MATH 308 — Lec 11 — Mathematical Induction

Lecture 11 — Mathematical Induction
MATH 308  Bridge to Advanced Mathematics  Chartrand §6.1–6.2

Goals for today. By the end of this lecture you should be able to:

• state the Principle of Mathematical Induction and explain the domino/ladder intuition behind it;

• structure an induction proof into a base case and an inductive step, naming the inductive hypothesis;

• prove a summation formula, a divisibility fact, and an inequality by induction;

• prove $\left\lvert \mathcal{P}(A) \right\rvert = 2^n$∣P(A)∣=2n for a set with $n$n elements — the fact we promised back in Lecture 1.

The Principle of Mathematical Induction (§6.1)

Suppose we have, for each $n \in \mathbb{N}$n∈N, a statement $P(n)$P(n) — one claim per natural number. For example, “$P(n)$P(n): the sum of the first $n$n odd numbers is $n^2$n2.” We want to prove that all of $P(1), P(2), P(3), \dots$P(1),P(2),P(3),… are true. We cannot check infinitely many statements one at a time. Induction gives us a way to capture them all with two finite checks.

The picture to keep in mind is a row of dominoes, or the rungs of an infinite ladder. To knock down every domino it is enough to know two things: (1) the first one falls, and (2) whenever one falls, it topples the next. Then domino $1$1 falls, so domino $2$2 falls, so domino $3$3 falls, … — and no domino is left standing.

Induction as falling dominoes: push the first (base case) and guarantee each topples its successor (inductive step); then every domino falls.

Theorem 1 (Principle of Mathematical Induction, PMI). Let $P(n)$P(n) be a statement for each $n \in \mathbb{N}$n∈N. Suppose:

1. (base case) $P(1)$P(1) is true; and

2. (inductive step) for every $k \in \mathbb{N}$k∈N,  if $P(k)$P(k) is true then $P(k+1)$P(k+1) is true.

Then $P(n)$P(n) is true for every $n \in \mathbb{N}$n∈N.

The assumption “$P(k)$P(k) is true” that we get to use inside the inductive step has a name: the inductive hypothesis. The whole art of an induction proof is to start from the case $k$k and reduce the case $k+1$k+1 back to it.

Remark 1. Induction need not start at $1$1. If the base case is proved at some integer $n_0$n0​ (possibly $0$0 or negative) and the inductive step runs for all $k \ge n_0$k≥n0​, then $P(n)$P(n) holds for all integers $n \ge n_0$n≥n0​. We use $n_0 = 0$n0​=0 in Theorem 5 below.

A summation formula (§6.1)

Proposition 2. *For every $n \in \mathbb{N}$n∈N,

$1 + 2 + 3 + \cdots + n \;=\; \frac{n(n+1)}{2}.$1+2+3+⋯+n=2n(n+1)​.

Proof. Let $P(n)$P(n) be the displayed equation.

Base case ($n=1$n=1): the left side is $1$1 and the right side is $\frac{1\cdot 2}{2}=1$21⋅2​=1. So $P(1)$P(1) holds.

Inductive step: fix $k \in \mathbb{N}$k∈N and assume $P(k)$P(k): that $1+2+\cdots+k = \frac{k(k+1)}{2}$1+2+⋯+k=2k(k+1)​ (this is the inductive hypothesis). We must deduce $P(k+1)$P(k+1). Adding $k+1$k+1 to both sides,

$\underbrace{1+2+\cdots+k}_{=\,\frac{k(k+1)}{2}} + (k+1) \;=\; \frac{k(k+1)}{2} + (k+1) \;=\; (k+1)\!\left(\frac{k}{2}+1\right) \;=\; \frac{(k+1)(k+2)}{2}.$=2k(k+1)​1+2+⋯+k​​+(k+1)=2k(k+1)​+(k+1)=(k+1)(2k​+1)=2(k+1)(k+2)​.

