epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 85
Indeed, we need to prove that each one of the possible conclusions is
true, because we want all of them to hold. If we have already completed the
proof that one of the two (or more) implications is true, we can use it to
prove the remaining ones (if needed).
EXAMPLE 3. The lines y = 2x-\-l and >; = -3x + 2 are not perpendicular,
and they intersect in exactly one point.
Proof:
Hypothesis:
A. The two lines have equations y = 2x-\-l and y = -3x-\-2.
{Implicit hypothesis: All the properties and relations between lines can be
used.)
Conclusion:
B. The lines are not perpendicular.
C. The lines intersect in exactly one point.
Parti. If A, then B.
Two lines are perpendicular if their slopes, m and mi, satisfy the equation
m = — 1/mi, unless one of them is horizontal and the other vertical, in which
case one slope is equal to zero and the other is undefined.
The first line has slope 2 and the second has slope —3, so the lines are
neither horizontal nor vertical. In addition, —3 7^—1/2. Thus, the lines are
not perpendicular.
Part 2. If A, then C.
This second part is an existence and uniqueness statement: There is one
and only one point belonging to both lines.
We can prove this part in two ways:
a. The given hues are distinct and nonparallel (as they have different
slopes); therefore, they have only one point in common.
b. We can find the coordinates of the point(s) in common by solving the
system:
' y = 2x-i-l
); = -.3x + 2.
By substitution we have:
2x + 1 = -3x + 2.
The only solution of this equation is x = 1/5.
The corresponding value of the y variable is y = 2(1/5) + 1 = 7/5.