Download 1. The sequence 2. Proof by contradiction

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notes & corrections
math 100
Friday, 4-7-17
1.
The sequence
Here is the beginning of a sequence: OTTFF... What’s next?
2.
Proof by contradiction
When trying to prove that P implies Q, where P is a collection of facts known (or assumed)
to be true and Q is a new fact whose truth is in question, it is sometimes easier to rely on the
logical equivalence
P −→ Q ⇐⇒ (P ∧ (∼ Q)) −→ (R ∧ (∼ R))
and prove that the negation ∼ (P −→ Q) leads to a contradiction.
A proof by contradiction is also called an indirect proof, in that it doesn’t explain directly
why P −→ Q, but shows instead that the opposite is impossible. Many mathematicians, though
happy to use indirect proofs when necessary, consider it something of a last resort. Whatever
your feelings may be on the subject,† it is a good idea to study any given proof by contradiction
to see if it can be reworked into a direct proof of one kind or another.
Examples...
Theorem 1. If x is a rational number, then x2 6= 2.
Proof. Suppose, to the contrary that x is rational and x2 = 2, then there exist integers a and
b such that x = a/b, where we may assume that a and b have no common factors. From the
assumption that x2 = 2 it follows that (a/b)2 = 2and so a2 = 2b2 .
It follows that 2 divides a2 , and since 2 is prime, this implies that 2 divides a or 2 divides a, so
2 divides a.‡ Hence there is an integer c such that a = 2c.
Substituting 2c for a in the equation a2 = 2b2 gives
4c2 = (2c)2 = 2b2
and dividing this equation through by 2 shows that b2 = 2c2 . Repeating the argument of the
previous paragraph, it follows that 2 divides b. But this contradicts the assumption that a and b
have no common factors, since 2 divides both.
This contradiction shows that the assumption that there is a rational solution to the equation
x2 = 2 must be false.
Comments: I didn’t assume that a and b are positive. Does it matter? Also, the symbol ‘’ (at
the end of the last line of a proof) is the same as ‘QED’—it signifies that we have completed the
proof.
Finally, the advantage of trying a proof by contradiction in this case is that it gives us a toehold: the assumption that x2 = 2 allows us to draw further conclusions that eventually lead to
an contradiction. Without this assumption, we don’t really know how to begin.
In the next example I will use some (well-known) facts and theorems from basic calculus.
Specifically...
(1) Polynomials are both continuous and differentiable on the entire real line.
(2) If x is a real number and n is an even integer, then xn ≥ 0.
†
Perhaps it is a little early in your mathematical education to have feelings on this particular subject.
Here we are using the theorem I stated on Wednesday: the integer n > 1 is prime if and only if for any two
integers a and b, if n divides ab, then n divides a or n divides b.
‡
1
2
(3) The intermediate value theorem: If f (x) is continuous on the interval [a, b] then for
every value y between f (a) and f (b), there is an c in (a, b) such that f (c) = y.
(4) Rolles’ theorem: If f (x) is differentiable on the interval [a, b] and f (a) = f (b), then
there is a point c in the interval (a, b) such that f 0 (c) = 0.
Theorem 2. The equation 2x7 + 3x3 + x − 5 = 0 has exactly one real solution.
Comment: This is an existence and uniqueness statement: we need to show that a solution to
the equation exists and that it is unique. This typically requires two arguments, one for each
claim.
Proof. Let f (x) = 2x7 + 3x3 + x − 5, then f (x) is a polynomial and therefore continuous and
differentiable on the entire real line (fact (1)). To show that a solution to the equation f (x) = 0
exists, we observe that f (0) = −5 and f (1) = 1 and since −5 < 0 < 1, it follows from the
intermediate value theorem (number (3) on the preceding list) that there is a point x0 in the
interval (0, 1), such that f (x0 ) = 0. To show that the solution is unique, we will use a proof-bycontradiction.
Suppose to the contrary that there is a point x1 6= x0 such that f (x1 ) = 0. Since f (x) is
differentiable on the whole real line and f (x0 ) = 0 = f (x1 ), it follows from Rolles’ theorem
(number (4) on the list above) that there is a point x2 between x1 and x0 such that f 0 (x2 ) = 0.
But f 0 (x) = 14x6 + 9x2 + 1, so
f 0 (x) ≥ 1 > 0
for all real numbers x, because the exponents in f 0 (x) are all even (fact (2)), so it is impossible
that f 0 (x2 ) = 0. This contradiction implies that the assumption that there is more than one
solution to the equation f (x) = 0 must be false.
Comment: In class I asked you to figure out what I didn’t like about this last proof. If you are
still thinking about that, then stop reading now.
As I mentioned before the examples, it is good practice to dissect proofs by contradiction to
see if they can be transformed into direct proofs. The proof above that x2 = 2 has no rational
solutions does not lend itself to such a transformation.§ Direct proofs do exist, but they use
different types of arguments,√and typically require more assumptions. For example, if we agree
that there is a real solution 2 to the equation x2 = 2, it is possible to show (directly) that this
number has a property that rational numbers do not have, and so it cannot be rational.
The proof by contradiction of the uniqueness of the solution of f (x) = 0 on the other hand can
be transformed to a direct proof rather easily. The key difference between the two cases is that
in the second one, we have more to work without making contradictory assumptions, in that the
fact that f 0 (x) ≥ 1 for all real numbers x does not come from the assumption that there are two
solutions, but from the nature of f (x).
Exercise: Rewrite the proof of the uniqueness of the solution in the second example as a direct
proof.
Hint: What can you say about a function ϕ(x) that is differentiable on the entire real line such
that ϕ0 (x) > 0 for all x? You may want to cite the more general version of Rolles’ theorem, the
mean value theorem, or you can just rely on one of its well-known consequences.
§
Though you can reach a different type of contradiction using much the same reasoning, by assuming for example
that you have found the rational solution with the smallest positive numerator, and then finding a solution with
an even smaller positive numerator, for example.