Abstract Mathematics : Proofs
Prepared by Dr Phil Stephenson (April 2015)
Additional Reading
1) Chapter 2 in the Subject Guide.
2) Chapters 1 – 3 in Biggs, N. L., Discrete Mathematics
3) Chapters 1 – 4 and 6 in Eccles, P. J., An Introduction to Mathematical Reasoning
In this document, we meet various strategies for proving or disproving mathematical
statements.
To show that a mathematical statement concerning all positive integers n is true, we must give
a general proof that shows that the statement is true for all such n.
To show that a mathematical statement concerning all positive integers n is false, we only have
to offer one counter-example that shows that the statement is false for one value of n.
If you are told that the non-negative integer n is even, then you can write n = 2k, for some nonnegative integer k.
If you are told that the positive integer n is odd, then you can write n = 2k + 1 for some nonnegative integer k. Alternatively, you can write n = 2k – 1 for some positive integer k.
If you are told that the non-negative integer n is divisible by 3, then you can write n = 3k, for
some non-negative integer k.
(Ex 1) Show that the following statement is false:
The number
1
2
n(n + 1)(n + 3) is a square number, for all odd positive integers n.
(Solution)
The statement is false. A counter-example follows:
If n = 5, then
1
2
n(n + 1)(n + 3) = 120, but 120 is not a square number.
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(Ex 2) Show that the following statement is true:
n2 is odd for all odd positive integers n.
(Solution)
The statement is true. A proof follows:
Let n be any odd positive integer. Then n = 2k + 1 for some integer k ≥ 0.
n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 = 2N + 1, where N = 2k2 + 2k.
N is an integer since k is an integer. Hence, n2 is odd for all odd positive integers n.
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Abstract Mathematics : Proofs
Number Systems and Sets
= {1, 2, 3, …}. This is the set of natural numbers (also called positive integers).
= {…, –3, –2, –1, 0, 1, 2, 3, …}. This is the set of integers.
The positive integers are 1, 2, 3, …, whereas the non-negative integers are 0, 1, 2, …
is the set of rational numbers. A rational number is a number that can be expressed in the
form
m
where m and n are integers and n ≠ 0.
n
For example, –3, –2/3, 0, 1.27, 2, 2.5 ∈ However, π, e,
a ∈ A means a is a member of the set A
2 , 51/3 ∉ The set of all numbers (e.g. all points on a line) is the set of real numbers, denoted .
So, ⊆ ⊆ ⊆ A ⊆ B means set A is contained in the set B
If a real number is not rational, then it is called irrational (e.g. 2 is irrational).
2 ∈ , but
2 ∉ .
3
4
∈ , but
3
∉ . –2 ∈ , but –2 ∉ .
4
Taking the intersection, we have {2, 3, 7, 9, 11} ∩ {3, 6, 9, 12} = {3, 9}
and taking the union, we have {2, 3, 7, 9, 11} ∪ {3, 6, 9, 12} = {2, 3, 6, 7, 9, 11, 12}.
Interval notation
The set {x ∈ : a ≤ x ≤ b} can be written in interval notation as [a, b]
The set {x ∈ : a < x < b} can be written in interval notation as (a, b)
The set {x ∈ : a ≤ x < b} can be written in interval notation as [a, b)
The set {x ∈ : a < x ≤ b} can be written in interval notation as (a, b]
The set {x ∈ : x ≤ a} can be written in interval notation as (– ∞, a]
The set {x ∈ : x ≥ a} can be written in interval notation as [a, ∞)
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Abstract Mathematics : Proofs
ln(x) and ex
y
y = ln x
1
0
1
e
x
ln x is only defined when x > 0, as is evident from the graph.
As x tends to 0 from the right, then ln x tends to –∞.
As x tends to ∞, then ln x tends to ∞.
The inverse of ln x is ex. The graph of ex can be obtained from the graph
of ln x by reflecting in the line y = x.
y
y = ex
1
0
ex > 0 for all x.
ex → ∞ as x → ∞, and ex → 0 as x → –∞.
(1) ln(fg) = ln(f) + ln(g)
f 
(2) ln   = ln(f) – ln(g)
g
(3) ln(gh) = hln(g)
(4) ln(eg) = g
(5) eln g = g
Page 3 of 34
x
Abstract Mathematics : Proofs
(Ex 3) Show that the following statement is true:
The function f, defined by f(x) = x2 – 3x where x ∈ [2, 6], is increasing.
(Solution)
The statement is true. A proof follows:
Assume x1, x2 ∈ [2, 6] with x1 < x2.
f(x2) – f(x1) = (x22 – 3x2) – (x12 – 3x1)
= x22 – 3x2 – x12 + 3x1
= x22 – x12 – 3x2 + 3x1
= (x2 – x1)(x2 + x1) – 3(x2 – x1)
= (x2 – x1)[(x2 + x1) – 3]
Now, x1 < x2, so that x2 – x1 > 0
Also, 2 ≤ x1 < x2 ≤ 6, so that x2 > 2 ∴ (x2 + x1) > 4 ∴ (x2 + x1) – 3 > 1 > 0
Hence, f(x2) – f(x1) is the product of two positive terms, and so is positive.
So, f(x2) – f(x1) > 0. So, f(x1) < f(x2).
It has been shown that if x1 < x2 then f(x1) < f(x2).
Hence, f is increasing.
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Division Theorem
We shall meet the Division Theorem formally in the Prime Numbers document.
However, we don’t need to know about prime numbers to use the theorem here.
We can write:
By the Division Theorem, any positive integer can be written in the form
2k or 2k + 1, where k ∈ .
We can also write:
By the Division Theorem, any positive integer can be written in the form
3k, 3k + 1 or 3k + 2, where k ∈ .
We can also write:
By the Division Theorem, any positive integer can be written in the form
5k, 5k + 1, 5k + 2, 5k + 3 or 5k + 4, where k ∈ .
I have given examples for the cases 2, 3 and 5, but we can use 2, 3, 4, 5, 6, etc.
Which case we choose depends on the question. Two examples follow.
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Abstract Mathematics : Proofs
(Ex 4) Show that the following statement is true:
For any prime p, it is not possible for both 8p – 1 and 8p + 1 to be prime.
(Solution)
The statement is true. A proof follows:
Let p be any prime number.
By the Division Theorem, p can be written in the form 3, 3k + 1 or 3k + 2, where k is a
non-negative integer (noting that if 3k is prime, then k = 1).
If p = 3, then 8p + 1 = 25 = 5 × 5, so 8p + 1 is not prime
If p = 3k + 1, then 8p + 1 = 8(3k + 1) + 1 = 24k + 9 = 3(8k + 3) so 8p + 1 is not prime
If p = 3k + 2, then 8p – 1 = 8(3k + 2) – 1 = 24k + 15 = 3(8k + 5) so 8p – 1 is not prime
Hence, for any prime p, it is not possible for both 8p – 1 and 8p + 1 to be prime.
