| (all letters are
integers) |
Comments |
| Division |
Definition |
a|b implies b=ac this is read "a divides b" |
if integer a goes into integer b evenly, it also goes into
bc. 4/2, 8/2,16/2 |
| Theorem 1a |
if a|b and a|c then a|(b+c) |
7 goes into 14 and 7 goes into 21. Therefore 7 goes into
14+21. Outline of Proof:
a|b implies b=as [from definition]
a|c = implies c=at
b +c = as + at = a(s + t)
So from the definition, we have
a | (b+c)
Example:
If 6|12 and 6|18 then 6|(12+18) |
| 1b |
if a|b then a|bc |
if a|b, then by the definition a=bc Example:
If 6|12 then 6 | 12*7 |
| 1c |
if a|b and b|c, a|c |
Outline of Proof: a|b implies b=as
b|c implies c=bt
c=bt=ast=(st)a st=an integer
By the definition, a|c since c=(st)a from
Theorem 1
if 6/3 goes evenly and 12/6 goes evenly
then 12/3 goes evenly. |
| Ex. 2.3 5 |
if a|b and b|a a=b or a=-b (other exercises follow the same
pattern.)
|
Outline of Proof: a|b implies a=bs [from the definition]
b|a implies b=at
b=at=bst
1=st
this means that s=1, t=1 or s=-1, t=-1
so a=-1*b or a=1*b
Example:
6 |6 and 6 |-6
|
| Prime |
Definition |
A prime has only itself
and 1 as factors |
This is often a very helpful restriction. See
prime numbers below for more information.Note: The concept
of prime is not restricted to numbers. Polynomials, for example, can be factored into
prime polynomials |
| |
Definition |
if gcd(a,b)=1, a and b are relatively prime |
If a and b have only 1 as a common factor, then they are
relatively prime. This is often helpful as a restriction. Example
gcd(15,7)=1 15 and 7 are relatively prime since no numbers divides both. |
| |
Definition |
if for a(1),a(2)a(3)..a(n)
gcd(a(i),a(j))=1
fot 1<=i<j<=n |
Every pair is relatively prime. This is useful in making sure
that each pair has a inverse. Example:
13,11,10 are relatively prime since
gcd(13,10)=1, gcd(13,11)=1 and gcd(10,11)=1 |
| Division Algorithm |
Definition |
a=dq+r |
d is the divisor, r is the remainder, q is the quotient. 6/5 =
1*5 + 1 |
| GCD |
Definition
gcd(a,b) |
The largest d such that d|a and
d|b |
d is the largest number that goes into a and b. Another way to look at this is to factor each number into a set of prime
factors. Those factors from each of the sets make up the gcd.
contrast this with lcm.
Example:
gcd(15,5)=5 since 5 |15 and 5 |5 and 5 is the largest number that does so. |
| Definition |
if gcd(a,b)=1, a and b are relatively prime |
If a and b have only 1 as a common factor, then
they are relatively prime. This is often helpful as a restriction. Example
gcd(15,7)=1 15 and 7 are relatively prime since no numbers divides both. |
| Theorem for Euclidean
algorithm. |
if a=bq+r, gcd(a,b)=gcd(b,r) |
Let d|a and d|b
then d | bq and d|(a-bq) and d |r see Theorem 1
This means that d divides a, b and r
gcd(a,b)=d=gcd(b,r)Example:
gcd(252,198)=gcd(198,54) since
252=198*1 +54
gcd(198,54)=gcd(54,36) since
198=54*3+36
gcd(54,36)=gcd(36,18) =18 since
36=18*2+0. [stop when a number goes in evenly.]
Therefore since gcd(36,18)=18,
gcd(36,18)=gcd(54,36)=gcd(198,54)
=gcd(252,198)=18
Hence gcd(252,198)=18.
This is the outline of the Euclidean algorithm.
See Euclidean Algorithms for example |
| Linear combinations
|
gcd(a,b)=sa + tb |
See Linear
Combination for mode detailed information on how to calculate t and s. This is a
generalization and example of the Euclidean Algorithm Let
d=gcd(a,b), then
d|a and d|b
d|a implies that d|sa
d|b implies that d|tb Theorem 1
d|sa and d|tb implies d|(sa+tb)
Therefore, since d is the greatest common divisor, d=sa+tb. |
| Lemma 1 |
if gcd(a,b)=1 (relatively prime) and
a|bc, a|c |
Outline of Proof:: sa+tb=1 [def of gcd; see relatively prime]
csa +ctb = c [multiply by c]
since a|bc a|tbc [Theorem 1]
a | csa [a | (cs)a ]
Since a |csa and a|tbc, a|(csa+ctb)
a | (c*(sa+tb))
but (sa+tb) is a integer so
a |c
What this says is that if a and b are relatively prime and a goes in b*c, then a goes
into c. If a and b were not relatively prime, there would not be a unique division in step
4.
Example:
gcd(13,6)=1 and 6 | 13*2, then 6|2. |
| Theorem 2 Law of Cancellation |
If ac~bc mod m and
gcd(c,m)=1, then a~b mod b If a,m are relatively prime then a has an inverse |
Outline of Proof: ac~bc mod m means that
m | (ac-bc) = m | c(a-b)
Since gcd(c,m)=1 and m |c(b-a)
from Lemma 1 we get
m | (b-a)
This is the definition of a~b mod m
Example (5 is prime)
14~ 4 mod 5
divide by 2
7 ~ 2 mod 5
If a is relatively prime to m, a has an inverse; if m
is prime, m is relatively prime to all a.
