CSE 311 Lecture 16: Induction

Topics

Mathematical induction

A method for proving statements about all natural numbers.

Using induction

Using induction in formal and English proofs.

Example proofs by induction

Example proofs about sums and divisibility.

Induction starting at any integer

Proving theorems about all integers $n\geq b$ for some $b\in\mathbb{Z}$.

Strong induction

Induction with a stronger hypothesis.

Using strong induction

An example proof and when to use strong induction.

How would you prove this theorem?

Mods and exponents

For all integers $a, b, m > 0$ and $k \geq 0$, $\congruent{a}{b}{m} \rightarrow \congruent{a^k}{b^k}{m}$.

We don’t have a proof rule to say “perform this step repeatedly.”

Perform a step repeatedly with induction!

Induction is a logical rule of inference that applies (only) over $\N$.

If we know that a property $P$ holds for 0, and

we know that $\forall k. P(k)\rightarrow P(k+1)$, then

we can conclude that $P$ holds for all natural numbers.

// f(x) = x for all x >= 0.
public int f(int x) {
  if (x == 0) { return 0; }
  else        { return f(x - 1) + 1; }
}

Induction: how does it work?

Suppose that we are given $P(0)$ and $\forall k. P(k)\rightarrow P(k+1)$.

How does that give us $P(k)$ for a concrete $k$ such as $5$?

Using the induction rule in a formal proof

Using the induction rule in a formal proof: key parts

Translating to an English proof: the template

What is $\sum_{i=0}^{n}2^i$ for an arbitrary $n\in\N$?

Recall that $\sum_{i=0}^{n}2^i = 2^0 + 2^1 + \ldots + 2^n$.

Prove $\sum_{i=0}^{n}2^i = 2^{n+1}-1$ for all $n\in\N$

Prove $\sum_{i=0}^{n}i = n(n+1)/2$ for all $n\in\N$

What number divides $2^{2n}-1$ for every $n\in\N$?

Prove $3 \vert (2^{2n}-1)$ for all $n\in\N$

Changing the start line

How can we prove $P(n)$ for all integers $n\geq b$ for some integer $b$?

Define a predicate $Q(n) = P(n+b)$ for all $n\geq0$.

Then $(\forall n. Q(n))\equiv(\forall n \geq b. P(n))$

Use ordinary induction to prove $Q$:

Prove $Q(0) \equiv P(b)$.

Prove $(\forall k. Q(k)\rightarrow Q(k+1)) \equiv (\forall k\geq b. P(k)\rightarrow P(k+1))$.

By convention, we don’t define $Q$ explicitly. Instead, we modify our proof template to account for the non-zero base case $b$.

Inductive proofs for any base case $b\in\Z$

① Let $P(n)$ be [ definition of $P(n)$ ].

We will show that $P(n)$ is true for every integer $n\geq b$ by induction.

② Base case ($n = b$):

[ Proof of $P(b)$. ]

③ Inductive hypothesis:

Suppose that $P(k)$ is true for an arbitrary integer $k\geq b$.

④ Inductive step:

We want to prove that $P(k+1)$ is true.

[ Proof of $P(k+1)$. This proof must invoke the inductive hypothesis. ]

⑤ The result follows for all $n\geq b$ by induction.

 

Example: prove $3^n \geq n^2 + 3$ for all $n\geq 2$

Recall how induction works

Note that we have $P(0), \ldots, P(k)$ when proving $k+1$.
So we can safely assume all of them, rather than just $P(k)$.

The strong induction rule of inference

Strong induction for $P$ follows from ordinary induction for $Q$ where

$Q(k) = P(0) \wedge P(1) \wedge P(2) \wedge \ldots \wedge P(k)$

To see why, note the following:

$Q(0) \equiv P(0)$

$Q(k+1) \equiv Q(k) \wedge P(k+1)$

$(\forall n. Q(n))\equiv(\forall n. P(n))$

Strong inductive proofs for any base case $b\in\Z$

① Let $P(n)$ be [ definition of $P(n)$ ].

We will show that $P(n)$ is true for every integer $n\geq b$ by strong induction.

② Base case ($n = b$):

[ Proof of $P(b)$. ]

③ Inductive hypothesis:

Suppose that for some arbitrary integer $k\geq b$, $P(j)$ is true for every integer $b \leq j \leq k$.

④ Inductive step:

We want to prove that $P(k+1)$ is true.

[ Proof of $P(k+1)$. The proof must invoke the strong inductive hypothesis. ]

⑤ The result follows for all $n\geq b$ by strong induction.

 

Example: the fundamental theorem of arithmetic

Examples

$48= 2\cdot 2\cdot 2\cdot 2\cdot 3$

$591 = 3 \cdot 197$

$45,523 = 45,523$

$321,950 = 2 \cdot 5 \cdot 5 \cdot 47 \cdot 137$

$1,234,567,890 = 2 \cdot 3 \cdot 3 \cdot 5 \cdot 3,607 \cdot 3,803$

We will use strong induction to prove that a factorization into primes exists (but not that it is unique).

Prove that every integer $\geq 2$ is a product of primes

Strong induction is particularly useful when …

We need to reason about procedures that given an input $k$ invoke themselves recursively on an input different from $k-1$.

Example:

Euclidean algorithm for computing $\gcd{a}{b}$.

// Assumes a >= b >= 0.
public static int gcd(int a, int b) {
  if (b == 0)
    return a;             // GCD(a, 0) = a
  else  
    return gcd(b, a % b); // GCD(a, b) = GCD(b, a mod b)
}

We use strong induction to reason about this algorithm and other functions with recursive definitions.