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140 The Nuts and Bolts of Proof, Third Edition
and P(-l) = 2. From the first condition, we obtain P(0) = a(0)^+
b{0)-{-c = c. Thus, c = —1. From the second condition we obtain
P(-l) = a{-lf + b(-l) + c = a - ft + c. Using the fact that c = -1,
the second equation yields a-b = 3. We have one equation and two
variables. Thus, one variable will be used as a parameter. As an
example, we can write a = ft + 3 and then choose any value we like for
ft. If ft = 0, we obtain the polynomial P{x) = 3x^-1. Clearly, this is
only one of infinitely many possibihties.
5. Because ft^ will be a negative number, ft must be negative. Indeed,
powers of positive numbers are always positive. Moreover, a cannot be
an even number, because an even exponent generates a positive result.
The numbers a and ft can be either fractions or integer numbers. Let us
try ft = -27 and a= 1/3. Then a^ = (1/3)"^^ = 3^^, which is a positive
integer number, and ft^ = (-27)^^^ = -3, which is a negative integer.
6. We can consider two cases: either a„ > 0 or a„ < 0. Let us assume that
an > 0. When the variable x has a very large positive value, the value of
P{x) will be positive, because the leading term, a„x", will overpower all
the other terms {i.e., limx->+oo ^W = +oo). When the variable x is
negative, and very large in absolute value, the value of P{x) will be
negative for the same reason {i.e., lim^-^-cx) P{^) = —oo). Polynomials
are continuous functions. Therefore, by the Intermediate Value
Theorem, there exists a value of x for which P(x) = 0. We can prove
in a similar way that the statement is true when «„ < 0.
7. This can be either a theoretical or constructive proof. Let us use the
constructive approach. We can write a = p/q and ft = n/m, where m, n,
p, and q are integer numbers, q^O, and m^O. Let c = (a H- ft)/2. Then
a < c <b, and c is a rational number because c = (mp + nq)/ (2qm).
Then consider d = (a-\- c)/2 and/ = (c + ft)/2. These two numbers are
both rational (write them explicitly in terms of m, n, p, q\ and
a < d < c <f<b (again use Example 3 in the Equivalence Theorems
section).
8. Consider k = -l. Then 2-i>4-^
UNIQUENESS THEOREMS
For completeness sake, we must prove that: (a) the polynomial p{x) has
a solution, and (b) the zero is unique.
a. Existence. Find the value(s) of the variable x for which p{x) = 0.
To do so, we have to solve the equation x~ft = 0. Using the
properties of real numbers we obtain x = b.