Download MAT 300 Mathematical Structures

MAT 300 Mathematical Structures
John Quigg
Spring 2012
revised February 20, 2012
Notes on: Section 3.1
1. Mathematical proofs — direct proofs of conditionals
A mathematical theorem is a precise statement about mathematical concepts, and a proof
of the theorem is a rigorous demonstration that the theorem is true, using logic, the definitions of the terms used in the statement of the theorem, and other theorems that have been
proven previously.
The best way to learn how to do proofs is to look at many examples. In each case we
analyze the statement of the theorem, determining what the hypotheses and the conclusion
are. The hypotheses are statements that we assume are true, and the conclusion is the
statement that we must prove.
Example 1. We will prove the following theorem:
Theorem 2. Let a, b ∈ R. If 0 < a < b then a2 < b2 .
Scratchwork. We will usually begin with a discussion, called scratch work, where we figure
out the proof, and then we will write up the formal proof. None of the scratch work can be
used in the proof itself. One consequence of this is that in our scratch work we are allowed
to state things informally, and even sometimes not quite completely correctly!
The proof must stand completely on its own, and doesn’t have to give any explanation of
how the arguments in the proof were figured out.
The first thing we have to do is carefully analyze the statement of the theorem. In this
case the theorem has two sentences. The first sentence specifies the universe of discourse
of the free variables a, b. The second sentence is the conclusion of the theorem, and is the
statement that we must prove.
We recognize the words “if” and “then” in the conclusion, and this lets us see that the
logical structure of the statement is a conditional P → Q, where P is “0 < a < b” and Q is
“a2 < b2 ”.
To prove the conditional P → Q, we assume P , and deduce Q. This already tells us how
the proof starts and ends:
Start of proof: “Assume that 0 < a < b.”
End of proof: “Therefore, a2 < b2 .”
However, we don’t have to use those exact words! What comes in between the start and
the end depends upon the properties of the mathematical concepts involved in the theorem.
In this case, the theorem involves real numbers and inequalities, and we are allowed to use
the basic properties of these things in our proof.
Specifically, in this proof we will need the following two properties of inequalities:
• transitivity: For all x, y, z ∈ R, if x < y < z then x < z. Here “x < y < z” is an
abbreviation for “x < y and y < z”.
• For all x, y ∈ R and t > 0, if x < y then tx < ty.
2
We don’t have to explicitly state these facts in our proof, but it should be clear how we’re
using them.
Note that once we assume 0 < a < b, by transitivity we’ll have 0 < b. So, we have two
positive numbers a > 0 and b > 0.
Multiplying both sides of the inequality a < b by a gives a2 < ab. On the other hand,
multiplying both sides of a < b by b gives ab < b2 .
Thus we have a2 < ab < b2 , so, again by transitivity, we get a2 < b2 , which is what we
wanted.
It seems that this will work, so let’s try the formal proof:
Proof. Assume that 0 < a < b. Then a > 0 and b > 0.
Multiplying both sides of a < b by a gives
a2 < ab,
and multiplying both sides of a < b by b gives
ab < b2 .
Thus a2 < ab < b2 , so
a2 < b2 .
Example 3. Consider the theorem written as Example 3.1.1 in the textbook:
Theorem 4. Suppose x > 3 and y < 2. Then x2 − 2y > 5.
Here the hypotheses are x > 3 and y < 2, and the conclusion is x2 − 2y > 5.
But: the free variables x and y have not really been very clearly “introduced”. The
hypotheses x > 3 and y < 2 could be interpreted as giving universes of discourse: (3, ∞) for
x and (−∞, 2) for y. But it is more common to infer from the context (inequalities) that
the universe of discourse for both x and y is R.
Problem 3.1 # 14 asks for the actual proof of this theorem, and would involve using the
inequalities x > 3 and y < 2 to deduce x2 − 2y > 5. As the hint in 3.1 # 14 states, the proof
would also use Theorem 2 to deduce x2 > 9 from x > 3.
Example 5. Consider the following:
Theorem 6. Let x ∈ R. If x < 2 then 1 − x > 0.
We’ll see that this theorem is actually false. Let’s analyze it: the first sentence gives the
universe of discourse for x, and the second sentence is the conclusion. The logical structure
of the conclusion is a conditional P → Q, where P is “x < 2” and Q is “1 − x > 0”. The
theorem states that P → Q is true for every x ∈ R.
3
But: if x = then P is true and Q is false. Thus the conditional P → Q is false in this
2
case.
3
Therefore the theorem itself is false. The false case x =
is a counterexample to the
2
theorem. Usually, and certainly in this case, the counterexample is not unique; there are
other counterexamples that we could have given. We must give a specific counterexample,
but we only need 1 — we must choose one.
Finding a counterexample to a theorem is one way to disprove it. If a theorem is true we
want to prove it, but it it’s false we want to disprove it.
3
2. Proof by contrapositive
Example 7. This time we’ll use a new strategy for proving a conditional.
Theorem 8. Let x ∈ R. If
x−5
=1
x+4
then x 6= 2.
Scratchwork. The universe of discourse of the free variable x is R. The conclusion is a
conditional P → Q.
Our earlier strategy for proving P → Q would be to assume P and deduce Q. This is a
direct proof of the conditional. However, this time we will give an indirect proof.
Recall that we are allowed to use any logical tautology in a proof. Here we’ll use the
tautology
(P → Q) ↔ (¬Q → ¬P ),
i.e., every conditional is equivalent to its contrapositive. Thus, if we can prove ¬Q → ¬P
then we’ll be able to immediately deduce that P → Q is true. This method is proof by
contrapositive, but more generally it’s an example of an indirect proof. Since indirect proofs
are harder to follow than direct proofs, to make the proof clearer we will indicate our strategy
in the proof. Then we’ll give a direct proof of the contrapositive. Finally, we’ll conclude that
the original conditional statement is true.
x−5
The contrapositive of our conclusion is “if x = 2 then
6= 1”.
x+4
Thus, we see the start and the end of the proof:
Start of proof: “We argue by contrapositive. Assume that x = 2.”
x−5
x−5
End of proof: “Thus
6= 1. Therefore, if
= 1 then x 6= 2.”
x+4
x+4
What comes in between? Presumably we want to plug the value x = 2 into the expression
x−5
, and show that the resulting value is not 1.
x+4
Let’s try the formal proof:
Proof of Theorem 8. We argue by contrapositive. Assume that x = 2. Then
x−5
2−5
=
x+4
2+4
−3
=
6
1
=−
2
6= 1.
Thus
x−5
x−5
6= 1. Therefore, if
= 1 then x 6= 2.
x+4
x+4