Download MAA245 NUMBERS 1 Natural Numbers, N

MAA245 NUMBERS
1
Natural Numbers, N
1.1
Peano’s Axioms
N is a set with the properties:(i) ∃ an element denoted “1” in N; so N is non-empty.
(ii) There is a function S : N → N called the Successor Function. Thus, ∀ n ∈ N,
∃ S(n) ∈ N. Write S(1) = 2, S(2) = 3, . . ..
(iii) 1 6= S(n) for any n ∈ N; i.e. 1 is not the successor of any element of N.
(iv) S(m) = S(n) only if m = n; i.e. S : N → N is an injection.
(v) The Principle of Induction
Suppose A ⊆ N. If: (a) 1 ∈ A, and (b) n ∈ A ⇒ S(n) ∈ A, then A = N.
Here, A is a set defined by some property (e.g. an equation to be satisfied). To
prove that every natural number has the property (i.e. that A = N), need to show:(a) 1 has the property (“anchor”);
(b) assuming that n has the property (the “inductive hypothesis”), it follows that
S(n) also has the property.
Note: (v) ⇒ every n ∈ N\{1} is the successor of some m ∈ N.
1.2
Addition of Natural Numbers
Define “+” inductively. ∀n, m ∈ N:
(a) n + 1 := S(n)
(b) n + S(m) := S(n + m)
Thus, n + 2 = n + S(1) = S(n + 1) = S(S(n)),
n + 3 = n + S(2) = S(n + 2),
etc.
This definition ensures that addition is closed on N, i.e. that if m, n ∈ N, then
n + m ∈ N (by induction, since S(n + m) ∈ N whenever n + m ∈ N).
A binary operation on a set A is a function f : A × A → A; i.e. “input” is
an ordered pair of members of A, and output is also a member of A; so the binary
operation is closed on A. Thus addition is a binary operation on N.
1
We now prove some properties of addition.
Lemma 1.2.1
1 + n = n + 1 ∀n ∈ N.
Proof (by induction)
(a) For n = 1: 1 + n = 1 + 1 = n + 1.
(b) Given the inductive hypothesis 1 + n = n + 1, must show: 1 + S(n) = S(n) + 1.
Now,
1 + S(n) = S(1 + n)
= S(n + 1)
= (n + 1) + 1
= S(n) + 1
[Definition of +, part (b)]
[Inductive hypothesis]
[Definition of +, part (a)]
[Definition of +, part (a)].
Lemma 1.2.2
(m + 1) + n = (m + n) + 1 ∀m, n ∈ N.
Proof (by induction on n)
(a) For n = 1: (m + 1) + n = (m + 1) + 1 = (m + n) + 1.
(b) Given the inductive hypothesis (m + 1) + n = (m + n) + 1, must show:
(m + 1) + S(n) = (m + S(n)) + 1.
Now,
(m + 1) + S(n) = S((m + 1) + n)
= S((m + n) + 1)
= S(S(m + n))
= S(m + S(n))
= (m + S(n)) + 1
[Definition of +, (b)]
[Inductive hypothesis]
[Definition of +, (a)]
[Definition of +, (b)]
[Definition of +, (a)].
Theorem 1.2.3
m + n = n + m ∀m, n ∈ N.
Proof (by induction on m)
(a) For m = 1: m + n = 1 + n = n + 1 [Lemma 1.2.1]
= n + m.
(b) Given the inductive hypothesis m+n = n+m, must show: S(m)+n = n+S(m).
Now,
S(m) + n = (m + 1) + n
= (m + n) + 1
= (n + m) + 1
= S(n + m)
= n + S(m)
[Definition of +, (a)]
[Lemma 1.2.2]
[Inductive hypothesis]
[Definition of +, (a)]
[Definition of +, (b)].
A binary operation ∗ on a set A is said to be commutative if
r ∗ s = s ∗ r ∀ r, s ∈ A,
2
i.e. if the order of the two elements doesn’t matter. Theorem 1.2.3 ⇒ Addition is
commutative on N.
Theorem 1.2.4
(m + n) + p = m + (n + p) ∀m, n, p ∈ N.
Proof: Exs. 1, Q.1.
A binary operation ∗ on a set A is said to be associative if
(r ∗ s) ∗ t = r ∗ (s ∗ t) ∀ r, s, t ∈ A,
i.e. if the order in which successive operations are done doesn’t matter. Theorem
1.2.4 ⇒ Addition is associative on N.
The commutative and associative properties mean that any number of additions
may be rearranged in any way, and brackets are not required.
Example
(m + n) + (p + q) = (n + (q + p)) + m = q + p + n + m
[See Exs. 1, Q. 2]
Theorem 1.2.5
Given any m ∈ N, 6 ∃ n ∈ N such that n + m = m
Proof (by induction on m)
(a) For m = 1:
n + 1 = S(n) [Definition of +, (a)]
6=
1
[Axiom (iii)]
i.e. n + m 6= m if m = 1.
(b) Given the inductive hypothesis, n + m 6= m, must show n + S(m) 6= S(m).
Now, n + S(m) = S(n + m) [Definition of +, (a)]
6=
S(m)
[Inductive hypothesis and Axiom (iv)].
Since + is commutative, we also have m + n 6= m for any m, n ∈ N.
A binary operation ∗ has an identity element i∗ ∈ A if i∗ ∗ a = a and a ∗ i∗ = a
∀a ∈ A, i.e. operating with i∗ leaves any element of A unchanged. Theorem 1.2.5
⇒ 6 ∃ i+ ∈ N: there is no additive identity element in N.
Theorem 1.2.6
If m + n = m + p, then n = p ∀ m, n, p ∈ N.
This is the Cancellation Law for addition.
Proof: Exs. 1, Q.3.
3
1.3
Multiplication of Natural Numbers
Define “×” inductively. ∀m, n ∈ N:
(a) n × 1 := n
(b) n × S(m) := (n × m) + n.
Since n ∈ N (on right hand side of (a)) and (b) uses addition which is closed on N,
this definition ensures that multiplication is closed on N; so × is a binary operation on
N.
Notation: write “mn” for “m × n” where convenient.
The following lemmas and theorems, apart from 1.3.2 and 1.3.5, are similar to
those for addition, and may be proved similarly.
Lemma 1.3.1
n × 1 = 1 × n ∀ n ∈ N.
We now have n × 1 = 1 × n = n, i.e. 1 is a multiplicative identity element in N.
Theorem 1.3.2
If an identity element exists for a binary operation ∗ on a set A, it is unique.
Proof
Supose i1 , i2 are both identity elements for ∗. Then
i1 = i1 ∗ i2 [i2 is an identity element]
=
i2
[i1 is an identity element]
i.e. i1 and i2 are actually the same.
Example: 1 is the unique multiplicative identity element in N: m×n = n or n×m = n
only if m = 1.
Lemma 1.3.3
(m + 1)n = mn + n ∀ m, n ∈ N.
Theorem 1.3.4
m × n = n × m ∀ m, n ∈ N,
i.e. multiplication is commutative on N.
Theorem 1.3.5
(m + n)p = mp + np ∀ m, n, p ∈ N.
Proof (by induction on n)
(a) For p = 1:
(m + n)p = (m + n) × 1 =
m+n
[Definition of +, (a)]
= m × 1 + n × 1 [Definition of +, (a)]
=
mp + np
(b) Given the inductive hypothesis, (m + n)p = mp + np:
(m + n) × S(p) = ((m + n)p) + (m + n) [Definition of ×, (b)]
4
= (mp + np) + (m + n) [Inductive hypothesis]
= (mp + m) + (np + n) [+ is associative & commutative]
=
mS(p) + nS(p)
[Definition of ×, (b)].
Because multiplication is commutative, we also have
m(n + p) = mn + mp
from Theorem 1.3.5.
Given two binary operations ∗ and ∨ on a set A, we say that ∗ distributes over ∨
if
r ∗ (s ∨ t) = (r ∗ s) ∨ (r ∗ t)
and
(r ∨ s) ∗ t = (r ∗ t) ∨ (s ∗ t)
∀ r, s, t ∈ A .
We have shown (Theorem 1.3.5) that multiplication distributes over addition on N.
Theorem 1.3.6
m × (n × p) = (m × n) × p ∀ m, n, p ∈ N,
i.e. multiplication is associative on N.
1.4
Exponentiation (Powers)
Define inductively: ∀m, n ∈ N:
(a) n1 := n
(b) nS(m) := n × nm .
Since n ∈ N and × is closed on N, exponentiation is closed on N; it is a binary
operation, and can be symbolised by ∧, i.e. write m ∧ n for mn .
The following theorems can be proved by induction:Theorem 1.4.1
1n = 1 ∀ n ∈ N.
Theorem 1.4.2
nm × np = nm+p ∀ m, n, p ∈ N.
Theorem 1.4.3
(np)m = nm × pm ∀ m, n, p ∈ N.
Note that ∧ is not commutative (mn 6= nm in general), and Theorem 1.4.3 is a
one-sided Distributive Law, i.e.
(n × p) ∧ m = (n ∧ m) × (p ∧ m)
but
n ∧ (p × m) 6= (n ∧ p) × (n ∧ m) .
Theorem 1.4.4
n(pm) = (np )m ∀ m, n, p ∈ N.
Note that ∧ is not associative: (np )m 6= n(p
5
m)
in general.
1.5
Order in N
“=” (equals) is an equivalence relation; it partitions N into the sets {1}, {2}, {3}, . . ..
We now define “>” (greater than) and “<” (less than).
Given m, n ∈ N:
m > n if ∃ k ∈ N such that m = n + k;
m < n if ∃ p ∈ N such that m + p = n.
It follows immediately that m > n if and only if n < m.
The following four theorems show that this is a “sensible” way to define > and <.
Theorem 1.5.1
S(n) > n ∀n ∈ N.
Proof
S(n) = n + 1, so let m = S(n), k = 1 in definition of >.
Theorem 1.5.2
n > 1 ∀n ∈ N\{1}.
Proof: Exs. 1, Q. 7
Theorem 1.5.3
Given m, n ∈ N, one and only one of the following is true: (i) m = n, or (ii) m > n,
or (iii) m < n. This property is called Trichotomy.
Proof
Firstly, show that not more than one of the three options can be true.
Theorem 1.2.5 ⇒ if m = n + k, then m 6= n; i.e. cannot have both m > n and
m = n. Similarly, cannot have both m < n and m = n.
Now suppose both m > n and m < n. Then
m = n+k
= (m + p) + k
= m + (p + k)
[+ is associative],
but (p + k) ∈ N [+ is closed on N], so Theorem 1.2.5 ⇒ this is impossible.
Secondly, show that one of the three options is true, by induction on n.
(a) For n = 1:
if m = 1, then m = n; if m 6= 1, then m > 1 [Theorem 1.5.2], i.e. m > n.
(b) Inductive hypothesis: one of these is true: (i) m = n, or (ii) m < n, or (iii) m > n.
Then:
(i) If m = n, then m + 1 = S(n), i.e. m < S(n).
(ii) If m < n, i.e. n = m + k, then
S(n) = (m + k) + 1
6
= m + (k + 1)
.
. . m < S(n) [since (k + 1) ∈ N].
(iii) If m > n, i.e. m = n + p:either: p = 1, so m = n + 1, i.e. m = S(n);
or: p 6= 1, so p = S(q) = q + 1 for some q ∈ N,
.. . m =
=
=
.
..m >
n + (q + 1)
(n + 1) + q
S(n) + q
S(n) .
In all cases (i), (ii) and (iii), have shown that one of these is true: m = S(n), or
m < S(n), or m > S(n), thus completing the induction.
The symbol “≥” means “> or =”; the symbol “≤” means “< or =”.
Trichotomy ⇒ : if m ≥ n and m ≤ n, then m = n.
Theorem 1.5.4
If m < n and n < p, then m < p, ∀ m, n, p ∈ N.
This property is called Transitivity.
Proof: Exs. 1, Q. 8.
Theorem 1.5.5
m>n⇔m+p>n+p
∀ m, n, p ∈ N,
i.e. addition, and also cancellation of addition, preserve ordering.
