Reading and Writing Mathematical Proofs

Spring 2014

Lecture 1: Proofs in Textual Form

Organization

Lecturer: Dr. Kevin Verbeek, MF 4.106,k.a.b.verbeek@tue.nl

Web page:Part of Data Structures:

http://www.win.tue.nl/~speckman/2IL50.html

Book: Daniel Solow.How to Read and Do Proofs (5th edition)

not mandatory

Schedule

Only 4 lectures Lectures usually on Monday 7&8 Also on Wednesday 11 February 3&4 (maybe…)

Prerequisites

This will not be covered

Logical inferences

Logical derivations (flag proofs)

In other words: Logic and Set Theory (2IT60)

This will be covered

Mathematical proofs in common English

Common proof techniques (proof by contradiction, induction, etc.)

Many proof examples

Proof

What is a proof?

What is a proof?

A proof is a method for establishing truth

What establishes truth depends on context

In physics: sufficient experimental evidence Correct until contradiction is observed

In courtroom: admissible evidence and witness testimony Evidence should be beyond reasonable doubt

Mathematical proof: there can be no doubt!

What is a proof?

A proof is a form of communication

Proof must convince reader (not the writer!) of correctness

Proof must be clearly written Should be easy to follow “Proving process” very different from written proof

Proof must be very precise No ambiguities

Proof can leave no doubts

Definition

Mathematical proof A convincing argument for the reader to establish the correctness of a mathematical statement without any doubt

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: 3 + 6 3 + 6 = 9

In what format should a proof be?

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

Proofs in Common English

Why not prove in 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!

Two Proof Formats

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. □

This is what you should write down

This should be in the back of your mind

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

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

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

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

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!

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

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?

A Simple Proof

Theorem

Two distinct circles can have at most two intersection points.

Does this even require a proof?

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 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! □

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 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! □

Proof Detail

When is a proof detailed enough?

Depends on what the reader will accept as true statements

A proof with more detail has “more obvious” statements…

… but is also much longer

A good proof must find the right balance

Axiomatic System

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”

Proving 2 + 2 = 4 requires >20,000 steps!

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

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! □

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. □

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 Or draw a figure

4. Don’t use complicated notation

5. Keep your proof as simple as possible

6. Make sure your axioms are actually “obvious” What is obvious to you may not be obvious to the reader

7. Finish your proof Connect everything with what you were trying to prove

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

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

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

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

Practice Proofs

Theorem

∀u∃v[P(u, v)] ⇒ ∃x∀y[P(x, y)]

State this theorem in common English

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. □

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.

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

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…

zx

y

z

z

z

z

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. □

z

z

z

z

Practice Proofs

Theorem (handshaking lemma)

For any graph G=(V, E) the number of vertices with odd degree is even.

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.

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. □