Download MATH 312H–FOUNDATIONS

MATH 312H–FOUNDATIONS
LOGIC
Logic is a separate (non-mathematical) discipline. We will use only basic facts from
logic which are (more or less) obvious by common sense.
We are dealing with statements (understood only as a mathematical statement)
which are true or false (there is no other possibility, expressed in Latin as the hypothesis “tertium non datur”). Thus every mathematical statement must be true or
false (decidable).
For example: Every number is an integer, arational or an irrational. True or False?
If a statement A is true we also say that “A holds” or “A is correct”, or a similar
expression.
Given a statement A we can form the statement “not A”, which can also be true
or false. Hence if A is true then “not A” is false and if A is false, then “not A” is
true. We write “¬A” for “not A”. Obviously this relation is symmetric interchanging
A and ¬A.
Now consider two statements A and B. We can form two new statements:
• “A and B” (conjunction)
• “A or B” (alternative), the ‘or’ does not mean exclusively “either...or”.
The following table shows when conjunction and alternative is true or false, whenever
A and B are each true or false:
A\ B true false
true true false
false false false
CONJUNCTION
A\ B true false
true true true
false true false
ALTERNATIVE
If B is the statement ¬A, then one finds that the statement “A and ¬A” is not
true while the statement “A or ¬A” is true.
For more than two statements similar “tables” can be constructed using an iteration
procedure for the two tables above.
Given the two statements A and B we also can form the new statement “if A holds
then B holds”. Equivalent saying is “if A is true then B holds” or “if A is true then
B is true”, or “A implies B”, or “A⇒B”, or “B⇐A”,or “B follows from A”, or “A
yields B”, or “B is necessary for A”, or “A is sufficient for B” and so on...
1
2
The statement “{“A⇒B”} and {“B⇒A”}” is expressed as “A is equivalent to B”,
or “A⇔B”, or “A is necessary and sufficient for B“, or “B holds if and only if A
holds”, and so on...
Proposition 0.1. The statement “A⇒B” is equivalent to the statement “¬B⇒ ¬A”.
Proof. Assume that “A⇒ B” holds. Then if A holds, so does B. So if ¬B holds, then
A cannot hold (since if A holds then B holds; we then are in the situation that B
and ¬ B would be true, a contradiction). The converse holds interchanging A and
¬A and also B and ¬B.
For more than two statements, say A,B and C, we have the logical rule that it
is always true that the statement “{“A⇒B”} and {“B⇒C”}” implies the statement
“A⇒C”. (transitive)
Example. A barman has four customers. He knows that the first one is under 18, the
second one over 18, the third is having a beverage and the fourth an alcoholic drink.
Which two questions he needs to ask to make sure that the law is not violated? The
answer is obvious.
Now the barman has four cards lying on his bar. He knows that each card has a
letter on one side and a number on the other side. He is asked to check whether the
rule “A card with a three can only have a D on it’s other side”. He sees 3, 7, A,and
D. Which two cards does he need to turn over in order to check the rule? The answer
is not so obvious.
You are invited to think it over, in particular, where the above implications A ⇒
B come into the play.
The moral is that the two problems are logical equivalent, although to many of us
it does not look alike. This is called the Watson test in phsychology.
Mathematical theorems mostly have the form “A⇒B”. A is called the assumption
(hypothesis, conditions) and B is called the claim (statement, assertion, conclusion
...). It says that if A is true, then B is true as well. Mathematics needs to make
assumptions what is supposed to be true (without proof). These assumptions are
called axioms. Later, we assume the rational numbers are given to us, which means
that the common rules of calculations are true. We would like to mention at this
point that these rules can be deduced from simpler axioms (the Peano axioms for the
natural numbers), but this is a topic left to special lectures and studies.
A mathematical proof consists of a sequence of statements A=A1 , A2 ,...,An =B for
which the axioms imply that Aj+1 is true whenever Aj is true. By transitivity, B is
true whenever A is true. Of course, we may show that A implies B by showing that
¬B implies ¬A. This is called an indirect proof, or proof by contradiction. You may
check with the above proof of the proposition how these rules apply here.
Finally we remark that if “A is not true then statement B is true” is a false statement (see the table of conjunction). This means that whenever you start with a
wrong assumptions, nothing is correct afterwards!
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In order to make statements easily accessible we need to find a new language
to express these statements in a simple form. In order to do this we introduce the
following terminology: Often, a statement A depends on a variable, say t. This means
that we are given a family of statements A(t) and A is the family of statements
(meaning here the conjunction of all statements A(t)). Then A is true if all the
statements A(t) are true, or for all t A(t) is true. This term for all we shall write as
“∀t”. On the other hand, if A is not true, then there exists a t for which A(t) is not
true. This existence is expressed as “∃t”. In case there is only one such t we shall
write “∃1 t” and call t unique.