Proof
m>n ⇒
⇒
⇒
⇒
m=n+k
m + p = (n + k) + p
m + p = (n + p) + k
m + p > (n + p)
Converse: Exs. 1, Q. 9.
Theorem 1.5.6
6 ∃k ∈ N such that n < k < S(n),
i.e. no natural number comes “between” any n ∈ N and its successor.
Proof (by induction on n)
(a) For n = 1: suppose 1 < k < S(1).
1 < k ⇒ k 6= 1, so k = S(j) = j + 1 for some j ∈ N.
7
Next,
k < S(1) ⇒ j + 1 < 1 + 1
⇒ j < 1 [Theorem 1.5.5],
contradicting Theorem 1.5.2; so 6 ∃k ∈ N such that 1 < k < S(1).
(b) Exs. 1, Q. 10.
Theorem 1.5.7
m > n ⇔ mp > np
∀ m, n, p ∈ N,
i.e. multiplication preserves ordering in N.
Corollary 1.5.8
If mp = np, then m = n.
This is the Cancellation Law for multiplication.
Proof
Unless m = n, Theorem 1.5.7 ⇒ mp 6= np (by Trichotomy).
Theorem 1.5.9
p > n ⇔ pm > nm
∀ m, n, p ∈ N,
i.e. exponentiation preserves ordering in N.
Theorem 1.5.10
m > p ⇔ nm > np
1.6
∀ n > 1; m, p ∈ N.
Finite and Infinite Sets
[See separate handout on Bounded Sets of Numbers for notes up to Theorem 1.6.4 ]
Theorem 1.6.5: The Well-Ordering of N
Every non-empty subset of N has a least member.
Proof
Let LA be the set of all lower bounds for A, where A ⊆ N and A 6= ∅. Then
1 ∈ LA [Theorem 1.5.2]. Also, LA 6= N: given any a ∈ A, S(a) 6∈ LA . Thus, ∃ l ∈ LA
with S(l) 6∈ LA ; otherwise, induction would give LA = N.
Because l ∈ LA , a ≥ l ∀ a ∈ A [Definition 1.6.1 (ii)]. So, if l 6∈ A, a > l ∀ a ∈ A.
Because 6 ∃k ∈ N with l < k < S(l) [Theorem 1.5.6], this implies a ≥ S(l) ∀ a ∈ A.
So S(l) is a lower bound for A, contradicting S(l) 6∈ LA .
We conclude that l ∈ A. Since l ∈ LA , Definition 1.6.2(ii) ⇒ l = min A.
Theorem 1.6.6
If a non-empty subset of N is bounded above, it has a greatest member.
Definitions 1.6.7
8
For each n ∈ N, let Cn = {k ∈ N; k ≤ n}. Let the symbol ↔ denote “has a
bijection to”. Then, for any set A:(i) If A ↔ Cn for some n ∈ N, then A is finite and has cardinality n; we count
the members of A by bijecting to Cn .
(ii) If A 6↔ Cn (A has no bijection to Cn ) for any n ∈ N, then A is an infinite set.
Notes:Can show: (i) n < m ⇔ Cn ⊂ Cm ; (ii) If n 6= m, Cn 6↔ Cm .
Hence: the cardinality of any finite set is unique; if A ⊂ B and B is finite, then A
is finite and A 6↔ B.
Examples
(i) A = {3, 6, 7} has cardinality 3: bijection to C3 is
3 ↔ 1,
6 ↔ 2,
7 ↔ 3.
(ii) N is infinite: for every n ∈ N, S(n) ∈ N but S(n) 6∈ Cn . Try a bijection from N to
Cn :
1 ↔ 1, 2 ↔ 2, . . . , n ↔ n, S(n) ↔?
Note that S is a bijection from N to N\{1}. Only infinite sets can have bijections to
proper subsets of themselves.
Definitions 1.6.8
(i) A set with a bijection to N is countably infinite.
(ii) A set is countable if it is either finite or countably infinite.
Theorem 1.6.9
Every finite subset of any ordered number system has a greatest and a least
member.
Proof
If A ↔ Cn , use the bijection to label members of A: A = {a1 , a2 , . . . , an }. Let
Ak = {a1 , a2 , . . . , ak } for any k ≤ n.
Firstly, a1 = max A1 . Then, if max Ak exists, either:aS(k) ≤ max Ak , so max AS(k) = max Ak ,
or:
aS(k) > max Ak , so max AS(k) = aS(k) .
In both cases, max AS(k) exists. When this induction reaches k = n, we have
Ak = An = A, so max A exists. Similarly, min An exists.
Note: If A is infinite, 6 ∃n for which k = n to “stop the induction”.
Corollary 1.6.10
A subset of N is finite iff it is bounded above.
Proof
Theorem 1.6.9 gives “Finite ⇒ Bounded above”.
9
If A ⊆ N is bounded above, Theorem 1.6.6 ⇒ max A exists. Let u = max A; then
A ⊆ Cu , i.e. A has cardinality ≤ u, and is finite.
1.7
N as an Algebraic Structure
An algebraic structure is a set, together with one or more binary operations (which
must be closed on the set). Algebraic structures are classified according to the properties of the binary operation(s). In particular, a semigroup hA, ∗i is a set A with
a binary operation ∗ which is associative. If ∗ is also commutative, then hA, ∗i is a
commutative semigroup.
Thus, hN, +i and hN, ×i are both commutative semigroups.
1.8
Inverting binary operations
If ∗ is a binary operation on a set A, we define its inverse ∗ as follows.
Given s, t ∈ A, if ∃ r ∈ A such that r ∗ s = t, then t ∗ s = r. If ∗ is commutative,
then
t ∗ s = r ⇔ t ∗ r = s.
If ∃ r ∈ A with r ∗ s = t for every s, t ∈ A, then ∗ is closed on A: t ∗ s ∈ A ∀s, t ∈ A.
The need to find sets of numbers on which the inverses of +, × and ∧ are closed
forces us to define new number systems. In particular, the inverse of addition, called
subtraction and denoted “−”, is not closed on N.
Proof
Given n, p ∈ N, suppose ∃ m ∈ N such that m + n = p. If subtraction is closed on
N, then also ∃ q ∈ N such that q + p = n. So
q + (m + n) = n
.. . (q + m) + n = n
with (q + m) ∈ N.
This is impossible by Theorem 1.2.5.
However, if p > n, then ∃ m ∈ N such that m + n = p, i.e. p − n = m. In this
case we have:
Theorem 1.8.1
p = q iff p − n = q − n.
Proof
Follows from Theorem 1.2.6 and definition of subtraction.
Theorem 1.8.2
p < q iff p − n < q − n.
Proof
Follows from Theorem 1.5.5 and definition of subtraction.
10
MAA245 NUMBERS
2
Integers, Z
2.1
Definition of Z
The set of ordered pairs of natural numbers is N × N. Define an equivalence relation
“≡” on N × N:
(p, n) ≡ (q, m)
iff
p+m=q+n
[p, n, q, m ∈ N].
[To verify that this is an equivalence relation, must show:
(a) Reflexive: (p, n) ≡ (p, n) ;
(b) Symmetric: If (p, n) ≡ (q, m), then (q, m) ≡ (p, n) ;
(c) Transitive: If (p, n) ≡ (q, m) and (q, m) ≡ (r, l), then (p, n) ≡ (r, l) :
see Exs. 2, Q.1.]
Example
(3, 1) ≡ (6, 4) since 3 + 4 = 6 + 1.
Definition 2.1.1
An integer is an equivalence class of ordered pairs (p, n) ∈ N×N under the relation
“≡” defined above. So two ordered pairs represent equal integers if they are members
of the same equivalence class.
Suppose p > n. Then ∃ s ∈ N such that p = n + s, so s = p − n. So, ∀k ∈ N,
p + k = (n + s) + k
= (k + s) + n [+ is associative & commutative on N],
.
. . (p, n) ≡ (k + s, k) [by definition of ≡]
Thus, if p > n, then (p, n) is a member of the equivalence class
Ks := {(k + s, k) : k ∈ N},
which is characterised by a single natural number s = p−n. Conversely, for each s ∈ N,
there is an equivalence class (i.e. an integer) Ks . The set of all such equivalence classes
is called the positive integers, Z+ :
Z+ = {Ks : s ∈ N}.
So Ks ↔ s is a bijection from Z+ to N.
Similarly, if p < n, ∃ t ∈ N such that n = p + t, so t = n − p; then
(p, n) ≡ (k, k + t) ∀ k ∈ N .
1
Thus, for each t ∈ N, there is an equivalence class
Lt := {(k, k + t) : k ∈ N}.
The set of all such equivalence classes is called the negative integers Z− :
Z− = {Lt : t ∈ N}.
Finally, if p = n, then
(p, n) ≡ (k, k) ∀ k ∈ N :
there is a single equivalence class, called zero:
0 = {(k, k) : k ∈ N}.
2.2
Arithmetic on Z
Define addition:(p, n) + (q, m) := (p + q, n + m)
(1)
Define multiplication:(p, n) × (q, m) := (pq + nm, qn + pm)
(2)
The binary operations + and × are closed on Z, since right-hand sides of (1) and (2)
are ordered pairs of natural numbers.
Suppose (p1 , n1 ) ≡ (p2 , n2 ) and (q1 , m1 ) ≡ (q2 , m2 ). Can show [Exs. 2, Q.2]:
(p1 , n1 ) + (q1 , m1 ) ≡ (p2 , n2 ) + (q2 , m2 ),
(p1 , n1 ) × (q1 , m1 ) ≡ (p2 , n2 ) × (q2 , m2 );
i.e. the result of adding or multiplying integers doesn’t depend on which members of
the equivalence classes are used in the calculations (1) or (2): the binary operations
are well-defined on Z.
Furthhermore,
(k + s, k) + (k + r, k) ≡ (k + (s + r), k)
and
(k + s, k) × (k + r, k) ≡ (k + sr, k)
[Exs. 2, Q. 3]; i.e. if we identify each positive integer Ks = {(k + s, k), k ∈ N} with
the natural number s, adding or multiplying the positive integers gives the same result
as adding or multiplying the natural numbers.
Definition: Two algebraic structures hA, ∗, . . .i and hB, ∗, . . .i are isomorphic if there
is a bijection ↔ between A and B which is preserved by each of the binary operations
2
∗, . . .; i.e. if, given a1 , a2 ∈ A and b1 , b2 ∈ B, with a1 ↔ b1 and a2 ↔ b2 , then
a1 ∗ a2 ↔ b1 ∗ b2 , . . ..
Thus hZ+ , +, ×i is isomorphic to hN, +, ×i under the bijection Ks ↔ s. Where
convenient, we can denote the positive integer Ks = {(k + s, k); k ∈ N} by s. Also,
denote the negative integer Lt = {(k, k + t), t ∈ N} by −t.
Using ordered-pair arithmetic (i.e. (1) and (2)), and applying properties of N
to each member of an ordered pair, we can show that + and × on Z have similar
properties to those on N:Theorem 2.2.1
Addition and multiplication are both commutative and associative on Z, and multiplication distributes over addition. There is a unique multiplicative identity element
in Z: it is 1 := {(k + 1, k), k ∈ N}.
However, addition on Z has further properties not found on N:Theorem 2.2.2
The integer 0 := {(k, k), k ∈ N} is the unique additive identity element in Z.
Proof
(p, n) + (k, k) = (p + k, n + k)
≡ (p, n) .
[Definition of + on Z]
Similarly, (k, k) + (p, n) = (p, n). So (k, k) is an additive identity, and is unique by
Theorem 1.3.2.
Definition 2.2.3
Suppose a binary operation ∗ on a set A has an identity element i∗ . Then a ∈ A
has an inverse element a if
a ∗ a = a ∗ a = i∗ .
Theorem 2.2.4
If a given set A has an inverse element under an associative binary operation ∗,
then the inverse element is unique.
Proof
Suppose a ∈ A has two inverses, a1 and a2 . Then
a1 =
=
=
=
=
a1 ∗ i ∗
[Definition of i∗ ]
a1 ∗ (a ∗ a2 ) [a2 is inverse]
(a1 ∗ a) ∗ a2 [∗ is associative]
i ∗ ∗ a2
[a1 is inverse]
a2 .
