reading and writing mathematical proofs autumn 2015 lecture 1: proofs in textual form
TRANSCRIPT
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Reading and Writing Mathematical
Proofs
Autumn 2015
Lecture 1: Proofs in Textual Form
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Organization
The required stuff…
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Organization
Lecturer: Arthur van Goethem MSc, MF 4.104B,[email protected]
Web page:Part of Data Structures
http://www.win.tue.nl/aga/teaching/2IL50/
Book: Daniel Solow.How to Read and Do Proofs (5th edition)
not mandatory
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Schedule
Only 4 lectures
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Prerequisites
Not covered
Logical inferences
Logical derivations (flag proofs)
In other words: Logic and Set Theory (2IT60)
Covered
Mathematical proofs in common English
Common proof techniques (proof by contradiction, induction, etc.)
Common errors
Many proof examples
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What is a proof?
Why explanations are not proofs…
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What is a proof?
A method for establishing truth
What establishes truth depends on context
Physics Sufficient experimental
evidence
Courtroom Admissible evidence
and witness testimony
Mathematical proof Not a doubt possible!
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What is a proof?
A form of communication
Proof must convince reader (not the writer!) of correctness
Proofs must be:
Clearly written Should be easy to follow Very different from “proving process”
Very precise No ambiguities!
Leaving no doubts
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Definition
Mathematical proof A convincing argument for the reader to establish the correctness of a mathematical statement without any doubt.
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3 + 6
3 + 6 = 9
Definition
Mathematical proof A convincing argument for the reader to establish the correctness of a mathematical statement without any doubt.
Statement must be true or false
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Definition
Mathematical proof A convincing argument for the reader to establish the correctness of a mathematical statement without any doubt.
In what format should a proof be?
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Logical derivation
Good Very systematic Hard to make mistakes
Bad Not convenient for statements
not stated in logical formulas Emphasis on logical reasoning
→ detract from crux argument Hard to read Cumbersome
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Common English
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ such that x = 2k + 1. Then, x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Thus, x2 is odd. □
Good Short and to the point Easy to read
Bad Logical reasoning
somewhat hidden Natural language can be
ambiguous
This is the kind of proof we expect
in Data Structures!
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Two Proof Formats
This is what you should write down
This should be in the back of your mind
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ such that x = 2k + 1. Then, x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Thus, x2 is odd. □
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
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Two Proof Formats
Theorem
If there is an x such that P(x,y) holds for all y, then for all v there is a u such that P(u, v) holds.
Proof
Let x* be the element such that P(x*,y) holds for all y. Then, for an arbitrary v there is a u, namely u = x*, such that P(u,v) = P(x*,v) holds. □
Must be able to translate:
Logic ↔ English
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Logic and Ambiguities
The stuff that is kind of doing that thing…
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Logic vs. English
Logic English
P ⋀ Q P and QBoth P and Q
P ⋁ Q P or Qcareful….
¬P
¬(P ⋁ Q)
not Pdoesn’t hold, cannot
neither P nor Q
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Logic vs. English
Logic English
P ⇒ Q P implies QIf P, then Q
P ⇔ Q P is equivalent to QP if and only if Q
P iff Q
For P ⇒ Q:¬Q ⇒ ¬P (equivalent)Q ⇒ P (not equivalent!)
contrapositiveconverse
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Logic vs. English
Logic English
∃x[x ϵ S: P(x)]∃x[x ϵ ℤ: P(x)]
• There is an x in S such that P(x) holds• There exists an integer x for which P(x)
holds• P(x) holds for some x
• There is at least one x s.t. …
∀x[x ϵ S: P(x)] • For all x in S it holds that P(x)• (careful) P(x) holds for any x in S• Let x be an arbitrary element of S,
then P(x) holds
∃x[x ϵ S: P(x)] ⋀ ∀x,y[x,y ϵ S: P(x) ⋀ P(y) ⇒ x=y]
• There is a unique x for which P(x) holds• There is exactly one x for which P(x) holds
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Ambiguities
Do you want beer or water?
