Induction is a way to prove that a statement is true for all $NN$. Formally, it
is introduced like this: say you have a set $S subset.eq NN$ such that $1 in
S$. Then, if $n in S => n + 1 in S$, $S = NN$. Usually we plug $n$ into a
proposition $P$ to make it more useful for proving statements. If $(forall n in
S)(P(n))$, and we can show $S = NN$ by induction, then $P(n)$ is true for all
$n in NN$.
Induction consists of two main parts. First, we have to show that $1 in S$.
This is the _base case_. Then, we assume the _inductive hypothesis_, $n in S$.
If we can prove, using the inductive hypothesis, that $n in S$ implies that $n
+ 1$ in $S$, then we'll have sufficient justification for $S = N$. When proving
a statement in practice, if $S$ is the set of all inputs to $P(n)$ that make it
true, we show that $P(1)$ is true ($1 in S$) and then show that $P(n) =>
P(n+1)$, which implies $(forall n in NN) P(n)$.
There are two equivalent formulations of induction, the principle of
mathematical induction (PMI, also called _weak induction_), and the principle
of complete induction (PCI, also called _strong induction_). They are
equivalent, but in some cases one is much more straightforward to show than the
other.
The previous strategy we discussed was _weak induction_, named such because we
need only show $P(1) and P(n) => P(n+1)$. _Strong induction_ requires us to
show a much stronger result, but also allows us to make a stronger assumption.
#definition[Strong induction][
If ${1,2,3,...,n-1} subset.eq S => n in S$, then $S = NN$.
]
Essentially in strong induction we assume that every natural up to $n$ is in
$S$, and if we can show that this implies $n$ is also in $S$, we have $S = NN$.
Note that in this case we do not have to show $1 in S$, but this is implied.
When proving a proposition $P$, we have to prove $P(1) and P(2) and ... and
P(n-1) => P(n)$, but we can assume the strong inductive hypothesis $P(1) and
P(2) and ... and P(n-1)$.
#example[Game theory of Nim][
In the game of _Nim_, two players take turns taking coins from two piles of
$n$ coins each. If there are $m$ coins in a pile, a player may choose to take
$1,2,3,...,m$ coins, *except* $m - 1$ coins. That is, a player may take any
nonzero amount, but may not leave exactly one coin left in the pile. The
player to take the last coin in play wins.
Example game: if there are 6 coins on the left and 8 coins on the right,
player 1 goes first and takes 4 coins from the left. Player 2 takes the
remaining 2 coins on the left. Player 1 takes all 8 coins from the right and
wins.
Prove that Player 2 has a winning strategy in all games of _Nim_.
This is actually a somewhat unexpected way in which strong induction can be
very useful. Let $S$ be a set of the natural numbers where Player 2 has a
winning strategy when there are $n in S$ coins left in both piles.
Suppose $m$ is a natural number and ${1,2,3,...,m-1} subset.eq S$. This is
our inductive hypothesis, we assume that Player 2 has a winning strategy for
when the game starts with $1,2,3,...,m-1$ coins in each pile. Now we seek to
show that this implies that Player 2 has a winning strategy when there are
$m$ coins in the pile. For the special case when $m = 1$, Player 1 takes 1
coin from either pile and Player 2 takes both, winning the game. Otherwise,
for $m > 1$, if Player 1 takes $m$ coins from a pile, Player 2 takes all the
coins in the other pile, winning. If Player 1 takes $j$ coins from either
pile where $1 <= j < m - 1$, leaving $m - j > 1$ coins in the pile, Player 2
can take $j$ coins from the other pile, leaving both piles with $m - j$
coins. This is essentially a new game of Nim with $m - j$ coins in each pile.
Then $m - j in {1,2,3,...,m-1}$ since $j <= 1 < m-1$, so by our inductive
hypothesis Player 2 has a winning strategy from this point on. Therefore
Player 2 has a winning strategy when $m$ coins are in each pile.
Since ${1,2,3,...,m-1} subset.eq S => m in S$, by strong induction, $S = NN$
and Player 2 has a winning strategy for all possible games of _Nim_.
]
#example[A hint for the fundamental theorem of arithmetic][
Prove that all natural numbers can be expressed as the product of primes.
