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Proving existential statements
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 Statements
o 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 Statements
o 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.
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
Counterexample
o 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
Counterexample
o 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 a2 = 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 Contradiction
o 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 Contradiction
o 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 Contradiction
o Many theorems in mathematics are conditional statements
(p→q). Now the negation of the implication p→q is
~(p→q) ≡ p Λ ~q
Clearly if the implication is true, then its negation must be false, i.e.,
leads to a contradiction.
Hencep→q ≡ (p Λ ~q) → c
where c is a contradiction.
Proofs by Contradiction
o 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 Contradiction
oWhen 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 Contradiction
o THEOREM: There is no greatest integer.
PROOF:
Suppose there is a greatest integer N. Then n ≤ N for every integer n.
Let M = N + 1
Now M is an integer since it is a sum of integers.
Also M > N since M = N + 1
Thus 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 Contradiction
o 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.

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Proving existential statements

  • 2. 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.
  • 3. Proving Existential Statements o 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.
  • 4. Proving Existential Statements o 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. o The disadvantage of a nonconstructive proof is that it may give virtually no clue about where or how x may be found.
  • 5. Disproving Universal Statements by Counterexample o 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.
  • 6. Disproving Universal Statements by Counterexample o 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 a2 = 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.
  • 7. Proofs by Contradiction o 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.
  • 8. Proofs by Contradiction o 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.
  • 9. Proofs by Contradiction o Many theorems in mathematics are conditional statements (p→q). Now the negation of the implication p→q is ~(p→q) ≡ p Λ ~q Clearly if the implication is true, then its negation must be false, i.e., leads to a contradiction. Hencep→q ≡ (p Λ ~q) → c where c is a contradiction.
  • 10. Proofs by Contradiction o 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.
  • 11. Proofs by Contradiction oWhen 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.
  • 12. Proofs by Contradiction o THEOREM: There is no greatest integer. PROOF: Suppose there is a greatest integer N. Then n ≤ N for every integer n. Let M = N + 1 Now M is an integer since it is a sum of integers. Also M > N since M = N + 1 Thus 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.
  • 13. Proofs by Contradiction o 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.
  • 14. 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.