203 - Chapter 4

An integer n is even if, and only if,

n equals twice some integer. n is even ⇔ ∃an integer k such that n = 2k.

An integer n is prime if, and only if,

n>1 and for all positive integers r and s, if n=rs, then either r or s equals n. n is prime ⇔ ∀positive integers r and s, if n = rs then either r = 1 and s = n or r = n and s = 1.

An integer n is composite if, and only if,

n>1 and n=rs for some integers r and s with 1<r<n and 1<s<n. n is composite ⇔ ∃positive integers r and s such that n = rs and 1 < r < n and 1 < s < n.

A nonconstructive proof of existence involves showing either

(a) that the existence of a value of x that makes Q(x) true is guaranteed by an axiom or a previously proved theorem or (b) that the assumption that there is no such x leads to a contradiction.

proofs of existence

According to the definition given in Section 3.1, a statement in the form ∃x ∈ D such that Q(x) is true if, and only if, Q(x) is true for at least one x in D. One way to prove this is to find an x in D that makes Q(x) true. Another way is to give a set of directions for finding such an x. Both of these methods are called constructive proofs of existence.

Statement: ∀ real numbers a and b, if a^2 = b^2, then a = b.

Counterexample: Let a = 1 and b = −1. Then a^2 = 1^2 = 1 and b^2 = (−1)^2 = 1, and so a^2 = b^2. But a = b since 1 != −1.

Existential Instantiation

If the existence of a certain kind of object is assumed or has been deduced then it can be given a name, as long as that name is not currently being used to denote something else

Use the method of exhaustion to prove the following statement: ∀n ∈ Z, if n is even and 4 ≤ n ≤ 26, then n can be written as a sum of two prime numbers.

Solution 4 = 2 +2 6= 3 +3 8= 3 +5 10= 5 + 5 12 = 5 +7 14= 11 +3 16= 5 + 11 18 = 7 + 11 20 = 7 + 13 22 = 5 + 17 24 = 5 + 19 26 = 7 + 19

counterexample

To disprove a statement of the form "∀x ∈ D, if P(x) then Q(x)," find a value of x in D for which the hypothesis P(x) is true and the conclusion Q(x) is false.

Method of Generalizing from the Generic Particular

To show that every element of a set satisfies a certain property, suppose x is a particular but arbitrarily chosen element of the set, and show that x satisfies the property.

An integer n is odd

if, and only if, n equals twice some integer plus 1. n is odd ⇔ ∃an integer k such that n = 2k + 1.

The disadvantage of a nonconstructive proof is that it

it may give virtually no clue about where or how x may be found.

To disprove a statement means

to show that it is false