Download 5. Integers

THE REAL NUMBER SYSTEM
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5. The Integers (10/6/10)
In this section, we axiomatically construct the set of integers, Z, starting with the elements
of N.
If we want to find x so that x + 4 = 3 using what we have so far, we are out of luck,
since we have no idea what negative numbers are. To rigorously define these, we will use the
time-honored method of equivalence classes of an equivalence relation, in this case a relation
on the set N × N.
Definition 5.1. Let (a, b), (c, d) ∈ N × N. Define (a, b) ∼ (c, d) provided a + d = b + c.
Theorem 5.1. The relation ∼ in Definition 5.1 is an equivalence relation on N × N.
For example, (5, 3) ∼ (3, 1) and (5, 11) ∼ (1, 7). (You should be thinking a − b when you
see (a, b).) Notice that in the first example, the difference between each first component
and second component is +2, while in the second example, the difference between each first
component and second component is −6. This is how we construct the integers.
Definition 5.2. The set of equivalence classes in Definition 5.1 is the set of integers,
denoted Z.
Consider the equivalence class
[(5, 3)] = {(2, 0), (3, 1), (4, 2), (5, 3), (6, 4), . . . }.
We identify this with the integer 2. Similarly, we identify the equivalence class [(5, 11)] with
the integer −6.
Now that we have a formal definition of Z, we want to be able to verify that the usual
operations of addition and multiplication hold. The problem is that we are now trying to
add/multiply entire sets (the equivalence classes), and we want to somehow use the numbers
in the ordered pairs in those sets to do this. Since each set contains infinitely many ordered
pairs, we must make a choice of some sort, and if our binary operations are to be “welldefined,” we have to get the same result no matter which ordered pair we choose.
And what should addition be anyway? Suppose we have the following identifications of
integers as ordered pairs inside their equivalence classes: x ↔ (a, b) and y ↔ (c, d). Then
when we compute x + y, we get
(a, b) + (c, d) ! (a − b) + (c − d)
= (a + c) − (b + d)
! (a + c, b + d).
(Notice the use of the undefined symbol !!)
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VAN WYK’S MATH 315
The following theorem shows that we get the same result no matter which ordered pairs we choose from each equivalence class.
Theorem 5.2. Assume (a1 , b1 ) ∼ (a2 , b2 ) and (c1 , d1 ) ∼ (c2 , d2 ), where each component is a
natural number. Then (a1 + c1 , b1 + d1 ) ∼ (a2 + c2 , b2 + d2 ).
You will see something similar to Theorem 5.2 every time you try to extend a binary
operation to a set of equivalence classes of elements that already have a binary operation ⊕
defined on them. Basically, you are trying to show the operation
[x] ⊕ [y] = [x ⊕ y]
makes sense, i.e., is independent of the choice of representative from [x] and [y]. Notice that
on left side of this equation, we are performing the operation ⊕ on two sets, while on the
right side of the equation, we are performing the operation ⊕ on two elements of those sets.
Anyway, we can now define addition of integers as [(a, b)] + [(c, d)] = [(a + c, b + d)]
(where each element x ∈ Z corresponds to the equivalence class [(a, b)] of elements of N × N
such that x = a−b). We can use the usual notation for integers if we want, but since addition
is defined by choosing representatives and adding as in Theorem 5.2, i.e. (a, b) + (c, d) =
(a + c, b + d), we’ll have to use that to prove various properties of addition.
By Theorem 5.2, we can choose any representative of any equivalence class and we’ll get
the same answer, which makes the following theorem very straightforward. Note: the integer
0 is the equivalence class [(1, 1)].
Theorem 5.3.
(1) For each a, b ∈ Z, a + b = b + a.
(2) For each a, b, c ∈ Z, a + (b + c) = (a + b) + c.
(3) For each a ∈ Z, a + 0 = a.
(4) For each a ∈ Z, there exists b ∈ Z such that a + b = 0.
Theorem 5.3 (1) and Theorem 5.3 (2) show is the commutative and associative laws of
addition extend to Z. Theorem 5.3 (3) shows Z contains an additive identity, and Theorem 5.3 (4) shows that every element of Z has an additive inverse.
Proof. We’ll prove Theorem 5.3 (1) and Theorem 5.3 (3) , and leave the other parts for the
exercises.
(1) Let (m, n) ∈ a and let (k, l) ∈ b. (That is, a = [(m, n)] and b = [(k, l)].) Then
a + b = [(m + k, n + l)] Definition of addition
= [(k + m, l + n)] Commutativity of addition of naturals
= b+a
Definition of addition.