The last expression is exactly the right-hand side of $P(k+1)$P(k+1) (namely $\frac{(k+1)\bigl((k+1)+1\bigr)}{2}$2(k+1)((k+1)+1)​). Thus $P(k)\Rightarrow P(k+1)$P(k)⇒P(k+1). By the PMI, $P(n)$P(n) holds for all $n \in \mathbb{N}$n∈N. ◻

There is also a wordless reason this formula is true. Take two copies of the staircase $1+2+\cdots+n$1+2+⋯+n and fit them together into an $n \times (n+1)$n×(n+1) rectangle of dots. The rectangle has $n(n+1)$n(n+1) dots, so one staircase has half that.

Two staircases (blue $=1{+}2{+}3{+}4$=1+2+3+4, red its mirror) tile a $4\times 5$4×5 grid: $2\!\sum_{i=1}^{n} i = n(n+1)$2∑i=1n​i=n(n+1).

A divisibility result (§6.2)

Recall from Lecture 7 that $d \mid m$d∣m (“$d$d divides $m$m”) means $m = d\,q$m=dq for some integer $q$q.

Proposition 3. For every $n \in \mathbb{N}$n∈N,  $3 \mid (4^n - 1)$3∣(4n−1).

Proof. Let $P(n)$P(n) be the statement “$3 \mid (4^n - 1)$3∣(4n−1).”

Base case ($n=1$n=1): $4^1 - 1 = 3 = 3\cdot 1$41−1=3=3⋅1, so $3 \mid (4^1-1)$3∣(41−1) and $P(1)$P(1) holds.

Inductive step: fix $k \in \mathbb{N}$k∈N and assume $P(k)$P(k), i.e. $4^k - 1 = 3q$4k−1=3q for some $q \in \mathbb{Z}$q∈Z. Then

$4^{k+1} - 1 \;=\; 4\cdot 4^k - 1 \;=\; 4(4^k - 1) + 3 \;=\; 4(3q) + 3 \;=\; 3\,(4q + 1).$4k+1−1=4⋅4k−1=4(4k−1)+3=4(3q)+3=3(4q+1).

Since $4q+1 \in \mathbb{Z}$4q+1∈Z, this shows $3 \mid (4^{k+1}-1)$3∣(4k+1−1), which is $P(k+1)$P(k+1). By the PMI, $P(n)$P(n) holds for all $n \in \mathbb{N}$n∈N. ◻

The trick to spot: rewrite the “$k+1$k+1” quantity so that the “$k$k” quantity (which the hypothesis controls) appears explicitly, with the leftover being visibly a multiple of $3$3. That single algebraic move — $4\cdot 4^k - 1 = 4(4^k-1)+3$4⋅4k−1=4(4k−1)+3 — is the entire inductive step.

An inequality (§6.2)

Proposition 4. For every $n \in \mathbb{N}$n∈N,  $2^n > n$2n>n.

Proof. Let $P(n)$P(n) be “$2^n > n$2n>n.”

Base case ($n=1$n=1): $2^1 = 2 > 1$21=2>1, so $P(1)$P(1) holds.

Inductive step: fix $k \in \mathbb{N}$k∈N and assume $P(k)$P(k): $2^k > k$2k>k. Then, doubling,

$2^{k+1} = 2\cdot 2^k > 2k = k + k \ge k + 1,$2k+1=2⋅2k>2k=k+k≥k+1,

where the last step uses $k \ge 1$k≥1. Hence $2^{k+1} > k+1$2k+1>k+1, which is $P(k+1)$P(k+1). By the PMI, $2^n > n$2n>n for all $n \in \mathbb{N}$n∈N. ◻

Remark 2. Notice we used $k \ge 1$k≥1 to turn $2k$2k into “at least $k+1$k+1.” Inequalities often need such slack from the domain; keeping the strict “$>$>” through the chain is what delivers a strict conclusion.

The size of a power set

We can finally settle a claim left open in Lecture 1: a set with $n$n elements has exactly $2^n$2n subsets.

Theorem 5. If $A$A is a finite set with $\left\lvert A \right\rvert = n$∣A∣=n, then $\left\lvert \mathcal{P}(A) \right\rvert = 2^n$∣P(A)∣=2n.