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(Ex 5) Show that the following statement is true:
The only possible remainders when n2 + 4n + 2 is divided by 5 are 2, 3 and 4,
for all n ∈ .
(Solution)
The statement is true. A proof follows:
Let n be any positive integer, and let N = n2 + 4n + 2.
By the Division Theorem, n can be written in the form 5k, 5k + 1, 5k + 2, 5k + 3 or
5k + 4, where k is a non-negative integer.
If n = 5k, then N = (5k)2 + 4(5k) + 2 = 25k2 + 20k + 2 = 5(5k2 + 4k) + 2
If n = 5k + 1 then N = (5k + 1)2 + 4(5k + 1) + 2 = 25k2 + 30k + 7 = 5(5k2 + 6k + 1) + 2
If n = 5k + 2 then N = (5k + 2)2 + 4(5k + 2) + 2 = 25k2 + 40k + 14 = 5(5k2 + 8k + 2) + 4
If n =5k + 3 then N = (5k + 3)2 + 4(5k + 3) + 2 = 25k2 + 50k + 23 = 5(5k2 + 10k + 4) + 3
If n =5k + 4 then N = (5k + 4)2 + 4(5k + 4) + 2 = 25k2 + 60k + 34 = 5(5k2 + 12k + 6) + 4
Hence, the only possible remainders when n2 + 4n + 2 is divided by 5 are 2, 3 and 4,
for all n ∈ .
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Abstract Mathematics : Proofs
In the next example, I demonstrate a method of proof called proof by exhaustion, in which we
check all possible cases. In my solution, I have used congruence notation, which we do not
meet formally until the ‘Congruences’ document.
In the meantime, let me just say that a ≡ b (mod n) when a – b is exactly divisible by n.
(Ex 6) Show that the following statement is true:
The equation x2 = 8 has no solutions in 9.
(Solution)
The statement is true. A proof by exhaustion follows:
9 = {0, 1, 2, 3, 4, 5, 6, 7, 8}
02 = 0 ≡ 0 (mod 9)
12 = 1 ≡ 1 (mod 9)
22 = 4 ≡ 4 (mod 9)
32 = 9 ≡ 0 (mod 9)
42 = 16 ≡ 7 (mod 9)
52 = 25 ≡ 7 (mod 9)
62 = 36 ≡ 0 (mod 9)
72 = 49 ≡ 4 (mod 9)
82 = 64 ≡ 1 (mod 9)
Hence, the equation x2 = 8 has no solutions in 9.
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(Ex 7) Show that the following statement is false:
Any number of the form 14k + 3, where k is a non-negative integer, must have a prime
divisor of this same form.
(Solution)
The statement is false. A counter-example follows:
Let k = 10. Then, 14k + 3 = 143 = 11 × 13.
Neither of the prime divisors (i.e. 11 and 13) is of the form 14k + 3 (k ≥ 0).
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Abstract Mathematics : Proofs
Notation: We write mn to mean m divides into n exactly. Also, n! is n factorial.
(Ex 8) Prove Euclid’s Theorem that there are infinitely many primes.
(Solution)
Proof by contradiction:
Assume there are only finitely many primes, with p the largest prime.
Let N = p! + 1 ………………………………………….. (1)
p is the largest prime and N > p, so N cannot be prime.
Any natural number > 1 that is not prime has at least one prime divisor.
So, N has a prime divisor q, say (so that qN). Note that q > 1 since q is prime.
p is the largest prime and q is also prime, so q ≤ p.
So, qp! (e.g. 3 ≤ 7 and 3 divides into 7⋅6⋅5⋅4⋅3⋅2⋅1).
So, qN and qp!
If ma and mb, then m divides any integer combination of a and b
∴ q(N – p!) ∴ q1 from (1)
However, q > 1, so we have a contradiction. Hence, the initial assumption is false.
Hence, there are infinitely many primes.
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Negation
The negation of a mathematical statement P is ‘not P’. We write ¬P to mean ‘not P’.
If the statement P is true then the statement ¬P must be false.
The truth table, with T for true and F for false, follows:
P
T
F
¬P
F
T
(Ex 9) State the negation of the following statement:
The number n4 – n2 is divisible by 12.
(Solution)
The negation of the given statement is:
The number n4 – n2 is not divisible by 12.
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Abstract Mathematics : Proofs
Notation: We can write ∀n ∈ to mean for all natural numbers n.
∀ is known as the universal quantifier, and means ‘for all’.
Notation: We can write ∃n ∈ to mean there exists a natural number n.
∃ is known as the existential quantifier, and means ‘there exists’.
We say that the statement p(n) is a predicate or variable proposition if p(n) must be true or
false once n is specified.
In general,
¬(∀n ∈ then p(n) is true) is (∃n ∈ such that ¬ p(n) is true)
We might write this more concisely as:
¬(∀n ∈ , p(n)) ⇔ (∃n ∈ , ¬ p(n))
Let p and q be propositions. A proposition is a statement that is either true or false, but not
both.
Compound
Proposition
p∧q
p∨q
p⇒q
p⇔q
Meaning
p and q
p or q
p implies q
p if and only if q
p ∧ q is true precisely when both p is true and q is true.
p ∨ q is true precisely when either p is true or q is true, or both are true.
Truth tables:
p
T
T
F
F
q
T
F
T
F
p∧q
T
F
F
F
p
T
T
F
F
q
T
F
T
F
p∨q
T
T
T
F
Suppose p is the statement ‘27 is divisible by 6’, and q is the statement ‘27 is divisible by 9’.
Then, p is false and q is true, so p ∧ q is false, and p ∨ q is true.
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Abstract Mathematics : Proofs
Truth tables:
p
T
T
F
F
q
T
F
T
F
p⇒q
T
F
T
T
p
T
T
F
F
q
T
F
T
F
p⇔q
T
F
F
T
p ⇒ q means that if p is true then q is true.
p ⇔ q means (p implies q) and (q implies p), sometimes written as p iff q.
(p ⇒ q) ⇔ [(not q) ⇒ (not p)]
To illustrate the previous statement, consider the definition of an injective function:
A function g is injective provided that:
For all a, b in the domain of g, then g(a) = g(b) ⇒ a = b
This is equivalent to saying the following:
A function g is injective provided that:
For all a, b in the domain of g, then a ≠ b ⇒ g(a) ≠ g(b)
Modus Ponens
If the propositions p and p ⇒ q are both true then q is true
If p ⇒ q and q ⇒ r then p ⇒ r
For example:
All squares are rectangles, and all rectangles are polygons. Hence, all squares are polygons.
