LECTURE 10: THE INTEGERS
Suppose you start with $5 and you pay $1
each day for a cup of coffee. Suppose your
credit is good at the store!
After the first day you have $4
After the second day you have $3
After the third day you have $2
After the fourth day you have $1
After the fifth day you have $0
After the sixth day you owe $1
After the seventh day you owe $2
(and so on.)
We can answer questions such as “if I owe
$2 and I find $5, how much do I have?”
In other words: “owe $2” acts like a number.
We can get to “owe $2” in any of the
following ways:
start with $0, spend $2
start with $1, spend $3
start with $2, spend $4 ...
These are ordered pairs: “start with $4,
spend $2” is not the same!
We call the ordered pairs
(0,2), (1,3), ..., (10,12),... equivalent.
How do we test this? One way:
(a,b)~(c,d) if a+d = b+c
Another way of looking at the same idea:
The natural numbers are not closed under
subtraction. If we wanted to make them
closed under subtraction, we would have to
add new numbers to “be” 1-3, 5-8, etc.
Also, if 1-3 and 2-4 mean anything, then they
should be equal, because
a-b = c-d
a+d = b+c
in the natural numbers, and we want to keep
the existing rules.
The relation ~ is reflexive:
(a,b) ~ (a,b)
a+b = a+b
And symmetric:
(a,b)~(c,d)
a+d = b+c
And transitive:
(a,b)~(c,d)~(e,f)
(c,d)~(a,b)
d+e = c+f
a+d = b+c
a+d + c+f = b+c + d+e
a+f = b+e
(a,b)~(e,f)
So it IS an “equivalence relation”. Therefore
all pairs are divided into equivalence classes.
In the same way that we represent an
equivalence class of finite sets (under the
“pairable” relation) by a whole number
=
0,1,2 ...
3
we represent an equivalence class of these
“credit-debit” pairs (under ~) by an integer.
(3,5) (10,12)
(0,2) ...
=
=
-2
We can label this as [(3,5)] where the square
brackets mean “the equivalence class
containing (3,5)” This is the same as [(0,2)];
each class has many names.
There is a natural embedding of the set N of
whole numbers into the set Z of integers.
(The symbol Z comes from the German “Zahlen”)
n
[(n,0)]
That is: “have $n” is the same as
“have $n, owe nothing”
(or as “have $n+1, owe $1, or...)
This embedding is 1-1, but it is not onto;
there are integers that cannot be reached in
this way.
These negative integers do not count
finite sets.
What is the arithmetic on integers?
Addition: [(a,b)]+[(c,d)] = (a+c,b+d). We
choose this definition, but need to verify that
it has the right properties. remember
doing this for
set addition?
(1) It’s well-defined.
This is not obvious! To add two integers, we
need to choose representative pairs. We must
show that this choice doesn’t affect the
result.
(2) It extends the addition operation on
natural numbers.
We also hope:
(3) It obeys the same laws that we’re used to:
commutativity, associativity, etc.
(1) Integer addition is well-defined:
If (a,b)~(a’,b’) and (c,d)~(c’,d’), then
(a,b)+(c,d) = (a’,b’)+(c’,d’).
Proof: By definition (a,b)+(c,d) = (a+c, b+d)
and (a’,b’)+(c’,d’) = (a’+c’,b’+d’). From our
assumptions we have
a+ b’= a’+b
c+d’ = c’+d.
Adding:
(a+b’)+(c+d’)=(a’+b)+(c’+d)
Rearranging: (a+c)+(c’+d’)=(a’+c’)+(b+d)
So
(a+c,b+d) ~ (a’+c’,b’+d’)
(a,b)+(c,d) ~ (a’,b’)+(c’,d’)
(2) Integer addition extends whole number
addition:
If a+b = c, then [(a,0)] + [(b,0)] = [(c,0)]
Proof: (a,0) + (b,0) = (a+b,0) = (c,0).
(3) Integer addition is commutative,
associative, and has a zero element [(0,0)]
such that [(a,b)] + [(0,0)] = [(a,b)].
Proofs: homework (answers will be on web)
EG: for commutativity, show
[(a,b)] + [(c,d)] = [(c,d)] + [(a,b)]
(4) A new property: Additive inverse
Every integer [(a,b)] has an opposite or
additive inverse [(b,a)]. Write this as
-[(a,b)] (this is a second, different use of the
subtraction sign!)
[(a,b)] + -[(a,b)] = [(a,b)] + [(b,a)]
= [(a+b,a+b)] = [(0,0)]
That is:
m + -m = 0 for any integer m
Also: --[(a,b)] = -[(b,a)] = [(a,b)]:
--m = m for any integer m.
