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28 The Nuts and Bolts of Proof, Third Edition
If we factor the denominator in the right-hand side, we obtain:
X 1_ X—1
6 4 (x- l)(x + 2)
We can divide by x-1 because x^l,
becomes:
so x-1^0, and the equation
X 1 1
6 4 x + 2
We can multiply by x + 2 because x / -2, so x + 2 ^^^ 0 to obtain:
2x^ + X + 6 = 0.
By hypothesis, this equation has at least one real solution, contradicting
the fact that a quadratic equation with a negative discriminant (A =
1 — 4(6)(2) = —47) has no real solutions (impHcit hypothesis). Therefore, the
two graphs have no points in common. •
EXAMPLE 11. Every positive number smaller than 1 is larger than its
square.
Discussion:
A. Consider the collection of all positive real numbers smaller than 1.
(Implicit hypothesis: We are familiar with all properties and operations
of real numbers.)
B. Every number in the collection described in A is larger than its square.
We will prove this statement in two ways and compare the proofs.
1. Direct Proof
Let X be a positive number smaller than 1; that is, x < 1. We can multiply
both sides of this inequality by x to obtain:
Therefore, x is larger than its square.
2. Proof by Contrapositive
x^ < X.
We need to construct the statement "not B." For "not B" there is at least
one positive real number smaller than 1 that is smaller than its square. Thus,