Hence a1 and a2 are the same.
3
Theorem 2.2.5
Every integer has a unique additive inverse element.
Proof
(p, n) + (n, p) = (p + n, n + p) ≡ (k, k) [= i+ ∈ Z];
similarly, (n, p) + (p, n) = (k, k). So (n, p) is an additive inverse of (p, n), and is unique
by Theorem 2.2.4.
Theorem 2.2.6
Subtraction is closed on Z
Proof
Since
(p + m, n + q) + (q, m) = ((p + m) + q, (n + q) + m)
= (p + (m + q), n + (m + q))
≡ (p, n) ,
definition of subtraction (Section 1.8) ⇒
(p, n) − (q, m) = (p + m, n + q) ∈ Z .
Thus
(p, n) − (q, m) = (p, n) + (m, q) ,
i.e. subtracting an integer (q, m) is the same as adding its additive inverse (m, q); i.e.
s − t = s + (−t).
Note: the symbol “−” has three uses:(i) Negative integers: if s ∈ N, −s ∈ Z−
(ii) Additive inverse
(iii) Subtraction
Example
5 − (−3) = −(−8) :
the four − signs here signify, respectively: subtraction; negative integer; additive
inverse; negative integer.
Rules for subtraction and additive inverses
Given a, b, c ∈ Z, the following may be proved by writing in ordered-pair form and
using definitions and theorems above:(i) −(a + b) = (−a) − b
(ii) −(a − b) = (−a) + b
(iii) (−a) × b = −(a × b) = a × (−b)
(iv) (−a) × (−b) = a × b
(v) a(b − c) = ab − ac
(vi) (a − b)c = ac − bc
4
(v) and (vi) mean that multiplication distributes over subtraction.
Also, a × 0 = 0 × a = 0.
Powers
For a ∈ Z and m ∈ N, we may define am inductively (in terms of multiplication)
as in Section 1.4.
2.3
Z as an Algebraic Structure
(i) An algebraic structure hA, ∗i is a group if:
(a) ∗ is associative on A
(b) ∗ has an identity element in A
(c) Every a ∈ A has an inverse element under ∗.
If ∗ is also commutative on A, then hA, ∗i is a commutative group (or Abelian
group).
Thus, hZ, +i is a commutative group. Also, hZ, ×i is a commutative semigroup.
(ii) An algebraic structure hA, ∗, ∨i (i.e. set A with two binary operations ∗ and ∨) is
a ring if:
(a) hA, ∗i is a commutative group
(b) hA, ∨i is a semigroup
(c) ∨ distributes over ∗
If ∨ is also commutative on A, then hA, ∗, ∨i is a commutative ring.
Thus, hZ, +, ×i is a commutative ring
2.4
Order in Z
Recall: (p, n) ≡ (q, m) iff p + m = q + n. Now define:
(p, n) > (q, m) iff p + m > q + n
(3)
(p, n) < (q, m) iff p + m < q + n
(4)
Notes:
(i) > and < are well-defined: choosing different members of same equivalence class
yields same ordering.
(ii) (3) ⇒ (k + s, k) > (k, k) and (4) ⇒ (k, k + s) < (k, k),
i.e. positive integers are greater than zero, negative integers are less than zero.
Theorem 2.4.1
The isomorphism between Z+ and N preserves order.
Proof
Given r, s ∈ N, the corresponding positive integers are represented by (k + r, k)
and (k + s, k). Now,
r < s ⇔ (k + r) + k < (k + s) + k
[Theorem 1.5.5, adding k twice]
⇔ (k + r, k) < (k + s, k)
[Definition of < on Z]
5
We can also prove the following properties of integers, which are similar to those of
natural numbers in Theorems 1.5.3 – 1.5.5 and 1.8.2:
(i) Trichotomy
(ii) Transitivity
(iii) Preservation of ordering on addition or subtraction of a number to/from both
sides.
Theorem 2.4.2
Suppose a < b (a, b ∈ Z);
then ac < bc if c > 0,
but ac > bc if c < 0.
Note the difference from Theorem 1.5.7; but consistent when c ∈ Z+ .
Example
7 > 4,
.. . 7 × 3 > 4 × 3 (i.e. 21 > 12),
but 7 × (−3) < 4 × (−3) (i.e. −21 < −12).
In particular, if c = −1,
a < b ⇒ a × (−1) > b × (−1)
⇒ −a > −b .
Theorem 2.4.3
(i) If ab > 0, then either : a > 0 and b > 0 or : a < 0 and b < 0
(ii) If ab < 0, then either : a > 0 and b < 0 or : a < 0 and b > 0
(iii) If ab = 0, then either a = 0 or b = 0.
Example
Find all n ∈ Z satisfying n2 − 3n − 4 > 0.
(n − 4)(n + 1) > 0.
By Theorem 2.4.3(i),
either:
n − 4 > 0 and n + 1 > 0
⇔ n > 4 and n > −1
[add 4 or −1 to both sides]
⇔ n>4
[since n > 4 ⇒ n > −1]
or:
n − 4 < 0 and n + 1 < 0
⇔ n < 4 and n < −1
⇔ n < −1
[since n < −1 ⇒ n < 4]
Solution set: {n ∈ Z : n > 4 or n < −1} = Z\{−1, 0, 1, 2, 3, 4}.
Theorem 2.4.4
6
6 ∃k ∈ Z such that n < k < n + 1 (for any n ∈ Z).
Proof
Follows from the same property in N, given bijections between N and Z+ and
between N and Z− , and existence of only one other integer, 0.
Henceforth, we do not distinguish between Z+ and N.
Theorem 2.4.5
If a non-empty subset of Z is bounded below [or above], then it has a least [or
greatest] member.
[Note: Z is not well-ordered; if A ⊆ Z, min A does not exist unless A is bounded
below.]
Proof
If a > 0 ∀a ∈ A, then A ⊆ Z+ = N, and Theorems 1.6.5, 1.6.6 apply.
Otherwise, find a lower bound l if bounded below. Then let
Al = {n ∈ Z : n = a − l + 1, a ∈ A}
⊆ Z+ = N.
Theorem 1.6.5 applies to Al , and min A = (min Al ) + l − 1.
Similarly for “greatest member”.
Theorem 2.4.6
A non-empty subset of Z is finite iff it is bounded above and below. [Cf. Corollary
1.6.10 for N.]
Theorem 2.4.7
Z is countably infinite.
Proof
A bijection f : Z → N is:
f (s) =
2s
if s ∈ Z+
1 − 2s if s ∈ Z− ∪ {0}
i.e. 0 ↔ 1, 1 ↔ 2, −1 ↔ 3, 2 ↔ 4, −2 ↔ 5, . . . .
7
MAA245 NUMBERS
3
3.1
Foundations of Number Theory
Division
Theorem 3.1.1: The Division Algorithm
Given n ∈ Z and d ∈ N, ∃ a unique quotient q ∈ Z and a unique remainder r ∈ Z such
that
n = qd + r with 0 ≤ r < d .
Proof
Note first: for any n ∈ Z and m ∈ N, the set
K = {k ∈ N : km > n}
is non-empty, since:
if
n ≤ 0, k = 1 ⇒ km > n ∀ m ∈ N ;
if
n > 0, k = n + 1 ⇒ km > n ∀ m ∈ N .
Thus, min K exists (by Well-Ordering of N)
Now consider three cases:(i) If 0 ≤ n < d, let q = 0 and r = n; then n = qd + r and 0 ≤ r < d.
(ii) If n ≥ d, let K = {k ∈ N : kd > n} and let l = min K.
1 × d 6> n, so 1 6∈ K, ∴ l > 1.
Let q = l − 1, so q ∈ N and q 6∈ K; so qd ≤ n. Let r = n − qd; then r ≥ 0. But
ld
∴ (q + 1)d
∴ qd + d
∴d
∴r
>
>
>
>
<
n
[l ∈ K]
n
n
n − qd
d.
(iii) If n < 0, let K = {k ∈ N : kd ≥ −n} and let q = − min K, so q ∈ Z− , and −q ∈ K.
Thus (−q)d ≥ −n, so n − qd ≥ 0. Let r = n − qd, so r ≥ 0. Also, −q = min K, so
−q − 1 6∈ K, i.e.
(−q − 1)d < −n
∴ n − qd < d
∴ r < d.
1
We have shown that q, r exist with the required properties; now show uniqueness.
Suppose n = qd + r and n = q 0 d + r0 ; then (q − q 0 )d = r0 − r.
Suppose r < r0 (without loss of generality); then
(q − q 0 )
∴ q − q0
∴ (q − q 0 )d
∴ r0 − r
>
≥
≥
≥
0
1
d
d.
(since d > 0)
[(q − q 0 ) ∈ Z]
But r0 < d and r ≥ 0, so r0 − r < d − 0, contradicting r0 − r ≥ d. Thus r 6< r0 (and r 6> r0
similarly), so r = r0 and hence q = q 0 .
Divisors
If r = 0, i.e. if n = qd, then d is a divisor or factor of n, and n is a multiple of d (and of q).
Notation: d | n (or d - n) if d is (or is not) a factor of n.
3.2
Prime Numbers and Prime Factorisation
For every n ∈ N, the numbers 1 and n are factors [1 × n = n]. If there are no other factors,
n is prime; otherwise, n is composite. 1 is considered neither prime nor composite.
The first few primes are: 2, 3, 5, 7, 11, 13, 17, 19, . . .. Note the irregular distribution of
primes; there is no formula to determine all primes, or even to yield only numbers that are
prime.
Theorem 3.2.1
Every n ∈ N\{1} can be expressed as the product of one or more prime factor(s).
Proof
If the theorem is false, ∃ a least number m0 which doesn’t satisfy it (by well-ordering).
Now, m0 is not prime (primes satisfy the theorem). So m0 = k0 × lo , with 1 < k0 , l0 < m0
and k0 , l0 ∈ N.
Either: k0 and l0 satisfy the theorem; but then m0 would satisfy it;
Or: k0 or l0 doesn’t satisfy it, contradicting m0 being the least such number.
Theorem 3.2.2: The Fundamental Theorem of Arithmetic
The prime factorisation of any n ∈ N\{1} is unique.
Proof
Suppose the theorem is false; let m be the smallest number with two prime factorisations,
m = p1 × p2 × . . . × pk = q 1 × q2 × . . . × ql .
None of the pi (i = 1, 2, . . . , k) can equal any of the qj (j = 1, 2, . . . , l); otherwise we could
write p1 = q1 and
m1 := p2 × p3 × . . . × pk = q2 × q3 × . . . × ql ,
2
i.e. ∃ m1 < m (since m = m1 p1 ) with two prime factorisations, contradicting the definition
of m.
Since p1 6= q1 , can suppose p1 < q1 .
Let
r :=
=
=
<
(q1 − p1 )q2 q3 . . . ql
(1)
q1 q2 q3 . . . ql − p1 q 2 q3 . . . ql
m − p1 q2 q3 . . . ql
(2)
m.
Now, p1 is not a factor of (q1 − p1 ); otherwise q1 − p1 = sp1 , so q1 = (1 + s)p1 , i.e. q1 would
not be prime. Thus (1) yields a prime factorisation of r not including p1 .
But (2) ⇒
r = p1 p2 p3 . . . pk − p1 q 2 q 3 . . . ql
= p1 (p2 p3 . . . pk − q2 q3 . . . ql )
i.e. a prime factorisation of r that does include p1 .
So r has two prime factorisations, and r < m, contradicting definition of m. Thus
6 ∃ m ∈ N with two (or more) prime factorisations.
To find the prime factorisation of any n ∈ N (or determine whether n is prime): test
successive primes (2, 3, 5, . . .) for divisibility into n. If divisible, divide and test the quotient
again with the same prime; otherwise, go to next prime. Stop after the largest prime p with
p2 ≤ n has been tested: if p2 > n and p is a factor, so pq = n, then q < p, so the factor q
will already have been divided out.
Example 1
Show that 137 is prime.