#proof[
Let $m$ be a natural number. Note that $m = 2$ is prime so its the product
of itself and 1. Now assume that all natural numbers $n$ such that $1 < n <
m$ can be written as the product of primes. Then $m$ is either prime, in
which case its prime factors are $1 dot m$, or $(exists s,t in
{1,2,...,m-1})(m = s t)$. Then by our inductive hypothesis both $s$ and $t$
can be written as the product of primes, so $m$ is also the product of
primes. By strong induction, we conclude all natural numbers can be written
as the product of primes.
]
]<fundamental-thm-arithmetic-hint>
== Well ordering principle
This is another property that characterizes $NN$.
#fact[Well ordering principle][
Every nonempty subset of $NN$ has a smallest element.
]
We demonstrate the WOP in an alternative proof of
@fundamental-thm-arithmetic-hint.
#example[
All natural numbers can be expressed as the product of primes.
#proof[
Suppose there exists some non-empty set $S subset.eq NN$ that contains all
of the natural numbers which cannot be expressed as a product of primes.
By the WOP, there is a least number in this set, $T in S$. $T$ cannot be
prime because it would be the product of $1 dot T$, so it must be
composite. Then $T = s t$ for some natural numbers $s,t$, which are less
than $T$. But $T$ is the least element of $S$ so neither $s$ nor $t$ are in
$S$, and therefore they both have prime factorizations. But this implies
that $T$ also has a prime factorization, a contradiction. Therefore no such
set $S$ can exist, and all natural numbers have a prime factorization.
]
]
#theorem[The division algorithm][
For all integers $a$ and $b$, with $a != 0$, there exist unique integers $q$
and $r$ such that $b = a q + r$ and $0 <= r < |a|$.
#proof[
Let us consider the case where $a > 0$. Then we have a set
$
S = {b - a k, k in ZZ, b - a k >= 0}
$
It follows that we should exclude 0 from $S$ because otherwise if $0 in S$,
$b - a k = 0$ for some integer $k$, and $a | b$, so $b = a dot b/a$.
In our restricted $S$, there is a least element, call it $r$. Then for some
integer $q$, $a q + r = b$. We know that $r > 0$ so let's show $r < |a|$.
Since we're considering only $a > 0$ for simplicity, we just show $r < a$.
If $r >= a$, then either $0 in S$ when $r = a$ because $a q + a = b$
implies that $a | b$, or, when $r > a$, there exists another element in
$S$, $b - a (q + 1)$. This is a contradiction since we assumed $b - a q$
was the smallest element in $S$ and clearly $b - a q - a$ is smaller.
Therefore $r < |a|$.
Now we show the uniqueness of $q, r$. Suppose there were other integers
satisfying the properties of $q$ and $r$, $q'$ and $r'$. Then we have
$
r' >= r
$
without loss of generality (otherwise just relabel). Then $a q + r = a q' +
r'$, so $a(q - q') = r' - r$. Then $a$ divides $r' - r$.
]
]
== Relations, partitions
#definition[
A relation on a set $A$ is a set of ordered pairs $(a,b)$ where $a,b in A$. A
relation from a set $A$ to a set $B$ is set of ordered pairs $(a,b)$ where $a
in A$ and $b in B$.
]
#abuse[
If $R$ is a relation from $A$ to $B$, for $a in A$ and $b in B$ where $(a,b)
in R$, we can abbreviate it $a R b$.
]
#definition[
For any set $A$, the identity relation on $A$ is the set
$
I_A = {(a,a) : a in A}
$
]
#definition[
The domain of the relation $R$ from $A$ to $B$ is the set
$
"Dom"(R) = {x in A : (exists y in B)(x R y)} \
$
The range is
$
"Rng"(R) = {y in B : (exists x in A)(x R y)}
$
]
So the domain of $R$ is the set of all first coordinates and the range is the
set of all second coordinates.
#theorem[
1. $"Dom"(R^(-1)) = "Rng"(R)$
2. $"Rng"(R^(-1)) = "Dom"(R)$
]
#definition[
Let $R$ be a relation from $A$ to $B$, let $S$ be a relation from $B$ to $C$.
The composite of $R$ and $S$ is
$
S compose R = {(a,c) : (exists b in B)((a,b) in R and (b,c) in S)}
$
]
== Equivalence relations
Let $A$ be a set and $R$ a relation on $A$.
