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 169
obtain, namely [(n + 1) + 1]^. If we evaluate [{n + 1) + 1]^ we
obtain [(n + 1) + lf= n^ -\-4n-\- 4. Therefore, we have to compare
the expressions 2n^ + 4n + 2 and n^ + 4n + 4. We will do so by
calculating their difference: (2n^ + 4n + 2) - {n^ + 4n + 4) = n^ - 2.
Because n^ > 36, n^ - 2>0.
This proves that (2n^ + 4n + 2) > {r? + 4^ + 4). If we now hst all
the steps just performed altogether, we have:
2"+^ = 2 X 2" > 2(n + 1)^ = 2^^ + 4n + 2
>n^ + 4n-\-4 =
[{n+l)+lf.
The inequahty is therefore true. So, by the principle of mathematical
induction, the inequahty will hold true for all integers fc > 6.
34. Because this is an existence statement, it is enough to find one number
(not necessarily an integer) such that 2^>(/c+ 1)^. Consider k = 6.
35. The statement does not seem to be true. We can try to prove that it is
false by providing a counterexample. Consider t = 1 and q = 1/2. Then
t-\-q = 3/2, and this is not an irrational number. (We can indeed prove
that the statement is false for every two rational numbers. We can
prove that the sum of two rational numbers is always a rational
number. Indeed, if t and q are two rational numbers, we can write
t = a/b, where a and b are relatively prime integers, and b/0; and
q = c/d, where c and d str relatively prime integers, and d^O. Then
bdy^O, and t-\-q = {ad-\-bc)/bd. Therefore, t-\-q is a well-defined
rational number because ad + be and bd are both integers and bd ^ 0.)
36. This is an existence statement. To prove that it is true we only need to
exhibit three consecutive integer numbers whose sum is a multiple of
3. Consider 3, 4, and 5. Then 3 + 4 + 5=12, which is a multiple of 3.
The fact that this statement is true for any three consecutive integer
numbers (see Exercise 28) is irrelevant. It just makes it very easy to
find an example.
37. The statement seems false; therefore, we will search for a counterexample.
Consider n = 5. Then 5 is a multiple of itself, but 5^ = 25,
which is not a multiple of 125.
38. (We cannot use proof by induction because we do not know what the
smallest number is that can be used to estabUsh the base case.) We are
going to prove that n^ -{-n = 2t for some integer number t. Using
factorization we can write n^ + n = n(n + 1). If n is an even number,
then n = 2q for some integer number q. Thus, n^ + n = n(n + 1) =
2[q{n + 1)]. Because the number q{n + 1) is an integer, this proves that