Page 9 of 34
Abstract Mathematics : Proofs
Negation revisited
p
T
T
F
F
q
T
F
T
F
p⇒q
T
F
T
T
¬(p ⇒ q)
F
T
F
F
From the above truth table, we see that ¬(p ⇒ q) is true only when p is true and q is false.
So, an important point to remember is that the negation of p ⇒ q is p ∧ (¬q)
“If p then q” ⇔ “p ⇒ q”.
(Ex 10) Suppose n belongs to the set of positive integers.
Write down the negation of the statement A.
A: If n > 100 then n can be expressed as a sum of two squares.
(Solution)
“If p then q” ⇔ “p ⇒ q”. Negation of p ⇒ q is p ∧ (¬q)
Negation of A: n > 100 and n cannot be expressed as a sum of two squares.
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(Ex 11) Suppose n belongs to the set of positive integers.
Write down the negation of the statement B.
B: (n is not odd) or (n4 is not odd).
(Solution)
p
T
T
F
F
q
T
F
T
F
p∨q
T
T
T
F
¬(p ∨ q)
F
F
F
T
From the above truth table, we see that ¬(p ∨ q) is true only when p is false and q is false.
So, the negation of (p ∨ q) is (¬p) ∧ (¬q)
Negation of B: (n is odd) and (n4 is odd).
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Abstract Mathematics : Proofs
Logical Equivalence
Two statements p and q are said to be logically equivalent provided that p ⇔ q is true.
To show that two statements p and q are logically equivalent, we can construct a truth table for
p and another truth table for q. In all cases, the two truth tables should take the same logical
truth-value (T or F).
The truth table will require (22 =) 4 rows if two statements are involved.
For example, use 4 rows to show ¬(p ∧ q) and ¬ p ∨ ¬ q are logically equivalent.
The truth table will require (23 =) 8 rows if three statements are involved.
For example, use 8 rows to show (p ⇒ q) ⇒ r and p ⇒ (q ⇒ r) are logically equivalent.
(Ex 12)
Prove that p ⇒ q and ¬ q ⇒ ¬ p are logically equivalent.
(Solution)
Truth table:
p
T
T
F
F
p⇒q
T
F
T
T
q
T
F
T
F
¬q
F
T
F
T
¬p
F
F
T
T
¬q⇒¬p
T
F
T
T
The truth-values match in all 4 cases, so p ⇒ q and ¬ q ⇒ ¬ p are logically equivalent.
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(Ex 13)
Prove that p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) are logically equivalent.
(Solution)
Truth table:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
q∨r
T
T
T
F
T
T
T
F
p ∧ (q ∨ r)
T
T
T
F
F
F
F
F
p∧q
T
T
F
F
F
F
F
F
p∧r
T
F
T
F
F
F
F
F
(p ∧ q) ∨ (p ∧ r)
T
T
T
F
F
F
F
F
The truth-values match in all 8 cases for the two highlighted columns.
Hence, p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) are logically equivalent.
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Page 11 of 34
Abstract Mathematics : Proofs
Converse statement
The converse of the implication statement p ⇒ q is q ⇒ p.
Note that the converse of a true implication is not necessarily true.
(Ex 14) Consider the following statement P, concerning n ∈ :
P: If n is divisible by 9 then n is divisible by 27.
(a) State the converse of P.
(b) Is P true? Justify answer.
(c) Is the converse of P true? Justify answer.
(Solution)
(a) Converse: If n is divisible by 27 then n is divisible by 9.
(b) P is not true. Counter-example: n = 9.
(c) The converse of P is true. A proof follows:
Let n be divisible by 27. Then n = 27k for some k ∈ .
So, n = 9(3k), where 3k ∈ . Hence, n is divisible by 9.
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(Ex 15) State and prove the converse of the following statement.
For all real numbers x, if x2 – 6x + 9 > 0, then x > 3.
(Solution)
The converse is: For all real numbers x, if x > 3, then x2 – 6x + 9 > 0.
Proof of converse:
Let x be any real number, such that x > 3.
x2 – 6x + 9 = (x – 3)2 > 0 since x ≠ 3.
Hence, if x > 3 then x2 – 6x + 9 > 0, for all real numbers x.
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Abstract Mathematics : Proofs
Contrapositive statement
The contrapositive of the implication statement p ⇒ q is ¬q ⇒ ¬p.
These two statements are logically equivalent, as we showed in an earlier example.
Occasionally it is easier to prove ¬q ⇒ ¬p than it is to prove p ⇒ q, so it is useful to be aware
of the fact that p ⇒ q is equivalent to ¬q ⇒ ¬p. We often use this in question 1 of the exam.
(Ex 16) Consider the following statement S, concerning n ∈ :
S: If 12n + 5 is not prime then n is even.
(a) Write down the contrapositive of S.
(b) Is the contrapositive of S true or false? Justify your answer.
(c) Is S true or false? Justify your answer.
(d) Write down the negation of S.
(e) Write down the converse S* of the statement S and show that S* is false.
(Solution)
(a) Contrapositive of S: If n is odd then 12n + 5 is prime.
(b) The contrapositive of S is false.
Counter-example: n = 5
If n = 5 then 12n + 5 = 65 = 5 × 13
That is, n is odd, but 12n + 5 is not prime.
(c) S is logically equivalent to its contrapositive.
Since the contrapositive of S is false, then so too is S.
(d) “If p then q” ⇔ “p ⇒ q”. Negation of p ⇒ q is p ∧ (¬q)
Negation of S: 12n + 5 is not prime and n is odd.
(e) Converse S*: If n is even then 12n + 5 is not prime.
The converse S* is false.
Counter-example: n = 2
If n = 2 then 12n + 5 = 29
That is, n is even, but 12n + 5 is prime.
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Abstract Mathematics : Proofs
Working backwards to obtain a proof
Sometimes it helps to work back from what you know will be the last line of a proof. Writing
down the last line, and then deducing what the second-last line must be, and then perhaps the
third-last line, may give you enough inspiration to deduce how to begin the proof.
In a later document, we shall meet congruences. If I am asked to prove a result from the
definition of congruence alone, then I often adopt the approach of working backwards.
(Ex 17) Suppose x, y, z ≥ 0 and x + y + z = c, for some positive constant c.
Suppose, also, that
1
1
1
9
+
+
≥ .
x
z
c
y
−1
 1  1 1 1 
Prove that (x + y + z) ≥   + +  .
3
 3  x y z 
1
(Solution)
I am not sure how to proceed, so I decide to work backwards. Once I know what to do, then I
shall re-write the proof.
