NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
DAMIEN PITMAN
Definitions & Theorems
Definition: A direct proof is a valid argument that verifies the truth of an impli-
cation by assuming that the premise is true and deducing that the
conclusion is true under that assumption.
Comment: A direct proof of a universally quantified statement usually begins
by assuming the premise for an arbitrary variable x in the universe
of quantification and then showing that we can deduce the conclusion no matter what value x takes in the universe.
Definition: We say that the proof of an implication is trivial when the conclusion
is always true.
Definition: We say that the proof of an implication is vacuous when the premise
is always false.
Definition: Sometimes it is convenient to prove different portions of a statement
separately. So long as all the proof taken in conjunction implies the
original statement, this is a valid proof technique called proof by
cases or proof by exhaustion.
Definition: Suppose we want to prove a statement concerning an arbitrary ele-
ment x of a domain S. When our proof proceeds by examining only
one subset of S at a time we say that we are doing a proof by cases.
Comment: When the proof of one case is (nearly) identical to the proof of an-
other case, it is common to prove both cases as one; but we preface
this proof with the phrase without loss of generality or simply the
acronym wlog.
Definition: Two integers x and y are said to be of the same parity if x and y are
both even or are both odd. When two integers are not of the same
parity they are said to be of opposite parity.
Definition: A proof by contrapositive is a direct proof of the contrapositive of
an implication.
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NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
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Comment: The proof of a biconditional P ⇔ Q must verify both an implication
P ⇒ Q and its converse Q ⇒ P. The most obvious way to do this
is with a direct proof for each of P ⇒ Q and Q ⇒ P. However,
it is often the case that one of the implications will be more easily
proved by contrapositive.
Definition: A proof by contradiction is a proof whereby we assume that the
negation of the statement to be proved were true, and then we deduce that a logical contradiction must follow from that negation.
Comment: Since any statement that is not false is true, proof by contradiction
is a valid proof technique for any statement.
Definition: Consider a universally quantified statement
∀ x ∈ S, R( x ).
This statement is true when the negation
¬(∀ x ∈ S, R( x )) ≡ ∃ x ∈ S, ¬ R( x )
is true. Thus, when we find an x ∈ S for which ¬ R( x ) is true, we
say that we have found a counterexample to the original statement
∀ x ∈ S, R( x ).
Definition: A disproof is a proof of the negation of a statement.
Definition: An existence theorem is a result that asserts the existence of at least
one object satisfying some specified properties.
Definition: The proof of an existence theorem is called an existence proof.
Definition: A constructive proof of an existentially quantified statement ex-
plains how such an element can be constructed or found. A nonconstructive proof demonstrates that such an element must exist
without actually explaining how to construct or find such an element.
Definition: An uniqueness proof is an argument that verifies there is only one
element satisfying a certain statement.
Definition: A corollary is a result that follows from another result.
Definition: A lemma is a result that helps to prove another result.
Definition: A conjecture is a statement with an unknown truth value
NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
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Examples of Direct and Contrapositive Proofs
Proposition: If n is an odd integer, then n2 is an odd integer.
Proof: We first suppose n is odd and show that it follows that n2 must be
odd. Since n is odd there is an integer x such that n = 2x + 1. . . .
Proposition: If n is an even integer, then n2 is an even integer.
Proposition: If n2 is an odd integer, then n is an odd integer.
Proof: We prove the contrapositive: If an integer n is even, then n2 is even.
Suppose n is even, . . .
Proposition: An integer n is odd if and only if n2 is odd.
Proof: We must prove two implications: 1) If n is an odd integer, then n2 is
an odd integer. 2) If n2 is an odd integer, then n is an odd integer. These
were proved above, so the biconditional is true.
Proposition: Let x ∈ Z. If 3x − 11 is odd, then x is even.
Proposition: Let x ∈ Z. If 3x − 11 is odd, then 5x + 2 is even.
Proposition: The sum and product of any two rational numbers is rational.
Examples of Proof by Cases
Proposition: For each integer n, n2 + n is even.
Proposition: Two integers x and y are of the same parity if and only if x + y is
even.
Proof: Case 1: Suppose x and y were of opposite parity. By concluding
that x + y is not even we will have proved the contrapositive of one
conditional. We can assume WLOG that x is even and y is odd. . . .
Case 2: Suppose x and y were even. . . .
Case 3: Supppose x and y were odd. . . .
The second conditional is covered by the second two cases, so
between the three cases, we have proved the biconditional.
Proposition: Two integers x and y are of the same parity if and only if 3x + 5y is
even.
Proof: Case 1:
Case 2:
Case 3:
Case 4:
Suppose
Suppose
Suppose
Suppose
x
x
x
x
and y are both even. . . .
and y are both odd. . . .
is even and y is odd. . . .
is odd and y is even. . . .
Comment: Because there is a lack of symmetry in x and y, Case 4 must be
proved separately from Case 3. In other words, it is not acceptable
to preface Case 3 with the phrase WLOG and then skip Case 4.
NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
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Proposition: Let a and b be integers. Then ab is even if and only if a is even or b
is even.
Lemma: If a or b is an even integer, then ab is even.
Proof: We assume WLOG that a is even. . . .
Lemma: Let a and b be integers. Then, if ab is even, then a is even or b is
even.
Proof: Suppose it is not the case that a or b is even, i.e., that a = 2x + 1 and
b = 2y + 1 for some integers x and y. . . .
Proposition: For any real numbers x and y, | x + y| ≤ | x | + |y|.
