Lecture Notes: Basic Proof Techniques

Preliminaries

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Tuesday Review

Questions:

Write the Extended Pigeonhole Principle, Alternate Version
Use the Pigeonhole Principle to prove that for a group of 25 people, there is some day of the week for which at least four people were born on that day.

Answers:

Extended Pigeonhole Principle, Alternate Version:
Let X and Y be any finite sets and let \( f: X \rightarrow Y \).
Then there is some \( y \in Y \) such that \( f(x)= y \) for at least \( \left\lceil \frac{\mid X \mid}{\mid Y \mid} \right\rceil \) values of x.
The people are set X
The days of the week are set Y
f is the function mapping people to the day of the week on which they were born.
By the Pigeonhole Principle, there is some day of the week such that at least \( \left\lceil \frac{\mid X \mid}{\mid Y \mid} \right\rceil = \left\lceil \frac{25}{7} \right\rceil = 4 \) people were born on that day.

Thursday Review

Sit with your assigned group
Review your answers to Written Assignment 1
Identify questions you would like to discuss as a class

The Importance of Mathematical Language

Write: Everybody loves somebody
But what does this really mean? Does it mean:
Write: For every person A, there is a person B such that A loves B.
Or perhaps: There is a person B such that for every person A, A loves B.
Notation: \( \forall \) means “for all,” “for any,” “for every”
Notation: \( \exists \) means “there exists” or “for some”
Using this notation, “For every person A, there is a person B such that A loves B” can be written succinctly as: \( \forall x \exists y L(x, y) \)
\( L(x, y) \) is the “loves” relationship, meaning that x loves y.
Or to express that “There is a person B such that for every person A, A loves B.” you would write: \( \exists y \forall x L(x, y) \)
We won’t use relationship notation until later in the course.
We will be using “for all” and “there exists” notation (e.g., to show that something is true for all integers).

Constructive Proofs

Write: Theorem: Every odd integer is equal to the difference between the squares of two integers.
An odd integer can be written as \( 2k + 1 \)
Give examples of squares from \( 0^2 \) to \( 4^2 \)
We call the squares of integers “perfect squares”
Give examples of odd integers: 1, 3, 5, 7 for \( k = 0, 1, 2, 3 \)
And show each square can be written as the difference of squares. For example:
\( 1 = 1 - 0 = 1^2 - 0^2 \)
\( 3 = 4 - 1 = 2^2 - 1^2 \)
\( 5 = 9 - 4 = 3^2 - 2^2 \)
\( 7 = 16 - 9 = 4^2 - 3^2 \)
Identify the pattern: \( (k+1)^2 - k^2 = k^2 + 2k + 1 - k^2 = 2k + 1 \)
Write: Proof: Any odd integer can be written as \( 2k + 1 \) for some integer k. We can rewrite that expression:
\( 2k + 1 = (k^2 + 2k + 1) - k^2 \) (adding and subtracting \( k^2 \))
\( = (k+1)^2 - k^2 \) (writing the first term as a square)
Now let \( m = k + 1 \) and \( n = k \).
Then \( 2k + 1 = m^2 - n^2 \), so we have identified integers m and n with the properties that we claimed.
Notes on style:
1. Argument is written in full sentences
2. Clear structure: works from the assumption to the conclusion
3. Rigorous: uses a mathematical definition of an odd number
4. Convincing: uses the appropriate amount of detail
Definition: An algorithm describes a series of steps in sufficient detail that it can be carried out mechanically by a machine or person.
Definition: A constructive proof describes an algorithm for finding the thing that the proof says exists.
Write pseudocode for finding the difference of squares for a particular odd integer.
In lab, you will implement this as a Python program.