 1  1 1 1 
(x + y + z) ≥   + + 
3
 3  x y z 
1
⇒
−1
 1  1 1 1 
⇒ c ≥   + + 
3
 3  x y z 
1
−1
⇒
11 1 1 3
 + + ≥
3  x y z  c
1
1
1
9
+
+ ≥
x
z
c
y
The proof follows: Let x, y, z ≥ 0 and x + y + z = c. Suppose that
1
1
9
1
+
+ ≥ .
x
z
c
y
−1
 1  1 1 1 
 1  1 1 1 
11 1 1 3
1
1
Then,  + +  ≥ , so c ≥   + +  , so (x + y + z) ≥   + + 
3
3
3 x y z c
 3  x y z 
 3  x y z 
−1
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Proof by contradiction
In an earlier example, I used a proof by contradiction to prove Euclid’s Theorem that there are
infinitely many primes. To prove p ⇒ q using a proof by contradiction, we begin by assuming
that p is true and q is false, and aim for a contradiction (i.e., a false statement). If we manage
to reach a contradiction, then we know the assumption that q is false is not true, so we deduce
that q is true. We can prove that 5 is irrational by using a proof by contradiction, but this
will be demonstrated in a later document (Chapter 8 in Subject Guide).
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Abstract Mathematics : Proofs
(Ex 18) Prove that there does not exist any prime p such that 23p + 1 is a square.
(Solution)
Assume that there does exist a prime p such that 23p + 1 is a square.
Then, 23p + 1 = n2, where n ≥ 7 since p ≥ 2.
23p = n2 – 1 = (n – 1)(n + 1)
p ≠ 23 since 23(23) + 1 = 530, which is not a square.
23p is a product of two distinct primes, so we have two possibilities:
either 23 = n – 1 and p = n + 1,
or 23 = n + 1 and p = n – 1.
So, either n = 24 so that p = 25,
or n = 22 so that p = 21.
However, neither 21 nor 25 is prime. So, we have reached a contradiction.
Hence, there does not exist any prime p such that 23p + 1 is a square.
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(Ex 19)
Let T be the statement
T: there are infinitely many odd numbers n such that 6n + 1 is not a prime.
Construct a proof of T.
[Hint: consider numbers of the form n = cm + 1 for a suitable value of c.]
(Solution)
T: there are infinitely many odd numbers n such that 6n + 1 is not a prime.
Proof of T:
Try n = cm + 1. If we choose c to be even, then n will be odd.
We want to find c such that 6n + 1 is not a prime and n is odd.
6n + 1 = 6(cm + 1) + 1 = 6cm + 7, which is not prime if c is a multiple of 7.
So, we want to find c such that c is a multiple of both 2 and 7. We can choose c = 14.
Then, 6n + 1 = 6(14m + 1) + 1 = 84m + 7 = 7(12m + 1), which is not prime.
Hence, there are infinitely many odd numbers n such that 6n + 1 is not a prime, as we
can choose n = 14m + 1, where m is a positive integer.
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Abstract Mathematics : Proofs
Sets
A set is a collection of objects (called elements or members). A set is denoted by a capital
letter (e.g. B), whereas an element of a set is denoted by a small (lower-case) letter (e.g. b).
If b is a member of a set B, then we write b ∈ B.
If b is not a member of a set B, then we write b ∉ B.
The elements of a set are listed within braces (curly brackets). For example, B = {b, c, d}.
The order in which elements are listed does not matter. For example, {b, c, d} = {b, d, c}.
The set B given by B = {1, 2, 3, 4} may be written as B = {n | n is a positive integer, n < 5}, or
as B = {n : n is a positive integer, n < 5}.
The vertical line ‘|’ and the colon ‘:’ are both read as “such that”, and the set is read as “the set
of all n such that n is a positive integer and n < 5”.
The empty set (also called null set) contains no elements, and is denoted by 0/ .
We write A ⊆ B to mean that all the elements of the set A belong to the set B.
If A ⊆ B, we can say “A is contained in B”, or that “A is a subset of B”, or that “B contains A”.
If A ⊆ B, then x ∈ A ⇒ x ∈ B.
If A ⊆ B and B ⊆ A then A = B.
The empty set is considered as a subset of every set. If A is a set then 0/ ⊆ A.
If there are n members in a set A, then the set A has 2n subsets, including 0/ and A. This
information has been needed in more than one past exam.
The universal set contains all elements being considered, and might be denoted by E.
Given a set A, which is a subset of the universal set E, we write the complement of the set A by
E \ A or A or Ac.
E \ A = {x ∈ E | x ∉ A}
A ⊆ B is equivalent to E \ B ⊆ E \ A.
Union of sets A and B :
A ∪ B = {x | x ∈ A or x ∈ B}.
x ∈ A ∪ B ⇔ {x ∈ A } ∨ {x ∈ B}.
Intersection of sets A and B: A ∩ B = {x | x ∈ A and x ∈ B}.
x ∈ A ∩ B ⇔ {x ∈ A } ∧ {x ∈ B}.
Page 16 of 34
Abstract Mathematics : Proofs
We can use Venn diagrams to illustrate the relationships between sets.
A
B
A
B
A
B
A A
A∪B
A∩B
E\A
(1)
A∩B = A ∪ B
(2)
A∪B = A ∩ B
(3)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(4)
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∩ (E \ B)
Power Sets
The set of all subsets of a set A is called the power set of A, and is denoted Ƥ (A).
(Ex 20) Let the universal set E be defined as E = {1, 2, 3, 4, 5, 6, 7, 8}.
Let the set A be defined as A = {1, 3, 5}. Let the set B be defined as B = {2, 3, 4, 5}.
(a) List the elements in the set E \ A.
(b) List the elements in the set A ∩ B.
(c) List the elements in the set A ∪ B.
(d) List the elements in the set B ∩ (E \ A).
(e) Write down the power set of the set A.
(f) What can the set A ∪ (E \ A) be written more concisely as?
(Solution)
(a) E \ A = {2, 4, 6, 7, 8}
(b) A ∩ B = {3, 5}
(c) A ∪ B = {1, 2, 3, 4, 5}
(d) B ∩ (E \ A) = {2, 3, 4, 5} ∩ {2, 4, 6, 7, 8} = {2, 4}
(e) 23 = 8, so 8 members. Ƥ (A) = { 0/ , {1}, {3}, {5}, {1, 3}, {1, 5}, {3, 5}, {1, 3, 5}}.
(f) A ∪ (E \ A) = E
***************************************************************
Page 17 of 34
Abstract Mathematics : Proofs
Here are two of the De-Morgan Laws:
(1)
A∩B = A ∪ B
(2)
A∪B = A ∩ B
Notice that the first law above looks similar to the rule ¬(p ∧ q) ⇔ ¬p ∨ ¬q
We are allowed to use the rules based in logic when proving results about sets.
The following table summarises the operations in sets and their corresponding operations in
logic.
