Download Basics in Mathematical Logic 1 Assertions

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Johann K. Brunner
Mathematics, October 2010
Basics in Mathematical Logic
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Assertions
In mathematics we generally deal with assertions (or statements), which are sentences
that are either true or wrong, like "The mathematics course is the …rst course in the
Innsbruck-Linz PhD-program in Economics".1
:A (= nonA) is the negation of an assertion A. :A is true if A is wrong and wrong if
A is true. By using the logical operators AND (^); OR (_), assertions can be combined
to a new assertion: Let A and B be assertions,
- then C = A ^ B is a new assertion, which is true if A and B are both true, otherwise
it is wrong.
- then D = A _ B is a new assertion, which is true if at least one of A and B is true,
it is wrong if both A and B are wrong.
Clearly, combined assertion like C or D can again be combined to more complex
statements.
A particularly important combination of two assertion A, B is
A ) B; which is de…ned as :A _ B:
A ) B should be read as "A implies B" or "If A then B". Note that if we know that the
assertion A ) B is true and that A is true, it follows that B must be true. On the other
hand, if we know that A ) B is true and that B is wrong, then A must be wrong as well
(i. e., :A must be true).
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Thus, we do not talk about phrases that express emotions ("What a wonderful day!") or requests
("Come with me!").
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In fact, the assertion A ) B is completely equivalent to the assertion :B ) :A
(which, by de…nition means B _ :A; which clearly is true whenever :A _ B is true).
If A ) B is true, we say that A is a su¢ cient condition for B, while B is a necessary
condition for A. (Note the di¤erence!)
Example: If a natural number is divisible by 6 (without rest), then it is divisible by 3
(without rest).
Two assertions A; B are called equivalent, if both A ) B and B ) A hold, which is
abbreviated by A , B: If A , B holds, A is true if and only if B is true (and, trivially,
the other way round).2
Example: A natural number is divisible by 6 (without rest) if and only if it is divisible
by 3 (without rest) and by 2 without rest.
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Theorems and proofs
An interesting mathematical statement (= assertion) is usually called a theorem or a
proposition. A lemma is a statement which is itself not that important, but is useful to
prove a theorem or a proposition. Statements typically have the form "A ) B"; where
A is a - possibly combined - statement (the assumption(s) or condition(s)) and B is a possibly combined - statement (the implications(s) or consequence). Frequently, not all
assumptions are mentioned explicitly, because it is understood that the statement is made
within a given framework.
To give an example:
Theorem 1: If good i is a normal good, it cannot be a Gi¤en good.
Here it is understood that we argue within the standard consumer theory, where households can consume n di¤erent goods, maximise utility and observe the budget constraint.
In order to prove a statement A ) B; we transform A and B until we arrive at a
formulation which is known to be true. That is, we try to …nd implications of A (or
equivalent statements of A) which are known to imply B, or to imply a known su¢ cient
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Instead of "if and only if" sometimes the abbreviation "i¤" is used: A , B means A is true i¤ B is
true.
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condition for B.
Example:
Proof of Theorem 1: Consider the demand function fi for good i, depending on
prices p1 ; :::; pn and the budget b, fi (p1 ; :::; pn ; b):
Good i is a normal good by assumption, which means that
@fi
@b
0:
(1)
By the Slutsky-equation, the (uncompensated) price-e¤ect can be separated into a substitution e¤ect and an income e¤ect:
@f c
@fi
= i
@pi
@pi
xi
@fi
;
@b
(2)
where fic denotes the compensated (or Hicksian) demand function and xi = fi (p1 ; :::; pn ; b)
denotes the quantity of good i.
We know from standard consumer theory that the own compensated price e¤ect is
always negative:
@fic
< 0:
@pi
If we use (1) and (3) in (2), we …nd
@fi
@pi
(3)
< 0; which means that good i is not a Gi¤en good.
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Note that equivalent ways to express the meaning of Theorem 1 would be:
A normal good is never a Gi¤ en good.
Let good i be a normal good. Then good i cannot be a Gi¤ en good.
Moreover, as :B ) :A is equivalent to A ) B; Theorem 1 can also be written as: If
good i is a Gi¤ en good, then it cannot be a normal good.
In fact, a statement A ) B is sometimes proved by contradiction, i. e. by showing
that if B is wrong it follows that A is also wrong (:A is implied by :B):
A speci…c technique to prove a statement is called proof by induction. This technique
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(or
a proof.
or the letters QED - "quod erat demonstrandum") is frequently used to designate the end of
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can be used, if natural numbers are involved in the statement. It consists of two steps:
Step 1: show that the statement is true for some (low) natural number n0 ; usually for 1.
Step 2: show that if the statement is true for some arbitrary number n, then it is also
true for the number n + 1:
These two steps allow the conclusion that the statement is true for all natural numbers.
To give an example:
Theorem 2: Let n be a natural number. Then sum of the …rst n odd numbers is
equal to n2 ; i. e.:
1 + 3 + 5 + ::: + (2n
1) = n2 :
(*)
Proof of Theorem 2:
Note …rst that the sum on the left-hand-side (LHS) of (*) consists indeed of n successive
odd numbers, starting with 1.
Step 1: Let n = 1. Then the statement is true because 2n
1 = 1 (the sum in (*) has
only 1 element), and 1 = 12 :
Step 2: Assume that the statement is true for some arbitrary (but odd) number n:
1 + 3 + 5 + ::: + (2n
1) = n2 :
Add the next odd number 2n + 1 on both sides to get
1 + 3 + 5 + ::: + (2n
1) + (2n + 1) = n2 + 2n + 1
Now we have the sum of the …rst n+1 odd numbers on the LHS. Clearly, the righthand side (RHS) is equal to (n + 1)2 : Thus the statement is true for n + 1; given
that it is true for n.
Conclusion: The statement is true for all natural numbers.
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