5.2 CONCLUSIONS INVOLVING V AND -+, BUT NOT 3 161EXAMPLE 5 (Some linear algebra background is helpful.) A set of vectors
{v,, v,,... , vnJ in a vector space V is said to be linearly independent if
and only if, for any n real numbers /I,, /I,,... , #In, if fllvl + #12v2 + -.. +
Pnvn = 0, then #I1 = P2 =. - = Bn = 0.
As indicated, Examples 2 and 4 were encountered earlier in the text, but
all except possibly Example 5 should be familiar to you. Also, you should
be able to recall other definitions from your mathematical experience that
have this logical form.
The problem we wish to confront now is how to go about proving a theo-
rem in which the conclusion is of the same logical form as we have just seen
in Examples 1 through 5, that is, the form (Vx)(p(x) -, q(x)).
EXAMPLE 6 Use the definition in Example 2 to prove that if I, and I, are
intervals, then I, n I, is an interval.
Solution The questions we must ask in approaching a proof of a statement
such as this one are. (1) "What is my desired conclusion?" (2) "According
to the relevant definitions, what must I do in order to arrive at this con-
clusion?" (3) "What do I have to work with? How can I bring the given
hypotheses to bear on the problem?" Implicit in the order of these ques-
tions is an important guiding rule, one that we had occasion to state and
use earlier in the text, in Article 4.1, after Example 7. The rule states that,
in setting up a proof of the type under consideration in this article, we
should at the outset focus on the desired conclusion, not on the hypoth-
eses. In the example at hand the desired conclusion is the statement
that I, n I, is an interval. The proof will be "set up" strictly in terms
of the definition of "I, n I, is an interval." The hypotheses, that is, the
assumptions that I, and I, are intervals, will be brought in and used in
the course of the proof. Our goal is to prove that I, n I, is an interval;
how is this to be done? By the definition in Example 2, we must show
that if a, b, and c are real numbers with a, c E I, n I, and a < b < c,
then b E I, n I,. Hence, to set up the proof, we begin by assuming that
a, b, and c are real numbers with a < b < c and a and c both elements
of I, n I, ("let a, b, and c be real numbers with.. ."). We must prove,
on the basis-of these assumptions and the given hypotheses, that b E
I, n I,. By definition of intersection, this means we must show that
b E I, and b E I,.
At this stage, the basic structure of the proof has been set; we must
now as$ how our hypotheses can be brought to bear on the problem.
We may reason as follows: Since I, is an interval, since a c b < c, and
since a and c are both in I, (since I, n I, c I ,), then b E I,. An identical
argument, with I, replacing I,, shows that b E 1,. Hence b E I, n I, and
I, n I, is an interval, as claimed.