Bridge to Abstract Mathematics: Mathematical Proof and Structures

(Dana P.) #1

300 PROPERTIES OF NUMBER SYSTEMS Chapter 9


Proof By the uniqueness of the additive inverse -x of a field element x
(Theorem I), we need only show that (- l)x behaves as -x is defined
to behave, namely, x + ( - l)x = 0. But
x + (-l)x = l(x) + (-l)x
= [1 + (- l)]x
=Osx
= 0
A result such as Theorem 3,
unremarkable as our familiarity


(Axiom 9)
(Axiom 11)
(Axiom 5)
(Theorem 2)
in a general field, is not so mundane or
with its application to R would have us
believe. Let us apply this result in the field Z, of integers modulo 7. The
theorem says that, in any field, the product of the additive inverse of the
multiplicative identity with any element equals the additive inverse of that
element. Let x = 5 in Z,. Note that in Z,, - 5 = 2 and - 1 = 6. Hence
our theorem boils down to the equation (6). (5) = 2. A quick check of the
Z, multiplication table indicates that this equation is true, as Theorem 3
predicts.


COROLLARY
Let (F, +, .) be a field with x, ye F. Then:

Proof (a) Using Theorem 3, we note the equations x(- y) = x[(- l)y] =
[(x)( - l)]y = ( - l)(xy) = - (xy). We may prove - (xy) = ( - x)y in a
similar manner.
(b) Our claim is that x is the additive inverse of -x. By uniqueness
of the additive inverse (Theorem I), we need show only that (-x) + x =



  1. But this holds since -x is the additive inverse of x.
    (c) (-x)(-y)= -[(x)(-y)].= -[-(xy)]=xy. The first two
    equalities follow from (a), the third from (b).
    Theorem 2 stated that, in a field, the product of any element with the
    additive identity zero equals zero. The following result indicates that the
    only way a product of two field elements can be zero is if one of the elements
    is zero.


THEOREM 4
Let (F, +, .) be a field with x, y E: F. If xy = 0, then either x = 0 or y = 0.

Proof We have seen both our general approach to this proof and this
specific proof earlier in the text (recall Example 1, Article 6.2). Given
xy = 0, suppose x # 0. We will try to conclude on this basis that y = 0.
Since x # 0, then x-I exists, by Axiom 10. Hence y = l(y) =
(x-lx)y = x-l(xy) = x-l(O) = 0, as desired.

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