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 147
COMPOSITE STATEMENTS
1. We can rewrite the equation x^ = y^ as x^ — y^ = 0. We can factor
the difference of two squares and rewrite the equation as
(^ — y){^ + y) = 0. This equahty imphes that either x - y = 0 or
x + y = 0. If both X and y are equal to zero, both equahties are
trivially true and x = y. Therefore, we will assume that x^Q and y#0.
From the previous equahties we obtain that either x = y or x = —y.
If x = —y, because neither x nor y is equal to zero one of them
would be a positive number and the other would be negative. But the
second part of the hypothesis states that the two numbers are
nonnegative. So, we have to reject this case.
Thus, the only conclusion we can accept is that x = y,
2. The function/is even. So/(x)=/(-x) for all x in its domain. Because
the function / is odd as well, /(—x) = —f{x) for all x in its domain.
Combining these two hypotheses yields /(x) =j{—x) = -f(x).
Thus, 2/(x) = 0 for all x in the domain off. This means that/(x) = 0 for
all X in the domain of the function.
3. Let n be a multiple of 3, and assume that n is not odd. Then it is even,
and therefore divisible by 2. Because n is divisible by 2 and by 3, it is
divisible by 6. (Even if n = 6 the statement is true as 6 is divisible by 6.)
Thus, the two choices listed in the conclusion are the only two possible
ones. The statement is therefore true.
4. Let us assume that x ^ y. We can rewrite the equality x^ = y^ as
(^ - y\^ + y){^ + J^) = 0. Therefore, x — y = ^ or x-\-y = ^ or
x^ + y^ = 0. The first equahty implies x = y, but we have excluded
this possibihty. The second equahty implies that x — —y. The statement
is true if there are no more possible choices but this. The last equality is
possible if and only if x = }^ = 0. Because we are working under the
hypothesis that x 7^ y, we cannot accept this conclusion. Therefore, if
X / }; the only possibihty left is that x = —y.
5. We will assume that A is nonempty. The set A — B is empty by
hypothesis. By definition of the set A — B, this means that there is no
element of A that does not belong to B. Then all the elements of A
belong to B. Therefore, A £B.
6. Case i. Let A = 0. By definition of union, AU B = {x|x £ A or x e B}.
Because the condition x G ^ is always false (i.e., there exists no x € A),
then AUB = {x\x G A oxx e B] = {x\x e B} = B.
Case 2. Let Ac.B.By definition of union, AU B = {x\x e A or x e B).
By hypothesis, A£B. Therefore, xeA implies x £ B. Thus, the