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CS104:Discrete Structures
C hap t er 2:
Proof Techniques
Introduction to Proofs
(In Book: Chapter 1-sec 1.8)
2
Introduction to Proofs
3
A proof : is a valid argument that establishes the truth of a mathematical
statement.
There are two types of proofs :
 Formal proof : where all steps are supplied and the rules for each step in
the argument are given
 Informal proof : where more than one rule of inference may be used in
each step, where steps may be skipped, where the axioms being assumed
and rule of inference used are not explicitly stated.
Introduction to Proofs (Cont.)
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 A theorem is a statement that can be shown to be true.
 Less important theorems sometimes are called propositions.
(Theorems can also be referred to as facts or results.)
 it may be the universal quantification of a conditional statement with one or
more premises and a conclusion, or it may be another type of logical statements.
 Proof is a valid argument (sequence of statements) that establishes the
truth of a theorem. And the statements to be used in proofs include:
 Axioms or postulates (statement assumed to be true
without proof)
o Ex: If x is positive integer then x+1 is positive integer.
 Hypothesis (premises) of the theorem
 Previously proven theorems
Proven
Axioms Hypothesis theorems
Rules of inference
 Rules of inference used to draw conclusions and
to move from one step to another
New Theorems
Introduction to Proofs (Cont.)
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 Lemma is a less important theorem that is helpful in the proof of other
results
 A corollary is a theorem that can be established directly from a theorem
that has been proved.
 A conjuncture
• is a statement that is being proposed to be a true statement, usually on
the basis of some partial evidence.
• When a proof of a conjuncture is found, the conjuncture becomes a
theorem
Introduction to Proofs (Cont.)
6
 Methods of proving theorems:
 Direct proofs
 Proof by Contraposition
 Vacuous Proofs
 Trivial Proofs
 Proofs by Contradiction
 Proof by cases
Methods of proving theorems – Direct proofs
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 A direct proof of a conditional statement pq is constructed when the
first step is the assumption that p is true; subsequent steps are constructed
using rules of inference, with the final step showing that q must also be
true.
• A direct proof shows that a conditional statement pq is true by showing
that if p is true then q must also be true.
• In a direct proof, we assume that p is true and use axioms, definitions,
and previously proven theorems, together with rules of inference, to show
that q must also be true.
Methods of proving theorems – Direct proofs
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 Definition : The integer n is even if there exists an integer k such that n =
2k, and n is odd if there exists an integer k such that n = 2k + 1.
 Note that every integer is either even or odd, and no integer is both
even and odd.
 Two integers have the same parity when both are even or both are odd;
they have opposite parity when one is even and the other is odd.
Methods of proving theorems – Direct proofs
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 Example Give a direct proof of the theorem :
“If n is an odd integer, then n2 is odd.”
 Proof:
We assume that n is odd. (hypothesis)
=> n = 2k + 1, where k is some integer. (definition of odd number)
We can square both sides of the equation n = 2k + 1
to obtain a new equation that expresses n2
=> n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2+2k)+1
Since r = 2k2+2k is an integer
(Axiom)
=> n2 = 2r+1 is odd.
 Example :Use a direct proof to show that :
“If n is an even integer, then n2 is even.”
(By squaring)
Methods of proving theorems – Direct proofs
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 Example Give a direct proof of the theorem :
“if m and n are both perfect squares, then nm is also perfect”
 Proof:
We assume that m and n are both perfect squares :
m = s2 and n = t2
m.n = s2.t2 = (st)2
m.n is also perfect
 Can we use this methodology to prove theorems of the form ∀x(P(x) → Q(x)) ?
Proofs of theorems of this type that are not direct proofs, that is, that do not
start with the premises and end with the conclusion are called indirect proof.
Methods of proving theorems – Proof by
Contraposition
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 An extremely useful type of indirect proof is known as proof by contraposition
( p → q equivalent to ￢q →￢p )
 Methodology of Proof by Contraposition : It is proving the conditional statement
p → q by showing that its contrapositive, ￢q →￢p, is true
to prove that
p→q
Begin with
is true
￢q
is true as a premise
￢p
must be also true
using (axioms,
definitions previously
proven theorems with rules of inference)
End with
It is showing the contrapositive
￢q →￢p
is true
Methods of proving theorems – Proof by
Contraposition
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 Example Prove that if n is an integer and 3n + 2 is odd, then n is odd.
 Proof:
We assume that the conclusion of the conditional statement
“if 3n + 2 is odd, then n is odd” is false .
=> We assume that n is even.
(hypothesis)
=> n = 2k , where k is some integer. (definition of even number)
( Substituting 2k for n in hypthesis : 3n+2 )
=> 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1)
(Substituting 2k for n)
Since r = 3k + 1 is an integer (Axiom)
=> 3n + 2 = 2r is even . (definition of even number)
Because the negation of the conclusion of the conditional statement implies that the
hypothesis is false, the original conditional statement is true.
