Lecture Notes: Proof by Mathematical Induction

Preliminaries

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Tuesday Review

Questions:

Write a real world example of an equivalence.
Write a real world example of an implication for which the converse is not true.
For each example: write the converse, inverse, and contrapositive

Possible answers:

Equivalence: The glass is half full
Implication: If it is sunny, the sun is up.

Thursday Review

Review Written Assignment 2.

Fibonacci Sequence

The Fibonacci numbers are defined as: \( F_n = F_{n-1} + F_{n-2} \) where \( F_0 = 0 \) and \( F_1 = 1 \)

Code for a recursive Python program might look like:
Write:

def fib(n):
  if (n == 0) or (n == 1):
      return n
  else:
      return fib(n - 1) + fib(n - 2)

Ask: Who has written a recursive implementation like this?
Notice the base case and the recursive case
We know the program won’t run forever because the recursive case calls smaller values of \( n \). Eventually, the base cases will be reached.
Today, we will learn about “inductive proofs,” which have a similar structure

Summation Notation

First, I will introduction some notation
Suppose we want to calculate:
Write: What is the sum of the first \( n \) powers of 2?
For example, for \( n = 5 \):
\( 2^0 + 2^1 + 2^2 + 2^3 + 2^4 = 1 + 2 + 4 + 8 + 16 = 31 \)
Or for any integer \( n \):
\( 2^0 + 2^1 + 2^2 + \ldots + 2^{n-1} \)
Notice the last term is \( 2^{n-1} \), because the first term was \( 2^0 \)
Using elipses is alright, but it isn’t totally unambiguous.
It is better to use sum notation.
Write: Sum notation: \[ \sum_{i=0}^{n-1} 2^i \]
This represents the sum of all terms between the lower limit and the upper limit, inclusive
We are using the capital Greek character “sigma”
A few examples: \( n = 0, 1, 2, 3, 4 \)

\[\begin{aligned} n &= 0 & &\sum_{i=0}^{0-1} 2^i = 0 \\ n &= 1 & &\sum_{i=0}^{1-1} 2^i = 2^0 = 1 \\ n &= 2 & &\sum_{i=0}^{2-1} 2^i = 2^0 + 2^1 = 1 + 2 = 3 \\ n &= 3 & &\sum_{i=0}^{3-1} 2^i = 2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7 \\ n &= 4 & &\sum_{i=0}^{4-1} 2^i = 2^0 + 2^1 + 2^2 + 2^3 = 1 + 2 + 4 + 8 = 15 \\ \end{aligned}\]

If the upper limit is smaller than the lower limit, then the sum is empty
If you are familiar with your powers of two, you might see the pattern:
Each of these sums is one less than a power of two
If the pattern holds for all n, then the sum can be expressed as a simple function: \[ \sum_{i=0}^{n-1} 2^i = 2^n - 1 \]
Let’s test this for each of \( n = 0, 1, 2, 3, 4 \)
Plug each value into the formula, and show it works
Let’s take a closer look at \( n = 4 \)

\[\begin{aligned} \sum_{i=0}^{4-1} 2^i &= 2^0 + 2^1 + 2^2 + 2^3 \\ &= (1 + 2 + 4) + 8 \\ &= 7 + 8 \\ &= (2^3 - 1) + 2^3 \\ &= 2(2^3) - 1 \\ &= 2^4 - 1 \\ \end{aligned}\]

By rearranging the terms, we see the sum equals \( 2^n - 1 \)
Also, notice that the current sum is formed from the previous sum \( (1+2+4) \) and a new term \( (8) \)
We will use these insights to build an inductive proof

Inductive Summation Proof 1

In programming, a recursive function has a base case and a recursive case
In an inductive proof, we have a base case and an induction step. In the induction step, we show that if the thing we’re trying to prove holds up to \( n \), then we can prove it holds for \( n + 1\)
Draw Ladder: Our base case is the bottom rung of the ladder. We show the base case is true. In the induction step, we show that if we are on any rung, we can reach the next rung. This will prove we can reach any rung!
Write: Theorem: For any \( n \geq 0 \), \[ \sum_{i=0}^{n-1} 2^i = 2^n - 1 \]
Write: Proof: We refer to this equality as predicate \( P(n) \).
We will show our base case, \( P(0) \), is true:

\[\begin{aligned} \sum_{i=0}^{0-1} 2^i &= 0 \\ 2^0 - 1 &= 0 \end{aligned}\]

Induction hypothesis: assume that \( P(k) \) is true for some \( k \geq 0 \) \[ \sum_{i=0}^{k-1} 2^i = 2^k - 1 \]
Induction step: we will prove that if \( P(k) \) is true, then \( P(k+1) \) must be true: \[ \sum_{i=0}^{(k+1)-1} 2^i = 2^{k+1} - 1 \]

\[\begin{aligned} \sum_{i=0}^k 2^i &= \left(\sum_{i=0}^{k-1} 2^i \right) + 2^k & &\text{(break off the sum's last term)} \\ &= \left(2^k - 1\right) + 2^k & &\text{(substitute induction hypothesis)} \\ &= 2\left(2^k \right) - 1 = 2^{k+1} - 1 \\ \end{aligned}\]

Having proven the base case and induction step, we have proved \( P(n) \) is true for all \( n \geq 0 \).
Ask: Does the structure of an inductive proof make sense? What three pieces does an inductive proof need?