Examples of Counterexamples
Example: For each open sentence below, attempt to find values for which the
statement is true and values for which the statement is false. One
statement is false for all real values, but the rest can realize both
truth values.
| a + b| = | a| + |b|
| a + b| = a + b
| a + b| ≤ a + b
| a + b| 6= a + b
p
a2 + b2 = a + b
p
a2 + b2 = | a | + | b |
√
√
√
a+b = a+ b
1
1 1
= +
a+b
a b
sin( x + y) = sin( x ) + sin(y)
√
√
2x + 6 = 2 x + 3
Example: Find a counterexample to the statement: All prime numbers are odd.
Example: Disprove the following statement: All rational numbers are integers.
Example: The sum of three consecutive integers is even.
Example: The sum of three consecutive integers is odd.
Example: Goldblach’s Conjecture, Twin Prime Conjecture
Example: Fermat’s last theorem, Four Color Theorem
NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
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Examples of Existence Proofs
Proposition: There exists an integer n such n2 is not positive.
Proposition: There exist integers that are neither prime nor even.
Proof: Any odd number that is not prime satisfies the given properties. For
example 3 · 5 = 15 or 5 · 7.
Proposition: There exist integers a and b such that a + b > ab.
Proposition: For any integer m, there exists an integer n such that n > m.
Proposition: There exists an integer solution to the equation x2 + 3x − 4 = 0.
Proposition: There exists a noninteger root of the polynomial f ( x ) = x2 + 5x − 7.
Proof: We notice that f (1) = −1 and that f (2) = 2. We also know that
polynomials are continuous at every real number. Thus, we can
apply the intermediate value theorem; which, in this case, asserts
that every value in [−1, 2] (between f (1) = −1 and f (2) = 2) is
attained by f within the interval [1, 2]. In particular, there exists a
number 1 < c < 2 such that f (c) = 0.
Example: Prove or Disprove: There exists an integer n such that n and n2 have
opposite parity.
Proposition: There exist irrational numbers a and b such that ab is rational.
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Example: Disprove the statement: There exists an even number n such that n+
2 is
an integer.
Proof: We need to show that for every even number
n +1
2
is not an integer.
Proposition: For each nonzero rational number r there exists a nonzero rational
number s such that rs = 1.
Example: Disprove the statement: There exists a nonzero rational number s, such
that for all nonzero rational numbers r, rs = 1
Comment: The following result is true. It can be proved by disproving its nega-
tion, which is what was done in the previous example.
Proposition: There is no rational number s, such that for all nonzero rational
numbers r, rs = 1.
NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
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Examples of Proof by Contradiction
Proposition: The sum of a rational number and an irrational number is irrational.
Proposition: No three consecutive integers sum to 100.
Proposition: Let x, y ∈ R+ . Then
p
x2 + y2 6= x + y.
Proposition: Let x, y ∈ R. The product xy = 0 if and only if x = 0 or y = 0.
Proposition: The square root of 2 is irrational.
Comment: The examples that follow are arranged by content instead of tech-
nique
Examples of Divisibility Proofs
Proposition: For each integer n, n3 + 2n is divisible by 3.
Proposition: Let n ∈ Z. If 3 6 |n, then 3|n2 − 1.
Proposition: Let n ∈ Z. If 3 6 |n, then 3|n3 + n + 1.
Proposition: Let n ∈ Z. If 5 6 |n, then 5 divides exactly one of n2 + 1 or n2 − 1.
Proof: Case 2: Let n = 5k + 2. Then n2 + 1 = 25k2 + 20k + 4 + 1
Case 3: Let n = 5k + 3. Then n2 + 1 = 25k2 + 30k + 9 + 1
NOTES ON PROOF TECHNIQUES (OTHER THAN INDUCTION)
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Examples of Set Proofs
Theorem: For any set A, ∅ ⊆ A.
Proof: This follows from the tautology: f ⇒ P.
Comment: Understanding this is somewhat subtle. There are no elements in
∅ to consider. The means that, no matter what set A is, there is no
element of ∅ that is not in A. This means that we do not violate
the definition of a subset if we say that ∅ ⊆ A and we are basically
forced to agree that the emptyset is a subset of every set - even itself!
Theorem: For any set A, A ⊆ A.
Proof: This follows from the tautology P ⇒ P.
Comment: If we carefully read the definition of a subset we notice that all we
need is for every element of A to be in A. Since this is true for any
set A, we do have that A ⊆ A for any set A.
Proposition: Let A and B be sets such that A ⊆ B. Then, if x ∈
/ B, then x ∈
/ A.
Proof: Consider the contrapositive statement. If x ∈ A, then x ∈ B. Since
this implication is equivalent stating that A is a subset of B, the
contrapositive is true.
Proof: Suppose to the contrary that there were an element x ∈
/ B such that
x ∈ A. Since x ∈ A and A ⊆ B we know that x ∈ B, which is a
contradiction.
Proposition: Let A and B be two sets in the same universal set U. Then A ⊆ B if
and only if B ⊆ A.
Proof: ( A ⊆ B) ⇔ ( x ∈ A ⇒ x ∈ B) ≡ ( x ∈
/B⇒x∈
/ A ) ⇔ ( B ⊆ A ).
Proposition: For any sets A and B, A ⊂ B ⇒ A ⊆ B.
Proof: This follows from the tautology ( P ∧ Q) ⇒ P.
Proposition: For any set X, ∅ ∩ X = ∅
and ∅ ∪ X = X.
Proof: The emptyset has no elements in common with any set, so the intersection of the emptyset with any other set is empty. Likewise, the
emptyset has no elements to add to any other set, so the union of
the emptyset with any other set X is still X.
Theorem: For any sets A, B and C,
(1) Distributive Laws
(a) A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )
(b) A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )
(2) De Morgan’s Laws
(a) ( A ∪ B) = A ∩ B
(b) ( A ∩ B) = A ∪ B.