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 133
Because/and g are odd functions,/(-x) = -f{x) and g(-x) =
Thus, we obtain:
-gix).
(/ + g)(-^) =f(-^) + di-^) = (-/W) + (-^W)
= -[fM + g{x)]==-(f + g){x).
Therefore, (/ + g)(-x) — -(/ + g)(x), and -(/ + g)(x) in general is
not equal to (f-\-g)(x). So, the equality if-\-g)(x) = (f-{-g){-x)
does not seem to be true for all real numbers. Can we find a
counterexample? The functions f(x) = x and g(x) = 2x are two odd
functions (check this claim). Their sum is the function
{f-{-g)(x) = 3x, which is not even. (Moreover, (f-\-g)(x) = 3x is
odd.)
11. The function ^^ is not going to be defined for all values of x, in
general. Indeed, it is not defined for all values of x that are zeros
for the function g. Thus, in general the statement is false. The fact
that f/g is either even or odd is not relevant. As an expUcit counterexample,
consider f{x) = x and g(x) = x^ - x^. Try to prove the
following statement: Let/and g be two odd functions defined for all
real numbers. Their quotient, the function ^/^ defined as (if/g){x)) =
(f(x)/g(x)), is an even function defined for all real numbers for which
g{x) / 0.
12. Let n = xyzzyx, with l<x<9,0<y<9, and 0<z<9. Then we can
write:
n = X + lOj + lOOz + l,000z + 10,000}; + 100,000x.
Therefore, n= l,100z+ 10,010};+ 100,001x= 11 x 100z+ 11 x 910}; +
11 x9091x. Thus, n=ll(100z + 910}; + 9091x). Because the number
lOOz + 910}; + 909 Ix is an integer, we conclude that n is divisible by 11.
13. It is true that, if two numbers are rational, then their sum is rational;
however, the converse of this statement is not true. If x is an irrational
number, its opposite, -x, is irrational as well. Their sum is 0, which is
a rational number. Thus, the sum of two numbers can be rational
without either one of them being rational.
14. To prove that g is even, we need to show that ^(x) = ^(-x) for all
real numbers x. Because / is an odd function, /(-x) = -/(x), and
g(x) = {f{x)f= {-fix)f= if{-x)f= gi-x). Thus, g is even.
15. This statement is false. Consider/(x) = x^ + 1. Then ^(x) = (x^ + 1)^.
If we use xi = -1 and X2 = 0, then xi <X2 but ^(xi)>^(x2).
Calculus approach: If we want to find the derivative of g, using the
chain rule, we obtain g\x) = 3{f(x)ff(^)- The factor 3(/(x))^ is