5.3 PROOF BY SPECIALIZATION AND DIVISION INTO CASES 177
- Let A = (aij), , , and B = (bij), , , be square matrices. Recall Exercise 11 from
Article 5.2 and prove:
(a) If A is diagonal, then A = At.
(b) If A is both upper and lower triangular, then A is diagonal. [What theorem
emerges from the combination of this fact with the results of (b) and (c) of
Exercise 11, Article 5.2?]
*(c) If A and B are diagonal matrices, then their product AB is a diagonal matrix.
(d) If A and B are upper (resp., lower) triangular matrices, then their product
AB is an upper (resp., lower) triangular matrix.
- Suppose that I, and I, are intervals on the real line such that I, n I, # 0.
Prove that I, u I, is an interval.
- Critique and complete (instructions in Exercise 11, Article 4.1):
(a) THEOREM For any real numbers x, y, and z, xv (yvz) = (xv y) vz (recall
Exercise 11).
"Proof" We divide the argument into the cases x < y < z and x > y > z. In the
first case xv(yvz) = xvz= z= yvz= (xvy)vz, so that xv(yvz) =
(xvy)vz, as desired. In the second case xv(yvz)=xvy=x=xvz=
(x v y) v z, so that the desired result holds in this case as well. In either case we
have the desired conclusion.
(b) THEOREM For any positive real numbers x and y, In (xly) = In x - In y.
[Note: Assume we already proved that (1) in (xy) = In x + In y for any positive
real numbers x and y and (2) In 1 = 0.1
"Proof" Let x be an arbitrary real number; we begin by using the special case
y = llx in (1). By (2), we have 0 = In 1 = In (x. llx) = In x + In (l/x). Hence
In (1/x) = -In x for any x > 0. Next, let x and y be arbitrary positive real num-
bers. Then In (xly) = In (x. lly) = In x + In (l/y) = In x - In y, as desired.
(c) THEOREM For any sets X and Y in U, if Y = X u (Y n XI), then X c Y.
Start of "Proof" Let x be an arbitrary element of U. Clearly either x E: X or x E X'.
We will divide the argument into these two cases....
(d) THEOREM If a function f, mapping reals to reals, is decreasing on an interval
I (i.e., if x, < x, implies f(x,) > f(x,) for any x,, x, E R), then f is one to one on I.
Start of "Proof" Assume f is decreasing on I. To show f is one to one on I, assume
that x, and x, are elements of I such that f(x,) = f(x,). We must prove x, = x,.
For suppose x, # x,. Then.... (Note: This approach anticipates indirect proof,
to be studied in Article 6.2.)
(e) FACT The function f(x) = ax (a > 0, a # 1) is one to one on R. [Note: Use the
results of (d), and Exercise 6(c), Article 5.2, together with the derivative for-
mula d/dx(ax) = ax In a where a > 0 and a # 1, and well-known facts about
the significance of the sign of the first derivative.]
Start of "Proof" We consider the sign of the derivative of the function f(x) = ax.
Since a>O and a#1, then either a> 1 or O< a< I....
