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 141
b. Uniqueness. We can prove this in at least three ways:
i. We can use the result stating that a polynomial of degree n has
at most n solutions. Therefore, a polynomial of degree 1 has
one solution. Because we found it, it must be the only one.
ii. The solution is unique because of the algebraic process used to
find it.
iii. We could assume that the number t is another zero of the
polynomial p(x). Thus, p{t) = 0. Because p{b) = 0 (from part a),
we have p{t) — p{b). This impHes that t-b = b-b. Adding b to
both sides of the equation yields t = b. This is the same solution
we found in part a. Thus, the solution is unique.
2. In this case, we have to prove the existence of the solution of the
equation cos ^ == ^ in the interval [0, n^. One way of achieving this goal
is to graph the functions f{x) = cos x and g(x) = x. If the two graphs
have only one intersection point in the interval [0,7r], the proof is
complete. In this case, the graph can be used as a proof because we are
interested in the situation on a finite interval; therefore, the graph
shows all the possibiUties. Another way to prove the statement is to
graph the function h{x) = (cos x)/x and to show that there is only one
value of X corresponding to y=l. (Be careful: This function is not
defined at x = 0. When x = 0, cos 0 = 1, so cos 0 ^ 0.)
3. We can start by finding a solution for the given equation and then
prove that it is unique. Through algebraic manipulation, we obtain
X = ^ as a solution. Because it is possible to evaluate the third root of
any real number, this expression is well defined for any value of b. Let y
be another solution of the same equation. Then x^ — b = 0 = y^ — b.
This implies that x^ — y^ = 0. Using factorization techniques for the
difference of two powers, this equation can be rewritten as (x — y)
(x^ -\- xy-\- y^) = 0. This product will equal zero only if either x — y — 0
or x^ -\- xy-\- y'^ = 0. The factor x^ -\-xy-\- y'^ is never equal to zero; it is
irreducible (you can try to solve for one variable in terms of the other
using the quadratic formula and check the sign of the discriminant.)
Thus, the only possibiHty is that x- y = 0. This implies that x = y and
the two solutions do indeed coincide. Therefore, the solution is unique.
4. A second-degree polynomial can be written as P(x) = ax^ + fex + c,
where a, b, and c are real numbers, and a^O. We will use the
conditions given in the statement to find a, b, and c.
P(0) = a(Of + b(0) + c = c
P(l) = a(lf -\-bil)-\- c = a-\-b-\- c
P(-l) = a(-lf + b(-l) -\-c = a-b-^c.