Sets
(E \ A)
A∩B
A∪B
Logic
¬p
p∧q
p∨q
(Ex 21) Prove that A ∩ B = A ∪ B
(Solution)
x ∈ A∩ B
⇔
x∉A∩B
⇔
¬ (x ∈ A ∩ B)
⇔
¬ ( (x ∈ A) ∧ (x ∈ B) )
⇔
¬ (x ∈ A) ∨ ¬ (x ∈ B)
⇔
(x ∈ A ) ∨ (x ∈ B )
⇔
x∈ A ∪ B
Thus, A ∩ B = A ∪ B
***************************************************************
For sets A and B in a universal set E, then A\ B means A – B. So, A\ B = {x ∈ A  x ∉ B}
(Ex 22) Prove that A\ B = A ∩ (E \ B)
(Solution)
x ∈ A\ B
⇔
(x ∈ A) ∧ (x ∉ B)
⇔
(x ∈ A) ∧ (x ∈ E\ B)
⇔
x ∈ A ∩ (E \ B)
Thus, A\ B = A ∩ (E \ B)
***************************************************************
Page 18 of 34
Abstract Mathematics : Proofs
Cartesian products
If A and B are sets then the Cartesian product A × B is the set of all ordered pairs (a, b), where
a ∈ A and b ∈ B.
× is the set of all ordered pairs of natural numbers.
× is the set of all ordered pairs of real numbers. We normally denote × by 2.
For example, the line with equation y = 2x + 3 could be described in set notation as follows:
{(x, y) ∈ 2 | y = 2x + 3}.
Quantifiers
The universal statement
∀x ∈ E, p(x)
means that for all x in the set E, the statement p(x) is true.
The existential statement
∃x ∈ E, p(x)
means that there exists an x in the set E such that the statement p(x) is true.
To show that a universal statement is false, we just have to show that p(x) is false for one value
of x. That one value of x is a counter-example.
To show that an existential statement is false, we have to show that p(x) is false for all values
of x.
¬(∀x ∈ E, p(x)) ⇔ (∃x ∈ E , ¬ p(x))
¬(∃x ∈ E, p(x)) ⇔ (∀x ∈ E , ¬ p(x))
(Ex 23) Suppose that P(x, y) is a predicate involving two variables x, y from a set E.
Find the negation of the statement
∃x ∈ E, ∀y ∈ E, P(x, y).
(Solution)
The negation of the given statement is
¬(∃x ∈ E, ∀y ∈ E, P(x, y))
⇔ ∀x ∈ E, ¬(∀y ∈ E, P(x, y))
⇔ ∀x ∈ E, ∃y ∈ E, ¬P(x, y)
***************************************************************
Page 19 of 34
Abstract Mathematics : Proofs
(Ex 24) Find the negation of the statement
∀x ∈ , ∃y ∈ , 3x + y = 28
(Solution)
The negation of the given statement is
¬(∀x ∈ , ∃y ∈ , 3x + y = 28)
⇔ ∃x ∈ , ¬(∃y ∈ , 3x + y = 28)
⇔ ∃x ∈ , ∀y ∈ , ¬(3x + y = 28)
⇔ ∃x ∈ , ∀y ∈ , 3x + y ≠ 28
***************************************************************
(Ex 25) The statement S is as follows:
S: for all sets A, B, C, (A ∪ B)C ∪ C = (A ∩ B)C ∩ C.
Indicate in a diagram the two sets whose equality is asserted by S, and prove that S is
false by constructing an explicit counter-example.
(Solution)
Venn diagrams:
A
B
A
C
B
C
(A ∪ B)C ∪ C
(A ∩ B)C ∩ C
Counter-example: Let A = {a}, B = {b}, C = {c}, E = {a, b, c, d}
(A ∪ B)C ∪ C = {a, b}C ∪ {c} = {c, d} ∪ {c} = {c, d}
(A ∩ B)C ∩ C = { 0/ }C ∩ {c} = E ∩ {c} = {c}.
Hence, S is false.
***************************************************************
Page 20 of 34
Abstract Mathematics : Proofs
Revision Problems
1.
Show that the following statement is false:
The number n2 + n + 17 is a prime number for all n ∈ .
2.
Show that the following statement is true:
The equation x3 – x = 2 has no solutions in 7.
3.
Show that the following statement is true:
The number n4 – n2 is divisible by 12 for all n ∈ .
4.
Show that the following statement is true:
If n is any positive integer that is not a multiple of 3, then n2 – 1 is divisible by 3.
5.
Show that the following statement is false:
If (2n + 7)2 – 1 is divisible by 24, then n is divisible by 6, for all n ∈ .
6.
Show that the following statement is true:
If n is divisible by 6, then (2n + 7)2 – 1 is divisible by 24, for all n ∈ .
7.
Show that the following statement is true:
If n is any odd positive integer, then n2 – 1 is divisible by 8.
8.
Consider three statements called p, q and r. Suppose the statement p ∨ q is
true and that the statement r ⇒ q is false. For each of the three statements
p, q and r, decide whether the statement is true, false or unknown.
9.
p and q are propositions such that p ⇒ q is true and q is false.
Is the proposition p ∨ q true or false?
10.
Let S be the following statement concerning n ∈ .
S: If n is an odd number then 6n + 1 is a prime.
(a) Show that S is false.
(b) Write down the converse S* of the statement S and show that S* is false.
(c) Write down the contrapositive of S.
Is the contrapositive of S true or false? Justify your answer.
(d) Write down the negation of S.
Page 21 of 34
Abstract Mathematics : Proofs
11.
Prove that p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r) are logically equivalent.
12.
Prove that ((¬p ⇔ q) ⇒ (q ∨ (r ⇒ p))) and ((p ⇒ ¬ (q ∨ ¬r)) ⇒ (p ⇒ r)) are logically
equivalent.
13.
Prove that ((p ∧ q) ⇒ (r ∨ (p ⇔ ¬r))) is always true, regardless of what truth-values p,
q and r take.
14.
Prove that ((¬p ⇒ ¬q) ⇒ (¬(q ∧ r) ∨ p)) is always true, regardless of what truthvalues p, q and r take.
15.
Let P be the following statement concerning n ∈ .
P: If n is divisible by 49 then n is divisible by 7.
(a)
State the contrapositive of P.
(b)
State the converse of P.
(c)
Is P true? Justify answer.
(d)
Is the converse of P true? Justify answer.
16.
Prove that there does not exist any prime p such that 37p + 1 is a square.
17.
What values can p take if p is a prime such that 7p + 1 is a square?
18.
Let the universal set E be defined as E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Let the set A be defined as A = {2, 6, 8}. Let the set B be defined as B = {1, 2, 6, 8, 9}.
(a) List the elements in the set E \ A.
(b) List the elements in the set A ∩ B.
(c) List the elements in the set A ∪ B.
(d) List the elements in the set A ∩ (E \ B).
(e) Write down the power set of the set A.
