Bridge to Abstract Mathematics: Mathematical Proof and Structures

ANSWERS AND SOLUTIONS TO SELECTED EXERCISES 377

y = x, then clearly ax = ay. So assume y - x E 8 and -a E 8. Then the

product (y - x)(-a) E 9. But (y - x)(-a) = (-a)(y) + (-a)(-x) = ax - ay,

so ax - ay E 8, as desired.

4. (b) (i) Proof Let a, b E F with a 2 0, b 2 0. Then a2 < b2 - 0 < b2 - a2 -

0 < (b - a)(b + a) - b - a > 0. The last step follows since b + a > 0, by (a)

of Definition 1, since, necessarily, either a > 0 or b > 0 (otherwise a = 0 and

b = 0, contradicting a2 < b2).

5. (a) (ii) Proof The result is clearly true if x = 0, so assume x # 0. Then
   JxI2 = Ix21, by Theorem 5(b) (using specialization, with x = y). But x2 > 0, by
   Theorem l(c), so that lx21 = x2, by Definition 4. Hence Ix12 = x2, as
   desired.


6. (a) (iii) Proof Given x, y E F with y # 0, we have Ixlyl = Ix(l/y)l [by
   Theorem 5(b)] = Ixl(l/lyl) [by (ii) of this exercise] = Ixl/lyl.

Article 9.3

1. (e) Proof Let S = (3,4) u (6). To prove 3 = glb S, note first that 3 I x for
   all x E S. For if x = 6, then 3 < 6 = x, whereas if x E (3,4), then 3 < x < 4 so
   that, in particular, 3 I x. For part (b) of Definition 3, let E > 0 be given. Let
   k = min (3, ~12). Then, 3 < 3 + k < 4 so that y = 3 + k E S and
   y = 3 + k I 3 + ~/2 < 3 + E, as required. To prove 6 = lub S, note first that if
   x E S, then either 3 < x < 4 (so that x < 6) or x = 6. Hence x I 6 for all x E S,
   so that (a) of Definition 3 is satisfied. For (b) of Definition 3, let E > 0 be given.
   Then 6 E S and 6 - E < 6, as desired.


2. (c) Proof Let y = lub T; then t I y for all t E T. Since S G T, then s I y
   for all s E S. Now let x = lub S. If x > y, then E = x - y > 0 and, by (b) of
   Definition 3, there must exist s' E S such that s' > s - E. But then
   s' > x - E = x - (X - y) = y. But s' E S and s' > y contradicts our earlier
   conclusion that s I y for all s E S. Hence our supposition x > y must be false,
   so that x I y; that is, lub S I lub T, as desired.


3. Proof Let S be a nonempty subset of F which is bounded below in F, say, by
   B. Then B I x for all x E S. Consider the set T = ( - x 1 x E S}. Letting y E T,
   wenote that y= -xfor somex~S, SO that y= -XI -Bfor allye T; that
   is, - B is an upper bound for T. By completeness of F, T has a least upper
   bound in F, call it u. Now -u E F, since F is a field, and by Exercise 5(a),
   
   
   • u = glb ( - x 1 x E T}. But clearly { - x 1 x E T} = S (you should verify this),
     so that -u = g!b S and S has the greatest lower bound in F.
   
   
   
   

Article 9.4

1. (g) 8i/17 (i-e., x = 0, y = A).


2. (a) z = (-8) + (-E)i.


3. (b) Proof Let z1 = x1 + yli, 2, = x, + y2i. Since z1 and z, are both real, then
   y1 = y2 = 0. Now z1z2 = (xl + yli)(x2 + y2i) = (x1x2 - yly2) + (xlyZ - x2y1)i =
   (xlxZ - (0 0)) + (xl -0 - x2 0)i = xlxZ = Re (2,) Re (z,).