3.2 Complete Induction and Well-Ordering Principle

Objectives:

•

Utilize the principle of complete induction in proving mathematical statements
•

Apply induction to a new theorem on the natural numbers

3.2.1 Building Complete Induction

Let’s revisit our infinite list of items to complete from last time:

1.

Prove that $P(1)$ is true. (Prove a base case is true.)
2.

Prove that, if $P(1)$ is true, then $P(2)$ is true (i.e., $P(1)\Rightarrow P(2)$ )
3.

Prove that, if $P(2)$ is true, then $P(3)$ is true (i.e., $P(2)\Rightarrow P(3)$ )…….
608981813029.

Prove that $P(608981813028)\Rightarrow P(608981813029)$ .

From Steps 2 onward we are proving that, if the statement from the previous step has been proven true, then the next step in the sequence should also be true. This is known as the domino effect: if one domino $P(n)$ gets knocked down $P(n)$ , then so does the next domino $P(n+1)$ .

Let’s say that we are on Step 3. We know that $P(1)$ is true from Step 1 and that $P(2)$ is true from Step 2. While we are proving the statement $P(3)$ , we come to an impasse: $P(2)$ does not necessarily imply $P(3)$ ! But then we start breathing easily again, because $P(1)$ implies $P(3)$ . Since at Step 3 we know both $P(1)$ and $P(2)$ are true we can use one or both of those statements to prove that $P(3)$ is true.

The same goes for every step: given our algorithm, if we are at Step 5 trying to show $P(5)$ is true, then we can use any combination of the statements $P(1)$ , $P(2)$ , $P(3)$ and $P(4)$ to prove this statement. This becomes incredibly helpful in some instances and leads us to our next theorem:

Theorem 3.2.1.

Let $P(n)$ be a predicate over the natural numbers. Assume the following:

1.

the statement $P(1)$ is true, and
2.

for all $k\in\mathbb{N}$ , if $P(j)$ is true for all $j\leq k$ , then $P(k+1)$ is true.

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

Example 3.2.2.

Define a sequence of numbers by $a_{1}=1$ , $a_{2}=3$ , and $a_{n}=3a_{n-1}-2a_{n-2}$ for $n$ being a natural number greater than 2. Then show that $a_{n}=2^{n}+1$ for all $n\in\mathbb{N}$ . (Consider how many base cases you have to verify before moving onto the inductive step.)

We can generalize this one step further. In an elegant solution like that displayed in the Mr. Beast video from the previous section, we only need to knock over one domino for the rest to get knocked over. Sometimes the dominoes don’t work as nicely and we have to knock over the first couple of dominoes ourselves. But our process totally allows for this!

Theorem 3.2.3 Complete Induction.

Let $P(n)$ be a predicate over the natural numbers. Assume the following:

1.

the statements $P(1),P(2),\dots,P(m)$ are true, and
2.

for all $k\geq m$ , if $P(j)$ is true for all $m\leq j\leq k$ , then $P(k+1)$ is true.

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

Example 3.2.4.

Prove that every amount of postage that is at least 12 cents can be made from 4-cent and 5-cent stamps. (Hint: first prove that the postage amounts of 12 cents, 13 cents, 14 cents, and 15 cents can be done. Then come up with an inductive step.)

To get to our next major example, we first need to make a definition that will be prevalent in Chapter 4:

Definition 3.2.5.

A set $A$ is a collection of elements.¹¹1This definition is quite simple - it turns out to be too simple due to a concept called Russell’s Paradox. We will not run into anything that makes this definition untenable for our working purposes, however. A subset $B$ of $A$ is a set containing some (or none, or all) the elements of $A$ . We write $B\subseteq A$ (LaTeX $\backslash$ subseteq) and say “ $B$ is a subset of $A$ ”.

Example 3.2.6.

Find two or three meaningfully different subsets of $\mathbb{N}$ .

Now pick two or three more proper subsets of $\mathbb{N}$ with infinitely many elements. (A proper subset of $A$ is a subset which is not equal to $A$ .)

Example 3.2.7.

Of the examples you gave above, what is the least element of each subset?

Can you come up with an example of a subset of $\mathbb{N}$ with no least element?

Example 3.2.8.

Perhaps we can convince ourselves in the following statement, keeping a healthy amount of skepticism about us: “Every nonempty subset of the natural numbers has a least element.” Why is the term “nonempty” important in this statement?

Example 3.2.9.

Throughout let’s call $S$ a subset of the natural numbers. Say that I were working with the empty set $S$ . Would the following statements be true or false? “ $1$ is an element of $S$ ”, “ $2$ is an element of $S$ ”, “ $3$ is an element of $S$ ”,…

Example 3.2.10.

Now say that we have a set $S$ with no least element? Would the following statements be true or false? This time they are displayed in order - answer each one having answered the previous one, and give your reasoning.

(a)

$1$ is an element of $S$ .

Blank space for you to write your work.
(b)

$2$ is an element of $S$ .

Blank space for you to write your work.
(c)

$3$ is an element of $S$ .

Blank space for you to write your work.

Example 3.2.11.

Look at the previous two examples. What must be true about a set with no least element? What elements can that set contain?

Example 3.2.12.

Write your conclusion from the previous statement as a conditional statement. Then take the contrapositive of that statement. What do you notice?

Example 3.2.13.

We are almost ready to formalize our proof. What would be a good method of proof? So far we have learned “direct”, “contrapositive”, “contradiction”, “induction”, and “complete induction”. Explain your reasoning for your choice - the correct answer is actually a combination of these methods, so more than one of the proof methods points toward the correct answer.

Theorem 3.2.14 Well-Ordering Principle.

Every nonempty subset of the natural numbers has a least element.

Example 3.2.15.

Use the Well-Ordering Principle to prove the following: Every positive integer greater than one has a prime factorization - i.e., it can be factored into a product of primes (see Section 2.3). This is the Fundamental Theorem of Arithmetic. A “proof with gaps” is given below.

Let $C$ be the set of positive integers greater than 1 that cannot be factored into a product of primes.

We would like to show that $C$ is empty, as this would prove our statement, so assume $C$ is non-empty and proceed by contradiction.

We apply the Well-Ordering Principle to $C$ in order to find a least element $n$ in $C$ .

Derive a contradiction if $n$ is prime with $n$ being in $C$ .

Derive a contradiction if $n$ is composite with $n$ being in $C$ . (Here you should use that $n$ is the least element in $C$ .) Then conclude the proof.