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 131
nonnegative: P(l) = -(1)^ + 2(1) - (3/4) = 1/4. Another way to
prove that the statement is false is to construct the graph of the
polynomial P(x), and to observe that the graph is not completely
located below the x-axis.
4. The statement seems to be true. But, if x = 1, then y=l. Thus, we
have found a counterexample. The statement becomes true if we
either change the hypothesis to "the reciprocal of a number x > 1" or
the conclusion to "0 < y < 1."
5. If n = 1, then 3^ + 2 = 5, which is a prime number.
If n = 2, then 3^ + 2 = 11, which is a prime number.
If n = 3, then 3^ + 2 = 29, which is a prime number.
If n = 4, then 3"* -h 2 = 83, which is a prime number.
If n = 5, then 3^ + 2 = 245, which is not a prime number. Therefore,
the statement is false.
6. (The functions fog and fob are equal if and only if
/ o ^(x) =/ o h{x) for all the values of the variable x. By definition
of composition of function, this equaUty can be rewritten as
f{g{x)) =f(h{x)). From this, can we conclude that ^(x) = h{x)l Or
could we find a function / such that f{g{x)) =f{h{x)) even if
g(x) ^ h{x)l The answer does not seem to be obvious. Let's look
for some functions that might provide a counterexample. Keep in
mind that counterexamples do not need to be "comphcated.") Let's
try to use ^(x) = x and h{x) = —x. Is it possible to choose a function/
such that/(^(x)) =f(h(x))l Given our choices of the functions g and
h, this equaUty becomes /(x) =/(—x). So we need a function that
assigns the same output to a number and its opposite. What about
f(x) = x^? We will now check to see if we really have found a
counterexample:
fog(x)=f(g(x))=f(x)
= x'
f o h(x) =f{h{x)) =fi-x) = {-xf = x\
So the equaUty/(^(x)) =/(/z(x)) holds, but the functions g and h are
not equal. Using the same choices for g and /i, we can use
f{x) = cosx, or/(x) = x"*, or/(x) = x^, or any other even function.
7. Let us try to prove this statement. Let n be the smallest of the five
consecutive integers we are going to add. So the other four numbers
can be written as n + 1, n + 2, n + 3, and n + 4. The sum of these five
numbers is S = n + (n + 1) + (n + 2) + (n 4- 3) + (n + 4) = 5n + 10.
This number is divisible by 5, so the statement is true.