Division of 137 by each of 2, 3, 5, 7, 11 leaves a non-zero remainder, i.e. 137 is not divisible
by these primes. 132 = 169 > 137, so no further testing needed; thus 137 is prime.
Example 2
Find prime factors of 1288.
1288 = 644 × 2 :
644 = 322 × 2 :
322 = 161 × 2 :
factor 2
factor 2
factor 2
161 is not divisible by 2, 3 or 5.
161 = 23 × 7:
factor 7
2
23 < 7 , so no further testing needed.
Prime factors are: 23 × 7 × 23.
3
Eratosthenes’ sieve
To list all the primes up to some given n: write down all natural numbers from 2 to n;
cross out all multiples of 2, except 2 itself; first remaining number after 2 is 3, so cross out
all multiples of 3, except 3 itself. Continue similarly: after crossing out all multiples of p,
cross out multiples of first remaining number after p, as long as p2 ≤ n.
E.g. for n = 15:
2
3
64
5
66
7
68
69
6 10
11
6 12
13
6 14
6 15
[4, 6, 8, 10, 12, 14 crossed out as multiples of 2; 9, 15 crossed out as multiples of 3.] 52 > 15,
so multiples of p ≥ 5 not required.
Theorem 3.2.3
There exist infinitely many primes.
Proof
First define n! (“n factorial”) inductively:
(a) 1! := 1
(b) S(n)! := (n + 1) × n!.
So n! = n × (n − 1) × (n − 2) × . . . × 2 × 1, and its prime factorisation includes all primes
≤ n, but no primes > n. Now suppose there is a finite number of primes; then ∃ a greatest
prime, pl , and pl ! is divisible by every prime. So (pl ! + 1) has remainder 1 when divided by
every prime, i.e. is not divisible by any prime ≤ pl . Thus (pl ! + 1) is either a prime (> pl ), or
it has prime factors all > pl , contradicting definition of pl as greatest prime.
Theorem 3.2.4
Suppose x and y have no common factors, i.e. no prime appears in the prime factorisations
of both x and y. Then, for any z ∈ N:(a) If x | yz, then x | z
(b) If x | z and y | z, then xy | z
Proof of (a)
Let the prime factorisations of y and z be
y = p1 p2 . . . pn ,
z = q1 q 2 . . . qm ;
so yz = p1 p2 . . . pn q1 q2 . . . qm
(1),
which includes all the prime factors of x (since x | yz). But the prime factorisation of x does
not contain any pi [it has no common factors with y]; so all factors of x must be qj ’s in (1);
but qj ’s are factors of z. So x | z.
3.3
Congruence Arithmetic
Given d ∈ N and a, b ∈ Z, we say “a is congruent to b, modulo d” and write
a ≡ b (mod d)
4
if a and b have the same remainder on division by d.
Theorem 3.3.1
Congruence is an equivalence relation.
Proof: Exs. 3, Q. 8
Theorem 3.3.2
a ≡ b (mod d)
Proof
iff d is a divisor of (a − b).
a ≡ b (mod d) ⇒ a = q1 d + r and b = q2 d + r
⇒ a − b = (q1 d + r) − (q2 d + r)
= (q1 − q2 )d ,
for some
r, q1 , q2
i.e. d is a divisor of (a − b).
Conversely, if (a − b) = qd, divide d into a: a = q1 d + r. Then
b = a − (a − b) = q1 d + r − qd
= (q1 − q)d + r ,
i.e. same remainder (r) for b as for a.
Examples
(i) Show that 137 ≡ 89 (mod 6).
Either:- divide: 137 = 22 × 6 + 5; 89 = 14 × 6 + 5:
or: 137 − 89 = 48, 48 = 8 × 6 (no remainder).
(ii) Show that 11 ≡ −94 (mod 5).
Either:- 11 = 2 × 5 + 1; −94 = −19 × 5 + 1
or: 11 − (−94) = 105 = 21 × 5.
both have remainder 5;
Theorem 3.3.3
Suppose a ≡ b (mod d) and m ≡ n (mod d). Then:(i) (a + m) ≡ (b + n) (mod d)
(ii) (a × m) ≡ (b × n) (mod d)
Proof of (i)
(i) a − b = q1 d and m − n = q2 d
[Theorem 3.3.2]
Then
(a + m) − (b + n) =
=
=
∴ (a + m) ≡
(a − b) + (m − n)
q1 d + q 2 d
(q1 + q2 )d
(b + n) (mod d)
5
[Theorem 3.3.2].
Example
Show that (137 × 89 − 1) is divisible by 12.
Since 137 ≡ 5 (mod 12) and 89 ≡ 5 (mod 12),
137 × 89 − 1 ≡ 5 × 5 − 1 (mod 12)
= 24
≡ 0 (mod 12) ,
i.e. (137 × 89 − 1) is divisible by 12.
Theorem 3.3.4
If a ≡ b (mod d), then ak ≡ bk (mod d) for any k ∈ N.
Proof (by induction on k).
(a) For k = 1: a1 ≡ b1 (mod d) (since a1 = a and b1 = b).
(b) Given inductive hypothesis, ak ≡ bk (mod d):
aS(k) = ak × a
≡ bk × b (mod d)
= bS(k) .
[Theorem 3.3.3 (ii) & inductive hypothesis]
Example
Show that 111,111 is divisible by 7.
9 × 111, 111 = 999, 999 = 106 − 1; since 7 and 9 have no common factors, 7 | 111, 111 iff
7 | 999, 999. [Theorem 3.2.4(a)]
Now, 106 − 1 ≡
=
=
≡
=
≡
36 − 1 (mod 7)
(32 )3 − 1
93 − 1
23 − 1 (mod 7)
7
0 (mod 7),
[10 ≡ 3 (mod 7)]
[9 ≡ 2 (mod 7)]
i.e. 7 | (106 − 1), so 111, 111 is divisible by 7.
Congruence Classes
The relation, “≡ (mod d)” partitions Z into d equivalence classes, members of which
have the same remainder on division by d. We define the congruence class of r (mod d) as
rbd := {a ∈ Z; a = qd + r, q ∈ Z},
e.g.
3b5 = {. . . , −12, −7, −2, 3, 8, 13, . . .};
0b2 is the set of all even integers;
6
1b2 is the set of all odd integers.
The set of all congruence classes (mod d) is a finite number system, denoted Zd , i.e.
Zd = {0bd , 1bd , . . . , (d\
− 1)d }.
Arithmetic on Zd (with each rbd treated as a single number): m
b +n
b, m
b −n
b and m
b ×n
b
(with the subscript d omitted when d fixed) are found by determining the congruence class
to which m + n, m − n and m × n belong. Theorem 3.3.3 ensures that these operations are
well-defined on Zd .
Examples
On Z12 :(i) b
2+b
5=b
7
b
b
(ii) 7 + 9 = b
4, since 7 + 9 = 16 ≡ 4 (mod 12)
(iii) b
2−b
5=b
9, since 2 − 5 = −3 ≡ 9 (mod 12)
b
b
b
(iv) 7 × 9 = 3, since 7 × 9 = 63 ≡ 3 (mod 12)
(v) b
6×b
6=b
0, since 6 × 6 = 36 ≡ 0 (mod 12)
3.4
Bases and Divisibility Tests
Given any b ∈ N\{1} and n ∈ N, use Division Algorithm repeatedly:
n =
=
=
=
q0 b + r0
(q1 b + r1 )b + r0 = q1 b2 + r1 b + r0
...
rk bk + rk−1 bk−1 + . . . + r1 b + r0
[i.e. with qk = 0], with 0 ≤ ri < b (i = 0, 1, 2, . . . , k); here k is the least integer for which
bk+1 > n.
We write the number n in base b as rk rk−1 . . . r1 r0 ; e.g. in base ten, “2503”= 2 × 103 +
5 × 102 + 0 × 101 + 3, but in base 6, “2503”= 2 × 63 + 5 × 62 + 0 × 61 + 3 (= 615 in base
ten).
Arithmetic in other bases
Set up addition and multiplication tables, e.g. for base 5:+ 1 2 3
1 2 3 4
2 3 4 10
3 4 10 11
4 10 11 12
4
10
11
12
13
×
1
2
3
4
1 2 3 4
1 2 3 4
2 4 11 13
3 11 14 22
4 13 22 31
Notes: tabulate from 1 + 1 to (b − 1) + (b − 1), similarly for ×; all entries in base b.
7
Example
24 × 43
20
4
40 1300 310
3 110 22
1
3 0 0
+
3 1 0
+
1 1 0
+
2 2
1
2 2 4 2
[Same procedure as in base 10.]
Divisibility tests
If s ≡ t (mod d), then:(i) d | s iff d | t [divisibility by d ⇔ “ ≡ 0 (mod d)00 ]
(ii) if c is a factor of d, c | s iff c | t [s ≡ t (mod d) ⇒ s ≡ t (mod c)
Note: in this context, c | 0 ∀c ∈ N.
Applications in base b
(a) Factors of b (including b itself)
n = rk bk + . . . + r1 b + r0
= (rk bk−1 + . . . + r1 )b + r0
i.e. n ≡ r0 (mod b) .
By (i) & (ii) above, divisibility of n by b or factors of b depends on the last digit (r0 ) of the
base-b representation of n. In particular, since 0 ≤ r0 < b, b | n iff last digit r0 = 0; if c is a
factor of b, c | n iff c | r0 . E.g. in base ten, 2 | n iff the last digit of n is 0, 2, 4, 6 or 8.
(b) Factors of (b − 1) (including (b − 1) itself)
Since b ≡ 1 (mod (b − 1)),
bi ≡ 1i = 1 (mod (b − 1)) ∀ i ∈ N,
so (bi − 1) ≡ 0 (mod (b − 1)) ∀ i ∈ N
(1).
Now,
n =
=
k
X
i=0
k
X
ri bi
(writing b0 = 1)
ri {(bi − 1) + 1}
i=0
8
≡
∴n ≡
k
X
i=0
k
X
ri (0 + 1) (mod (b − 1))
[using (1) and Theorem 3.3.3]
ri .
i=0
Thus n is divisible by (b−1) or factors of (b−1) iff the sum of digits in the base-b representation
of n is divisible.
Examples
(i) 4728 in base ten:
4 + 7 + 2 + 8 = 21; 2 + 1 = 3 which is a factor of 9 = 10 − 1.
So 3 | 4728 (but 9 - 4728).
Also, 2 | 4728 (from (a) above).
Since 2 and 3 have no common factors, Theorem 3.2.4(b) ⇒ 6 is also a divisor of 4728.
(ii) 354 in base 7:
3 + 5 + 4 = 15 (base 7); 1 + 5 = 6 = b − 1 [b = 7], so 6 | 354 (base 7).
[Note: 3547 = 18610 (subscripts denote bases).]
(c) Factors of (b + 1) (including (b + 1) itself)
Since b ≡ −1 (mod (b + 1)),
bi ≡ (−1)i (mod (b + 1)).
But (−1)i = 1 for even i, and (−1)i = −1 for odd i.
Thus bi − 1 ≡ 0 (mod (b + 1)) for even i,
and bi + 1 ≡ 0 (mod (b + 1)) for odd i.
Write
X
X
n =
r i bi +
ri bi
even i
=
X
odd i
i
ri {(b − 1) + 1} +
even i
≡
X
∴n ≡
even i
ri {(bi + 1) − 1}
odd i
ri (0 + 1) +
even i
X
X
X
ri (0 − 1) (mod (b + 1))
odd i
ri −
X
ri .
odd i
Thus n is divisible by (b + 1) or factors of (b + 1) if (sum of 1st, 3rd, 5th, . . . , digits) − (sum
of 2nd, 4th, 6th, . . . , digits) is divisible.
Example
27093 in base ten:
X
ri = 2 + 0 + 3 = 5,
even i
X
ri = 7 + 9 = 16 ;
odd i
9
5 − 16 = −11, divisible by 11; so 11 | 27093.
Also, 2 + 7 + 0 + 9 + 3 = 21, 2 + 1 = 3, so 3 | 27093.
Since 11 and 3 have no common factors, Theorem 3.2.4(b) ⇒ 11 × 3 = 33 is also a divisor
of 27093.