19.
Let the universal set E be defined as E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Define the set A as A = {2, 4, 6, 8}. Let the set B be defined as B = {1, 4, 6, 9, 10}.
(a) List the elements in the set E \ A.
(b) List the elements in the set A ∩ B.
(c) List the elements in the set A ∪ B.
(d) List the elements in the set B ∩ (E \ A).
(e) Write down the power set of the set A.
Page 22 of 34
Abstract Mathematics : Proofs
20.
Show that the following statement is false:
4224 – 1 is a prime number.
21.
Show that the following statement is true:
No number of the form n2 – 1 (where n is an integer with n ≥ 3) is a prime number.
22.
Show that the following statement is true:
The number n2 + 8n + 15 is not a prime number for any positive integer n.
23.
Show that the following statement is true:
The number n2 + 9n + 20 is an even number for all positive integers n.
24.
Show that the following statement is true:
The number n2 + 7n + 13 is an odd number for all positive integers n.
25.
Find an unlimited supply of counter-examples to the following statement:
The number 8n + 1 is a prime number for all positive integers n.
26.
Show that the following statement is false by constructing a counter-example:
A ∩ (B ∪ C) = A ∪ (B ∩ C).
27.
Find the negation of the statement
∃m ∈ , ∀n ∈ , m ≥ n.
28.
Find the negation of the statement
∀m ∈ , ∃n ∈ , m + n is prime.
29.
Prove, without the use of diagrams, that A ∪ B = A ∩ B
30.
Let q denote the following statement: ∀n ∈ , 3n2 + 31 is a prime.
(a) Find the negation of q.
(b) Prove that ¬q is true.
31.
Let q denote the following statement concerning n ∈ .
If n is the square of a positive even integer,
then n is the sum of two successive odd positive integers.
(a) Prove that q is true.
(b) State the converse of q.
(c) Prove that the converse of q is false.
Page 23 of 34
Abstract Mathematics : Proofs
32.
The Fibonacci sequence (F0, F1, F2, F3, …) is defined by
F0 = 0
F1 = 1
Fn = Fn – 1 + Fn – 2,
for n ≥ 2.
Show that Fn + 3 = 4Fn + Fn – 3 for all n ≥ 0.
33.
The statement P is as follows:
P: For all sets A, B and C, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Prove that P is true.
[Hint: This is a set-equality question, so you must show that each set is contained in the
other.]
34.
Find an unlimited supply of counter-examples to the following statement:
The number 6n + 5 is a prime number for all positive integers n.
Page 24 of 34
Abstract Mathematics : Proofs
Solutions to Revision Problems
1.
The statement is false. A counter-example follows:
Let n = 17. Then n2 + n + 17 = 172 + 17 + 17 = 17(17 + 1 + 1) = 17 × 19
Therefore, n2 + n + 17 is not a prime number when n = 17.
****************************************************************
2.
The statement is true. A proof by exhaustion follows:
7 = {0, 1, 2, 3, 4, 5, 6}
03 – 0 = 0 ≡ 0 (mod 7)
13 – 1 = 0 ≡ 0 (mod 7)
23 – 2 = 6 ≡ 6 (mod 7)
33 – 3 = 24 ≡ 3 (mod 7)
43 – 4 = 60 ≡ 4 (mod 7)
53 – 5 = 120 ≡ 1 (mod 7)
63 – 6 = 210 ≡ 0 (mod 7)
Hence, the equation x3 – x = 2 has no solutions in 7.
****************************************************************
3.
The statement is true. A proof follows:
Let N = n4 – n2, where n ∈ .
N = n4 – n2 = n2(n2 – 1) = n2(n – 1)(n + 1)
N = n[(n – 1)n(n + 1)], so N is divisible by the product of three consecutive
integers, one of which must be divisible by 3. So, N is divisible by 3.
N = [(n – 1)n][n(n + 1)], where each of (n – 1)n and n(n + 1) is the product of two
consecutive integers, one of which must be divisible by 2. So, N is divisible by 4.
N is divisible by both 3 and 4 so N is divisible by LCM(3, 4).
LCM(3, 4) = 12.
Hence, N is divisible by 12 for all n ∈ .
****************************************************************
Page 25 of 34
Abstract Mathematics : Proofs
4.
The statement is true. A proof follows:
Let n be any positive integer that is not a multiple of 3.
By the Division Theorem, n can be written in the form 3k + 1 or 3k + 2, where k is a
non-negative integer.
If n = 3k + 1 then n2 – 1 = (3k + 1)2 – 1 = 9k2 + 6k = 3(3k2 + 2k) which is divisible by 3.
If n = 3k + 2 then n2 – 1 = (3k + 2)2 – 1 = 9k2 + 12k + 3 = 3(3k2 + 4k + 1), which is
divisible by 3.
Hence if n is any positive integer that is not a multiple of 3, then n2 – 1 is divisible by 3.
****************************************************************
5.
The statement is false. A counter-example follows:
Let n = 2. Then (2n + 7)2 – 1 = 120, which is divisible by 24.
However, n is not divisible by 6.
****************************************************************
6.
The statement is true. A proof follows:
Let n ∈ be divisible by 6, so that n = 6k where k ∈ .
(2n + 7)2 – 1 = (12k + 7)2 – 1 = 144k2 + 168k + 48 = 24(6k2 + 7k + 2)
6k2 + 7k + 2 ∈ , since k ∈ .
Hence, (2n + 7)2 – 1 is divisible by 24, for all n ∈ .
****************************************************************
7.
The statement is true. A proof follows:
Let n be an odd positive integer. Then, n = 2k + 1 for some non-negative integer k.
N = n2 – 1 = (2k + 1)2 – 1 = 4k2 + 4k = 4k(k + 1)
k(k + 1) is the product of two consecutive numbers, one of which must be even.
So, k(k + 1) is divisible by 2.
N is also divisible by 4, independently of k(k + 1).
2 × 4 = 8, so N is divisible by 8.
8.
****************************************************************
r ⇒ q is false, so r is true and q is false
p ∨ q is true and q is false, so p is true
9.
****************************************************************
p ⇒ q is true and q is false ∴ p is false.
So, both p and q are false. Hence, p ∨ q is false.
Page 26 of 34
Abstract Mathematics : Proofs
10. (a) Counter-example for S: n = 9, since 9 is odd and 6n + 1 = 55, which is not prime.
(b) Converse S*: If 6n + 1 is a prime then n is an odd number.
Counter-example for S*: n = 2, since 6n + 1 = 13, which is prime, but n is even.
(c) The contrapositive of S is: If 6n + 1 is not prime then n is an even number.
The contrapositive of S is logically equivalent to S.
Since S is false then so too is the contrapositive of S.
(d) The negation of S is: n is an odd number and 6n + 1 is not a prime.
