Bridge to Abstract Mathematics: Mathematical Proof and Structures

ANSWERS AND SOLUTIONS TO SELECTED EXERCISES 369

Article 5.3

(a) Proof Let X, A, and B be sets such that X c B. To prove

X u (A n B) E (X u A) n B, let x E X u (A n B). To prove x E (X u A) n B,

we must prove x E (X u A) and x E B. Since x E X u (A r\ B), then either

x E X or x E A n B. We divide the argument into cases. Case I: If x E X,

then x EX u A since X G X u A. Since X c B and XEX, then x E B, so

that our desired conclusion holds in this case. Case II: If x E A n B, then

x E A so that x E X u A. Furthermore, x E A n B implies x E: B. Again, we

have our desired conclusion in this case. Since these cases are exhaustive,

our theorem is proved.

(b) Proof Let A, B, and X be arbitrary sets. Then A = A n U =

A n (X u X') = (A n X) u (A n X') = (B n X) u (B n X') =

Bn(XuX')=BnU=B.

(a) Proof Let x, y E R. Let x' = (x + y)/2 and y' = (x - y)/2. By Exercise

9(g, h), using specialization, we have sin (x' + y') - sin (x' - y') = 2 cos x' cos y'.

Resubstituting, this becomes sin x - sin y = 2 cos ((x + y)/2) sin ((x - y)/2), as

desired. 0

(6) (ii) Proof Let x, y E R and consider the cases x < y, x = y, and x > y. If

x c y, then (+)(x + y + Ix - yl) = (&(x + y + y - x) = (4)(2y) = y = x v y.

If x = y, then (&x + y + Ix - yl) = (i)(x + x) = (#(2x) = x = x v y. If

x > y, then (i)(x + y + Ix - y() = @)(x + y + x - y) = (4)(2x)=n = x = x v y. 0

(c) Proof Given diagonal matrices A and B, to prove AB = C = (cijln is

diagonal, we must prove that if i # j, then cij = 0. So let i and j be arbitrary,

distinct integers between 1 and n, inclusive; note that cij = z=, aikbkj.

Now since i # j, then for each k = 1,2,... , n, either k # i or k # j. If k # i,

then a& = 0 (since A is diagonal), whereas if k # j, then bkj = 0 (since B is

diagonal). In either case we have that the product aikbkj = 0, so that the sum

used to compute cij is the sum of n zeros and so equals zero, as desired.

Article 5.4

3. (b) Proof Define S in the usual manner. (i) Clearly 1 E S since (1)(2) =
   2 = (1)(2)(3)/3. (ii) Assume m E S. To prove m +^1 E S, note that

Lm2f k(k + 1) = z=, k(k + 1) + (m + l)(m + 2) = ($)dm + l)(m + 2) +

(m + l)(m + 2) = (i)[m(m + l)(m + 2) + 3(m + l)(m + 2)] =

(+)(m + l)(m + 2)(m + 3), as desired. 0.

7. (d) Proof Define S in the usual manner. (i) 1 E S since x - y divides
   x - y. (ii) Assume m E S. To prove m +^1 E S, we must prove x - y
   divides xm+' -y"+'. Now xm+' - ym+' = x(xm - y") + (xym - ym+') =
   x(xm - ym) + y"(x - y). Since x - y divides xm - ym (by induction hypothesis),
   then x - y divides x(xm - y"). Also, x - y clearly divides ym(x - y). Hence
   x - y divides the indicated sum, so that x - y divides xm+' - y"+', as
   required. O


8. (b) Proof Let S = (n E N 1 n 2 10 and n3 < 2"). Clearly 10 E S, since
   I 1000 < 1024. So assume m E S. To prove m + 1 E S, we must prove
   (m + < 2m+1. Now (m + = m3 + 3m2 + 3m + 1. Since m^2 10, then