Do you want beer or water or both? (or)
OR
Do you want either beer or water? (exclusive-or)
If you can solve any problem we pose, then you ace the course
If you can solve some problem we pose, then you ace the course
OR
If you can solve all problems we pose, then you ace the course
Always avoid ambiguities!
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Practice
1. There exists an element x of S such that x is at least as large as all other elements of S
∃x[x ϵ S: ∀y[y ϵ S: x ≥ y]]∃x[x ϵ S: ∀y[y ϵ S ⋀ y ≠ x: x ≥
y]]
2. There are no positive integers a, b, and c such that an + bn = cn for any integer n > 2
¬∃a,b,c[a,b,c ϵ ℕ+: ∃n[n ϵ ℤ ⋀ n > 2:
an + bn = cn]]∀a,b,c[a,b,c ϵ ℕ+: ∀n[n ϵ ℤ ⋀ n > 2:
an + bn ≠ cn]]3. ∀x,y[x,y ϵ ℤ: P(x) ⋀ Q(y) ⇒ ¬Q(x) ⋁
¬P(y)]
For all integers x and y it holds that, if P(x) and Q(y) hold, then Q(x) does
not hold or P(y) does not hold
4. ∀x,y[x,y ϵ S: ∃T[T ⊆ S : P(T) ⇔ x ϵ T ⋀ y ϵ
T]]
For all elements x and y of S, there exists a subset T of S for which P(T)
holds iff x and y are both in T
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Detail
What to describe and what not
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Definition
Mathematical proof A convincing argument for the reader to establish the correctness of a mathematical statement without any doubt.
How detailed should a proof be?
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A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Does this even require a proof?
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A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1 and C2 have (at least) three intersection points p1, p2, and p3. Note that p1, p2, and p3 must lie on both C1 and C2. But, three points on a circle uniquely define the circle. Thus, C1 = C2. Contradiction! □
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A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1 and C2 have (at least) three intersection points p1, p2, and p3. Note that p1, p2, and p3 must lie on both C1 and C2. But then the center of C1 (and C2) must be at the intersection of the bisector of p1 and p2, and the bisector of p2 and p3. Since the bisectors are lines, and two lines intersect only once, this uniquely defines the center (and radius) of both C1 and C2.Thus, C1 = C2. Contradiction! □
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Proof Detail
When is a proof detailed enough?
What will the reader accept as true statements
More detail results in “more obvious” statements…
… but is also much longer
A good proof must find the right balance
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Axiomatic System
Axiom A simple statement assumed to be true
Mathematical theory Set of axioms Theorems derived from axioms using logical inferences Note: Definitions are generally used like axioms (but are different)
Examples Euclid’s 5 postulates for geometry Peano axioms for natural numbers Zermelo–Fraenkel set theory with the axiom of choice
“Two sets are equal if they have the same elements”
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Axioms in this Course
Which axioms are you allowed to use?
Anything from high school math (middelbare school) Within reason, must be covered by any high school
Anything covered in class
Your proofs should be written as if intended for a fellow student The axioms you use should be accepted by all students
If you are unsure, just ask!
Exception: If you are asked to prove a statement, the statement itself can never be used as an axiom
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Axioms in this Course
Exception:
If you are asked to prove a statement, the statement itself can never be used as an axiom
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Practice
What are the axioms?
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1 and C2 have three intersection points p1, p2, and p3. Note that p1, p2, and p3 must lie on both C1 and C2. But, then the center of C1 (and C2) must be at the intersection of the bisector of p1 and p2, and the bisector of p2 and p3. Since the bisectors are lines, and two lines intersect only once, this uniquely defines the center (and radius) of both C1 and C2.Thus, C1 = C2. Contradiction! □
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Practice
What are the axioms?
