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142 The Nuts and Bolts of Proof, Third Edition
Thus,
c = -\
a — fe + c = 2.
We can simplify this system of three equations and obtain
c = -\
a + b = 4
a-b = 3.
So, a = 7/2, b — Ijl, and c = -1. Thus, the polynomial that satisfies the
given requirements is:
We have just proved that a polynomial satisfying the given
requirements exists. Is this polynomial unique? The values of a, b,
and c we obtained are the only solutions to the equations generated by
the three conditions given in the statement, as we can see from the
calculations performed to obtain them. (You might want to consult a
linear algebra book for a more theoretical proof of this statement.)
Therefore, the polynomial we obtained is the only one that satisfies the
requirements.
5. To find the coordinates of the intersection points, sety(x) = ^(x). Then
x^ — —x^ — 2x. This equation is equivalent to x{x^ + x + 2) = 0. This
product is zero if either x = 0 or x^ + x -\- 2 = 0. The quadratic
equation has no solution because its discriminant is negative. Thus,
the equation J{x) = g{x) has a unique solution: x = 0. The corresponding
value of the }^-coordinate is y = 0. Thus, the two graphs have a
unique intersection point—namely, (0,0).
EQUALITY OF SETS
1. First part: (AUB)U(AU C) CAU{BU C). Let x € (^ U B) U (^ U C).
Then either x e {AU B) or x e (AU C). Thus, either (x € .4 or x € B)
or (x € A or X € C). If we eliminate the redundant part of this
sentence (the repeated information), we can rewrite it as xeAov
X e B or X e C. This implies that either x e Aor x e {BU C). Thus,
xeAU(BUC).