Lecture Notes: Strong Induction
Preliminaries
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Course Content
Tuesday Review
Questions:
- Use summation notation to write: “The sum of the first n positive even numbers.”
- Write a closed-form solution for this sum.
Answers:
\[ \sum_{i=1}^{n} 2i \]
\[ = 2 \sum_{i=1}^{n} i = 2 \cdot \frac{n(n+1)}{2} = n(n+1)\]
Note that we substituted \( \sum_{i=1}^{n} i \) for its closed form, which we proved last class.
Thursday Review
Review Written Assignment 3.
Motivation for Strong Induction
- Last class, we proved theorems related to the Thue Sequence.
- Each element in the Thue Sequence was defined in terms of the immediately preceding element in the sequence.
- We used Mathematical Induction in our proofs:
- For the base cases, we proved the predicate for the first element of the sequence.
- For the induction hypothesis, we assumed the predicate was true for was true for \(T_k\) (a particular element).
- For the induction steps, we used the induction hypothesis to prove the predicate for \(T_{k+1}\) (the next element).
- Note that our induction step relied on just one preceding element.
- Next, consider a different sequence:
\( a_1 = 1\), \(a_2 = 3\), and for all \(n > 2\), \(a_n = a_{n-2} + 2 a_{n-1} \)
- Ask: What are some ways this sequence differs from the Thue Sequence?
- It is significant that \(a_n\) is defined in terms of two preceding elements.
- If we want to prove properties of this sequence, our proofs will need to consider multiple preceding elements.
- We will use an extension of Mathematical Induction, called Strong Mathematical Induction.
- For regular induction, we used the metaphor of climbing a ladder. We proved we could reach the next rung by using the previous rung. In strong induction, we will prove we can reach the next rung by using any of the previous rungs. The textbook has a terrifying illustration of this, which they liken to centipede using many legs to climb a ladder. Terrifying, but memorable!
Sequence Proof 1
- Let’s write out a few elements of this sequence:
\[\begin{aligned}
a_3 &= a_1 + 2 a_2 = 1 + 2 \cdot 3 = 7 \\
a_4 &= a_2 + 2 a_3 = 3 + 2 \cdot 7 = 17 \\
a_5 &= a_3 + 2 a_4 = 7 + 2 \cdot 17 = 41 \\
\end{aligned}\]
- Notice that so far, all the elements of the sequence are odd.
- Suppose we want to prove this fact:
- Write: Theorem: \(a_n\) is odd, for all \(n \geq 1\).
- Proof: We refer to the predicate “\(a_n\) is odd” as \(P(n)\).
- Base cases: \(P(1)\) is true, since \(a_1 = 1\), which is odd. \(P(2)\) is also true, since \(a_2 = 3\), which is odd.
- Induction hypothesis: Let \(n \geq 2\) be arbitrary but fixed, and assume that \(P(m)\) holds whenever \( 1 \leq m \leq n \).
- That is, all elements of the sequence are odd from \(a_1\) to \(a_n\).
- Induction step: We will prove that if our induction hypothesis is true, then \(P(n+1)\) must be true, that is, \(a_{n+1}\) must be odd.
\[\begin{aligned}
a_{n+1} &= a_{n-1} + 2 a_n \\
&= (2p + 1) + 2(2q + 1) & \text{ (substitute by I.H.)} \\
&= 2(p + 2q + 1) + 1
\end{aligned}\]
- Thus, \(a_{n+1}\) is odd.
- Having proven the base cases and induction step, we have shown that \(P(n)\) is true for all \(n \geq 1\).
- Credit: Dr. Trefor Bazett via YouTube
Sequence Proof 2
- Let’s consider another sequence:
\(a_1 = 3\), \(a_2 = 5\), and for all \(n > 2\), \(a_n = 3a_{n-1} - 2a_{n-2}\)
- Let’s write a few elements of the sequence:
\[\begin{aligned}
a_3 &= 3 a_2 - 2 a_1 = 3 \cdot 5 - 2 \cdot 3 = 9 \\
a_4 &= 3 a_3 - 2 a_2 = 3 \cdot 9 - 2 \cdot 5 = 17 \\
a_5 &= 3 a_4 - 2 a_3 = 3 \cdot 17 - 2 \cdot 9 = 33 \\
\end{aligned}\]
- Ask: Does anyone see the pattern here? Think about powers of 2.
- Perhaps a closed form solution is: \(a_n = 2^n + 1\).