10
MAA245 NUMBERS
4
4.1
Rational Numbers, Q
Definition of Q
The inverse of multiplication is (exact) division, denoted “÷”:
p÷n=m
iff
m × n = p.
Division is not closed on Z; if it was closed, the Division Algorithm would always give
remainder = 0.
Definition
A rational number is an equivalence class of ordered pairs (p, n) ∈ Z × Z+ under
the new equivalence relation ≡ defined by
(p, n) ≡ (q, m)
iff
p × m = q × n.
Example
(−8, 4) ≡ (−6, 3) since (−8) × 3 = (−6) × 4 .
For any k ∈ Z+ , (p, n) ≡ (kp, kn), since p × kn = kp × n.
Similarly, if p and n have a common factor d, i.e. p = sd, n = td (s ∈ Z, t ∈ Z+ ), then
(p, n) = (sd, td) ≡ (s, t) .
For every (p, n) ∈ Z × Z+ (except when p = 0), there is a unique ordered pair
(s, t) ≡ (p, n) such that s and t have no common factors. [Proof requires Fundamental Theorem of Arithmetic.] A rational number is expressed in lowest terms when
represented by a pair (s, t) with no common factors. If p = 0, (0, n) ≡ (0, 1) ∀n ∈ Z+ ,
so take (0, 1) as “lowest terms” form.
Example
Express (−30, 18) in lowest terms.
−30 = (−5) × 6; 18 = 3 × 6; so (−30, 18) ≡ (−5, 3) and −5, 3 have no common
factors.
Note: rationals are equal (=) when they are represented by equivalent (≡) ordered
pairs.
Henceforth we represent rationals by their lowest-terms form.
A rational number (p, n) is positive or negative according to whether p ∈ Z+ or
p ∈ Z− .
1
4.2
Arithmetic on Q
Define addition:(p, n) + (q, m) := (pm + qn, nm)
(1)
Define multiplication:(p, n) × (q, m) := (pq, nm)
(2)
Note that nm ∈ Z+ , pq ∈ Z and (pm + qn) ∈ Z; so + and × are closed on Q.
Also, addition (1) and multiplication (2) are well-defined on Q. Proof: Exs. 4, Q. 2.
Define QZ ⊂ Q by
QZ := {(p, n) ∈ Q : n = 1} .
Note the bijection between QZ and Z: (p, 1) ↔ p. Furthermore,
(p, 1) + (q, 1) = (p + q, 1) ,
(p, 1) × (q, 1) = (pq, 1) .
Hence hQZ , +, ×i is isomorphic to hZ, +, ×i.
Also, (p, n) × (n, 1) ≡ (p, 1), so (p, n) = (p, 1) ÷ (n, 1) [÷ is inverse operation of ×].
Hence, rationals in QZ may be represented by the corresponding integer, while other
rationals may be written in fraction notation,
(p, n) =
p
.
n
The following theorems for + and × on Q are similar to those on Z, and may be
proved using properties of Z and definitions (1) and (2).
Theorem 4.2.1
Addition and multiplication on Q are both commutative and associative, and multiplication distributes over addition.
Theorem 4.2.2
The unique additive identity in Q is (0, 1). Every (p, n) ∈ Q has a unique additive
may be written − np . Subtraction is
inverse, −(p, n) = (−p, n); so the fraction −p
n
closed on Q:
(p, n) − (q, m) = (p, n) + (−q, m)
Theorem 4.2.3
The unique multiplicative identity in Q is (1, 1).
However, the following property of multiplication on Q is not found on Z:-
2
Theorem 4.2.4
Every rational number except (0, 1) has a multiplicative inverse.
Proof
Given (p, n) ∈ Q:(i) If p ∈ Z+ :
(p, n) × (n, p) = (pn, np) ≡ (1, 1)
and
(n, p) × (p, n) = (1, 1)
similarly;
so (n, p) is a multiplicative inverse for (p, n).
In fraction notation:
p n
n p
× = × =1
n p
p n
(ii) If p ∈ Z− :
(−n, −p) ∈ Q is a multiplicative inverse for (p, n);
p −n
×
=1
n −p
(iii) If p = 0:
(0, n) × (q, m) = (0, nm) 6≡ (1, 1) for any n, n,
i.e. (0, n) has no multiplicative inverse in Q.
Note: inverses in (i) and (ii) are unique by Theorem 2.2.4.
Theorem 4.2.5
Division is closed on Q\{(0, 1)}: given (q, m) and (r, l) ∈ Q (with q 6= 0), ∃ (p, n) ∈
Q such that (p, n) × (q, m) = (r, l).
Proof
If q ∈ Z+ :Let (p, n) = (rm, lq). Then
(rm, lq) × (q, m) = ((rm)q, (lq)m)
= (r(mq), l(mq))
≡ (r, l) .
Similarly, if q ∈ Z− , (p, n) = (−rm, −lq).
Thus
(r, l) ÷ (q, m) =
(r, l) × (m, q)
if q ∈ Z+
(r, l) × (−m, −q) if q ∈ Z− ,
i.e. division by (q, m) is the same as multiplication by the multiplicative inverse.
The usual rules for arithmetic (+, −, ×, ÷) follow from the above definitions and
theorems.
3
4.3
Q as an algebraic structure
An algebraic structure hA, ∗, ∨i is a field if it is a commutative ring and also:
(a) ∨ has an identity element in A;
(b) Every a ∈ A, except the identity element for ∗, has an inverse element under
∨.
Thus hQ, +, ×i is a field, as is hZd , +, ×i when d is a prime (see Exs. 3, Q. 13 (iii)).
Also, hQ, +i is a commutative group, as is hQ\{(0, 1)}, ×i ; note that × is closed on
Q\{(0, 1)}, since (p, n) × (q, m) 6= (0, 1) if p 6= 0 and q 6= 0.
4.4
Countability of Q
Lemma 4.4.1
Let {A1 , A2 , . . .} be a countable set of sets, with each Ai (i = 1, 2, . . .) being a
countable set. Let aij be the j’th member of Ai (i, j ∈ N). Then:
(i) A1 × A2 × . . . × An is countable (for any n ∈ N);
(ii) A1 ∪ A2 ∪ . . . is countable.
Proof
Let pi be the i’th prime number. Define f : A1 × A2 × . . . × An → N by
f ((a1j1 , a2j2 , . . . , anjn )) = pj11 pj22 × . . . × pjnn
and define g : A1 ∪ A2 ∪ . . . → N by
g(aij ) = pij
(noting that A1 ∪ A2 ∪ . . . = {aij ; i, j ∈ N} ).
By the Fundamental Theorem of Arithmetic, f and g are bijections to subsets of N.
Theorem 4.4.2
Q is countably infinite.
Proof
From definition of Q, Q ⊂ Z × Z+ . Z and Z+ are countable; so Lemma 4.4.1(i) ⇒ Q
is countable.
Z ⊂ Q and Z is infinite, so Q is infinite.
4
4.5
Base-b notation for rationals
Given n ∈ Z+ and p ∈ Z+ , ∃ unique q0 , r0 ∈ Z, 0 ≤ r0 < n, such that
p = nq0 + r0
(1)
(by the Division Algorithm). q0 is the integer part of the rational number np .
, so work with −p
.]
[If p ∈ Z− , then −p ∈ Z+ and np = − −p
n
n
Working in base b, use the Division Algorithm repeatedly, multiplying the remainder
by b and dividing by n:r0 b = nq1 + r1 ,
so (1) ⇒
q1
r1
p
= q0 + +
;
n
b
bn
for k > 1:
rk−1 b = nqk + rk ,
rk−1
qk
rk
∴ k−1
= k+ k
b n
b
b n
q1
qk
rk
p
= q0 + + . . . + k + k
∴
n
b
b
b n
(by induction on k).
Set out calculation in “short division” style:
q 0 · q1 q2 q3 . . .
n )p · r0 0r1 0r2 0 . . .
[Notes: p = p · 000 . . .; “rk 0” is base-b form of (rk × b).]
We call q0 · q1 q2 q3 . . . the base-b expansion of np .
Consider two cases:(i) rk = 0 (for some k) iff
pbk = n(q0 bk + . . . + qk )
i.e. iff n | bk [by Theorem 3.2.4(a)] (given that np is in lowest terms). In this case,
there is a terminating base-b expansion of np : qk is the last non-zero quotient.
Example
Express 129
as a base-10 (i.e. decimal) expansion.
80
Since 80 = 24 × 5 and 10k = 2k × 5k , 80 is a factor of 104 ; so the decimal expansion
129
of 80 will terminate 4 places after the point.
1 ·
80 )129 ·
49
6 1 2 5
0 10 0 20 0 40 0
5
i.e.
129
80
= 1 · 6125.
(ii) If the prime factorisation of n contains factors not in the prime factorisation of
b, then n is not a factor of bk (for any k). ∃ (n − 1) possible values of remainder
rk , so some value must reappear after at most (n − 1) iterations of the Division
Algorithm. Thus the sequence of quotients will repeat with a period t ≤ n − 1, i.e.
qk+t = qk ∀ k > s, where s is number of non-repeating digits between the point and
the periodic part (0 ≤ s ≤ n − 1).
Example
Express
19
66
as a decimal expansion.
0·
66 )19 ·
2 8 7 8
0 58 0 52 0 58 0
19
i.e. 19
= 0 · 2878787 . . ., written 0 · 287 or 0 · 28̇7̇: the decimal expansion has period
66
t = 2, with s = 1 non-repeating digits.
Conversion from base-b expansion to fraction form
Given a terminating expansion q0 · q1 q2 . . . qk , i.e.
q0 +
qk
q1 q2
+ 2 + ... + k ,
b
b
b
multiply by bk to yield an integer; then
q0 · q1 q2 . . . qk =
q0 bk + q1 bk−1 + . . . + qk
bk
in fraction form, i.e.
q0 q1 q2 . . . qk
100 . . . 0
[k zeroes in denominator]
with numerator and denonimator in base b.
Example
In base 12 (with A, B as digits for base-10 numbers 10, 11 respectively),
3B · 59 =
[In base 10, this is
3B59
.
100
3 × 123 + 11 × 122 + 5 × 12 + 9
6837
=
.]
2
12
144
6
Given a repeating expansion q0 · q1 . . . qs qs+1 . . . qs+t , i.e.
q1
qs qs+1 bt−1 + . . . + qs+t
1
1
q0 + + . . . + s +
1 + t + 2t + . . . ,
b
b
bs+t
b
b
(where the terms up to qbss are the non-repeating part, while the terms in parentheses
represent the repetitions of digits from qs+1 to qs+t ):Note that
1
1
1
1
t
t
(b − 1) 1 + t + 2t + . . .
=
b + 1 + t + ... − 1 + t + ...
b
b
b
b
t
= b .
Thus
q1
qs o
(b − 1) × (q0 · q1 . . . qs qs+1 . . . qs+t ) = (b − 1) q0 + + . . . + s
b
b
qs+1 bt−1 + . . . + qs+t
+
bs
t
1
(b − 1){q0 bs + q1 bs−1 + . . . + qs }
∴ q0 · q1 . . . qs qs+1 . . . qs+t = s t
b (b − 1)
+qs+1 bt−1 + . . . + qs+t .
t
t
n
Example
Express the base-7 number 3 · 354 in fraction form in base ten.
Note: s = 0, t = 3.
73 × 3 · 3547 = 3354 · 3547
∴ (73 − 1) × 3 · 3547 = [3354 · 354 − 3 · 354]7
= 33517
33517
6667
1212
=
342 ten
202
=
57 ten
∴ 3 · 3547 =
in lowest terms.
7
Special case
If t = 1 with qs+1 = (b − 1) (note: qs+1 is the single repeating digit), then bt − 1 =
b − 1; we have
1
{(b − 1)(q0 bs + . . . + qs ) + (b − 1)}
− 1)
s
q0 b + . . . + (qs + 1)
,
=
bs
q0 · q1 . . . qs (b − 1) =
bs (b
which also has a terminating expansion.
Example
In base 10, consider 44 · 519.
Note: s = 2, t = 1, qs+1 = 9 = b − 1, qs = 1; so
44 · 519 =
4452
= 44 · 52 .