Example (4 and 9 are relatively prime)
9~ 21 mod 4
divide by 3
3 ~ 7 mod 4 |
| LCD/GCD |
Theorem |
ab= gcd(a,b)*lcm(a,b) |
When all the factors are written out, they form two sets, GCD
and LCM. |
| LCM |
Definition lcm(a,b) |
The smallest d such that a|d
and b|d. |
d is the smallest number than can be divided by a and b. Another
way to look at this is to factor each number into a set of prime factors. Those factors
that are the union of the sets make up the lcd.Contrast this with
gcd. |
| Modular Arithmetic |
Definition |
a mod m =r such that a=qm+r |
When a is divided by m q times, the remainder is r. This
definition is used in many proofs in the chapter.
Example
4 mod 5 means 5*0+4 or 5 goes into 4 zero times with 4 left over.
NOTE: The trig functions are a mod like function. The sine 0 = sine 360. They, like the
mod are periodic functions. In this case, q= the number of periods and r is the angle
remaining.
To determine the sun's position in the sky, it is important to know the angle with the
horizon, not the number of revolutions. This is a use of a mod function. |
| Definition |
Congruence(~): a ~ b mod m
means that m | (a-b)
or
a = b + qm
Also, it can be seen as a and b have the same remainder when divided by m. |
~ is the same as the 3 equal signs (I don't have
the symbol on my key board.) Outline of Poof:
Given a and b have the same remainders when divided by m:
a mod m=r
b mod m=r
Then,
a=qm+r
b=pm+r
a-b=qm-pm=(q-p)m
q-p is an integer so
m | (a-b)
or a~b mod m
Example
4 mod 5 means 5*0+4 or 5 goes into 4 zero times with 4 left over.
6 mod 3 means 3*2+0 or 3 goes into 6 with zero left over |
| Inverses |
if a*b~ 1 mod m,
and m is prime, then for any a there is an inverse bf b. If a,m are relatively prime
(gcd(a,m)=1) then a has an inverse b. |
For example 3*2~1 mod 5 therefore 3 and 2 are
inverses of each other.see table of inverses mod 5 and mod
4. 2 does not have an inverse mod for since
2*b ~ 1 mod 4 does not have a solution. Look at the tables and you will see that there is
not an onto / 1-1 relation in the mod 4 tables. There is in the mod 5 table.
NOTE: 3*b~1 mod 4 where b=3 Since 3 is relatively prime
to 4, 3 has an inverse, in this case 1. While the system does not have an inverse,
inverses do exist for those values of a that are relatively prime to m.
This is why the restriction a,m is relatively prime or equivalently, gcd(a,m)=1 is
added to the restrictions if we need to take the inverse of a. |
| Addition by a constant |
If a mod m~b mod m then a+c mod m ~ b+c mod m If a mod m =r, then
a+c mod m = c+r mod m |
This allows us to do math using just the
remainders.
Example:
5 ~ 1 mod 4 =1
add 7
5+7 ~ 1+7 mod 4
12 ~ 8 mod 4 = 1+7 mod 8 = 0 mod 4
1+7=8 or 0 mod 4
This means that we can add across a congruency and not change the relationship. If we
add an hour to 1am and an hour to 1pm, we get 2am ~ 2pm. We haven't changed the
relationship, but each clock is advanced one hour. |
| Multiplication by a constant |
If a mod m~b mod m then ac mod m ~ bc mod m If a mod m =r, then
ac mod m = rc mod m
Multiplication always works; division only works if m is prime. |
Multiplication by a constant. Example
5 ~ 1 mod 4 =1
multiply by 6
5*6 ~ 1*6 mod 4
30 ~ 6 mod 4 =2
1*6=6 mod 4 =2
Note unless m is prime, ac mod m~bc mod m does not equal a mod m ~ b mod m.
Example: 14 ~ 8 mod 6
divide by 2 and we get 7 ~ 4 mod 6
which is incorrect. |
| Addition |
if a~b mod m and c~d mod m, a+c (mod m) ~ b +d (mod m) |
Outline of proof: b=a+sm
d=c+tm
b+d = a + sm + c + tm
b+d = a+c + sm + tm|
b+d = a +c + m(s+t)
b+d = a+c +mq (q=st)
By definition, b+d(mod m) ~ a+c (mod m)
6~1 mod 5, 7 ~ 2 mod 5
6+7 mod 5 ~ 1+2 mod 5 =3 mod 5
Since 6+7=13 mod 5= 3 mod 5 |
| Multiplication |
if a~b mod m and c~d mod m, a*c (mod m) ~ b *d (mod m) |
Outline of proof: b=a+sm
d=c+tm
b*d = (a + sm) * (c + tm)
b*d = ac +smc + atm + stm*m
b*d = ac + m(sc+at+stm)
b*d = a*c +mq (q=sc+at+stm)
By definition, b*d(mod m) ~ a*c (mod m)
6~1 mod 5, 7 ~ 2 mod 5
6*7 mod 5 ~ 1*2 mod 5 =2 mod 5
Since 6*7=42 mod 5 = 2 mod 5 |
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