We say $R$ is reflexive on $A$ if $forall x in A, x R x$. $R$ is symmetric if
$forall x,y in A$, if $x R y$, then $y R x$. $R$ is transitive if $forall x,y,z
in A$, if $x R y$ and $y R z$, then $x R z$.
#theorem[
Let $A$ be a set. For the power set $cal(P)(A)$, the relation "is a subset of"
is reflexive on $cal(P)(A)$ and transitive but not symmetric.
]
#definition[
A relation $R$ on a set $A$ is an equivalence relation on $A$ if $R$ is
reflexive on $A$, symmetric, and transitive.
]
We make an equivalence relation when we think of objects as being related by
having the same property.
An equivalence relation on a set divides the set into subsets of related
elements.
#definition[
Let $R$ be an equivalence class on a set $A$. For $x in A$, the equivalence
class of modulo $R$ (or $x mod R$) is the set
$
overline(x) = {y in A : x R y}
$
Each element of $overline(x)$ is a *representative* of the class. The set
$
A \/ R = {overline(x) : x in A}
$
of all equivalence classes is called $A "modulo" R$.
]
#theorem[
Let $R$ be an equivalence relation on a nonempty set $A$. For all $x$ and $y$
in $A$,
1. $x in overline(x)$ and $overline(x) subset.eq A$
2. $x R y$ if and only if $overline(x) = overline(y)$
3. $x cancel(R) y$ if and only if $overline(x) sect overline(y) = emptyset$
]
== Congruence relations
Now we define congruence relations on $ZZ$ and show that congruence mod $m$ is
an equivalence relation. We describe its equivalence classes.
#definition[
Let $m$ be a fixed positive integer. For $x,y in ZZ$, we say $x$ is congruent
to $y$ modulo $m$ and write $x = y (mod m)$ if $m$ divides $(x-y)$. The number
$m$ is called the modulus of the congruence.
]
#theorem[
For every fixed positive integer $m$, congruence modulo $m$ is an equivalence
relation on $ZZ$.
]
#proof[
1. Reflexivity. Let $x$ be an integer. We show $x = x (mod m)$. $m dot 0 = 0 = x - x$, $m | x - x$.
2. For symmetry, suppose $x = y (mod m)$. Then $m | x - y$. Thus $(exists k)(x - y = k m)$. But $-(x-y) = -(k m)$, or $y - x = (-k) m$. So $m | y - x$ so $y = x (mod m)$
3. Suppose $x = y (mod m)$ and $y = z (mod m)$. Thus $m | x - y$ and $m | y - z$. Thus $m | x - y + y - z <=> m | x - z$ so $x = z (mod m)$. So congruence modulo $m$ is transitive.
]
#definition[
The set of equivalence classes for the relation congruence modulo $m$ is
denoted $ZZ_m$.
]
#theorem[
Let $m$ be a fixed positive integer. Then
1. for integers $x$ and $y$, $x = y (mod m)$ if and only if the remainder when $x$ is divided by $m$ equals the remainder when $y$ is divided by $m$
2. $ZZ_m$ consists of $m$ distinct equivalence classes.
]
The equivalence classes of $ZZ_m$ $overline(0), overline(1), ...,
overline(m-1)$ which are exactly all of the possible remainders when integers
are divided by $m$. For this reason the elements of $ZZ_m$ are sometimes called
the residue classes modulo $m$.
== Partitions
#definition[
Let $A$ be a nonempty set. $cal(P)$ is a partition of $A$ if $cal(P)$ is a set of subsets of $A$ such that
1. if $X in cal(P)$, then $X != emptyset$
2. if $X in cal(P)$ and $Y in cal(P)$, then $X = Y$ or $X sect Y = emptyset$
3. $union.big_(X in cal(P)) X = A$
]
In other words a partition of a set $A$ is a pairwise disjoint collection of
nonempty subsets of $A$ whose union is $A$.
#theorem[
If $R$ is an equivalence relation on a nonempty set $A$, then $A\/R$ is a
partition of $A$.
]
#proof[
Every equivalence class $overline(x)$ is a subset of $A$ and is nonempty
because it contains $x$. Any two equivalence classes are either equal or
disjoint. Also,
$
union.big_(x in A) overline(x) subset.eq A
$
because each $overline(x) subset.eq A$. To prove $A subset.eq union.big _(x in A) overline(x)$, suppose $y in A$. Because $y in overline(y)$, then