Inductive Summation Proof 2

Write: Theorem: For any \( n \geq 0 \), \[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]
Write: Proof: We refer to this equality as predicate \( P(n) \).
We will show our base case, \( P(0) \), is true:

\[\begin{aligned} \sum_{i=1}^{0} i &= 0 \\ \frac{0(0+1)}{2} &= 0 \end{aligned}\]

Induction hypothesis: assume that \( P(k) \) is true for some \( k \geq 0 \) \[ \sum_{i=1}^{k} i = \frac{k(k+1)}{2} \]
Induction step: we will prove that if \( P(k) \) is true, then \( P(k+1) \) must be true: \[ \sum_{i=1}^{k+1} i = \frac{(k+1)(k+2)}{2} \]

\[\begin{aligned} \sum_{i=1}^{k+1} i &= \left(\sum_{i=1}^{k} i \right) + (k+1) & &\text{(break off the sum's last term)} \\ &= \left(\frac{k(k+1)}{2}\right) + (k+1) & &\text{(substitute induction hypothesis)} \\ &= \frac{k(k+1)}{2} + \frac{2(k+1)}{2} & &\text{(common denominator)} \\ &= \frac{k^2 + 3k + 2}{2} & &\text{(simplify)} \\ &= \frac{(k+1)(k+2)}{2} & &\text{(factor)} \\ \end{aligned}\]

Having proven the base case and induction step, we have proved \( P(n) \) is true for all \( n \geq 0 \).
See Figure 3.3 and 3.4 in the textbook for a geometric explanation of this sum

Product Notation

Next, we will see some inductive proofs for products
Similar to sums, we will use special notation to denote a product of an arbitrary number of factors
Write: What is the product of the first \( n \) positive integers? \[ 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n = \prod_{i=1}^n i \]
We are using the capital Greek character “pi”
With sums, the sum of no terms is zero
With products, the product of no factors is one

Inductive Product Proof

Write: Theorem: For any \( n \geq 1 \), \[ \prod_{i=1}^{n} \left( 1 + \frac{1}{i} \right) = n + 1 \]
For example, for \(n = 3\), the theorem says that:

\[\begin{aligned} \left(1 + \frac{1}{1} \right) \left(1 + \frac{1}{2} \right) \left(1 + \frac{1}{3} \right) &= 4 \\ \frac{1+1}{1} \cdot \frac{2+1}{2} \cdot \frac{3+1}{3} &= 4 \\ \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} &= 4 \end{aligned}\]

With the numerators and denominators cancelling, only the numerator of the last factor will remain
The same thing will happen in our inductive proof
Write: Proof: Let \(P(n)\) denote the predicate that \[ \prod_{i=1}^{n} \left( 1 + \frac{1}{i} \right) = n + 1 \]
We will show our base case, \( P(1) \), is true:

\[\begin{aligned} \prod_{i=1}^{1} \left( 1 + \frac{1}{i} \right) = 1 + \frac{1}{1} &= 2 \\ 1 + 1 &= 2 \end{aligned}\]

Induction hypothesis: assume that \( P(k) \) is true for some \( k \geq 1 \) \[ \prod_{i=1}^{k} \left( 1 + \frac{1}{i} \right) = k + 1 \]
Induction step: we will prove that if \( P(k) \) is true, then \( P(k+1) \) must be true: \[ \prod_{i=1}^{k+1} \left( 1 + \frac{1}{i} \right) = (k + 1) + 1 \]

\[\begin{aligned} \prod_{i=1}^{k+1} \left( 1 + \frac{1}{i} \right) &= \left( \prod_{i=1}^{k} \left( 1 + \frac{1}{i} \right) \right) \left( 1 + \frac{1}{k+1} \right) & &\text{(break off the last factor)} \\ &= (k+1) \left( 1 + \frac{1}{k+1} \right) & &\text{(substitute induction hypothesis)} \\ &= (k+1) + \frac{k+1}{k+1} & &\text{(distributing)} \\ &= (k+1) + 1 \\ \end{aligned}\]

The Thue Sequence (Needed for Lab)

Internally, computers represent information in binary: 0s and 1s
I will refer to a sequence of bits as a “binary string”
Write: For example: 10001001
This string as length 8
I will define the complement of a binary string as replacing all 0s with 1s, and all 1s with 0s
The complement of our earlier example is: 01110110
Concatenation of binary strings works the same as concatenating regular strings when programming
For example, concatenating 111 and 000 yields: 111000
The Thue sequence describes a sequence of binary strings.
Write: The Thue sequence is defined as \(T_0 = 0\) and for any \(n \geq 0\), \[ T_{n+1} = \text{ the concatenation of } T_n \text{ and the complement of } T_n\]
For example:

\[\begin{aligned} T_0 &= 0 \\ T_1 &= 01 \\ T_2 &= 0110 \\ \end{aligned}\]

