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The Natural Numbers N
• For this course, N = {0, 1, 2, 3, . . . }.
• Each element has a unique successor (“next” element).
• Each nonzero element has a unique predecessor (“previous” element).
Zero has no predecessor.
• Addition can be defined in terms of successors/predecessors. In this
sense addition is “repeated successors.”
• Similarly, multiplication is “repeated addition.”
• Similarly, exponentiation is “repeated multiplication.”
Algebraic properties of N (addition/multiplication):
1. N is closed under addition: if a, b ∈ N, then a + b ∈ N.
2. Addition is associative: (a + b) + c = a + (b + c).
3. There is an additive identity, 0, such that a + 0 = 0 + a = a.
4. Addition is commutative: a + b = b + a.
Properties 1-2 give the abstract definition of a semigroup. With properties
1-3, we have the abstract definition of a monoid.
If we replace addition with multiplication and 0 with 1, the above properties
hold. However, the additive and multiplicative structures are very different.
Subtraction/Division in N:
Subtraction can be defined as an “inverse operation” to addition:
a − b = c ⇐⇒ a = c + b.
Division can be defined as an “inverse operation” to multiplication:
a ÷ b = c ⇐⇒ a = cb.
A major shortcoming of N is that subtraction and division are not always
defined. For example, there is no x ∈ N such that x + 5 = 3, so we cannot
define x = 3 − 5 (at least not within N).
Note: Division with remainder will be discussed within the set of integers.
Cancellation Laws
Even without subtraction, we still have (for any a, b, c ∈ N):
• If a + b = a + c, then b = c.
• If x + y = z + y, then x = z.
Since addition in N is commutative, either law above implies the other.
There are similar laws for multiplication, with an additional restriction:
• If ab = ac and a 6= 0, then b = c.
• If xy = zy and y 6= 0, then x = z.
We’ll discuss how to prove these after introducing a few more ideas.
Order Relations
The concept of ordering is less important in abstract algebra, but there a
few key concepts that are best understood in terms of ordering.
A binary relation ⊆ on the set S is a partial ordering if:
• For all a ∈ S, a ⊆ a. (Reflexive Property)
• If a ⊆ b and b ⊆ a, then a = b. (Antisymmetric Property)
• If a ⊆ b and b ⊆ c, then a ⊆ c. (Transitive Property)
The subset relation (A ⊆ B means x ∈ A =⇒ x ∈ B) satisfies these
properties, hence the notation.
Total Orderings
In a partial ordering, it might be the case that two objects are
“incomparable.” For example, take the sets A = {1, 2} and B = {2, 3}.
Neither is a subset of the other.
A binary relation ≤ on the set S is a total ordering if:
• The relation ≤ is a partial ordering on the set S.
• For any a, b ∈ S, a ≤ b or b ≤ a (Trichotomy Law).
Note that if both a ≤ b and b ≤ a, then Antisymmetry implies a = b. Hence
”Trichotomy” rather than ”Dichotomy.” You might also consider how to
modify the partial/total ordering properties to get similar properties of the
proper subset relation ⊂, or of < (these are strict partial/total orderings).
Ordering of N
The usual ≤ relation on N (a total ordering) can be defined completely in
terms of successors and predecessors. Recall that each element of N has a
unique successor, and each nonzero element has a unique predecessor. In
more familiar notation, x − 1 is the predecessor of x.
• For any x ∈ N, we set 0 ≤ x.
• To satisfy antisymmetry, x ≤ 0 must be false whenever x 6= 0.
• If x and y are both nonzero, we set x ≤ y if and only if x − 1 ≤ y − 1.
Well-Ordered Sets
A key property of N is that (using the usual ordering) every nonempty set
contains a minimal element. This is known as the Well-Ordering Principle.
In general a total ordering ≤ on a set S is a well-ordering if:
• For any nonempty subset A of S, there exists x ∈ A such that x ≤ y for
any y ∈ A. Such an x is called a minimal element of A.
