calculus book - Mathematics and Computer Science

1.3. LOGIC 21
(m/n) 2 = 2 is impossible, that is, there is no rational square root of 2.
Proof by contradiction tends to be awkward logically, and does not
generally build a logical link between hypothesis and conclusion. For
this reason, proof by contradiction should be regarded as a last resort.
Happily, most proofs by contradiction can easily be re-written as proofs
by contraposition, though as in Theorem 1.3 it may be necessary to
reformulate the implication appropriately.
It is a common mistake, especially under exam pressure, to start a
proof by assuming the conclusion. This amounts to assuming what is
to be proved, and is clearly wrong. To emphasize:
When proving a logical implication, the conclusion is never
assumed.
Let us return for a moment to Theorem 1.2, which asserts (loosely)
the impossibility of squaring the circle in Euclidean geometry. Here
are the basic ideas of the proof: First it is shown that if a segment of
length π could be constructed in finitely many steps with a straightedge
and compass, then the real number π would satisfy a polynomial
relation with rational coefficients—something like π 2 − 10 = 0 or
1 − π 2 /6 + π 4 /120 = 0 (neither of these is correct). But for purely analytic
reasons (that are, unfortunately, beyond the scope of this book),
such a relation would imply existence of an integer between 0 and 1.
Since no such integer exists, the purported construction is impossible,
in the sense of being incompatible with basic properties of numbers.
Counterexamples The previous items deal with establishing the
truth of a logical implication. The dual task, proving the falsity of
a logical implication, is accomplished by means of counterexamples. A
counterexample to the assertion H =⇒ C is an object x that both
satisfies the hypothesis and fails to satisfy the conclusion. Existence
of such an x proves that the intersection H ∩ C c is non-empty, so that
H ⊂ C is false, see Figure 1.5.
While the falsity of the assertion H =⇒ C can be proven by
finding an object that both satisfies the hypothesis and fails to satisfy
the conclusion, the statement H =⇒ C cannot be proven by finding
an example y satisfying both the hypothesis and the conclusion; in
Figure 1.6, such an example exists but the statement “H =⇒ C”
is false. To prove a logical implication, it must be shown that every
object satisfying the hypothesis satisfies the conclusion.