(a) -0 = O;
(b) 't:/x E F, -(-x) = x;
(c) 1-^1 = 1 and (-1)-^1 = -1;
( d) 't:/x E F, x · 0 = O;
(e) xy=O<=:? eitherx=O ory=O;(f) if x -j. 0, then x-^1 -j. 0 and (x-^1 )-^1 = x;
1.1 The Field Properties 7(g) if x, y -j. 0, then xy -j. 0 and (xy)-^1 = x -^1 y-^1 ;
(h) 't:/x E F, (-l)x = -x;
(i) 't:/x, y E F, (-x)y = -(xy) = x(-y);
(j) (-1)(-1) = 1;
(k) 't:/x, y E F, (-x)(-y) = xy.
Proof. (a) Observe that 0 + 0 = 0. This says that 0 is an additive inverse
of 0. Theorem 1.1.3 (c) says that there is only one additive inverse of O; namely,
-0. Thus, -0 = 0.
(b) Observe that (-x) + x = x + (-x) = 0 by (Al) and (A4). This says
that xis an additive inverse of (-x). Theorem 1.1.3 (c) says that -x has only
one additive inverse; namely, -(-x). Thus, -(-x) = x.
( c) Exercise 4.
(d) Let x E F. Then (give reasons)x. 0 + x. 0 = x. (0 + 0) = x. 0 = x. 0 + 0.
That is, x · 0 + x · 0 = x · 0 + 0. Applying Theorem 1.1.2 (a), (cancellation),
we have x · 0 = 0.
( e) This is a two-part proof.
Part 1: First we prove the "=>" part. Let xy = 0. Suppose x -j. 0. Then by
(M4), :ix-^1 E F, and
By the associative law, this implies
(x-^1 x)y = 0, i.e.,
1·y=0
y = o.