is a
2
Proposition 12: If x − 5 x + 6 ≥ 0, then x ≤ 2 or x ≥ 3.
maximum of the function ax + bx + c.
2
Solution
Using elimination method, we have
A1: x − 5 x + 6 ≥ 0.
A2: ( NOT C ) x > 2.
2
B1: x ≥ 3.
A3: ( x − 2 )( x − 3 ) ≥ 0.
Solution
Because there is no keyword in the proposition, we first use the
forward-backward method.
A4: ( x − 3 ) ≥ 0.
B1: For every real number x, ay + by + c ≥ ax + bx + c.
2
Since A4 = B1, the proof is completed.
14
2
Lesson 4-1
Since “For every” appears, we use the choose method.
1.15 Max/Min Method
There are two max/min methods: max/min 1 and max/min 2. The
max/min 1 method is used when we want to prove
1) All of set S is to the right of x. min {s : s ∈ S } ≥ x.
A1: Choose a real number x.
B2: Such that ay + by + c ≥ ax + bx + c.
2
B3:
2
( y − x )  a ( y + x ) + b  ≥ 0.
2) All of set S is to the left of x. max {s : s ∈ S } ≤ x.
A2: If y − x = 0, the proposition is true. So, we then concentrate when
y − x ≠ 0.
A3: Therefore, either y − x > 0 or y − x < 0.
S
x
0
Since “either/or” appears, we now use cases method.
A4: Case 1: assume that y − x > 0.
S
B4: a ( y + x ) + b ≥ 0.
x
0
A5: From the fact that y = −b / 2a and a < 0, working from A4, we have
2ax + b > 0.
A6: And so a ( y + x ) + b = ax + b / 2 = ( 2ax + b ) / 2 > 0.
The max/min 2 method is used when we want to prove
1) Some of set S is to the right of x. min {s : s ∈ S } ≤ x.
Since A6 = B4, the proof of case 1 is completed.
2) Some of set S is to the left of x. max {s : s ∈ S } ≥ x.
A4: Case 2: assume that y − x < 0.
B4: a ( y + x ) + b ≤ 0.
S
A5: From the fact that y = −b / 2a and a < 0, working from A4, we have
0
x
2ax + b < 0.
A6: And so a ( y + x ) + b = ax + b / 2 = ( 2ax + b ) / 2 < 0.
Since A6 = B4, the proof of case 2 is completed.
Since both cases are proved, the proof is now completed.
We can see that the proof of case 2 is almost identical to that of
the first case. We can therefore omit the proof of case 2 by saying that
“Assume, without loss of generality, that case 1 occurs…,” which means
that the proof of case 1 can in general be applied with case 2 or any other
cases.
15
Lesson 4-1
Proposition 14: If R is the set of all real numbers, then
The max/min 1 method can be summarized as follows.
When to use it
min { x ( x − 2 ) : x ∈ R} ≥ −1.
When B has the form
“ min {s : s ∈ S } ≥ x ” or
“ max {s : s ∈ S } ≤ x. ”
What to assume
Choose an s ∈ S , and A
What to conclude
How to do it
s ≤ x or s ≥ x
Convert B to “For all” statement.
Use choose method.
Work forward from A and the fact that
s ∈ S . Also work backward.
The max/min 2 method can be summarized as follows.
When to use it
When B has the form
“ min {s : s ∈ S } ≤ x ” or
“ max {s : s ∈ S } ≥ x. ”
A
What to assume
What to conclude
That there is an s ∈ S for which s ≥ x or
How to do it
Convert B to “There is” statement.
Solution
We use max/min 1 method. Converting B to “for all” statement,
we have
s≤x
B1: For all real numbers x, x ( x − 2 ) ≥ −1.
Use construction method to produce the
desired s ∈ S .
A1: Choose a real number x.
B2: x ( x − 2 ) ≥ −1.
B3: x − 2 x + 1 ≥ 0.
2
B4: ( x − 1) ≥ 0, which is always true for all real numbers.
2
Therefore, A1 = B4 and the proof is completed.
16
Lesson 4-1
Summary of proof methods
Proof Method
Forward - Backward
When to Use It
As a first attempt, or when B does not have a
recognizable form.
Contrapositive
When
NOT A
and
NOT B
give useful
What to Assume
What to Conclude
A
B
NOT B
NOT A
NOT A.
information.
When the statement A and
two possible alternatives.
Construction
How to Do It
Work forward using the facts given in A and apply the
backward process to B by asking “How or when can I
