The elements of the set of integers:
Z = {...,−5,−4,−3,−2,−1, 0, 1, 2, 3, 4, 5, ...}
consist of three types of numbers:
I. The (positive) natural numbers {1, 2, 3, 4, 5, ...},
II. The negative integers {−1,−2,−3,−4,−5, ...}, and
III. The number 0.
The ordering on the natural numbers extends to an ordering of the integers:
... < −5 < −4 < −3 < −2 < −1 < 0 < 1 < 2 < 3 < 4 < 5 < ...
but there is no well-ordered principle for the integers since many subsets of Z
(including Z itself) have no smallest element.
Negative numbers may seem obvious today, but there was a long period of
time when only positive numbers were used. The introduction of 0 is often cited
as evidence of the scientific superiority of the eastern cultures during the
middle
ages. It is remarkable that we can easily (if tediously) extend the operations
of
addition and multiplication of natural numbers to include 0 and the negative
integers, maintaining all the fundamental laws of arithmetic . Negative numbers
are necessary today because, in our society based upon the right to the pursuit
of happiness through credit card debt, we need to be able to subtract!
Definition of Addition. a + b is defined on a case by case basis:
Case I. b = n is positive. Then a + n is defined by induction.
(i) a + 1 is the next number after a,
(ii) Each a + (n + 1) is the next number after a + n
Case II. b = −n is negative. Then a + (−n) is also defined by induction.
(i') a + (−1) is the number immediately before a, and
(ii') Each a + (−(n + 1)) is the number immediately before a + (−n)
Case III. b = 0. By definition,
a + 0 = a
The associative and commutative laws of addition can now be proved for
this new definition of addition by the same proof-by-induction strategy we used
in §1.1 (but it is tedious, involving lots of different cases , so we won’t do
it!)
Since the first case of the definition of addition is identical to the
definition of
addition of natural numbers from §1.1, the two additions give the same result
when applied to two natural numbers.
Before we tackle multiplication, we introduce:
The Negation Transformation: Negation is the function:
− : Z → Z
defined by: −(n) = −n, −(0) = 0 and −(−n) = n. It is clear that
−(−a) = a for all integers a
(double negatives cancel) and that taking negatives reverses order:
if a < b then − b < −a
Negation has the fol lowing three important properties, too:
Proposition 1.2.1. For all integers a,
a + (−a) = 0
(and because of this, we say that −a is an additive inverse of a).
Proof: This is clearly true for a = 0 since 0+(−0) = 0+0 = 0. Otherwise,
either a or −a must be a natural number, so it is enough by the commutativity
of addition to prove the “additive inverse” sentence n+(−n) = 0 for all n, which
we will now do by induction:
(i) 1 + (−1) is the number before 1, by addition definition (i'), which is 0.
(ii) For each n, once we know n + (−n) = 0, then since −(n + 1) is the
number before −n, we also know that:
(n + 1) + (−(n + 1)) = (n + 1) + (−n + (−1))
by addition definition (i'), and then
(n + 1) + (−n + (−1)) = (1 + (−1)) + (n + (−n))
by the commutative and associative laws of addition. But now:
1 + (−1) = 0, n + (−n) = 0 and 0 + 0 = 0
allow us to conclude that (n + 1) + (−(n + 1)) = 0, hence the induction.
Proposition 1.2.2. −a is the only additive inverse of a.
Proof: Suppose b is another additive inverse of a. Then:
−a + (a + b) = −a + 0 = −a
but using the associative law of addition, we also have:
−a + (a + b) = (−a + a) + b = 0 + b = b
so −a = b. Thus any other additive inverse of a is equal to −a, which is the
same thing as saying that −a is the only additive inverse of a.
Proposition 1.2.3. Negation is a linear
transformation , meaning:
−(a + b) = −a + (−b)
Proof: By the laws of addition and Proposition 1.2.1:
(a + b) + (−a + (−b)) = (a + (−a)) + (b + (−b)) = 0
so −a + (−b) is an additive inverse of a + b. From Proposition 1.2.2, we know
that there is only one additive inverse of a + b, so −a + (−b) = −(a + b).