Now if in the above situation we consider the family as alternative, then we can
say that A is true if there is one t so that A(t) is true, that is ∃t such that A(t) is
true, ¬ A to be true requires that there is no t such that A(t) is true, or for all t
¬A(t) holds.
A special logical instrument to conlude the truth of a statement is given by Mathematical Induction. We assume that we are given the natural numbers 1,2,3,...
Principle of mathematical induction. Let for each natural number n be given a
statement An . Then if
• Base of induction: the statement A1 is true.
• Induction step: If the statement An is true then An+1 holds as well.
the conclusion of the mathematical induction is that all statements An are true.
This principle has several variants known as recursive definition, inductive definition, finite induction... their meaning is clear whenever they appear in the course.
To make it clear at this point by an example: We can say: If n is an even integer,
then it can be written in the form 2m where m is a natural number. The proof could
use a finite induction, once n is fixed. Then, given n the statement “n can be written
as a multiple of a natural number” is correct. Mathematical induction provides the
conclusion that the statement “Every even ineger can be written as a multiple of 2
and another integer” is correct. Of course, there is only a small difference in the two
aspects. Both are saying the same using mathematical induction (a trivial aspect of
the principle). The last statement can be proven directly without induction of course.
The important point in using mathematical induction is that in order to prove that
a statement holds for all n you only need to show that it holds for n = 1 and it holds
for n + 1 provided it holds for n. This makes the proof much easier in many cases.
Mathematical induction has turned out to be a very useful instrument for proving
correctness of statements. We give some examples.
Example. The sum of the first 10 integers is 55. Gauss found this as a school boy at
the age of 6 or so. He detected a general principle how to calculate the sum.
Theorem 0.2. For all integers n it holds that
1 + 2 + ... + n =
n(n + 1)
.
2
Proof. We proceed by induction. The statements are
.
An : 1 + 2 + ... + n = n(n+1)
2
4
Base of induction: For n = 1
1 + 2 + ...n = 1
1(1 + 1)
=1
2
hence statement A1 is correct.
Induction step: Let An be true. Then
1+2+...+n+(n+1) = (1+2+...+n)+(n+1) =
(n + 1) + ((n + 1) + 1)
n(n + 1)
+n+1 =
2
2
shows that An+1 holds as well.
The conclusion is that An holds for all n.
Of course there is a proof without mathematical induction: If n is even
n
n
n
1 + 2 + +... + n = (1 + n) + (2 + (n − 1)) + (3 + (n − 2)) + ... + ( + ( + 1)) = (n + 1) .
2
2
2
The case of an odd natural number is left as an exercise. This is Gauss’ proof. No
induction needed.
Example
Theorem 0.3. For all positive integers n 7n − 2n is divisible by 5.
Proof. For n = 0 the statement is correct. Let it be true for n. Then
7n+1 − 2n+1 = 7(7n − 2n ) + 2n (7 − 2)
is clearly divisible by 5. By induction, the theorem is proved.
Example The binomial formula is an important one in analysis. It is based on the
factorial
n! = n(n − 1)(n − 2)...3 · 2 · 1
for an integer n ≥ 1 and the binomial coefficients given by
n!
n
=
k
k!(n − k)!
for k = 0, 1, ..., n and n = 0, 1, 2, .... Here we put 0! = 1. Thus we get
Theorem 0.4. (Pascal triangle) For each n
n+1
n
n
k = 1, 2, ..., n.
=
+
k
k−1
k
Moreover,
n
n
= 1.
=
n
0
Proof. We prove the first formula only, since the second one is true trivially by
inspecting the left hand sides.
So let k = 1, ..., n be arbitrary. Then
n!
n
n
n!
=
+
+
k−1
k
k!(n − k)! (k − 1)!(n − k + 1)!
n+1
n!
.
n + 1 − k)!(n + 1 − k + k) =
=
k!
k
5
Theorem 0.5. (Binomial Theorem) For each natural number n and all rational numbers a and b we have
n X
n k n−k
n
a b .
(a + b) =
k
k=0
Proof. We use induction over n.
Base of induction: For n = 1 nothing has to be shown.
Induction step: Let the formula be true for n. We shall show it for n + 1. We have
(a + b)n+1 = (a + b)n (a + b)
n n X
n k+1 n−k X n k n+1−k
a b
a b
+
=
k
k
k=0
k=0
n n−1 X
n k+1 n−k X n k n+1−k
n+1
a b
+ bn+1
a b
+
= a
+
k
k
k=1
k=0
n X
n
n
n+1
n+1
ak bn+1−k
+
= a
+b
+
k−1
k
k=1
n+1
X n+1
ak bn+1−k
=
k
k=0
using the previous Pascal’s theorem.
For later use: The numbers a and b can be replaced by any real or complex numbers,
and other algebraic objects allowing powers and interchanging multiplications.