CSE 311 Lecture 16: Induction

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Topics

Mathematical induction: A brief review of Lecture 15.
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 $b\in\mathbb{Z}$.
Induction with a stricter hypothesis: Proving theorems by induction on a stricter hypothesis.

Mathematical induction

A brief review of Lecture 15.

What is induction?

Induction$\rule{P(0); \forall k. P(k)\rightarrow P(k+1)}{\forall n. P(n)}$

Domain: natural numbers ($\N$).

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.

Induction: how does it work?

Induction$\rule{P(0); \forall k. P(k)\rightarrow P(k+1)}{\forall n. P(n)}$

Domain: natural numbers ($\N$).

How do we get $P(5)$ from $P(0)$ and $\forall k. P(k)\rightarrow P(k+1)$?

1.	First, we have $P(0)$.	$P(0)$
2.	Since $P(k)\rightarrow P(k+1)$ for all $k$, we have $P(0)\rightarrow P(1)$.	$\ \Downarrow_{\ P(0)\rightarrow P(1)}$
3.	Applying Modus Ponens to 1 and 2, we get $P(1)$.	$P(1)$
4.	Since $P(k)\rightarrow P(k+1)$ for all $k$, we have $P(1)\rightarrow P(2)$.	$\ \Downarrow_{\ P(1)\rightarrow P(2)}$
5.	Applying Modus Ponens to 3 and 4, we get $P(2)$.	$P(2)$
$\vdots$		$\ \Downarrow_{\ P(k)\rightarrow P(k+1)}$
11.	Applying Modus Ponens to 9 and 10, we get $P(5)$.	$P(5)$

Using the induction rule in a formal proof: key parts

Induction$\rule{P(0); \forall k. P(k)\rightarrow P(k+1)}{\forall n. P(n)}$

Prove $P(0)$

Base case

Let $k\geq0$ be an arbitrary integer

3.1.

Assume that $P(k)$ is true

Inductive
hypothesis

3.2.	$\ldots$
3.3.	Prove $P(k+1)$ is true

Inductive
step

4.	$P(k) \rightarrow P(k+1)$	Direct Proof Rule
5.	$\forall k. P(k) \rightarrow P(k+1)$	Intro $\forall$: 2, 4
6.	$\forall n. P(n)$	Induction: 1, 5

Conclusion

Translating to an English proof: the template

① Let $P(n)$ be [ definition of $P(n)$ ].: We will show that $P(n)$ is true for every integer $n\geq 0$ by induction.
② Base case ($n=0$):: [ Proof of $P(0)$. ]
③ Inductive hypothesis:: Suppose that $P(k)$ is true for an arbitrary integer $k\geq 0$.
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 somewhere. ]
⑤ The result follows for all $n\geq 0$ by induction.

Prove $P(0)$

Base case

Let $k\geq0$ be an arbitrary integer

3.1.

Assume that $P(k)$ is true

Inductive
hypothesis

3.2.	$\ldots$
3.3.	Prove $P(k+1)$ is true

Inductive
step

4.	$P(k) \rightarrow P(k+1)$	Direct Proof Rule
5.	$\forall k. P(k) \rightarrow P(k+1)$	Intro $\forall$: 2, 4
6.	$\forall n. P(n)$	Induction: 1, 5

Conclusion

Example proofs by induction

Example proofs about sums and divisibility.

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

① Let $P(n)$ be $\sum_{i=0}^{n}i = 0 + 1 + \ldots + n = n(n+1)/2$.: We will show that $P(n)$ is true for every integer $n\geq 0$ by induction.
② Base case ($n = 0$):: $\sum_{i=0}^{n}i = 0 = 0(0+1)/2$ so $P(0)$ is true.
③ Inductive hypothesis:: Suppose that $P(k)$ is true for an arbitrary integer $k\geq 0$.
④ Inductive step:: We want to prove that $P(k+1)$ is true, i.e., $\sum_{i=0}^{k+1}i = (k+1)(k+2)/2$. Note that $\sum_{i=0}^{k+1}i = ($$\sum_{i=0}^{k}i$$) + (k + 1) = ($$k(k+1)/2$$) + (k+1)$ by the inductive hypothesis. From this, we have that $(k(k+1)/2) + (k+1) =$ $(k+1)(k/2 + 1) =$ $(k+1)(k+2)/2$, which is exactly $P(k+1)$.
⑤ The result follows for all $n\geq 0$ by induction.

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

Let’s look at a few examples:: $2^{2*0} - 1 = 1 - 1 = 0 = 3 * 0$; $2^{2*1} - 1 = 4 - 1 = 3 = 3 * 1$; $2^{2*2} - 1 = 16 - 1 = 15 = 3 * 5$; $2^{2*3} - 1 = 64 - 1 = 63 = 3 * 21$; $2^{2*4} - 1 = 256 - 1 = 255 = 3 * 85$

It looks like $3 \vert (2^{2n}-1)$.: Let’s use induction to prove it!

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

① Let $P(n)$ be $3 \vert (2^{2n}-1)$.: We will show that $P(n)$ is true for every integer $n\geq 0$ by induction.
② Base case ($n = 0$):: $2^{2 * 0} - 1 = 1 - 1 = 0 = 3*0$ so $P(0)$ is true.
③ Inductive hypothesis:: Suppose that $P(k)$ is true for an arbitrary integer $k\geq 0$.
④ Inductive step:: We want to prove that $P(k+1)$ is true, i.e., $3 \vert (2^{2(k+1)}-1)$. By inductive hypothesis, $3 \vert (2^{2k}-1)$ so $2^{2k}-1 = 3j$ for some integer $j$. We therefore have that $2^{2(k+1)} - 1 $ $=$ $2^{2k+2} - 1$ $=$ $4($$2^{2k}$$) - 1$ $=$ $4($$3j+1$$) - 1$ $=$ $12j + 3 = 3(4j + 1)$. So $3 \vert (2^{2(k+1)}-1)$, which is exactly $P(k+1)$.
⑤ The result follows for all $n\geq 0$ by induction.

Induction starting at any integer

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

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 $n\geq 2$

① Let $P(n)$ be $3^n \geq n^2 + 3$.: We will show that $P(n)$ is true for every integer $n\geq 2$ by induction.
② Base case ($n = 2$):: $3^2 = 9 \geq 7 = 4 + 3 = 2^2 + 3$ so $P(2)$ is true.
③ Inductive hypothesis:: Suppose that $P(k)$ is true for an arbitrary integer $k\geq 2$.
④ Inductive step:: We want to prove that $P(k+1)$ is true, i.e., $3^{(k+1)} \geq (k+1)^2 + 3$ $=$ $k^2 + 2k + 4$. Note that $3^{(k+1)} = 3($$3^k$$)$ $\geq$ $3($$k^2+3$$)$ by the inductive hypothesis. From this we have $3(k^2+3) = 2k^2 + k^2 + 9$ $\geq$ $k^2 + 2k + 4$ $=$ $(k+1)^2 + 3$ since $k\geq 2$. Therefore $P(k+1)$ is true.
⑤ The result follows for all $n\geq 2$ by induction.

Induction with a stricter hypothesis

Proving theorems by induction on a stricter hypothesis.

Changing the hypothesis

Can we use induction to prove $P(n)$ if we can’t show that $P(k) \to P(k+1)$ in the inductive step?: Usually, but it requires a creative leap!
Find a predicate $Q(n)$ such that: $Q(n) \to P(n)$, $Q(0)$, and $Q(k) \to Q(k+1)$.
Then, we can prove $Q(n)$ by induction and get $P(n)$ by MP.: We call this using a stricter (or stronger) hypothesis.

Example: prove $\sum_{i=1}^{n}1/i^2 < 2$ for all $n\geq 1$

① Let $P(n)$ be $\sum_{i=1}^{n}1/i^2 < 2$.: We will try to show that $P(n)$ is true for every integer $n\geq 1$ by induction.
② Base case ($n = 1$):: $\sum_{i=1}^{1}1/i^2 = 1/1^2 = 1 < 2$ so $P(1)$ is true.
③ Inductive hypothesis:: Suppose that $P(k)$ is true for an arbitrary integer $k\geq 1$.
④ Inductive step:: We want to prove that $P(k+1)$ is true, i.e, $\sum_{i=1}^{k+1}1/i^2 < 2$. Note that $\sum_{i=1}^{k+1}1/i^2 = $$\sum_{i=1}^{k}1/i^2$$ + 1/(k+1)^2$. By the inductive hypothesis, we know that $\sum_{i=1}^{k}1/i^2 < 2$, and from this we can conclude that $\sum_{i=1}^{k}1/i^2 + 1/(k+1)^2 < 2 + 1/(k+1)^2$. But this is not $P(k+1)$!

We need a stricter hypothesis! Let’s try $Q(n) := \sum_{i=1}^{n}1/i^2 \leq 2 - 1/n$, noting that $Q(n) \to P(n)$ for all $n >= 1$.

Prove instead $\sum_{i=1}^{n}1/i^2 \leq 2 - 1/n$ for all $n\geq 1$

① Let $Q(n)$ be $\sum_{i=1}^{n}1/i^2 \leq 2 - 1/n$.: We will show that $Q(n)$ is true for every integer $n\geq 1$ by induction.
② Base case ($n = 1$):: $\sum_{i=1}^{1}1/i^2 = 1/1^2 = 1 \leq 2 - 1/1 = 1$ so $Q(1)$ is true.
③ Inductive hypothesis:: Suppose that $Q(k)$ is true for an arbitrary integer $k\geq 1$.
④ Inductive step:: We want to prove that $Q(k+1)$ is true, i.e, $\sum_{i=1}^{k+1}1/i^2 \leq 2 - 1/(k+1)$. Note that $\sum_{i=1}^{k+1}1/i^2 = $$\sum_{i=1}^{k}1/i^2$$ + 1/(k+1)^2$. By the inductive hypothesis, we know that $\sum_{i=1}^{k}1/i^2 \leq 2 - 1/k$. Adding $ 1/(k+1)^2$ to both sides, we get $\sum_{i=1}^{k}\frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k} + \frac{1}{(k+1)^2} = 2 - \frac{(k+1)^2-k}{k(k+1)^2} = 2 - \frac{k^2 + k + 1}{k(k+1)^2} = 2 - \frac{k(k + 1) + 1}{k(k+1)^2}$. Note that $\frac{k(k + 1)}{k(k+1)^2} < \frac{k(k + 1)+1}{k(k+1)^2}$ when $k\geq1$, so $\sum_{i=1}^{k}\frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{k(k + 1) + 1}{k(k+1)^2} < 2 - \frac{k(k + 1)}{k(k+1)^2}$ $= 2 - \frac{k+1}{(k+1)^2} = 2 - \frac{1}{k+1}$. This shows that $Q(k+1)$ holds.
⑤ The result follows for all $n\geq 1$ by induction.

Summary

Induction lets us prove statements about all natural numbers.: A proof by induction must show that $P(0)$ is true (base case).; And it must use the inductive hypothesis $P(k)$ to show that $P(k+1)$ is true (inductive step).
Induction also lets us prove theorems about integers $n\geq b$ for $b\in\Z$.: Adjust all parts of the proof to use $n\geq b$ instead of $n\geq 0$.
Induction proofs can use a hypothesis that implies $P(k)$.: It is not always possible to use $P(k)$ as the induction hypothesis.; In those cases, we use induction to prove $Q(n)$, where $Q(n) \to P(n)$.