Now we are finally ready for the:
Definition of Subtraction: For all integers a and b:
a − b = a + (−b)
(that is, subtraction of b is defined to be addition of the additive inverse of
b)
Finally (for the integers), we use negatives to define multiplication:
Definition of Multiplication. ab is defined on a case-by-case basis.
Case I. b = n is a positive. Then a × n is defined by induction:
(i) a × 1 = a,
(ii) Each a × (n + 1) = a × n + a.
Case II. b = −n is negative. Then a × (−n) is defined to be −(a × n).
Case III. b = 0. Then a × 0 = 0
Remark: The definitions in Cases II and III are forced upon us, if we want
multiplication to satisfy the distributive law ! (see the exercises) Notice also
that
a × (−1) = −a
so the negation transformation is the same as multiplication by −1.
Again, we will not go through the tedious exercise of proving the rest of
the basic laws of arithmetic, but it can be done with only induction and these
definitions, if you are willing to work through all the cases.
Recap: The new number 0 is the additive identity, meaning that:
a + 0 = a
for all integers a. Negation takes an integer to its additive inverse, allowing
us to define subtraction as addition of the additive inverse. Note that 1 is the
multiplicative identity, meaning that a×1 = a for all integers a, but
integer
multiplicative inverses only exist for the integers 1 and −1.
The rational numbers can be thought of geometrically as
slopes of lines :
Q = {(slopes of) lines that pass through (0, 0) and a point (b, a)}
where a, b ∈ Z and b ≠ 0 (so the line isn’t vertical.)
The line L passing through (0, 0) and (b, a) has equation:
by = ax
and the slope is also the y- coordinate of the intersection of L with the
(vertical)
line x = 1. In particular, different lines through the origin have different
slopes.
Many different points with integer coordinates will lie on the same line L!
If (b', a') is another point with integer coordinates, then by the equation
above
for the line L, we see that (b', a') is also on L exactly when:
ba' = ab'
Definition: An integer fraction is a symbol of the form:
where a, b ∈ Z and b ≠ 0. Two integer fractions are equivalent,
written:
if (b, a) and (b', a') are on the same line through the origin.
Note: The symbol “ ~” is called a relation, and it is easy to see that:
(i) ~ is reflexive, meaning that 
(ii) ~ is symmetric meaning that if 
then
(iii) ~ is transitive meaning that if a and
then 
(A relation satisfying (i)-(iii) is called an equivalence relation.)
Definition: The equivalence class
is the set of all fractions that are equivalent to
. Notice that:
whenever , that is,
whenever (b, a) and (b', a') are on the same line
through the origin, or, as we noticed above, whenever ba' = ab'.
Thus we can reinterpret the rational numbers as:
Q = {equivalence classes of integer fractions}
and this reinterpretation is very useful for seeing the arithmetic of Q.
Before we do this, let’s notice that the rational numbers are still ordered:
if the line through (0, 0) and (b, a) intersects the vertical line x = 1 at a
point
that is below the intersection of the line through (0, 0) and (d, c).
Un like the integers , there is no such thing as the next rational number after
a rational number
, so there is no way to use
induction to define addition.
Instead, we use the rule for adding fractions that we learned in gradeschool.
Definition of Addition: Given rational numbers
and 
,
then:
(using the arithmetic of integers to define the numerator and denominator ).
But we need to check something!
Is addition is well-defined? Here’s the problem. If
and
how do we know that:
This isn’t obvious, and it needs to be checked, because if it weren’t true, then
this would be bad definition of addition, because it would only be an
addition of
integer fractions, and not of rational numbers, which are equivalence classes of
integer fractions. This problem will arise whenever we try to make a definition
involving equivalence classes of fractions (or other things). So beware!
Proof that addition is well-defined: Suppose:
and
This means that ba' = ab' and dc' = cd'. But then:
(substituting, and applying the laws of arithmetic for
integers). So:
as desired.
Now that the definition is OK, it is very useful!
Addition is associative. Proof:
and
and these are the same because of the associative laws for integer arithmetic!
Similarly, you can prove that addition is commutative.
Definition of Multiplication:
Proof that multiplication is well-defined: If
and
, then
ba' = ab' and dc' = cd', so 
. But
this gives us:
which is just what we needed to check.
It is now easy to prove the associative and commutative laws for the
multiplication
of rational numbers. To see that multiplication distributes with addition,
though, we need an extra:
