Proving Existential Statements∃x D such that Q(x)∈

is true if, and only if,Q(x) is true for at least one x in D.

1. find an x in D that makes Q(x) true.2. Give a set of directions for finding such an x.

o Both of these methods are called constructive proofs of existence.

Proving Existential Statementso Prove the following: an even integer n that can be written ∃

in two ways as a sum of two prime numbers.o Let n = 10. Then 10 = 5 + 5 = 3 + 7 and 3, 5, and 7 are all prime

numbers.

o Suppose that r and s are integers. Prove the following: an ∃integer k such that 22r + 18s = 2k.o Let k = 11r + 9s. o Then k is an integer because it is a sum of products of integers;

o and by substitution, 2k = 2(11r + 9s), which equals 22r + 18s by the distributive law of algebra.

Proving Existential Statementso A nonconstructive proof of existence involves showing

eithera) 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.

o The disadvantage of a nonconstructive proof is that it may give virtually no clue about where or how x may be found.

Disproving Universal Statements by Counterexampleo To disprove a statement means to show that it is false.

∀x in D, if P(x) then Q(x).

– Showing that this statement is false is equivalent to showing that its negation is true

∃x in D such that P(x) and not Q(x).

– Example is given to show that statement is true and actual statement is false.

– Such as example is called counterexample.

Disproving Universal Statements by Counterexampleo 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. Such an x is called a counterexample.

o Statement: real numbers a and b, if a∀ 2 = b2, then a = b.o Counterexample:o Let a = 1 and b = −1. Then a2 = 12 = 1 and b2 = (−1)2 = 1, and

so a2 = b2. But a b since 1 −1.

Proofs by Contradictiono We will assume that the claim we are trying to prove is wrong

and reach a contradiction.o If all the derivations along the way are correct, then the only

thing that can be wrong is the assumption, which was that the claim we are trying to prove does not hold.

o This proves that the claim does hold.o This method of proof is also known as reductio ad absurdum

because it relies on reducing a given assumption to an absurdity.

Proofs by Contradictiono For example, if a man accused of holding up a bank can prove that he

was at some place else at the time the crime was committed, he will certainly be cleared. The logic of his defense is as follows:

Suppose I did commit the crime. Then at the time of the crime, I would have had to be at the scene of the crime. In fact, at the time of the crime I was in a meeting with 20 people far from the crime scene, as they will testify. This contradicts the assumption that I committed the crime since it is impossible to be in two places at one time. Hence that assumption is false.

Proofs by Contradictiono Many theorems in mathematics are conditional statements

(p→q). Now the negation of the implication p→q is

~(p→q) p ~q

Clearly if the implication is true, then its negation must be false, i.e., leads to a contradiction.

Hencep→q (p ~q) → c

where c is a contradiction.

Proofs by Contradictiono The method of proof by contradiction, may be summarized as

follows:

1. Suppose the statement to be proved is false.

2. Show that this supposition leads logically to a contradiction.

3. Conclude that the statement to be proved is true.

Proofs by ContradictionoWhen to use proof by contradiction

– Proof by contradiction is indicated– If you want to show that there is no object with a certain

property,

– Or if you want to show that a certain object does not have a certain property.

Proofs by Contradictiono THEOREM: There is no greatest integer.

PROOF:

Suppose there is a greatest integer N. Then n ≤ N for every integer n.

Let M = N + 1Now M is an integer since it is a sum of integers.Also M > N since M = N + 1Thus M is an integer that is greater than the greatest integer, which is a contradiction. Hence our supposition is not true and so there is no greatest integer.

Proofs by Contradictiono THEOREM:

Give a proof by contradiction for the statement:

“If n2 is an even integer then n is an even integer.”

PROOF:Suppose n2 is an even integer and n is not even, so that n is odd.

Hence n = 2k + 1 for some integer k.

Now n2 = (2k + 1) 2

= 4k2 + 4k + 1

= 2•(2k2 + 2k) + 1

= 2r + 1 where r = (2k2 + 2k)

This shows that n2 is odd, which is a contradiction to our supposition that n2 is even. Hence the given statement is true.

Proofs by Contradiction-Home work

o Prove by contradiction method, the statement: If n and m are odd integers, then n + m is an even integer.