Logical Equivalence

Two statements are equivalent if one statement is true whenever the second is true, and vice versa. That is, either both statements are true, or both are false.
For example:
- “The sun is up” and “it is daytime” are equivalent
  A mathematician would write: Write: “The sun is up if and only if it is daytime.”
- “A number is odd” and “a number is not even.”
  A mathematician would write: “An integer \( n \) is odd if and only if \( n \) is not even.”
Ask: Other examples?
iff means “if and only if,” meaning the statements are equivalent.
Let’s refer to the different parts of the equivalence with symbols:
p is “the sun is up”
q is “it is daytime”
p and q are propositions. Propositions can be true or false.
p iff q is an equivalence. An equivalence is a shorter way of writing these two statements:
if p then q: “if the sun is up, then it is daytime”
if q then p: “if it is daytime, then the sun is up”
Furthermore, each of these statements can be rewritten as their contrapositive without changing their meaning:
if not q then not p: “if it is not daytime, then the sun is not up”
if not p then not q: “if the sun is not up, then it is not daytime”
Take a minute to read the original statements and their contrapositives, and convince yourself they are logically equivalent.

Direct Proof

Next, we will prove an equivalence:
Write: Theorem: For any integer \( n \), \( n^2 \) is odd iff \( n \) is odd.
To prove an equivalence, we must prove the statements in both directions.
Write: Proof, part 1: Prove that if \( n \) is odd, then \( n^2 \) is odd.
If \( n \) is odd, then \( n = 2k + 1 \), where \( k \) is some other integer.
\( n^2 = (2k + 1)^2 \)
\( = 4k^2 + 4k + 1 \)
\( = 2(2k^2 + 2k) + 1 \)
Let \( j = 2k^2 + 2k \). Then \( n^2 = 2j + 1 \), meaning that \( n^2 \) is odd.
Next, we prove the other statement.
Write: Proof, part 2: Prove that if \( n^2 \) is odd, then \( n \) is odd.
First, we rewrite the statement as its contrapositive:
“if \( n \) is not odd, then \( n^2 \) is not odd,” or equivalently:
“if \( n \) is even, then \( n^2 \) is even.”
If \( n \) is even, then it can be written as \( n = 2k \), where \( k \) is some other integer. Then:
\( n^2 = (2k)^2 \)
\( = 4k^2 \)
\( = 2(2k^2) \)
Thus, \( n^2 \) is also even.
Using this theorem, it is easy to prove the following:
Write: Corollary: If \( n \) is odd, then \( n^4 \) is odd.
Write: Proof: Since \( n \) is odd, our theorem says \( n^2 \) is odd.
Using the theorem again, since \( n^2 \) is odd, \( (n^2)^2 = n^4 \) is also odd.

Inverse, Converse, Contrapositive

Let’s take a closer look at the theorem we proved
The statements we proved are implications:
Write: Implications have the form: “if p then q”
Example: “if you live in MA, then you live in the USA”
We chose to prove the contrapositive of an implication, since contrapositives are logically equivalent
Write: Contrapositive: “if not q then not p”
“if you do not live in the USA, then you do not live in MA”
There are two related ways to change implications, but these are not logically equivalent to the original implication
Write: Inverse: “if not p then not q”
“if you do not live in MA, then you do not live in the USA”
We negate both parts of the implication
Ask: Can someone give an example that disproves the inverse?
Write: Converse: “if q then p”
“if you live in the USA, then you live in MA”
We change the order of the implication
Ask: Can someone give an example that disproves the converse?

Rational and Irrational Numbers

Write: Definition: A rational number can be expressed as the ratio of two integers.
For example: \( 1.25 = \frac{5}{4}, 0.\bar3 = \frac{1}{3}, \ldots \)
Write: Definition: An irrational number cannot be expressed as the ratio of two integers.
For example: \( \pi \approx 3.14, \sqrt{2} \approx 1.414, \ldots \)

False Proof

Let \( a = b \). Then:

\[\begin{aligned} a^2 &= ab &\text{ (multiplying both sizes by a)} \\ a^2 - b^2 &= ab - b^2 &\text{ (subtracting } b^2 \text{ from both sides)} \\ (a+b)(a-b) &= b(a-b) &\text{ (factoring both sides)} \\ (a+b) &= b &\text{ (dividing both sides by } a - b \text{)} \\ 2b &= b &\text{ (substituting a for b, since a = b)} \\ 2 &= 1 &\text{ (dividing both sides by b)} \\ \end{aligned}\]

Read each step, and see if you can catch the step where we made a mistake
Answer: \( a - b = 0 \), and dividing by 0 is invalid!