conclude that the statement B is true?”
Work forward from NOT B and backward from
B
are one of the
When the statement A and B contain the key
word “no” or “not.”
When B has the quantifiers “there is, there are,
there exists.”
A
That there is the
desired object.
Construct the object, then show that it has the certain
property and that the something happens.
A, and choose a
general object with
the certain
property.
That the something
happens.
Work forward from A and the fact that the general object
has the certain property to conclude that something
happens.
B is in the form:
Choose
Specialization
There is an “object” with a “certain property”
such that “something happens.”
When B has the quantifiers “for all, for each, for
every, for any.”
B is in the form:
When A has the quantifiers “for all, for each,
for every, for any.”
A
B
Look for one particular object to apply specialization to.
(Usually it is the same object as when the choose method
is used in the backward process.)
A
is in the form:
Direct Uniqueness
When
B
has the word “unique” in it.
Be sure that the general object can be replaced with any
object to satisfy the “for all” quantifier.
Identify, in the for-all statement, the object, the certain
property, and the something that happens.
There are two such
objects, and A.
17
That the two
objects are equal.
Verify that this object has the certain property and that
something happens for this particular object.
Work forward using A and the properties of the objects.
Also work backward to show the objects are equal.
Lesson 4-1
Proof Method
Elimination
Cases
When to Use It
When B has the form “ C OR
“ A implies C OR D. ”
D ”, such as,
When A has the form “ C OR D ”, such as,
“ C OR D implies B .”
What to Assume
A and NOT C
A and
NOT D )
Case 1:
Case 2:
When
or “ max
Max/Min 2
Contradiction
has the form “ min
B
{s : s ∈ S } ≥ x ”
{s : s ∈ S} ≤ x. ”
B has the form
“ min { s : s ∈ S } ≤ x ” or
{s : s ∈ S} ≥ x. ”
When
NOT B
s ∈ S,
gives useful information.
A and
and
(or work forward from
NOT C , and backward from
A and NOT D, and
Both cases must
imply B .
C implies B;
then prove that D implies B.
s≤x
Convert
or
First prove that
s≥x
A
B
to “For all” statement.
Use choose method.
A
When
“ max
and
A
backward from C.)
C
D
Choose an
How to Do It
Work forward from
D.
(or
Max/Min 1
What to Conclude
D (or C )
That there is an
s ∈ S for which
s ≥ x or s ≤ x
NOT B
Some contradiction
Work forward from A and the fact that
backward.
Convert B to “There is” statement.
s ∈ S.
Also work
Use construction method to produce the desired
Work forward from
A
and
NOT B
s ∈ S.
to reach a
contradiction.
When the statement
possible alternatives.
B
is one of the two
Indirect Uniqueness
When the statement B contains the key word
“no” or “not.”
Induction
When a statement
integer
n
P (n)
beginning with
is true for each
n0 .
There are two
different objects,
and A.
P (n)
n.
is true for
Some contradiction.
That
P ( n + 1)
true; also show
that
true.
P ( n0 )
is
Work forward from A using the properties of the two
objects and the fact that they are different.
First show that
is true by replacing
n
in
P (n)
by
n0 .
is
Write
P ( n + 1)
the fact that
18
P ( n0 )
and relate
P (n)
P ( n + 1)
to
is true to show that
P (n)
then use
P ( n + 1)
is true.
Lesson 4-1
Diagram of proof methods
Forward-Backward
Contrapositive
Contradiction
Induction
Indirect Uniqueness
Construction
Choose
Specialization
Direct Uniqueness
Elimination
Cases
Max/Min 1
Max/Min 2
References
[1]
How to Read and Do Proofs, by Daniel Solow, Wiley, 2002.
19

1 How to Read and Do Proofs [1]

Transcription

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