Bridge to Abstract Mathematics: Mathematical Proof and Structures

194 METHODS OF MATHEMATICAL PROOF, PART II Chapter 6

Solution Let A and B be given invertible n x n matrices. We must find a

matrix X, ., such that (AB)X = X(AB) = I,. Now since A is invertible,

there exists an n x n matrix C such that AC = CA = I,. Since B is in-

vertible, there corresponds an n x n matrix D such that BD = DB = I,.

We get the desired X from C and D, namely, by letting X = DC. Note

that (AB)X = (AB)(DC) = A(BD)C = A(I,)C = AC = I,. Verify that

X(AB) = I,. 0

Obviously, the key to the preceding proof, once the setting-up is com-

pleted, is the choice X = DC of X in terms of D and C. If you had never

seen this proof before, how might you have discovered that X = DC was

the proper choice? Back in Chapter 1 we emphasized two main methods

of discovery that mathematicians use and that are worth recalling here.

One was drawing a picture (a method we will use in the next example);

the other was carrying out computations in specific examples. Assuming

that you have some computational familiarity with matrices, you might

look at a particular pair of invertible matrices, say A =

(; :) and

B = (-i

:). Computations yield

9

D=(-:, i), and X=(-! -- 8.

After some experimentation, you might finally notice the relationship X =

DC, and this observation should lead to the speculation that the choice

X = DC of X will work in general.

In the next example we use geometric motivation for our selection of the

existentially quantified unknown, as we apply definitions from Example 4.

' EXAMPLE 9 Suppose S and T are both open subsets of R. Prove that

S n T and S u T are both open.

Solution We deal first with intersection. Let S and T be arbitrary open

subsets of R. According to the definition of "open," in order to prove

that S n T is open, we must show that each point of S n T is an interior

point. So begin by letting x be an arbitrary point in S n T. (Note: We

are using here, almost unconsciously, proof-writing techniques over

which we labored hard in Article 5.2. In particular, we are setting up

our proof in terms of the desired conclusion, not the hypotheses.) To

prove that x is an interior point of S n T, we must produce a positive

real number 6 such that the 6 neighborhood N(x; 6) of x is a subset of