We call this an “inductive definition” or a “recursive definition”
The definition of \(T_0\) is the base case
The definition of \(T_{n+1}\) is the constructor case
Let’s see how we can calculate \(T_1\) and \(T_2\) using the constructor case
(Walk through the process)
Notice that the first half of any string in the sequence is the same as the previous string in the sequence

Thue Sequence Induction Proof 1

Write: Theorem: For every \(n \geq 0\), the length of \(T_n\) is \(2^n\).
Proof: We will prove this by induction. Let \(P(n)\) denote the predicate we are proving.
First, we show that our base case \(P(0)\) is true. \(T_0 = “0”\), which has a length of 1. \(2^0 = 1\). So our base case is true.
Induction hypothesis: assume that \( P(k) \) is true for some \( k \geq 0 \). That is, \(T_k\) has length \(2^k\).
Induction step: we will prove that if \( P(k) \) is true, then \( P(k+1) \) must be true. \(T_{k+1}\) is formed by concatenating \(T_k\) with its complement. By the induction hypothesis, we know that \(T_k\) has length \(2^k\). Also, \(T_k\)’s complement has the same length. So \(T_{k+1}\) will have length: \(2\left(2^k\right) = 2^{k+1}\)

Thue Sequence Induction Proof 2

Write: Theorem: For every \(n \geq 1\), \(T_n\) begins with 01, and ends with 01 if n is odd or with 10 if n is even.
Proof: \(T_1=01\), and the first two bits of \(T_n\) are the same for every \(n \geq 1\).
We will use induction to prove how \(T_n\) ends.
P(n) denotes the predicate that \(T_n\) ends with 01 if n is odd or with 10 if n is even.
Base case: For our base case \(P(1)\), 1 is odd and \(T_1=01\), which ends in 01.
Induction hypothesis: assume that \(P(k)\) is true for some \( k \geq 1 \). That is, \(T_k\) ends with 01 if k is odd or 10 if k is even.
Induction step: \(T_{k+1}\) ends with the complement of the last two bits of \(T_k\).
- If \(k+1\) is even, then \(k\) is odd, and \(T_{k+1}\) will end with 10 (complement of \(T_k\)).
- If \(k+1\) is odd, then \(k\) is even, and \(T_{k+1}\) will end with 01 (complement of \(T_k\)).

Thue Sequence Induction Proof 3

Write: Theorem: For every \(n \geq 0\), \(T_n\) never has more than two 0s or two 1s in a row.
Proof: See Example 3.18 in the textbook.

Pigeonhole Principle Proof

Next, we will use induction to prove a theorem we have used before.
Tell me if you recognize this theorem.
Write: Theorem: If \(f: X \rightarrow Y\) and \(\vert X \vert > \vert Y \vert \), then there are elements \(x_1, x_2 \in X\) such that \(x_1 \neq x_2\) and \(f(x_1) = f(x_2)\).
Proof: We restate the principle as claiming that predicate \(P(n)\) is true for every \(n \geq 2\), where \(P(n)\) states that:
For any finite sets \(X\) and \(Y\) such that \(\vert X \vert > \vert Y \vert \) and \(\vert X \vert = n\), if \(f: X \rightarrow Y\), then there are distinct elements \(x_1, x_2 \in X\) such that \(f(x_1) = f(x_2)\).
Note that we have introduced \(n\), an induction variable. In this proof, we will perform induction using the size of X. Alternatively, we could have performed induction using the size of Y, but the proof would be written differently.
Base case: For our base case \(P(2)\), \(\vert X \vert = 2\). Since \(\vert X \vert > \vert Y \vert \) and we have a function from X to Y, then \(\vert Y \vert = 1\).
This is because Y must have fewer than 2 elements, and must have at least one element for the function to map to.
In our base case, the two elements of X must both map to the unique element of Y.
Induction hypothesis: Suppose that \(P(k)\) is true for some fixed \(k \geq 2\).
Induction step: Prove that if \(P(k)\) is true, then \(P(k+1)\) is true. Pick an arbitrary element \( x \in X \). There are two possibilities: either there is another element \( x’ \in X \) such that \( x’ \neq x \) but \( f(x’) = f(x) \), or there is not.
In the first case, the induction step is proven: we have found two distinct elements of X that map to the same element of Y.
In the second case, x is the only element of X for which the value of f is equal to f(x).
Let X’ be the result of removing x from X and let Y’ be the result of removing f(x) from Y.
Now \(\vert X’ \vert = k \) and \( \vert Y’ \vert = \vert Y \vert - 1 < n \), so the induction hypothesis applies.
The function \(f’: X’ \rightarrow Y’\) is identical to f on the elements of X’.
By the induction hypothesis, there are distinct \( x_1, x_2 \in X’ \) such that \(f’(x_1) = f’(x_2) \in Y’ \). Then \( x_1 \) and \( x_2 \) are also distinct members of X for which f has identical values.