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 139
10. a. Check the statement for n = 2:
b. Assume it is true for a generic number n > 2; that is.
-"=(» :>
c. Is the statement true for n +1? Using the associative property of
multipHcation of matrices, we can write A^^^ = Ax A^. Thus, by
the inductive hypothesis we obtain:
Therefore, the statement is true for all n > 2 by the principle of
mathematical induction.
EXISTENCE THEOREMS
1. (We are trying to find a function defined for all real numbers. Usually
polynomials are good candidates. But, in general, the range of a
polynomial is much larger than the interval [0,1]. We could try to
construct a rational function, because it is possible to use the
denominator to control the growth of the function. But often rational
functions are not defined for all real numbers. We could try using
transcendental functions.) The functions sin x and cos x are bounded,
because — l<sinx<l and — l<cosx<l, but their ranges include
negative values as well. We could try to square them, or to consider
their absolute values to obtain bounded and nonnegative functions.
The functions sin^x, cos^x, |sinx|, and |cosx| are functions defined
for all real numbers and whose ranges are in the interval [0,1]. You can
graph them to check this claim, if you wish. Note: The functions with
the second power are differentiable; the ones with the absolute value
are not differentiable.
2. Just use n = 2. Then 2^ + 7^ = 53, which is a prime number.
3. We are searching for a number b such that ab = n, where n is an integer.
Because 0 is a rational number, we know that a / 0 and fc ^ 0. Then we
can solve the equation and use b = na~^. So, for example, b = a~^
satisfies the requirement. Because a is irrational, a~^, 2a~^, 3a~\... are
irrational as well.
4. A second-degree polynomial P(x) can be written as P(x) = ax^-ibx
+ c, where a, b, and c are real numbers and a^/^O. We are looking
for a second-degree polynomial satisfying the requirements P(0) = —1