- Write: Theorem: \(a_n = 2^n + 1\) for all integers \(n \geq 1\).
- Write: Proof: Let \(P(n)\) be the predicate that \(a_n = 2^n + 1\). We will prove \(P(n)\) for all \(n \geq 1\).
- Base cases:
\[\begin{aligned}
P(1): a_1 &= 2^1 + 1 = 3 \\
P(2): a_2 &= 2^2 + 1 = 5 \\
\end{aligned}\]
- Induction hypothesis: Fix \(n \geq 2\), and assume that for all \(m\), \(1 \leq m \leq n\), \(P(m)\) holds. That is, \(a_m = 2^m + 1\).
- Induction step: Show \(P(n+1)\). That is, \(a_{n+1} = 2^{n+1} + 1\).
\[\begin{aligned}
a_{n+1} &= 3a_n - 2a_{n-1} \\
&= 3(2^n + 1) - 2(2^{n-1} + 1) & \text{ (by the I.H.)} \\
&= 3 \cdot 2^n + 3 - 2^n - 2 \\
&= 2 \cdot 2^n + 1 \\
&= 2^{n+1} + 1 \\
\end{aligned}\]
Measuring Cup Proof (Needed for Lab)
- Next, we will write an inductive proof where our base cases don’t start at 0 or 1.
- Write: Theorem: A set of measuring cups includes a 4-cup, 9-cup, 11-cup, and 14-cup measure. Show that this set can be used to measure out any number of cups greater than or equal to 11.
- Proof: Let \(P(n)\) be the predicate that \(n\) cups can be measured using the available cup sizes. We will show \(P(n)\) is true for any \(n \geq n_0 = 11\).
- Base Cases:
\[\begin{aligned}
P(11): 11 &= 11 \cdot 1 \\
P(12): 12 &= 4 \cdot 3 \\
P(13): 13 &= 4 \cdot 1 + 9 \cdot 1 \\
P(14): 14 &= 14 \cdot 1 \\
\end{aligned}\]
- Induction hypothesis: For some fixed but arbitrary \(n \geq n_1 = 14\), \(P(m)\) holds for all \(m\) such that \(n_0 \leq m \leq n\).
- Induction step: We will show \(P(n+1)\) given that \(P(m)\) is true for all \(m\) between \(n_0 = 11\) and \(n\) inclusive.
- By the I.H., \(P(n-3)\) is true, since \(n - 3 \geq n_0 = 11\). This means we can measure \(n - 3\) cups using our measures. Using an additional 4-cup scoop, we can measure \((n - 3) + 4 = n + 1\) cups. Thus, \(P(n+1)\) is true.
Coin Game
- Next, let’s use strong induction in a very different context.
- A simple game involves two players (Ali and Brad) and a pile of coins.
- Write: Ali moves first, and then the two players alternate.
- There are only two other rules.
- On each move, a player picks up one or two coins.
- The player who picks up the last coin loses.
- Who wins, if each player plays as well as possible?
- The winner depends on the number of coins in the pile at the beginning of the game: \(n\)
- Examples (draw coins and losing/winning for Ali):
- If \(n=1\), Ali loses, because she must take the last coin. \(n=1\) is a losing position for Ali.
- If \(n=2\), Ali can take one coin, and so force Brad into a losing position. \(n=2\) is a winning position for Ali.
- Note: If \(n\) is a losing position for Ali, then having that number of coins on the table at the start of your turn means you are in a losing position, whether you are Ali or Brad. And vice versa, for winning positions.
- If \(n=3\), Ali can take two coins, and so force Brad into a losing position. \(n=3\) is a winning position for Ali.
- If \(n=4\), either move Ali makes will cause her to lose. Regardless of whether she takes one coin or two coins, Brad is given a winning position for his next turn (three or two coins at the start of his next turn, respectively). \(n=4\) is a losing position for Ali.
- If \(n=5\), Ali can take one coin, and so force Brad into a losing position. \(n=5\) is a winning position for Ali.
- Ask: What is the pattern?
- Hint: Calculate the remainder when n is divided by 3.
- If n divided by 3 has remainder 0 or 2, n coins is a winning position.
- If n divided by 3 has remainder 1, n coins is a losing position.
- In a winning position, taking the correct number of coins will put the other player into the losing position.
- In the losing position, taking either number of coins will put the other player into the winning position.
Coin Game Proof
- Next, we will describe winning or losing positions in a theorem.