The Ordering of the Integers
We define [(a,b)] > [(a’,b’)] if a+b’ > b+a’.
This is well-defined.
If m>n, then m+p > n+p
If m>n, then n<m.
Homework
questions:
prove these
statements!
For integers m,n, either m>n, m<n, or m=n.
...
[(2,4)] [(2,3)] [(2,2)] [(3,2)] [(4,2)] ...
An Integer Number Line
(5) “Classification theorem for integers”:
Every integer is either the embedding of a
natural number or the opposite of the
embedding of a natural number. Only 0 is
both.
Proof: [(a,a)] = [(0,0)] is the embedding of 0.
If a>b, [(a,b)] = [(a-b,0)], the embedding of
a-b.
If a<b, [(a,b)] = [(0,b-a)], the opposite of the
embedding of b-a.
Because of this, we can write any integer as n
or as –n, where n is a natural number.
Whole Number Line
0
1
2
3
4
5
3
4
5
greater -->
Integer Number Line
-5 -4 -3 -2 -1 0
<-- smaller
1
2
An integer is (the embedding of) a whole
number if and only if it is greater than or
equal to 0.
Integers greater than 0 are called positive.
Integers less than 0 are called negative.
Integers greater than or equal to 0 can also
be called non-negative
We can combine this with the Prime
Factorization Theorem to get a unique
representation for any nonzero integer as
 p1 p2 ... p j
n1
n2
nj
EXAMPLE: -650 = - 2 x 52 x 13
NOTE: Don’t say “integer” when you mean
“negative integer”.
An integer is negative if it is –m where m
is a positive whole number.
Being written as –m does NOT make
an integer negative. 5 = -(-5).
INTEGER SUBTRACTION (this is why we
are doing all this!)
Define [(a,b)] –[(c,d)] = [(a+d, b+c)].
This is well-defined in the sense that it
doesn’t depend on our choice of pairs.
And it restricts to whole number subtraction
where defined.
Proof: homework; similar to the ones for
addition!
INTEGER SUBTRACTION:
[(a,b)] –[(c,d)] = [(a+d, b+c)].
As the right-hand side can always be
computed, the integers are closed under
subtraction!
But we don’t need subtraction at all anymore:
[(a,b)] –[(c,d)] = [(a+d, b+c)]
=[(a,b)]+[(d,c)]
=[(a,b)] + -[(d,c)]
That is:
m-n = m+(-n)
“Subtraction is just adding the opposite!”
MULTIPLICATION:
We want to keep as many of the old rules as
we can. If we want the distributive law, then
0 = (a-a) x b = (a + -a) x b
= axb + (-a)xb
So (-a)xb is the opposite of axb.
That is: (-a)xb = -(axb).
Also, ax(-b) = -(axb)
and (-a)x(-b) = -(ax(-b))
= --(axb)
= axb.
By the Classification Theorem, every
product of integers is of the form
ab, (-a)b, a(-b) or (-a)(-b)
for whole numbers a,b. This lets us find the
product of any pair of integers.
EXAMPLES: (-5)x6 = -(5x6) = -30
(-3)x(-2) = (3x2) = 6
If we have a product of three or more
integers, we just “count minus signs.”
(-a)(-b)(-c) = -abc
(-u)(-v)(-w)(-x) = uvwx
(-1)n = 1 if n is even, -1 if n is odd.
(-a)n = an if n is even, -an if n is odd.
use repeated multiplication: (-a) (-a)... (-a)
DIVISION:
If a/b exists, then (-a)/b and a/(-b) also
exist and equal –(a/b).
Also, (-a)/(-b) exists and equals a/b.
The same basic idea: “is the total number of
minus signs in the product odd or even?”
works for both multiplication and division.
EXAMPLE: Is (-4)2 x3x-5 positive?
-20 x –(3/2)
Answer: no, there are a total of 5 minus
signs (two in the (-4)2 !) So it’s negative.
COMPUTING:
Some calculators let you press the
key before entering digits to enter a negative
number.
Others have a +/- key that you press
after a number to change its sign.
Many have both.
HAND/MENTAL CALCULATION:
Multiplication, division, and exponents are
easy.
Addition and subtraction: Use identities
to change the problem and get an addition or
a subtraction of a smaller whole number from
a larger. Then compute as for whole numbers.
23 – 42 = -(42 – 23) = -19
45 – (-68) = 45 + 68 = 113
-45 + 68 = -(45 – 68)
= --(68-45)
= 68-45 = 23