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 125
13. (a) If there is at least one i with 0 < i < n for which Ui :^ bu then there
is at least one number x for which P(x) and Q{x) are not equal, (b) If
ai = bi for all i, with 0 < i < n, then P{x) and Q{x) are equal for all real
numbers, (c) If there exists at least one real number for which P{x)
and Q{x) are not equal, then there is at least one i with 0 < i < n for
which at i=- bt.
14. (a) If the product of two integer numbers is not odd, then at least one
of the integer numbers is not odd (or: the two integer numbers are not
both odd), (b) If the product of two integer numbers is odd, the two
integer numbers are odd. (c) If two integer numbers are not both odd,
their product is not odd.
15. (a) If two numbers are both not even, then their product is not even.
(b) If at least one of two integer numbers is even, their product is even.
(c) If the product of two integer numbers is not even, then both (all)
numbers are not even.
16. Let X and y be any two real numbers such that x<y. Can we
prove that/ o g(x) </ o g{y)l Because g is nondecreasing, s = g(x) <
g{y) = t. Because / is nondecreasing f(s) <f(t); that is, f{g{x)) =
f(g{y)). So, the statement is true.
17. (Because we cannot directly check all products between rational and
irrational numbers, we could consider using the contrapositive of
the original statement.) Let's assume that xy is rational. Then xy = a/b
where a and b are integers and b^O. Because x is a nonzero rational
number, x = c/d, where with c and d integers, ^ ^ 0, and c^O. What
can we find out about yl Because of our assumption, (c/d)y = a/b.
Because c/d 7^ 0, we can multiply both sides of the equation by its
inverse, d/c, and we obtain y = (ad)/(bc). The numbers a, b, c, and d
are integers, and bc^O because b^^O and c^O. Therefore, >; is a
rational number. Thus, we proved that the contrapositive of our
original statement is true, so the original statement is true as well.
18. For the sake of simphcity, let us assume that n is positive. Because
n has at least three digits, we can write n = rst...cba, where
r, s, t,..., b, a represent the digits; therefore, they are all numbers
between 0 and 9, and r^O. Because we have some information about
the number formed by the two rightmost digits, we will isolate them
and write n = {rst... c) x 100 + fca. By hypothesis, ba is divisible by 4,
so ba = At for some integer number t. Thus, n = {rst ...c)x 100 + 4t =
4[25(rst...c) + t]- The number 25{rst.. .c)-ht is an integer. This
proves that n is divisible by 4. Repeat the proof for the case in which n
is a negative number.