Bridge to Abstract Mathematics: Mathematical Proof and Structures

9.1 FIELDS 299

of R is analogous to that of our other examples of fields now; in particular,
we must verify (i.e., prove) the 11 field properties for R in that approach.
We now wish to prove some elementary theorems about fields. Any
theorem we can prove about fields in general is, of course, true in any of the
specific examples of fields considered in Examples 1 through 7. On the other
hand, there are familiar properties of the real number field that do not
appear in our list of general field theorems, agd indeed, are not true in an
arbitrary field. Several such results are found in Articles 9.2 and 9.3. An
analysis of their proofs reveals that they depend for their validity on prop-
erties of R other than pure field properties, and so are true in less generality
than the theorems we are about to state. In particular, some may fail to
hold in our various specific examples of fields.
Throughout the following discussion we use freely the basic principles
"equals added to (respectively, multiplied by) equals yield equals." In
symbols, if (F, +, .) is a field with x, y, z E F and if x = y, then x + z =
y + z and xz = yz. An easy consequence of Axiom 4 and this property is
additive cancellation, namely, if a, b, c E F and a + b = a + c, then b = c
(see Exercise 1).

THEOREM 1

Let (F, +, .) be a field. Then the additive and multiplicative identities of F are

unique. The additive and multiplicative inverses of a given element x E F (x # 0 in

the multiplicative case) are uniquely determined by x.

Proof First, suppose 0 and 0' are both additive identities for F. Then
0 = 0 + 0' = 0', where the first equation uses the fact that 0' is an addi-
tive identity, with the second using that property in reference to 0. Thus
0 = 0' and the uniqueness of the additive identity is established. Second,
suppose y and z are both additive inverses for a given x E F. Then
y=O+y=(y+x)+y=(z+x)+y=z+(x+y)=z+O=z,sothat
y = z, and the uniqueness of additive inverses is proved. The key step
(y + x) + y = (z + x) + y in the preceding sequence follows from the
assumption y + x = 0 = z + x.
The arguments in the multiplicative case are identical to those just
given and are left as an exercise (Exercise 2).

THEOREM -2

Let(F,+;) beafieldandx~F. Thenx.O=O.x=O.

Proof We claim first that x 0 + x 0 = x 0 + 0. This follows from the
sequence of steps x -0 + 0 = x 0 = x. (0 + 0) = x 0 + x 0. By ad-
ditive cancellation, we conclude 0 = x - 0. The equation 0 = 0. x follows
immediately from Axiom 8.

THEOREM 3

Let (F, +, .) be a field and x E F. Then (-l)x = -x.