****************************************************************
11.
Truth table:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
q∧r
T
F
F
F
T
F
F
F
p ∨ ( q ∧ r)
T
T
T
T
T
F
F
F
p∨q
T
T
T
T
T
T
F
F
p∨r
T
T
T
T
T
F
T
F
(p ∨ q) ∧ (p ∨ r)
T
T
T
T
T
F
F
F
The truth-values match in all 8 cases for the two highlighted columns.
So, p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r) are logically equivalent.
****************************************************************
12.
Truth table 1:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
¬p
F
F
F
F
T
T
T
T
¬p ⇔ q
F
F
T
T
T
T
F
F
r⇒p
T
T
T
T
F
T
F
T
q ∨ (r ⇒ p) ((¬p ⇔ q) ⇒ (q ∨ (r ⇒ p)))
T
T
T
T
T
T
T
T
T
T
T
T
F
T
T
T
Page 27 of 34
Abstract Mathematics : Proofs
A
B
p ⇒ ¬ (q ∨ ¬r)
F
F
T
F
T
T
T
T
p⇒r
T
F
T
F
T
T
T
T
Truth table 2:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
¬r
F
T
F
T
F
T
F
T
q ∨ ¬r
T
T
F
T
T
T
F
T
¬ (q ∨ ¬r))
F
F
T
F
F
F
T
F
A⇒B
T
T
T
T
T
T
T
T
((¬p ⇔ q) ⇒ (q ∨ (r ⇒ p))) and ((p ⇒ ¬ (q ∨ ¬r)) ⇒ (p ⇒ r)) are logically equivalent since
the truth-values match in all 8 cases.
13.
****************************************************************
Truth table:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
p∧q
T
T
F
F
F
F
F
F
¬r
F
T
F
T
F
T
F
T
p ⇔ ¬r
F
T
F
T
T
F
T
F
r ∨ (p ⇔ ¬r)
T
T
T
T
T
F
T
F
(p ∧ q) ⇒ (r ∨ (p ⇔ ¬r))
T
T
T
T
T
T
T
T
The truth-value is T in all 8 cases, so ((p ∧ q) ⇒ (r ∨ (p ⇔ ¬r))) is always true.
14.
****************************************************************
Truth table:
A
B
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
¬p
F
F
F
F
T
T
T
T
¬q
F
F
T
T
F
F
T
T
¬p ⇒ ¬q
T
T
T
T
F
F
T
T
q∧r
T
F
F
F
T
F
F
F
¬ (q ∧ r)
F
T
T
T
F
T
T
T
¬ (q ∧ r) ∨ p
T
T
T
T
F
T
T
T
A⇒B
T
T
T
T
T
T
T
T
The truth-value is T in all 8 cases, so ((¬p ⇒ ¬q) ⇒ (¬ (q ∧ r) ∨ p)) is always true.
Page 28 of 34
Abstract Mathematics : Proofs
15. (a) Contrapositive is: If n is not divisible by 7 then n is not divisible by 49.
(b) Converse: If n is divisible by 7 then n is divisible by 49.
(c) P is true. A proof follows:
Let n be divisible by 49. Then n = 49k for some k ∈ .
So, n = 7(7k), where 7k ∈ . Hence, n is divisible by 7.
(d) The converse of P is not true. Counter-example: n = 7.
****************************************************************
16.
Assume that there does exist a prime p such that 37p + 1 is a square.
Then, 37p + 1 = n2, where n ≥ 9 since p ≥ 2.
37p = n2 – 1 = (n – 1)(n + 1)
p ≠ 37 since 37(37) + 1 = 1370, which is not a square.
37p is a product of two distinct primes, so we have two possibilities:
either 37 = n – 1 and p = n + 1,
or 37 = n + 1 and p = n – 1.
So, either n = 38 so that p = 39,
or n = 36 so that p = 35.
However, neither 35 nor 39 is prime. So, we have reached a contradiction.
Hence, there does not exist any prime p such that 37p + 1 is a square.
****************************************************************
17.
Assume that p is a prime such that 7p + 1 is a square.
Then, 7p + 1 = n2, where n ≥ 4 since p ≥ 2.
7p = n2 – 1 = (n – 1)(n + 1)
p ≠ 7 since 7(7) + 1 = 50, which is not a square.
7p is a product of two distinct primes, so we have two possibilities:
either 7 = n – 1 and p = n + 1,
or 7 = n + 1 and p = n – 1.
So, either n = 8 so that p = 9,
or n = 6 so that p = 5.
However, 9 is not prime.
Hence, the only possibility is p = 5.
****************************************************************
Page 29 of 34
Abstract Mathematics : Proofs
18.
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 6, 8}, B = {1, 2, 6, 8, 9}.
(a) E \ A = {1, 3, 4, 5, 7, 9, 10}
(b) A ∩ B = {2, 6, 8}
(c) A ∪ B = {1, 2, 6, 8, 9}
(d) A ∩ (E \ B) = {2, 6, 8} ∩ {3, 4, 5, 7, 10} = 0/
(e) A = 3 and 23 = 8, so 8 members.
Ƥ (A) = { 0/ , {2}, {6}, {8}, {2, 6}, {2, 8}, {6, 8}, {2, 6, 8}}.
***************************************************************
19.
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 4, 6, 8}, B = {1, 4, 6, 9, 10}.
(a) E \ A = {1, 3, 5, 7, 9, 10}
(b) A ∩ B = {4, 6}
(c) A ∪ B = {1, 2, 4, 6, 8, 9, 10}
(d) B ∩ (E \ A) = {1, 4, 6, 9, 10} ∩ {1, 3, 5, 7, 9, 10} = {1, 9, 10}
(e) A = 4 and 24 = 16, so 16 members. Ƥ (A) = { 0/ , {2}, {4}, {6}, {8}, {2, 4}, {2, 6},
{2, 8}, {4, 6}, {4, 8}, {6, 8}, {2, 4, 6}, {2, 4, 8}, {2, 6, 8}, {4, 6, 8}, {2, 4, 6, 8}}
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20.
The statement is false. A proof follows:
n2 – 1 = (n – 1)(n + 1)
So, 4224 – 1 = (4212)2 – 12 = (4212 – 1)(4212 + 1)
4212 – 1 > 1 and 4212 + 1 > 1
So, 4224 – 1 is the product of two integers, both exceeding 1.
Hence, 4224 – 1 is not a prime number
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21.
The statement is true. A proof follows:
Let n be an integer with n ≥ 3.
n2 – 1 = (n – 1)(n + 1)
n – 1 > 1 and n + 1 > 1 since n ≥ 3
So, n2 – 1 is the product of two integers, both exceeding 1.
So, n2 – 1 is not a prime number for n ≥ 3.
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Page 30 of 34
Abstract Mathematics : Proofs
22.