100
Summary
We have shown that a number is rational iff it has a terminating or periodic expansion in any base b; also, any terminating expansion can be written in repeating form,
e.g. in base 10,
300 = 299 · 9 .
44 · 52 = 44 · 519,
4.6
Exponentiation in Q
For a ∈ Q and n ∈ N, an is defined inductively (as in section 1.4), and Theorems
1.4.1 – 1.4.4 apply; also:
Theorem 4.6.1
n
p
pn
= n
q
q
∀n ∈ N;
p, q ∈ Q
(provable by induction).
Now define an for n ∈ Q\N by letting these theorems apply for all n ∈ Q.
Theorem 4.6.2
a0 = 1
8
Proof
an × a0 = an+0
∴ an a0 = an
∴ a0 = 1 .
[Theorem 1.4.2]
Note: 00 is undefined.
Theorem 4.6.3
a−n =
1
an
(for a 6= 0) .
Proof
an × a−n = an+(−n)
= a0 = 1 ,
1
∴ a−n = n .
a
[Theorem 1.4.2]
The inverse operation of exponentiation is called root-extraction. From the definition of inverse operation in Section 1.8,
√
(a, c ∈ Q, n ∈ N) .
a = n c iff c = an
√
If n is even, (−a)n = an , so n c is defined as the positive number satisfying c = an
(if such a ∈ Q exists) for even n 6= 0.
Theorem 4.6.4
√
√
c1/n = n c whenever n c exists in Q.
Proof
(c1/n )n = c(1/n)×n
[Theorem 1.4.4]
1
= c = c,
√
√
1/n
∴c
= nc
[set a = c1/n in definition of n c] .
√ p
∀p ∈ Z
[Theorem 1.4.4].
Corollary: cp/n = ( n c)
We have now defined ck for all k ∈ Q, c ∈ Q\{0}, whenever ck exists in Q.
Theorem 4.6.5
Exponentiation is not closed on Q\{0}.
Proof
It is sufficient to show that ∃ c, k ∈ Q\{0} for which ck 6∈ Q.
9
Suppose c ∈ N(⊂ Q) is a prime number and k = 12 . If c1/2 ∈ Q, then
p
c1/2 =
(p ∈ Z, n ∈ Z+ ) ,
n
p2
∴c =
[(c1/2 )2 = c]
2
n
∴ p2 = cn2 .
The prime factorisation of n2 must contain an even power (possibly 0) of c; so the
prime factorisation of cn2 contains an odd power of c. But the prime factorisation of
p2 contains an even power of c, contradicting p2 = cn2 (by the Fundamental Theorem
of Arithmetic). Hence c1/2 6∈ Q; we say, c1/2 is irrational. A new number system will
be required to accommodate irrational numbers.
4.7
Order in Q
Recall: (p, n) = (q, m)
Now define:
(p, n) > (q, m)
(p, n) < (q, m)
iff
iff
iff
pm = qn.
pm > qn
pm < qn
With m = n = 1, order in QZ is preserved in the isomorphism with Z.
For fractions,
q
pm
qn
p
>
iff
>
,
n
m
nm
nm
cf. “pm > qn” in definition of >; so put over common denominator and compare
numerators.
Example
To determine order between
4
4 × 16
64
=
=
;
5
5 × 16
80
4
5
and
13
:
16
13
13 × 5
65
=
=
;
16
16 × 5
80
65 > 64,
so
13
4
>
16
5
Properties of order in Z also apply in Q: trichotomy; transitivity; preservation of
inequality on addition or subtraction of a number to both sides; preservation/reversal
of inequality on multiplication of a positive/negative number by both sides. Furthermore, division of both sides by a positive/negative rational number preserves/reverses
inequality.
Example
4
13
> ;
16
5
but
13
3
13
÷ −
=− ,
16
4
12
4
3
16
÷ −
=− ,
5
4
15
10
and
−
13
16
<− .
12
15
4.8
Bounded sets in Q
The following properties of Q are in contrast to theorems 2.4.4 and 2.4.5 for Z:Theorem 4.8.1
The rationals are dense, i.e. there is a countably infinite set of rationals between
any a ∈ Q and b ∈ Q, b > a.
Proof
For every n ∈ Z+ \{1}, a < a + n1 (b − a) < b.
Theorem 4.8.2
Not every subset of Q that is bounded below [or above] has a least [or greatest]
member.
Proof (by counterexample)
The set
Q+ := {a ∈ Q; a > 0}
is bounded below: 0 is a lower bound.
Suppose Q+ has a least member, l; so l > 0 (since l ∈ Q+ ). Then 2l > 0, so
l
∈ Q+ ; but 2l < l, contradicting definition of l as least member of Q+ . So Q+ has
2
no least member.
However, Q+ has an infimum: inf Q+ = 0, since 0 is a lower bound but any l > 0
is not a lower bound (because 2l ∈ Q+ and 2l < l).
Further examples
(i) Let Q0+ = Q+ ∪ {0} = {a ∈ Q; a ≥ 0}. Then min Q0+ = 0.
(ii) Let P = {a ∈ Q; a = 1 − 21n , n ∈ N}. P is bounded above by 1, but max P does
not exist.
Proof
1
.
Suppose max P = g. Then g ∈ P , so g = 1 − 21k for some k ∈ N. Let h = 1 − 2k+1
Then h ∈ P and h > g. So g 6= max P : contradiction, ⇒ max P does not exist.
However, sup P = 1.
Proof
1 is an upper bound for P , so sup P ≤ 1 (if it exists). If c ∈ Q and c < 1, can
write c = 1 − mq with q ∈ Z+ and m ∈ Z+ .
Now, 2m > m ∀ m ∈ Z+ and q ≥ 1, so 1 − 21m > 1 − mq . But 1 − 21m ∈ P ; so c is
not an upper bound for P , for any c < 1. Hence sup P = 1.
Theorem 4.8.3
Not every subset of Q that is bounded below [or above] has an infimum [or supremum] in Q.
11
Proof
We show that
C := {c ∈ Q; c2 < 3} ⊂ Q ,
which is bounded above (e.g. by 2), has no supremum in Q.
[Note: Theorem 1.6.4 ⇒6 ∃ max C ∈ Q if 6 ∃ sup C ∈ Q.]
Need to show that
UC := {q ∈ Q; q ≥ c ∀c ∈ C}
has no least member.
Note first that, if a > 0, then a > b if a2 > b2 .
Suppose r is the least member of UC . Clearly r > 0. Let
s=
3r + 3
r+3
(so s > 0). Then
3r + 3
−r
r+3
3 − r2
=
.
r+3
s−r =
(1)
Also
(3r + 3)2
−3
(r + 3)2
6(r2 − 3)
=
.
(r + 3)2
s2 − 3 =
(2)
Three possibilities:(a) If r2 < 3 :
s2 < 3 from (2)
[(r + 3)2 > 0, r2 − 3 < 0]
and s > r from (1);
so ∃s ∈ C with s > r, i.e. r 6∈ UC .
(b) If r2 > 3 :
s2 > 3 from (2), so s ∈ UC
[s2 > 3 > c2 ∀ c ∈ C, so s > c ∀ c ∈ C].
But s < r from (1), contradicting r being the least member of UC .
(c) If r2 = 3, r 6∈ Q (see Section 4.6), so r 6∈ UC .
Similarly, C is bounded below by −2, but has no infimum.
Summary
Q is incomplete in two senses:(a) Algebraically: equations like q 2 = 3 have no solution in Q.
(b) Order: there exist bounded subsets of Q with no supremum or infimum in Q.
12
MAA245 NUMBERS
5
5.1
Real Numbers, R
Completeness
We seek to define a new number system,
R, satisfying the following:-
The Axiom of Completeness
“Every subset of R that is bounded above [below] has a supremum [infimum] in
R.”
Furthermore, R must have all the properties of Q. Specifically:Binary operations + and × must be defined such that hR, +, ×i is a field.
There must be a subset RQ ⊂ R such that hRQ , +, ×i is isomorphic to hQ, +, ×i.
Order relations > and < on R must be defined to have the same properties as on
Q. These relations are also needed to define sup and inf on R, so that the Axiom of
Completeness can be checked.
R from Q
Given a set α ⊂ Q, let α := Q\α. α is a Dedekind cut in the rationals (or simply a
5.2
Construction of
“cut”) if it has the following 3 properties:(a) α and α are both non-empty;
(b) If r ∈ α and s ∈ α, then r < s;
(c) α has no greatest member.
Note: from (b) and (c), α = Uα , the set of all upper bounds of α.
Examples
(i) The set A := {a ∈ Q; a ≤ 3} is not a cut: it has a greatest member.
(ii) The set B := {b ∈ Q; b < 3} is a cut.
Proof
(a) 0 ∈ B, 4 ∈ B, so B and B are non-empty.
(b) B = {q ∈ Q; q ≥ 3}; so ∀b ∈ B and ∀q ∈ B, b < 3 ≤ q, .. . b < q.
(c) B has no greatest member (cf. Exs. in Section 4.8).
(iii) The set C := {c ∈ Q; c2 < 3} is not a cut; 0 ∈ C and −2 ∈ C, and 0 > −2
contrary to (b).
However, let D := C ∪ Q− (where Q− := {q ∈ Q; q < 0}). Then D is a cut.
Proof
(a) 0 ∈ D, 2 ∈ D, so D and D are non-empty.
(b) If q ∈ D, then q 2 ≥ 3 and q ≥ 0. Thus q > d ∀ d ∈ Q− and q 2 ≥ 3 > c2 ∀ c ∈ C,
.. . q > c.
1
(c) C has no greatest member (see Section 4.8), and all members of
than 0 ∈ C.
Q− are less
Interpretation of cuts: we partition Q into a “left set” α and a “right set” α (where
rational numbers are plotted on a line, with s to the right of r if s > r). If sup α
exists in Q, then the partition point is at sup α, and α is a rational cut. E.g. in ex.
(ii) above, B is a rational cut with sup B = 3. If α has no supremum in Q (as in ex.
(iii) above), then α is an irrational cut; 6 ∃q ∈ Q at the partition point.
Definition
A real number is a Dedekind cut in the rationals.
Thus R has a subset RQ , the set of rational cuts, with a bijection between members
of RQ and Q: ∀ c ∈ Q, sup αc = c where αc ∈ RQ is defined as αc = {q ∈ Q; q < c}.
By analogy, we can think of irrational real numbers as suprema of irrational cuts;
so every partition point on the rational line is located at a (rational or irrational) real
number. In this sense, real numbers form a continuum (i.e. no gaps) on the number
line.
Roots as real numbers
√
Given c ∈ Q and
n ∈ N: if n c exists in
√
Otherwise, define n c as the irrational cut
{a ∈ Q; an < c}
{a ∈ Q; an < c} ∪ Q−
so
√
n
c ∈ R ∀c ∈ Q
√
n
c ∈ R ∀ c ∈ Q+
Q,
then
√
n
c = sup{a ∈
Q; an
< c}.
if n is odd
if n is even and c > 0 ;
if n is odd,
if n is even .
[{a ∈ Q; an < c} is empty if c < 0 and n is even.]
Positive and negative real numbers
The zero cut is α0 := {q ∈ Q; q < 0} .
For any cut α 6= α0 :
α ∈ R+
α ∈ R−
5.3
if
if
0∈α
0 6∈ α
Arithmetic in
(positive real number);
(negative real number).
R
For α, β ∈ R, the sum α + β and the product α × β may be defined as cuts, such that
hR, +, ×i is a field, with hRQ , +, ×i isomorphic to hQ, +, ×i.
2
5.4
Order in
R
Define < and > for cuts:
α < β if α ⊂ β
α > β if β ⊂ α
Theorem 5.4.1
Trichotomy applies in R: either α = β or α < β or α > β
Proof Suppose α 6= β. Then either α ⊂ β (i.e. α < β) or ∃ x ∈ α such that x 6∈ β
(from definition of ⊂). In the latter case, x ∈ β, so x > y ∀ y ∈ β [definition of cut,
(b)]; but if y < x and x ∈ α (a cut), then y ∈ α; so y ∈ β only if y ∈ α, i.e. β ⊂ α
(i.e. α > β).