We have proved the theorem “if 3n + 2 is odd, then n is odd”
Methods of proving theorems – Proof by
Contraposition
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 Example Use an indirect proof to show that “if n2 is odd then n is odd”
 Proof:
The contraposition is “if n is not odd then n2 is not odd ”
=> We Assume that n is not odd i.e., n is even (hypothesis)
=> n = 2k , where k is some integer. (definition of even number)
=> n2 = (2k)2 = 4k2 = 2(2k2)
Since r = 2k2 is an integer
=> n2 = 2r
(Axiom)
(definition of not odd number)
Thus , if n2 is odd then n is odd is true
Methods of proving theorems – Proof by
Contraposition
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 Example Prove that if n = ab where a and b are positive integers then a  n
or b  n
 Proof: Let p=“an”, q=“bn” and r=“n=ab”
We want to prove that r pq
=> The contraposition is (pq )   r (By definition)
=> p  q   r (De Morgan’s law)
Now, assume that an and bn (p  q)
=> a.b  n.n = n (by multiplying above twos)
=> ab  n
=>  r which contradicts the statement n = ab
Because the negation of the conclusion of the conditional statement implies that the
hypothesis is false, the original conditional statement is true.
Methods of proving theorems – Vacuous Proofs
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 Vacuous Proofs A conditional statement p  q is TRUE if p
is FALSE. If we can show that p is False, then we have a proof,
called vacuous proof, of the conditional statement p  q .
p
q
pq
F
F
T
T
F
T
F
T
T
T
F
T
 Example Show that the proposition P(0) is true, where P(n) is “If n > 1, then n2 > n”
and the domain consists of all integers.
 Proof: Note that P(0) is “If 0 > 1, then 02 > 0.” We can show P(0) using a vacuous proof.
Indeed, the hypothesis 0 > 1 is false. This tells us that P(0) is automatically true.
 Example Prove that if he is alive and he is dead then the sun is ice cold
 Proof: Since the hypothesis is always false the implication is vacuously true.
Methods of proving theorems – Trivial Proofs
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 Trivial Proofs A conditional statement p  q is TRUE if q is
True . If we can show that q is True, then we have a proof, called
Trivial proof, of the conditional statement p  q .
p
q
pq
F
F
T
T
F
T
F
T
T
T
F
T
 Example Let P(n) be “If a and b are positive integers with a ≥ b, then an ≥ bn where the
domain consists of all nonnegative integers. Show that P(0) is true.
 Proof: The proposition P(0) is “If a ≥ b, then a0 ≥ b0.” Because a0 = b0 = 1, the conclusion
of the conditional statement “If a ≥ b, then a0 ≥ b0” is true. Hence, this conditional
statement, which is P(0), is true
 Example Prove that if x=2 then x2  0 for all real numbers
 Proof: Since x2  0 is true then the implication is trivially true. (we didn’t use
the fact x=2)
 Example Use a trivial proof to show that if n > 1 then n2 ≥ n for all integers
Methods of proving theorems – Proofs by Contradiction
 Proofs by Contradiction
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Let p be the statement we want to prove is true, assuming not p (p )and show a
contradiction ¬p → q
Definition: The real number r is rational if there exist integers p and q with q  0 such
that r = p/q and p and q have no common factors A real number that is not rational is
called irrational
 Example Prove that 2 is irrational by giving a proof by contradiction.
 Proof: Let p is “2 is irrational “
To start a proof by contradiction, we suppose that ￢p is true , "It is not the case that is
irrational”
Let p is 2 is rational (hypothesis)
2 = a/b for some integers a and b (b0) and a and b have no common factors
We can square both sides of the equation 2 = a/b
=> 2 = a2/b2
Methods of proving theorems – Proofs by Contradiction
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=> 2b2 = a2
By the definition of an even integer it follows that a2 is even, so we
can substituting with a = 2c for some integer
=> 2b2 = (2c)2 = 4c2
(Dividing both sides of this equation by 2)
 b2 = 2c2 By the definition of an even integer it follows that b2 is even , i.e. b is even
We have now shown that the assumption of ￢p leads to the equation √2 = a/b, where
 a and b have no common factors,
 but both a and b are even ( that is, 2 divides both a and b)
That is our assumption of ￢p leads to the contradiction, ￢p must be false.
That is, the statement p, “ √2 is irrational,” is true. We have proved that √2 is irrational.
Methods of proving theorems – Proofs by Contradiction
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 To prove a conditional statement p  q by contradiction we prove that:
( p  q F is true ) which is equivalent to p  q .
 Example Give a proof by contradiction of the theorem :
“If 3n + 2 is odd, then n is odd.”
 Proof: Suppose that 3n+2 is odd and n is even [p  q]
=> n = 2r
(hypothesis q, definition of even numbers)
=>3n = 6r
(multiply both sides by 3)
=> 3n+2=2+6r
(add 2 to both sides)
=> 3n+2=2(1+3r)
=> 3n+2=2k
(Let k=1+3r)
=> Thus 3n+2 is even which is false (a contradiction !)