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ such that x = 2k + 1. Then, x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Thus, x2 is odd. □
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Proving Tips
Tips for writing proofs
1. State the proof techniques you’re using (see next lecture)
2. Keep a linear flow Proving process different from written proof
3. Describe every step clearly in words
4. Don’t use complicated notation
5. Make sure your axioms are actually “obvious” What is obvious to you may not be obvious to the reader
6. Finish your proof Connect everything with what you were trying to prove
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Analogy
The Maze Analogy
entrance/premise
exit/goal
Proving process Find path between
entrance and exit Erratic, creative
Written proof Give instructions for
others to find path Stick to essentials
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Analogy
Proving process
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Analogy
Written proof
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Practice
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Practice Proofs
Theorem
If a right triangle with side lengths x, y, and z (x ≤ y ≤ z) has area z2/4, then the triangle is isosceles (two sides have the same length).
y
xz
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Practice Proofs
Theorem
If a right triangle with side lengths x, y, and z (x ≤ y ≤ z) has area z2/4, then the triangle is isosceles (two sides have the same length).
Proof
Let T be a right triangle with side lengths x, y, and z (x ≤ y ≤ z) and area z2/4. The area of T can also be computed as xy/2, so we obtain xy/2 = z2/4. By Pythagoras’ theorem we know that z2 = x2 + y2. We hence get the equation xy/2 = (x2 + y2)/4. By multiplying both sides by 4 and moving the terms to one side, we obtain
x2 + y2 – 2xy = 0. By factoring the polynomial we obtain (x - y)2 = 0. Thus, x = y and T is isosceles. □
y
xz
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Practice Proofs
Theorem
If a right triangle with side lengths x, y, and z (x ≤ y ≤ z) has area z2/4, then the triangle is isosceles (two sides have the same length).
Proof
Let T be a right triangle with side lengths x, y, and z (x ≤ y ≤ z) and area z2/4. The area of T can also be computed as xy/2, so we obtain xy/2 = z2/4. By Pythagoras’ theorem we know that z2 = x2 + y2.
We get the following:
xy/2 = (x2 + y2)/4
2xy = x2 + y2
x2 + y2 - 2xy = 0
(x - y)2 = 0
x = y
Thus T is isosceles. □y
xz
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Practice Proofs
Theorem (handshaking lemma)
For any graph G=(V, E) the number of vertices with odd degree is even.
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Practice Proofs
Theorem (handshaking lemma)
Consider a group of people shaking hands with each other (but not necessarily with everyone). The number of people that shook hands with an odd number of people is even.
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Practice Proofs
Theorem (handshaking lemma)
For any graph G=(V, E) the number of vertices with odd degree is even.
Proof
Consider the sum of the degrees of all vertices: Σv d(v). Since this sum counts every edge exactly twice, the sum must be even. Note that vertices with even degree do not affect the parity of this sum. Hence, if V’ is the set of vertices with odd degree, then Σv ϵ V’ d(v) must also be even. Since the sum of two odd numbers is even, the sum Σv ϵ V’ d(v) is even if and only if |V’| is even. □
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Practice Proofs
Theorem
∀u∃v[P(u, v)] ⇒ ∃x∀y[P(x, y)]
State this theorem in common English
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Practice Proofs
Theorem
If for all u there exists a v such that P(u, v) holds, then there exists an x such that P(x, y) holds for all y.
Proof
Is it even true?
Let the domain be ℕ and P(u, v) = u < v. Then, for all u there exists a v, namely v = u + 1, such that u < v = u + 1. On the other hand, for any x, x < x never holds and hence x < y does not hold for all y. So the theorem does not hold. □
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Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds that x2 + y2 = z2.
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Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds that x2 + y2 = z2.
Proof
Seems tricky… Let’s try to prove something simpler. What if x = y?
Then the area of the triangle is x2/2. We need to prove that 2x2 = z2.
So 4 triangles should together have area z2. How about this configuration?
zx
y
z
zz
z
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Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds that x2 + y2 = z2.
Proof
Let’s return to the general case. What happens if we again try to form a square with side length z? We get a hole in the middle which seems to be a square. We may have enough for a proof here…
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Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds that x2 + y2 = z2.
Proof
Consider the configuration of four copies of T shown in the figure. Note that, for a right triangle, the sum of the non-right angles is 90°, since the sum of all angles is 180°. Hence, the outer shape is a square with side length z and has area z2. Thearea of a single triangle is xy/2. Because T is aright triangle, the middle hole is also a squarewith side length (y - x) and area (y - x)2. Wethus obtain: z2 = 2xy + (y - x)2 = x2 + y2. □
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