Transitivity then follows from properties of ⊂ in Set Theory. Properties of arithmetic operations (+, −, ×, ÷) with respect to inequalities are the same in R as in Q.
Order in Q is preserved in RQ . The definitions of R+ and R− (Section 5.2) correspond
to α > α0 and α < α0 respectively.
Theorem 5.4.2
Given a subset of
R,
A = {α1 , α2 , α3 , . . .}.
bounded above, let
γ := α1 ∪ α2 ∪ α3 ∪ . . . ,
i.e. γ contains all rationals in the cuts α1 , α2 , α3 , . . . . Then either A has a greatest
member (and hence a supremum) in R (i.e. one of the α’s), or γ is a cut (i.e. γ ∈ R)
and γ = sup A.
Proof (of γ = sup A)
By definition of ∪, αi ⊆ γ (i = 1, 2, 3, . . .), so γ ≥ αi ∀ αi [definition of > in R);
so γ is an upper bound for A.
For any δ < γ (δ ∈ R), ∃ r ∈ Q with r ∈ γ but r 6∈ δ. If r ∈ γ, then r ∈ αj for
some j; so δ < αj [by trichotomy] so δ is not an upper bound of A.
We have shown that defining real numbers as cuts satisfies the Axiom of Completeness. In particular, sup α = α, confirming identification of cuts with suprema (or
partition points).
Theorem 5.4.3
The Archimedean Property for R: for every α ∈ R+ and β ∈ R, ∃ n ∈ N such that
nα > β.
Proof Suppose ∃ β ∈ R such that nα ≤ β ∀ n ∈ N, so β is an upper bound for the set
B := {γ ∈ R; γ = nα, n ∈ N} .
Then B has a supremum [axiom of completeness]; let sup B = σ. Then σ − α < σ
[since α > 0], so (σ − α) is not an upper bound for B, i.e. ∃ m ∈ N with mα > σ − α;
3
thus σ < (m + 1)α, but (m + 1) ∈ N, so (m + 1)α ∈ B, contradicting σ being an
upper bound. Hence 6 ∃ β ∈ R with β ≥ nα ∀ n ∈ N.
5.5
Base-b Expansion and Uncountability of
R
Given α ∈ R+ and any base b, ∃ qk ∈ N ∪ {0} such that
qk ≤ αbk < qk + 1
for each k ∈ N ∪ {0} [Archimedean property]. Then
qk
bk
gives the base-b expansion of α, to k digits after the point.
[For α ∈ R− , work with −α ∈ R+ and finally include minus sign.]
Example √
Expand 3: (a) to 2 places in base 10; (b) to 3 places in base 7.
Note that qk ≤ αbk < qk + 1 iff qk2 ≤ α2 b2k < (qk + 1)2 (for α > 0).
(a) With b = 10 and k = 2:
√
1732 = 29929 < 30000 (= ( 3)2 × 102×2 ) < 30276 = 1742 ,
√
i.e. q2 = 173, so 3 = 1 · 73 . . .
(b) With b = 7 and k = 3 (but calculating in base 10):
√
( 3)2 × 72×3 = 352947;
5942 = 352836 < 352947 < 354025 = 5952 ,
so q3√= 594 (base 10) = 1 × 73 + 5 × 72 + 0 × 7 + 6
i.e. 3 = 1 · 506 . . . in base 7.
Notes:(i) Practical methods for finding qk only exist for special classes of real numbers.
(ii) Base-b expansions of irrational numbers do not terminate or repeat.
(iii) α = sup{qk /bk ; k ∈ N} with qk determined as above. Conversely, any base-b
expansion is a set of rational numbers qk /bk whose supremum is a real number [axiom
of completeness].
Theorem 5.5.1
R is uncountable.
Proof (by Cantor’s “diagonal argument”)
Suppose R is countable, so every real number can be denoted αn for some n ∈ N.
Choose base b and write each αn in base-b expansion:
α1 = q10 · r11 r12 r13 . . .
α2 = q20 · r21 r22 r23 . . . ,
4
i.e. rij is j’th digit after point in expansion of αi . Now construct a base-b expansion
0 · c1 c2 c3 . . .
where c1 6= r11 , c2 6= r22 , . . ., ck 6= rkk , . . . . Then 0 · c1 c2 c3 . . . is not in the list of αn ,
but it does represent a real number [note (iii) above]. Hence R is not countable.
5
MAA245 NUMBERS
EXAMPLES 1: NATURAL NUMBERS
When required to prove a theorem from the notes, you are allowed to use any definition,
lemma or theorem appearing in the notes before the required theorem.
1. Prove that addition on N is associative, i.e. that
(m + n) + p = m + (n + p)
∀ m, n, p ∈ N .
[Hint: Use induction on p.]
2. By applying the commutative and associative properties of addition several times,
show that
(m + n) + (p + q) = (n + (q + p)) + m
∀ m, n, p, q ∈ N .
3. Prove the Cancellation Law for addition, i.e. that
if
m + n = m + p,
then
n=p
(∀ m, n, p ∈ N) .
[Hint: You will need the axiom that S(m) = S(n) only if m = n.]
4. Using the definition of multiplication and the properties of addition:n×1=1×n
(i) Prove that
(ii) Hence prove that
∀n ∈ N.
(m + 1)n = mn + n
∀ m, n ∈ N .
[Hint: Use induction on n.]
(iii) Using the lemmas proved in (i) and (ii) above, prove that multiplication is
commutative on N, i.e. that
n×m=m×n
∀ m, n ∈ N .
5. Prove that
1m = 1
6. Prove that
nm × np = n(m+p)
∀m ∈ N.
∀m, n, p ∈ N.
[Hint: Use induction on p.]
7. Prove that
n>1
∀ n ∈ N\{1} ,
i.e. that every natural number, apart from 1 itself, is greater than 1.
[Hint: you may assume the theorem that S(n) > n.]
8. Using the definition of the symbol <, prove the transitivity property, i.e. that
if
m<n
and
n < p,
then
m<p
(∀ m, n, p ∈ N) .
9. Using the definition of the symbol > and the cancellation law, prove that
if
m + p > n + p,
then
m > n.
1
10. Complete the proof of the theorem that
6 ∃ k ∈ N such that n < k < S(n) .
[It was shown in lecture notes that the theorem is true for n = 1: to complete
the induction, you need to show that if the theorem is true for some n, then
6 ∃ l ∈ N such that S(n) < l < S(S(n)) .]
11. Prove that
n < p ⇔ nm < pm
∀m, n, p ∈ N.
[Hint: Once you have proved the implication in one direction, you can use trichotomy to prove the converse.]
12. For each of the following, either: state that it is empty, or state that it is finite
and find its cardinality, or prove that it is countably infinite by finding a bijection to
N. [Note that since a bijection always has an inverse bijection, you can either find
a bijection from the given set to N, or from N to the given set. Write down your
bijection as a formula, just as you would for any function.] Rigorous proofs are not
required.
(i)
{n ∈ N : n + n = 8}
(ii)
{n ∈ N : n + n = 7}
(iii)
{n ∈ N : n + m = 8
for some
m ∈ N}
(iv)
{n ∈ N : m + 8 = n
for some
m ∈ N}
(v)
{n ∈ N : m + m = n
for some
m ∈ N}
13. We can define 2N, a subset of N, as follows:2N := {m ∈ N : m = 2n, n ∈ N} .
Thus it is the set of all natural numbers which can be expressed as 2 times a natural
number.
(i) What is the usual name given to the set 2N, and which of the sets in Q.13 is it
identical to?
(ii) Prove that 2N is closed under each of the operations of addition, multiplication
and exponentiation. You may use any of the definitions, lemmas and theorems in
the lecture notes relating to these operations on natural numbers.
(iii) Prove that 1 ∈
/ 2N.
(iv) Is the set 2N ∪ {1} also closed under each of the operations of addition,
multiplication and exponentiation? Give proofs to justify your answers.
14. Subtraction on N is defined as the inverse operation of addition: if q > p, then
q−p = m ⇔ m+p = q.
Using only this definition and the distributive law, prove that
n(q − p) = nq − np .
[Hint: it may be useful to let q − p = m.]
2
MAA245 NUMBERS
EXAMPLES 2: INTEGERS
When required to prove a theorem from the notes, you are allowed to use any definition,
lemma or theorem appearing in the notes before the required theorem.
1. Show that the relation “≡” , defined by
(p, n) ≡ (q, m)
if and only if
p + m = q + n,
is reflexive, symmetric and transitive.
2. Show that addition is a well-defined operation on the integers. To do this, write
down the results of the additions,
(p1 , n1 ) + (q1 , m1 )
and
(p2 , n2 ) + (q2 , m2 ) ;
and
(q1 , m1 ) ≡ (q2 , m2 )
then suppose
(p1 , n1 ) ≡ (p2 , n2 )
and use the equation defining equivalence to show that the results of the two
additions are in the same equivalence class.
3. Show that hZ+ , +, ×i is isomorphic to hN, +, ×i . To do this, you need to verify
that
(k + s, k) + (k + r, k) ≡ (k + (s + r), k)
and
(k + s, k) × (k + r, k) ≡ (k + sr, k) .
Z is associative.
Prove that multiplication on Z is commutative.
4. Prove that addition on
5.
6. Show that the unique multiplicative identity element in Z is the integer 1 (defined as {(k + 1, k), k ∈ N} ). To do this, suppose ∃ (q, m) ∈ N × N which
leaves any other ordered-pair integer unchanged when multiplying it; then prove
that (q, m) ≡ (k + 1, k).
7. Use the arithmetic of ordered-pair integers to verify the following rules (noting
that in some instances the minus sign indicates subtraction, whereas in others it
indicates an additive inverse):(i) − (a − b) = (−a) + b
(ii) (−a) × (−b) = a × b
1
(iii) a × 0 = 0
8. Using the definition of the symbol < for ordered-pair integers, show that the following properties apply:(i) Transitivity: If
(p, n) < (q, m)
and
(q, m) < (r, l),
then
(p, n) < (r, l) .
(ii) Preservation of order on subtraction of a number from both sides:
If
(p, n) < (q, m),
then
(p, n) − (r, l) < (q, m) − (r, l) .
9. Writing a = (p, n), b = (q, m), c = (r, l) , and noting that c > 0 if r > l and
c < 0 if r < l, use ordered-pair arithmetic to show that:
If
If
a<b
a<b
and
and
c > 0,
c < 0,
then
then
ac < bc;
ac > bc.
10. Using the results of Q.9, Q. 7 (iii) and Trichotomy, show that:ab > 0
ab < 0
ab = 0
iff
iff
iff
(a > 0 and b > 0) or
(a > 0 and b < 0) or
a = 0 or b = 0 .
(a < 0 and b < 0);
(a < 0 and b > 0) ;
11. (a) Determine whether each of the following subsets of Z are closed under each of
the operations of addition, subtraction and multiplication:(i) {−1, 0, 1}
(ii) Z−
(iii) Z\{0}
(b) What kind of algebraic structure is hZ− , +i? Explain why there is an isomorphism between hZ− , +i and hN, +i, but not between hZ− , ×i and hN, ×i.
(c) Let S = {−1, 0, 1} and T = {−1, 1}. What kinds of algebraic structure are
hS, ×i and hT, ×i?
12. Prove that every non-empty subset of
member.
Z
that is bounded above has a greatest
13. Refer to Examples 1, Q.13. Do any of the answers change if the sets are considered as subsets of Z (i.e. if the definitions are changed from {n ∈ N : . . .} to
{n ∈ Z : . . .})?
2
MAA245 NUMBERS
EXAMPLES 3: FOUNDATIONS OF NUMBER THEORY
1. Show that if d is a divisor of both m and n (for some m, n ∈ Z), then d is also a
divisor of am + bn, ∀a, b ∈ Z.
2. Investigate the distribution of primes as follows.
(a) Use Eratosthenes’ sieve to list all the prime numbers up to 130.
(b) We define the k’th decade Dk by
Dk := {n ∈ N; 10(k − 1) < n ≤ 10k},
e.g. D3 = {21, 22, . . . , 29, 30} .