Therefore the implication is true.
Methods of proving theorems – Proofs of Equivalence
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 Proofs of Equivalence To prove a theorem that is a biconditional statement,
that is, a statement of the form p ↔ q, we show that p → q and q → p are both
true.
 Example Prove the theorem:
“If n is an integer, then n is odd if and only if n2 is odd.”
“if n is odd then n2 is odd” proved in Example 5
“ n2 is odd then n is odd” proved in Example 7
 Therefore, n is even if and only if n2 is even
 Proving equivalence of several propositions:
If we want to prove that p1 p2  p3… pn
 Then it is sufficient to prove p1  p2,, p2  p3 … pn  p1
Methods of proving theorems – Counterexample
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 Counterexample : To show that a statement of the form ∀xP(x) is false, we need only
find a counterexample, that is, an example x for which P(x) is false.
 E.g. Prove that “For all real numbers x2 > x” is false
X=0.5 is a counterexample since 0.52 > 0.5 is not true
 Example Show that the statement: “Every positive integer is the sum of the
squares of two integers” is false.
 To show that this statement is false, we look for a counterexample which is 3
because we cannot write 3 as the sum of the squares of two integers.
To show this is the case, note that the only perfect squares not exceeding 3 are:
(02 = 0 & 12 = 1), but 3  1+0. Therefore the statement is false
Methods of proving theorems – proof by cases
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 Proof by cases : A proof by cases must cover all possible cases that
arise in a theorem. Each case may cover an infinite number of instances
 Example Prove "if n is an integer then n2 ≥ n".

Proof: We can prove that n2 ≥ n for every integer by considering three cases,
when n = 0, when n ≥ 1, and when n ≤ -1.
Case 1: When n = 0, since 02 = 0, 02 >= 0. So, n2 ≥ n is true.
Case 2: When n ≥ 1, From n ≥ 1, we get n2 ≥ n (by multiplying with n).
Case 3: When n ≤ -1, since n is negative, n2 is positive, so n2 ≥ n.
Hence, if n is an integer then n2 ≥ n.
Methods of proving theorems – proof by cases
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 Example Use a proof by cases to prove that |xy| = |x||y|, where x and y real
numbers. Hint: Consider all 4 cases: x and y positive or negative

Proof: using the fact that |a| = a if a ≥ 0 and |a| = -a if a < 0 , there are four cases
for x and y :
Case 1: x and y both are nonnegative: xy ≥ 0 so that |xy| = xy = |x||y|
Case 2 : x is nonnegative and y is negative: xy ≤ 0 so that |xy| = - xy = x(-y) = |x||y|
Case 3 : x is negative and y is nonnegative: xy ≤ 0 so that |xy| = - xy = (-x)y = |x||y|
Case 4 : x and y both are negative xy > 0 so that |xy| = xy = (-x)(-y) = |x||y|
Mistakes in Proofs
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 There are many common errors made in constructing mathematical and the most
common errors are mistakes in arithmetic and basic algebra.
 Each step of a mathematical proof needs to be correct and the conclusion needs to
follow logically from the steps that precede it.
Be careful!
Many mistakes result from the introduction of steps that
do not logically follow from those that precede it
Mistakes in Proofs
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 Example What is wrong with this famous supposed “proof” that 1 = 2 ?
 Proof: Suppose a and b are two equal positive integers. The error is that a − b equals

zero; division of both sides
of an equation by the same
quantity is valid as long as
this quantity is not zero.
Steps :
Reason
1.
a=b
( Given)
2.
a2 = ab
(Multiply both sides of (1) by a)
3.
a2 - b2 = ab - b2
(Subtract b2 from both sides of (2))
4.
(a − b)(a + b) = b(a − b)
5.
a+b=b
(Divide both sides of (4) by a − b )
6.
2b = b
(Replace a by b in (5) because a = b & simplify)
7.
2= 1
(Divide both sides of (6) by b)
(Factor both sides of (3))
Can you find the error 
Every step is valid except for one, step 5 where we divided both sides by a − b
Mistakes in Proofs
26
 Example What is wrong with this “proof” ?
“Theorem:” If n2 is positive, then n is positive.
 Proof: Suppose that n2 is positive. Because the conditional statement :
“If n is positive, then n2 is positive” is true, we can conclude that n is positive.
 Solution : Let P(n) be “n is positive” and Q(n) be “n2 is positive.”
Then our hypothesis is Q(n).
(1)
This statement “If n is positive, then n2 is positive” is the statement:
∀n(P (n) → Q(n)).
(2)
From the hypothesis (1) and the statement (2):
because we are not
using a valid rule of
inference.
We cannot conclude P(n)
Instead, we can prove that by using counterexample
A counterexample
supplied by n = −1 for
which n2 = 1 is positive,
but n is negative.
End of Chapter 2