Proof Methods, Schemes and Mind Maps of Communication

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Proof Methods

What is a proof?

Proofing as a social process, a communication art.

Theoretically, a proof of a mathematical statement is no

different than a logically valid argument starting with

some premises and ending with the statement. However,

in the real world such logically valid arguments can get

so long and involved that they lose their "punch" and

require too much time to verify.

In mathematics, the purpose of a proof is to convince

the reader of the proof that there is a logically valid

argument in the background. Both the writer and the

reader must be convinced that such an argument can be

produced – if needed.

What is a Proof?

That is, you are writing to convince me that you could

drop down to the logic level and provide all the details, if I

asked you to do so.

Rigor in proofs.

The above remarks should not be construed to mean that

you can get sloppy with your proofs – your audience

requires clarity, precision and, above all, correctness.

Phrases such as "clearly" or "it is easy to see that" are

neither clear nor easy for this audience.

When you say something follows from a definition, I

want to know "the definition of what?"

General Hints

The importance of definitions.

It can not be overemphasized how important definitions

are. Without a clear and crisp understanding of a

definition, you will not be able to use it in a proof. You

have to be able to recall a definition precisely when it is

needed – vague familiarity will not work for you.

Working backwards.

There is a big difference between discovering a proof

and presenting a proof. In presenting a proof you must be

convincing, and things need to follow in a logical order.

To discover a proof, you are under no such restrictions

and often the best procedure is to work the problem

backwards.

Direct Proof

In a direct proof one starts with the premise (hypothesis) and proceed directly to the conclusion with a chain of implications. Most simple proofs are of this kind. Definitions: An integer n is odd iff there exists an integer k so that n = 2k+1. An integer n is even iff there exists an integer s so that n = 2s. Example of a direct proof: If n is an odd integer then n 2 is odd. Pf: Let n be an odd integer. There exists an integer k so that n = 2k+1. n 2 = (2k+1) 2 = 4k 2

• 4k + 1 ⇒ n 2 = 2(2k 2
• 2k) + 1 Since 2k 2
• 2k is an integer, n 2 is odd.

The Empty Set

Thm : The empty set is unique. Pf : Suppose that A and B are empty sets. Since A is an empty set, the statement x∈A is false for all x, so (∀x)( x∈A ⇒ x∈B ) is true! That is, A ⊆ B. Since B is an empty set, the statement x∈B is false for all x, so (∀x)( x∈Β ⇒ x∈Α ) is also true. Thus, B ⊆ Α. Since A ⊆ B and B ⊆ Α we have A = B.  We will use a direct proof here ... but later we will use another technique to prove this.

The Universal Subset

Pf: Let A be a set. Since x ∈ ∅ is false for all x, (∀x)( x ∈ ∅ ⇒ x∈Α ) is true. Thus, ∅ ⊆ A. Since A was arbitrary, ∅ is a subset of every set.  Thm : The empty set is a subset of every set .* ∗ Observe that the statement is a quantified statement: (∀A, A a set)( ∅ ⊆ A) To prove a "for all" statement, one starts with an arbitrary element of the universe for the variable ... assuming no properties other than membership, and show that the statement is true for it.

Another For All Theorem

Thm : For every set A, A ⊆ A.

Pf : Let A be an arbitrary set. Since x∈A ⇒ x∈A for all x (whether x∈A is true or not) (∀x)(x∈A ⇒ x∈A ) is true. So, A ⊆ A for all sets A. 

Contradiction Proofs

This proof method is based on the Law of the Excluded Middle. Essentially, if you can show that a statement can not be false, then it must be true. In practice, you assume that the statement you are trying to prove is false and then show that this leads to a contradiction (any contradiction). This method can be applied to any type of statement , not just conditional statements. There is no way to predict what the contradiction will be.

Contradiction Proof

Example :

The √2 is irrational.

Pf : BWOC assume that √2 is rational.

There exist integers p and q so that √2 = p/q.

We may assume that the fraction is reduced,

i.e. no integer divides both p and q.

2 = p

2

/q

2

⇒ 2q

2

= p

2

, so p

2

is even.

Thus, p is even.

Definition : A real number r is rational iff it can be written as r = a/b with a and b integers and b ≠ 0. A real number is irrational if it is not rational.

The Empty Set

Thm : The empty set is unique. Pf : BWOC suppose that A and B are distinct empty sets. Since A is an empty set, the statement x∈A is false for all x, so (∀x)( x∈A ⇒ x∈B ) is true! That is, A ⊆ B. Since B is an empty set, the statement x∈B is false for all x, so (∀x)( x∈Β ⇒ x∈Α ) is also true. Thus, B ⊆ Α. Since A ⊆ B and B ⊆ Α we have A = B. →←  The contradiction method in this case does not add anything to the argument, it just puts an unnecessary layer over the basic direct method ... this is not wrong, its just in bad taste!!

Proofs of Biconditionals

Example :

Integer a is odd if and only if a+1 is even.

Pf : (Sufficiency, if a is odd then a+1 is even)

Suppose a is an odd integer.

There exists an integer k so that a = 2k + 1.

a+1 = (2k+1) + 1 = 2k+2 = 2(k+1)

Since k+1 is an integer, a+1 is even.

A proof of a P ⇔ Q statement usually uses the tautology P ⇔ Q ≡ (P ⇒ Q) ∧ (Q ⇒ P) That is, we prove an iff statement by seperately proving the "if" part and the "only if" part.

Power Sets

Thm : If a set S has n elements then P (S) has 2 n elements. We will provide at least two proofs of this important result in later sections when we talk about the techniques used. Examples : If S = ∅ then P (∅) = {∅} which is not the empty set! S contains 0 elements, and P (∅) has 2 0 = 1 element. If S = {a}, then P (S) = {∅,{a}} and has 2 1 = 2 elements. If S = {a,b} then P (S) = {∅,{a},{b},{a,b}} and has 2 2 = 4 elements.

Power Sets

Thm : Let A and B be sets. Then A ⊆ B iff P (A) ⊆ P (B). Pf : (⇒ Sufficiency) Let C ∈ P (A), then C ⊆ A. Since A ⊆ Β we have by transitivity that C ⊆ B. Thus, C ∈ P (B). Since C was arbitrary, P (A) ⊆ P (B). (⇐ Necessity) Let x ∈ A. Then {x} ⊆ A. Since P (A) ⊆ P (B), {x} ∈ P (B), so {x} ⊆ B. Thus x ∈ B, and since x was arbitrary, A ⊆ B.  Both parts of this proof illustrate the method of " element chasing " used frequently in proofs involving sets.