- Then, we will prove the theorem using strong induction.
- Write: Theorem: For any \(n \geq 1\), starting position with \(n\) coins is a losing position for Ali if and only if \(n = 3k + 1\) for some integer \(k\).
- That is, \(n\) is one more than a multiple of 3.
- Write: Proof: Let \(P(n)\) denote the predicate:
\(n\) coins is a losing position iff \(n = 3k + 1\)
- Base cases:
\(P(1)\): With one coin left, you must take it, so you will lose. \(1 = 3 \cdot 0 + 1 \).
\(P(2)\): With two coins left, you can take one, forcing your opponent to lose. \(2 \neq 3k + 1 \).
- Induction hypothesis: Fix \(n \geq 2\), and assume that for all \(m\) such that \( 1 \leq m \leq n\), \(P(m)\) holds. That is, \(m\) coins is a losing position for Ali iff \(m = 3k + 1\) (i.e., \(m\) leaves a remainder of 1 when divided by 3).
- Induction step: We will prove that if our I.H. is true, then \(P(n+1)\) must be true. That is, \(n+1\) coins is a losing position iff \(n + 1 = 3k + 1\).
- Suppose Ali has \(n+1\) coins where \(n \geq 2\). Ali can pick up either one or two coins, leaving \(n\) or \(n-1\) coins, respectively. In either case, the number of coins left will be in the range where our I.H. applies.
- Case 1: \(n + 1 = 3k + 1\). Ali taking one or two coins will leave remainders 0 or 2, respectively. By the I.H., these are winning positions for Brad. Thus, Case 1 is a losing position for Ali.
- Case 2: \(n + 1 = 3k + 0\). Ali taking two coins will leave remainder 1. By the I.H., this is a losing position for Brad. Thus, Case 2 is a winning position for Ali.
- Case 3: \(n + 1 = 3k + 2\). Ali taking one coin will leave remainder 1. By the I.H., this is a losing position for Brad. Thus, Case 3 is a winning position for Ali.
- \(P(n+1)\) is true in each case, so we have proven it.
- Having proven the base cases and induction step, we have shown that \(P(n)\) is true for all \(n \geq 1\).
Chocolate Game
I recommend reading the textbook’s description of another game and strong induction proof. This game involves breaking a chocolate bar, and trying to avoid getting left with the last piece.
Well-Ordering Principle
- Next, I will introduce an important mathematical principle
- Write: Well-Ordering Principle: Any nonempty set of nonnegative integers has a smallest element.
- This is related to the principle of mathematical induction, but that is beyond the scope of this course.
- We will use the Well-Ordering Principle to prove the FTA.
Fundamental Theorem of Arithmetic Proof
- Write: Fundamental Theorem of Arithmetic: Every integer \(n > 1\) has a unique prime decomposition.
- Proof: We will prove by induction on \(n\).
- Base case: \(n_0 = 2\). \(2 = 2^1\), and no other product of primes with positive exponents can equal 2.
- Induction hypothesis: Fix \(n \geq 2\), and suppose we know that for every \(m\), \(2 \leq m \leq n\) has a unique prime decomposition.
- Induction step: Consider \(n+1\).
- If \(n+1\) is prime, then \(n+1 = (n+1)^1\) is its unique prime decomposition.
- If \(n+1\) is not prime, let \(S\) be the set of all its factors greater than 1. By the Well-Ordering Principle, \(S\) has a smallest element \(p\), which must be prime; if it were not prime, any factor of \(p\) would be a smaller factor of \(n+1\).
- Let \(q = (n+1)/p\). Since \(q < n\), the induction hypothesis says \(q\) has a unique prime decomposition:
\[ q = p_1^{e_1} \cdot \ldots \cdot p_k^{e_k} \]
- \(p\) is either \(p_1\) or is a prime smaller than \(p_1\), which we call \(p_0\).
- Case 1: \(p = p_1\).
\[ n+1 = p_1^{e_1+1} \cdot \ldots \cdot p_k^{e_k} \]
- Case 2: \(p = p_0\).
\[ n+1 = p_0^1 \cdot p_1^{e_1} \cdot \ldots \cdot p_k^{e_k} \]
- Having proven the base case and induction step, we have shown all integers have prime decompositions.
- We haven’t shown the prime decompositions are unique, so this proof is incomplete.
Closing Thoughts
- Anything that can be proved using strong induction could also be proved using ordinary induction.