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:

• 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.

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.

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 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