Proof by Contradiction

In the false proof, we arrived at a contradiction due to a logical error
In a proof by contradiction, we can prove a statement by arriving at a contradiction without making logical errors
We will use proof by contradiction to prove:

Write: Theorem: \( \sqrt{2} \) is irrational
Proof: To derive a contradiction, assume \( \sqrt{2} \) is rational, meaning it can be represented as a ratio of integers.
Suppose that \( \sqrt{2} = \frac{a}{b} \), where \( \frac{a}{b} \) is the simplest form of the fraction (explain fraction simplification).
This means that at most one of a and b is even, or else we could simplify further.

\[\begin{aligned} \sqrt{2} &= \frac{a}{b} \\ b \cdot \sqrt{2} &= a &\text{ (multiplying both sides by b)} \\ 2b^2 &= a^2 &\text{ (squaring both sides)} \\ \end{aligned}\]

Thus, \( a^2 \) is divisible by 2 (i.e., even).
Write: The “Direct Proof” theorem from earlier.
By our earlier theorem this means \( a \) is even, too.
So we can write \( a = 2k \), where \( k \) is some other integer.

\[\begin{aligned} 2b^2 &= a^2 \\ 2b^2 &= (2k)^2 &\text{ (substituting for a)} \\ b^2 &= 2k^2 &\text{ (dividing both sides by 2)} \\ \end{aligned}\]

Thus, \( b^2 \) is divisible by 2 (i.e., even). By our earlier theorem this means \( b \) is even, too.

This is a contradiction, since we said that at most one of a and b is even!
Since we encountered a contradiction, \( \sqrt{2} \) must be irrational.

Case Analysis Proof

Earlier, we proved an equivalence (if and only if) by proving two implications
We will see something similar with this proof, which uses case analysis
In case analysis, you prove a general statement by proving it for subclasses
Write: In any group of 6 people, there are either at least 3 who all know each other, or at least 3 who are all strangers to each other. “Knowing” is symmetric: if A knows B, then B knows A.
Draw: Six vertices, representing people: A, B, C, D, E, F. Connect them in a triangle, showing three people who all know each other. Use a different color to show that nobody else knows each other.
We need to prove our statement for all possible combinations.
Write: Proof: Pick a person X. Of the remaining 5 people, there must be at least 3 whom X knows, or at least 3 whom X does not know.
We will prove either subcase.
Write: Case 1: If X knows at least 3 people.
If none of those 3 know each other: they form a set of 3 who don’t know each other. ✓
If at least 2 know each other: since they also know X, with X they form a set of 3 who know each other. ✓
Write: Case 2: If X does not know at least 3 people.
If all of those 3 know each other: they form a set of 3 who know each other. ✓
If at least 2 don’t know each other: since they also don’t know X, with X they form a set of 3 who do not know each other. ✓
Note that the logic of the cases is symmetric. So Case 2 could be rewritten:
Write: Case 2: If X does not know at least 3 people, then the argument is symmetric to Case 1, reversing the roles of “knowing” and “not knowing.”
Note that my illustration is an example of a graph
The people are “vertices” and the connections between them are “edges”
The statement we proved could be rephrased as a graph coloring problem: in a fully connected graph with six vertices and edges of two colors, we will always find a triangle with a single color
We will learn more about graphs later in the semester
Graphs are very relevant to Computer Science, since computer networks are naturally modeled as graphs