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 (c,d)~(a,b)
And transitive:
d+e = c+f
(a,b)~(c,d)~(e,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 ...
we represent an equivalence class of these “credit-debit” pairs (under ~) by an integer.
(3,5) (10,12) =
(0,2) ... =
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.
(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.
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.
0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
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
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