THE REAL NUMBER SYSTEM
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(3) Let (m, n) ∈ a and let (1, 1) ∈ 0. Then
a + 0 = [(m + 1, n + 1)] Definition of addition
= [(m, n)]
(m + 1, n + 1) ∼ (m, n)
= a
(m, n) ∈ a.
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The element b in Theorem 5.3 (4) is unique, for if a + b = a + b = 0, then b = 0 + b =
(b + a) + b = b + (a + b) = b + 0 = b. Following convention, we’ll denote the additive inverse
of the integer a by −a.
So, how do we multiply integers? Suppose we have the following identifications of integers
as ordered pairs inside their equivalence classes: x ↔ (a, b) and y ↔ (c, d). Then when we
compute x · y, we get
(a, b) · (c, d) ! (a − b) · (c − d)
= ac − ad − bc + bd
= (ac + bd) − (ad + bc)
! (ac + bd, ad + bc).
As with addition, we must verify that multiplying by choosing representatives is welldefined; that is the point of Theorem 5.4.
Theorem 5.4. Assume (a1 , b1 ) ∼ (a2 , b2 ) and (c1 , d1 ) ∼ (c2 , d2 ), where each component is a
natural number. Then (a1 c1 + b1 d1 , a1 d1 + b1 c1 ) ∼ (a2 c2 + b2 d2 , a2 d2 + b2 c2 ).
Proof. By hypothesis, we know
a1 + b2 = b1 + a2 and
c1 + d 2 = d 1 + c2 .
We want to show
(a1 c1 + b1 d1 ) + (a2 d2 + b2 c2 ) = (a1 d1 + b1 c1 ) + (a2 c2 + b2 d2 ).
First,
(a1 c1 + b1 d1 ) + (b2 c1 + a2 d1 ) = c1 (a1 + b2 ) + d1 (b1 + a2 )
= c1 (b1 + a2 ) + d1 (a1 + b2 )
= (b1 c1 + a1 d1 ) + (a2 c1 + b2 d1 ).
Secondly,
(a2 c1 + b2 d1 ) + (a2 d2 + b2 c2 ) = a2 (c1 + d2 ) + b2 (c2 + d1 )
= a2 (c2 + d1 ) + b2 (c1 + d2 )
= (a2 c2 + b2 d2 ) + (b2 c1 + a2 d1 ).
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VAN WYK’S MATH 315
Adding these two equations together and canceling the (b2 c1 + a2 d1 ) and (a2 c1 + b2 d1 ) terms
from each side yields the desired equation.
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So we can now define multiplication of integers as [(a, b)] · [(c, d)] = [(ac + bd, ad + bc)].
Time for some more properties!
Theorem 5.5.
(1) For each a, b ∈ Z, ab = ba.
(2) For each a, b, c ∈ Z, a(bc) = (ab)c.
(3) For each a, b, c ∈ Z, a(b + c) = ab + ac.
This theorem shows the various multiplicative properties (commutativity, associativity,
left distributive law) on N from Theorem 4.3 extend to Z. The proof of each part is similar
to those in Theorem 5.3: pick ordered pairs for each equivalence class (i.e., element of Z)
and use the appropriate definitions/properties to derive the desired result.
In light of the various theorems in this section, and using the function φ in Problem ??, we
may identify every natural number n with the integer [(n + 1, 1)] = [(n, 0)] and consider the
set of integers as a superset of the set of natural numbers. Every integer is either a natural
number or the additive inverse of a natural number, with the familiar algebraic structure
given by the rules of addition and multiplication we learned as children.
The only property of the integers which we still haven’t developed is that of order; that
will come in a later section.
Problems. 1. Find all the mistakes in this section.
2. Prove Theorem 5.1.
3. Draw a section of the lattice points of N × N in the Cartesian plane. Connect points that
are equivalent under the relation ∼ of Definition 5.1. Show that each equivalence class is
identified with the integer that is the x-intercept of the line you get.
4. Prove Theorem 5.2.
5. Prove Theorem 5.3 (2).
6. Prove Theorem 5.3 (4) .
7. Prove Theorem 5.5 (1) .
8. Define the function φ : N → Z by φ(n) = [(n + 1, 1)].
(a) Convince me that you don’t have to prove φ is well-defined.
(b) Prove φ is an injection.
(c) Figure out what φ has to do with Problem 3.
(d) Prove φ(m) + φ(n) = φ(m + n). (Note: It is also true that φ(m)φ(n) = φ(mn), so
you can prove that too if you are ambitious.)