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

Solutions for the Exercises at the End of the Sections and the Review Exercises 157
11. (The statement has only impUcit hypotheses. Before proceeding we
must be sure that we are famihar with the definition of rational
numbers and their operations and properties. We can reformulate the
statement in the following way: If ^ is a rational number, then
q 7^ V2. This statement is equivalent to: If ^ is a rational number,
then q^ i^ 2. Because we cannot directly check that the square of each
rational number is not equal to 2, we will try to prove the
contrapositive of the statement.) The statement to be proved is "If q
is a rational number, then c^ / 2." We will assume that there exists a
rational number ^ such that (^ — 2. Because g is a rational number, it
can be written as ^ = a/h, where a and h are relatively prime integer
numbers, b7«^ 0, and a 7^ 0 (because ^ / 0 as 0^ / 2). Because c^ — 2,
we have fi = 2. This is equivalent to:
0^ = 2b^ (*)
Thus, a^ is a multiple of 2. This implies that a is a multiple of 2.
Therefore, a = 2/c for some integer number k. If we substitute into
equation (*), we obtain 4/c^ = 2b^. Thus, 2fc^ = b^. This implies that b^
is a multiple of 2. Therefore, b is a multiple of 2; that is, b = 2s for
some integer number s. Our calculations show that a and b have at
least 2 as a common factor. However, by hypothesis, a and b are
relatively prime integer numbers. Because it is impossible to find two
numbers that satisfy both these conditions at the same time, we
cannot find a rational number such that q^ = 2.
12. Let y=zax-\-b and y=:cx'{-d be the equations of the two lines.
Because the lines are distinct by hypothesis, we know that either a^c
oxb^d. The coordinates of the intersection point are the solutions of
the system:
3; = ax 4- b
y = ex + d.
Therefore, we obtain ax + b = ex + d or (a - c)x = d - fc. If a — c = 0,
the system has no solutions because b^d, (We can explain this result
geometrically. The two lines have the same slope; therefore, they are
parallel and distinct. Thus, they have no points in common.) If
a - c7»^ 0, we obtain the only solution, x = ^. Therefore, the unique
intersection point is the one having coordinates (^,^frf^). (This
statement can be proved using its contrapositive as well. In this case,
start by assuming that the two lines have two points in common, and
use algebraic steps to obtain the conclusion that a = c and b = d)