The statement is true. A proof follows:
Let n be a positive integer.
n2 + 8n + 15 = (n + 3)(n + 5)
n + 3 > 1 and n + 5 > 1 since n ≥ 1
So, n2 + 8n + 15 is the product of two integers, both exceeding 1.
Hence, n2 + 8n + 15 is not a prime number for any positive integer n.
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23.
The statement is true. A proof follows:
Let n be a positive integer.
n2 + 9n + 20 = (n + 4)(n + 5)
n + 4 and n + 5 are consecutive positive integers.
So, one of n + 4 and n + 5 must be even, while the other must be odd, for any
positive integer n. So, n2 + 9n + 20 is divisible by an even number.
Hence, n2 + 9n + 20 is an even number for all positive integers n.
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24.
The statement is true. A proof follows:
Let n be a positive integer.
n2 + 7n + 12 = (n + 3)(n + 4)
n + 3 and n + 4 are consecutive positive integers.
So, one of n + 3 and n + 4 must be even, while the other must be odd, for any
positive integer n. So, n2 + 7n + 12 is divisible by an even number.
Thus, n2 + 7n + 12 is an even number for all positive integers n.
n2 + 7n + 13 = (n2 + 7n + 12) + 1.
Hence, n2 + 7n + 13 is an odd number for all positive integers n.
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25.
Let n = 3m + r, for some positive integers m and r.
Then, 8n + 1 = 8(3m + r) + 1 = 24m + 8r + 1, which we want to be not prime.
Choose r = 1. Then, 8n + 1 = 24m + 9 = 3(8m + 3), which is not prime.
Hence, an unlimited supply of counter-examples to the statement that 8n + 1 is a prime
number is the set {n = 3m + 1: m ∈ }.
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Page 31 of 34
Abstract Mathematics : Proofs
26.
The statement is false. A counter-example follows:
Let A = {a}, B = {b} and C = {c}. Then,
A ∩ (B ∪ C) = {a} ∩ {b, c} = 0/ .
A ∪ (B ∩ C) = {a} ∪ 0/ = {a} ≠ A ∩ (B ∪ C).
27.
****************************************************************
The negation of the given statement is
¬(∃m ∈ , ∀n ∈ , m ≥ n)
⇔ ∀m ∈ , ¬(∀n ∈ , m ≥ n)
⇔ ∀m ∈ , ∃n ∈ , ¬(m ≥ n)
⇔ ∀m ∈ , ∃n ∈ , m < n.
****************************************************************
28.
The negation of the given statement is
¬(∀m ∈ , ∃n ∈ , m + n is prime)
⇔ ∃m ∈ , ¬(∃n ∈ , m + n is prime)
⇔ ∃m ∈ , ∀n ∈ , ¬(m + n is prime)
⇔ ∃m ∈ , ∀n ∈ , m + n is not prime.
****************************************************************
29.
x ∈ A∪ B
⇔
x∉A∪B
⇔
¬ (x ∈ A ∪ B)
⇔
¬ ( (x ∈ A) ∨ (x ∈ B) )
⇔
¬ (x ∈ A) ∧ ¬ (x ∈ B)
⇔
(x ∈ A ) ∧ (x ∈ B )
⇔
x∈ A ∩ B
Thus, A ∪ B = A ∩ B
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Page 32 of 34
Abstract Mathematics : Proofs
30. (a) The negation of q is
¬(∀n ∈ , 3n2 + 31 is a prime)
⇔ ∃n ∈ , ¬(3n2 + 31 is a prime)
⇔ ∃n ∈ , 3n2 + 31 is not a prime.
(b) ¬q is true. For example, if n = 31 then 3(31)2 + 31 = 31(93 + 1) = 31 × 94.
Thus, if n = 31 then 3n2 + 31 is not a prime.
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31. (a) Examples: 42 = 16 = 8 + 8 = 7 + 9, 62 = 36 = 18 + 18 = 17 + 19.
Proof:
Let n be the square of a positive even integer m.
m is even and positive, so m = 2k, for some positive integer k.
Then, n = m2 = (2k)2 = 4k2 = (2k2 – 1) + (2k2 + 1)
2k2 is multiple of 2, so 2k2 is even, and so both (2k2 – 1) and (2k2 + 1) are odd.
Hence, n is the sum of two successive odd positive integers. Thus, q is true.
(b) The converse of q is as follows:
If n is the sum of two successive odd positive integers,
then n is the square of a positive even integer.
(c) The converse of q is false. A counter-example follows:
Let n = 5 + 7 = 12. Then n is not the square of a positive even integer since 32 = 9 and
42 = 16, and since 9 < 12 < 16.
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32.
For all n ≥ 0,
Fn + 3 = Fn + 2 + Fn + 1
= (Fn + 1 + Fn) + (Fn + Fn – 1)
= Fn + 1 + 2Fn + Fn – 1
= (Fn + Fn – 1) + 2Fn + (Fn – 2 + Fn – 3)
= 3Fn + (Fn – 1 + Fn – 2) + Fn – 3
= 3Fn + Fn + Fn – 3
= 4Fn + Fn – 3
****************************************************************
Page 33 of 34
Abstract Mathematics : Proofs
33.
x ∈ A ∪ (B ∩ C)
⇒ x ∈ A or x ∈ (B ∩ C)
⇒ x ∈ A or (x ∈ B and x ∈ C)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)
⇒ (x ∈ A ∪ B) and (x ∈ A ∪ C)
⇒ x ∈ (A ∪ B) ∩ (A ∪ C)
So, A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C) ………………………….. (1)
x ∈ (A ∪ B) ∩ (A ∪ C)
⇒ (x ∈ A ∪ B) and (x ∈ A ∪ C)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)
⇒ (x ∈ A and x ∈ A) or (x ∈ A and x ∈ C) or (x ∈ B and x ∈ A) or (x ∈ B and x ∈ C)
⇒ x ∈ A or (x ∈ A ∩ C) or (x ∈ A ∩ B) or (x ∈ B ∩ C)
⇒ x ∈ A or x ∈ (B ∩ C) since (A ∩ C) ⊆ A and (A ∩ B) ⊆ A
⇒ x ∈ A ∪ (B ∩ C)
So, (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C) ………………………….. (2)
From (1) and (2), it follows that
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) for all sets A, B and C.
Hence statement P is true.
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34.
Let n = 7m + r, for some positive integers m and r.
Then, 6n + 5 = 6(7m + r) + 5 = 42m + 6r + 5, which we want to be not prime.
Choose r = 5. Then, 6n + 5 = 42m + 35 = 7(6m + 5), which is not prime.
Hence, an unlimited supply of counter-examples to the statement that 6n + 5 is a prime
number is the set {n = 7m + 5: m ∈ }.
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Page 34 of 34