2 1. From the Real Numbers to the Complex Numbers
inverse -n such that
(1) n + (-n) = (-n) + n = 0.
Of course 0 is the only number whose additive inverse is itself.
Let a,b be given integers. As usual we write a-b for the sum a+(-b). Consider
the equation a + x = b for an unknown x. We learn to solve this equation at a
young age; the idea is that subtraction is the inverse operation to addition. To solve
a + x = b for x, we first add -a to both sides and use (1). We can then substitute
b for a + x to obtain the solution
x = 0 + x = (-a) + a + x = (-a) + b = b + (-a) = b - a.
This simple principle becomes a little more difficult when we work with multi-
plication. It is not always possible, for example, to divide a collection of n objects
into two groups of equal size. In other words, the equation 2 * a = b does not have
a solution in Z unless b is an even number. Within Z, most integers (±1 are the
only exceptions) do not have multiplicative inverses.
To allow for division, we enlarge Z into the larger system Q of rational numbers.
We think of elements of Q as fractions, but the definition of Q is a bit subtle. One
reason for the subtlety is that we want
, and
all to represent the same
rational number, yet the expressions as fractions differ. Several approaches enable
us to make this point precise. One way is to introduce the notion of equivalence
class and then to define a rational number to be an equivalence class of pairs of
integers. See [4] or [8] for this approach. A second way is to think of the rational
number system as known to us; we then write elements of Q as letters, x,y,u,v, and
so on, without worrying that each rational number can be written as a fraction in
infinitely many ways. We will proceed in this second fashion. A third way appears
in Exercise 1.2 below. Finally we emphasize that we cannot divide by 0. Surely
the reader has seen alleged proofs that, for example, 1 = 2, where the only error is
a cleverly disguised division by 0.
Exercise 1.1. Find an invalid argument that 1 = 2 in which the only invalid
step is a division by 0. Try to obscure the division by 0.
Exercise 1.2. Show that there is a one-to-one correspondence between the set
Q of rational numbers and the following set L of lines. The set L consists of all
lines through the origin, except the vertical line x = 0, that pass through a nonzero
point (a,b) where a and b are integers. (This problem sounds sophisticated, but
one word gives the solution!)
The rational number system forms a field. A field consists of objects which can
be added and multiplied; these operations satisfy the laws we expect. We begin
our development by giving the precise definition of a field.
Definition 2.1. A field F is a mathematical system consisting of a collection of
objects and two operations, addition and multiplication, subject to the following
1) For all x,y in F, we have x + y = y + x and x * y = y * x (the commutative
laws for addition and multiplication).
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