Proof. We induct on $n = \left\lvert A \right\rvert$n=∣A∣, starting the base case at $n = 0$n=0. Let $P(n)$P(n) be: “every set with $n$n elements has exactly $2^n$2n subsets.”

Base case ($n=0$n=0): the only set with $0$0 elements is $\varnothing$∅, and its only subset is $\varnothing$∅ itself, so $\mathcal{P}(\varnothing) = \{\varnothing\}$P(∅)={∅} and $\left\lvert \mathcal{P}(\varnothing) \right\rvert = 1 = 2^0$∣P(∅)∣=1=20. Thus $P(0)$P(0) holds.

Inductive step: fix $k \ge 0$k≥0 and assume $P(k)$P(k): every $k$k-element set has $2^k$2k subsets. Let $A$A be any set with $\left\lvert A \right\rvert = k+1$∣A∣=k+1. Pick one element $a \in A$a∈A and set $A' = A - \{a\}$A′=A−{a}, so $\left\lvert A' \right\rvert = k$∣A′∣=k. Now split the subsets of $A$A into two kinds:

• those that do not contain $a$a: these are exactly the subsets of $A'$A′, and there are $2^k$2k of them by the inductive hypothesis;

• those that do contain $a$a: each is of the form $X \cup \{a\}$X∪{a} for a unique $X \subseteq A'$X⊆A′, so there are again $2^k$2k of them.

Every subset of $A$A is of exactly one kind (it either omits $a$a or contains it), so these two families are disjoint and together exhaust $\mathcal{P}(A)$P(A). Therefore

$\left\lvert \mathcal{P}(A) \right\rvert = 2^k + 2^k = 2\cdot 2^k = 2^{k+1},$∣P(A)∣=2k+2k=2⋅2k=2k+1,

which is $P(k+1)$P(k+1). By the PMI, $\left\lvert \mathcal{P}(A) \right\rvert = 2^{\left\lvert A \right\rvert}$∣P(A)∣=2∣A∣ for every finite set $A$A. ◻

The doubling “$2^k + 2^k = 2^{k+1}$2k+2k=2k+1” is the heart of it: adding one new element $a$a to a set gives every old subset a choice — take $a$a or leave it — so the number of subsets doubles at each step.

Building $\mathcal{P}(A)$P(A) by adding one element at a time. Each new element doubles the count: $1 \to 2 \to 4 = 2^2$1→2→4=22 for $A=\{a,b\}$A={a,b}.

Remark 3. Two common pitfalls. (1) No base case. A valid inductive step alone proves nothing: dominoes that each topple the next still all stand if none is pushed. (2) Using what you’re proving. In the inductive step you may assume only $P(k)$P(k), never $P(k+1)$P(k+1) — that would be assuming the conclusion. We sharpen the hypothesis (assume all of $P(1),\dots,P(k)$P(1),…,P(k)) in strong induction, Lecture 12.

Homework problems

1. Prove by induction that $1 + 3 + 5 + \cdots + (2n-1) = n^2$1+3+5+⋯+(2n−1)=n2 for all $n \in \mathbb{N}$n∈N.

2. Prove that $3 \mid (n^3 - n)$3∣(n3−n) for all $n \in \mathbb{N}$n∈N. (Rewrite $(k+1)^3-(k+1)$(k+1)3−(k+1) so that $k^3-k$k3−k appears.)

3. Prove that $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$i=1∑n​i2=6n(n+1)(2n+1)​ for all $n \in \mathbb{N}$n∈N.

4. For which $n \in \mathbb{N}$n∈N is $2^n > n^2$2n>n2? Make a conjecture from small cases, then prove it by induction (mind the base case).

5. Where exactly does the “proof” that all horses are the same color break down? (It inducts on the size of a set of horses.) Which step fails, and for which $k$k?

Suggested textbook exercises (Chartrand §6.1–6.2): 6.2, 6.6, 6.10, 6.14, 6.20, 6.26.