(i) How many primes are there in each of the decades, D3 to D9 . Do you see
a pattern emerging? Explain this pattern. Now write down the number of
primes in each of the decades, D10 , D11 , D12 and D13 , and comment on the
results.
(ii) How many primes are there in each of the decades, D2 , D11 and D20 ? Explain
your observations. What is the obvious next member of this sequence of
decades? Find how many primes there are in this next decade in the sequence.
(iii) Give up trying to find regular patterns in the distribution of primes. All such
attempts are doomed to failure.
3. To determine whether 10001 is prime, what is the largest prime which you might
need to test for divisibility into 10001? [You are not asked to actually determine
whether 10001 is prime.]
4. Prove that no polynomial formula will yield only primes, as follows.
(i) Consider first linear formulae,
P (n) = a0 + a1 n
(a0 , a1 ∈ Z)
This is supposed to yield only primes, i.e. P (n) is a prime for every n ∈ N. In
particular, P (1) is prime. Let p1 = P (1) and then show that p1 is a factor of
P (1 + p1 ) (so that the formula fails to yield a prime when n = 1 + p1 ).
1
(ii) Now extend the proof to general polynomials,
P (n) = a0 + a1 n + a2 n2 + . . . + ak nk
(k ∈ N; ai ∈ Z(i = 0, 1, 2, . . . , k)) .
You should again let p1 = P (1) and show that p1 is a factor of P (1 + p1 ). If you
cannot immediately do the proof for general polynomials, try quadratic functions
first.
5. In the proof that there are infinitely many primes (Theorem 3.2.3 in the lecture
notes), we used the number (pl ! + 1), which can only have prime factors larger than
the supposed last prime pl . Another obvious choice of a number which only has
prime factors larger than pl is (pl ! − 1). Can you think of a number much smaller
than pl ! which also has the same property?
6. A prime gap is the difference between two successive prime numbers. For instance,
the largest prime gap you found in Q.2(a) was 14, between the successive primes
113 and 127.
Prove that there exist arbitrarily large prime gaps, i.e. that for any given g ∈
is always possible to find a prime gap larger than g.
N it
[Hint: consider all the numbers between n! and (n! + n).]
7. (i) Consider the proposition, “If x | yz, then x | z”. If x = 6, find values of y and z
for which the proposition is false.
(ii) Prove that the proposition is always true if we attach the condition that x
and y do not have any common factors. [Note that this means that the prime
factorisations of x and y can be written x = p1 p2 . . . pn and y = q1 q2 . . . qm where
none of the pi are equal to any of the qj . No conditions are attached to z, so its
prime factorisation may include some of the pi and some of the qj , as well as some
factors not in the prime factorisations of x or y.]
8. Show that congruence is an equivalence relation.
9. Show that if a ≡ b (mod d) and m ≡ n (mod d), then
a × m ≡ b × n (mod d)
10. (i) Show that a congruence class r̂d is closed under addition iff r = 0.
(ii) Show that a congruence class r̂d is closed under multiplication if r = 0 or r = 1.
Show that for r > 1 a congruence class r̂d is closed under multiplication iff (r2 − r)
is divisible by d. For d = 6, give examples of congruence classes r̂6 with r > 1
which are (a) closed, and (b) not closed under multiplication.
2
(iii) Show that if n2 is even, then n is even, and that if n2 is odd, then n is odd.
11. Show that (79 × 94 + 53 × 107) is divisible by 7 (without doing any multiplication
of numbers larger than 7).
12. Show that 9, 999, 999, 999, 999 (that’s 13 nines) is divisible by 53.
[Hints: Start by writing 1013 as (102 )6 × 10. At each stage of the process, you are
looking to simplify the calculation by finding congruences to small (positive or negative) numbers (mod 53): for instance, since 106 = 2 × 53, 100 ≡ −6 (mod 53).]
13. (i) The commutative and associative properties of addition on Z carry over to the
finite system of congruence classes Zd , as does the existence of an additive identity
element, which is 0̂d . Find the additive inverse of every element of Z7 . Write down
a formula for the additive inverses of the elements of Zd for general d. What kind
of algebraic structure is hZd , +i?
(ii) The commutative and associative properties of multiplication on Z carry over
to Zd , as does the distributive law. What kind of algebraic structure is hZd , +, ×i?
(iii) The element 1̂d is a multiplicative identity element for Zd . For each element
of Z6 and for each element of Z7 , find its multiplicative inverse if such an inverse
exists. Generalise your results: prove that every element of Zd except 0̂d has a
multiplicative inverse iff d is prime, but that there will be elements other than 0̂d
which do not have a multiplicative inverse iff d is composite.
14. Using only the 7-times-table and the table below, and without a calculator (assuming that you can subtract without a calculator), (i) convert the base-10 number
9376 to base 7, and (ii) convert the base-10 number 14660 to base 49.
2 × 72 = 98
3 × 72 = 147 4 × 72 = 196 5 × 72 = 245
6 × 72 = 294
72 = 49
73 = 343 2 × 73 = 686 3 × 73 = 1029 4 × 73 = 1372 5 × 73 = 1715 6 × 73 = 2058
74 = 2401 2 × 74 = 4802 3 × 74 = 7203 4 × 74 = 9604 5 × 74 = 12005 6 × 74 = 14406
15. Write out addition and multiplication tables for base 6 and do the following calculations in base 6 (all numbers being written in base 6).
(i) 544 + 53
(ii) 544 × 53
[You can check your answers by converting to base 10.]
3
16. When using bases larger than 10, it is customary to use capital letters for the extra
digits needed beyond 9. For instance, in base 16 we use A, B, C, D, E, F as digits
to represent the numbers written in base 10 as 10, 11, 12, 13, 14, 15, respectively.
Example: the base-16 number B5E converts to base 10 as 11 × 162 + 5 × 16 + 14 =
2910.
Without doing any division or converting to base 10, determine whether the following base-16 numbers are divisible by each of 5, 8, C and 11 (where these base-16
divisors are the numbers that would be written 5, 8, 12 and 17 in base 10):(i) 57F8
(ii) ACD2D5
(iii) 91FFEC
(iv) 34EAE
For your convenience, a base-16 addition table is provided below.
+ 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 A
A B
B C
C D
D E
E F
F 10
2
3
4
5
6
7
8
9
A
B
C
D
E
F
10
11
3
4
5
6
7
8
9
A
B
C
D
E
F
10
11
12
4
5
6
7
8
9
A
B
C
D
E
F
10
11
12
13
5
6
7
8
9
A
B
C
D
E
F
10
11
12
13
14
6
7
8
9
A
B
C
D
E
F
10
11
12
13
14
15
7
8
9
A
B
C
D
E
F
10
11
12
13
14
15
16
8
9
A
B
C
D
E
F
10
11
12
13
14
15
16
17
4
9
A
B
C
D
E
F
10
11
12
13
14
15
16
17
18
A
B
C
D
E
F
10
11
12
13
14
15
16
17
18
19
B
C
D
E
F
10
11
12
13
14
15
16
17
18
19
1A
C
D
E
F
10
11
12
13
14
15
16
17
18
19
1A
1B
D
E
F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
E
F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
MAA245 NUMBERS
EXAMPLES 4: RATIONAL NUMBERS
When required to prove a theorem from the notes, you are allowed to use any definition,
lemma or theorem appearing in the notes before the required theorem.
1. Show that the relation “≡” , defined by
(p, n) ≡ (q, m)
if and only if
p × m = q × n,
is reflexive, symmetric and transitive.
2. Show that addition is a well-defined operation on
Q.
3. Prove that multiplication distributes over addition on
Q.
4. Prove that every (p, n) ∈ Q has a unique additive inverse, (−p, n).
5. A number system, known as the algebraic numbers
equations of the form
a0 + a1 x + a2 x2 + . . . + an xn = 0,
A, consists of all solutions x of
n ∈ N;
a0 , a1 , a2 , . . . , an ∈ Z ,
(i.e. all roots of polynomials of any degree with integer coefficients). You do not
need to know how to construct this number system; you only need to know that
there are at most n solutions of any such equation.
Using the lemma that “A countable union of countable sets is countable”, show
that A is countable.
6. Find the base-10 (decimal) expansion of 7917/2500.
7. Find the base-10 (decimal) expansion of 31/72. Find a base b < 10 in which this
fraction would have a terminating base-b expansion, and find this expansion. List
all the other bases less than 72 in which it would have a terminating expansion; for
each such base, state the number of places after the point in the expansion.
8. Find the base-10 (decimal) expansions of 1/7, 2/7, 3/7, 4/7, 5/7 and 6/7. Describe
and explain any symmetries you observe within and between the 6 periodic decimal
expansions.
9. Find the base-7 expansions of 1/5, 2/5, 3/5 and 4/5. Do these expansions display
similar symmetries to those in the previous question?
[Hint: first find all multiples of 5 up to 6 × 5 in base 7.]
1
10. For each of the following numbers written as a decimal expansion, find the fraction
form in lowest terms:(i) 37.575
(ii) 2.837.
11. Prove that the following numbers are not members of
1/2
1/2
(i) 6
(ii) 27
(iii) c
1/3
Q:-
where c is any prime number
(iv) 721/3
12. For each of the following pairs of rational numbers, determine which is greater:(i)
22 355
,
7 113
(ii) −
38
17
, −
27
12
13. Prove from the definition of order on
then
Q that if (p, n) < (q, m) and (r, l) < (0, 1),
(p, n) ÷ (r, l) > (q, m) ÷ (r, l) .
14. Find a number c ∈ Q between 22/7 and 355/113. Then find two more rational
numbers between c and 355/113.
15. Give two examples of subsets of Q that are bounded above, one of which has a
greatest member and the other does not have a greatest member.
16. For each of the following subsets of Q, find (a) an upper bound, (b) the greatest
member, (c) a lower bound, (d) the least member, if these bounds and members
exist:1
a ∈ Q; a = n + , n ∈ N
n
(i)
(ii)
1
a ∈ Q; a = 1 + , n ∈ N, n < 1000
n
(iii)
1
a ∈ Q; a = 1 + , n ∈ N
n
17. Write down the supremum and infimum of the set
A := {a ∈ Q; a2 < 9}
Then prove that these numbers are the supremum and infimum of A.
18. Prove that the set
B := {b ∈ Q; b2 < 7}
has neither a supremum nor an infimum in Q, even though it is bounded both above
and below.
2
MAA245 NUMBERS
EXAMPLES 5: REAL NUMBERS
When required to prove a theorem from the notes, you are allowed to use any definition,
lemma or theorem appearing in the notes before the required theorem.
1. Prove that if α is a cut, then α is the set of all upper bounds of α.
2. For each of the following subsets of Q, determine whether the set is a cut; if so,
prove that it has the three properties that define a cut; if not, explain why it is not
a cut:(i) {a ∈ Q; a < −1}
(ii) {a ∈ Q; a2 < 1}
(iv) {a ∈ Q; a < 7} ∪ Q−
2
(iii) {a ∈ Q; a2 < 1} ∪ Q−
(
(v)
1
a ∈ Q; a = , q ∈ Q−
q
)
(vi) {a ∈ Q; a = −q 2 , q ∈ Q}
3. Write down the definition (in the form of a cut) of each of the following irrational
numbers:(i)
√
5
18
(ii)
√
6
7
(iii) −
√
6
7
(iv)
√
7
−6
4. Let αr and αs be rational cuts corresponding to the rational numbers r and s. Using
the definition of < in R, prove that αr < αs iff r < s.
√
5. Find the expansions of 3:(a) to 2 places in base 11;
(b) to 5 places in base 3.
Is the base-3 expansion in (b) repeating? Check by finding the 6th place.
√
6. Find the expansions of 3 11:(a) to 2 places in base 10;
(b) to 3 places in base 7.
[Note: part (a) is trivial, since you will need a calculator (working in base 10) in
any case.]
7. Suppose that an attempt is made to use Cantor’s proof to show that the rationals
are uncountable, i.e. by listing the base-b expansions of all the rationals and trying
to construct a base-b expansion of a rational number that is not in the list. At what
point does the attempted proof break down?
1