Note that “minimal element” can be defined even for a partial ordering. A
set A cannot have more than one minimal element: if x and y were both
minimal, then x ≤ y and y ≤ x, so x = y by antisymmetry.
Exercises
Assume that ≤ and < have their usual meanings.
1. Is ≤ a partial/total/well-ordering on the set of integers?
2. Is ≤ a partial/total/well-ordering on the set of rational numbers?
3. Is ≤ a partial/total/well-ordering on the set of real numbers?
4. Is ≥ a partial/total/well-ordering on the set of natural numbers?
5. Redo the previous exercise for the set of integers, rational numbers, and
real numbers.
More Exercises
6. If a and b are positive integers, define a | b to mean that b is a multiple
of a (there is some positive integer c such that b = ac). Is this a
partial/total/well-ordering?
7. Redo the previous exercise, using “nonzero” in place of “positive.”
8. Consider the set of all ordered pairs (x, y), where both entries are natural
numbers. Define (a, b) E (x, y) to mean either a < x or (a = x and
b ≤ y). Is this a partial/total/well-ordering?
9. Redo the previous exercise, replacing the natural numbers by the integers,
rational numbers, and real numbers.
Mathematical Induction
The importance of Well-Ordering in N is that it is equivalent to the following:
Principle of Mathematical Induction (PMI): Let A be a subset of N
with the following properties:
1. 0 ∈ A
2. For any k ∈ N, k ∈ A implies k + 1 ∈ A.
Then A = N.
Note: Condition (2) can be stated abstractly in terms of successors (k + 1
is the successor of k), but the additive notation should be more familiar.
Well-Ordering of N implies PMI for N
This is a useful proof that you may not have seen before.
• Let A be a subset of N satisfying conditions (1) and (2) of the previous
slide, but assume (for contradiction) that A 6= N.
• Let B be the subset of N containing all elements not in A. By our
assumption, B is a nonempty set.
• By Well-Ordering, B contains a minimal element, call it x. Note that
x∈
/ A. We cannot have x = 0 (why?), so x − 1 is still in N.
• x − 1 is smaller than x, so x − 1 ∈
/ B, and thus x − 1 ∈ A.
• By condition (2), x = (x − 1) + 1 ∈ A. This is our contradiction.
Proof by Induction
The PMI allows use to prove that a given statement (which may depend on
the value of a natural number n) is actually true for every n ∈ N.
Suppose that we have a statement (depending on the value of n) such that:
1. The statement is true for n = 0 (the “base case”)
2. For any integer k, if the statement is true for n = k, then the statement
is true for n = k + 1 (the “inductive step”)
Then the statement is true for any n ∈ N.
Example: Cancellation Law for Addition
We’ll show that for all natural numbers a, b, n, n + a = n + b implies a = b,
using induction (on the value of n). We will need to use the fact that
a + 1 = b + 1 if and only if a = b, which follows from that fact that
successors/predecessors in N are unique.
• Base Case: If n = 0, then a = 0 + a = n + a = n + b = 0 + b = n.
• Inductive Step: Suppose the result is true when n = k. If (k + 1) + a =
(k +1)+b, then we have (by the associative law) k +(a+1) = k +(b+1).
However, (a + 1) and (b + 1) are still in N, so we may apply the n = k
case to conclude that a + 1 = b + 1. As noted above, this implies a = b,
and we’ve thus shown that the result is true for n = k + 1.
Exercises: Proof by Induction
Prove each of the following, using induction on the value of n:
1. 0 + 1 + 2 + 3 + 4 + ... + n =
n(n+1)
.
2
2. 1 + 3 + 5 + 7 + 9 + ... + (2n + 1) = (n + 1)2.
3. If y > −1, then (1 + y)n ≥ 1 + yn.
4. If A is a set with exactly n elements, there are exactly 2n subsets of A
(including the empty set and A itself).
5. If A is a subset of N and n ∈ A, then A has a minimal element. Do not
use the Well-Ordering Principle in your proof.