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Lecture 1 An Introduction to Numbers
Summary
We started today by getting to know the policies and expectations in the course. All of
this is available already on the syllabus, but if you have any questions don't be shy
about emailing Andy. We also spent some time introducing ourselves briefly; this will
be continued as you post your own profiles for Homework 0. Afterwards, we started
talking about the basics in number theory, starting with the axioms. We finished by
introducing the notion of divisibility for the integers.
The Axioms of Number Theory
When trying to build a mathematical discipline from the ground up, one needs to
describe the fundamental objects and operations in the discipline and then define the
basic properties these objects will obey. These properties are called axioms, and they
are the "ground rules" the objects and operations must satisfy. With axioms in place,
one can then start proving theorems by manipulating the axioms.
In number theory, the basic objects of interest are integers. You might know these
objects as whole numbers. In this class we'll denote the set of all integers as :
(1)
The basic operations we have on the integers are addition, subtraction and
multiplication. We've avoided division since division doesn't behave very well on the
integers: the quotient of two integers is frequently not another integer. We also have
basic tools for comparing integers, namely equality and inequality.
With the fundamental objects, operations and comparisons in place, we can start
writing down the basic rules they all satisfy. Here's the list that we were able to come
up with in class:
for any (reflexivity of equality);
for any , if and then (transitivity of equality);
for any , implies (symmetry of equality);
for any , the sum is an integer (closure under addition) [note:
we didn't mention this in class];
1
for any , the product is an integer (closure under multiplication)
[note: we didn't mention this in class];
for any , (associativity of addition);
for any , (associativity of multiplication);
for any , (distributivity);
for any , (commutativity of addition);
for any , (commutativity of multiplication);
for any , (additive identity) [note: we didn't mention this in
class];
for any , (multiplicative identity) [note: we didn't mention this
in class];
for any there exists so that (additive identity);
for any with , then implies (cancellation of
multiplication);
for any , implies (substitution for multiplication);
for any , if and only if (substitution for a
cancellation of addition);
ddition;
for any , exactly one of the following is true (1) (2) (3)
(Trichotomy law);
for any , if then ;
for any , if and , then ;
for any , if and , then ; and
if is a nonempty set of positive integers, then has a least element (Well
ordering principle).
The last axiom — the well ordering principle — probably sticks out as the ugly
duckling of the bunch, and it certainly isn't one which most people think of when
rattling off basic properties of the integers. It is, however, essential to what we'll be
doing in class, as it is logically equivalent to mathematical induction — a tool that
we'll be using with some frequency in this course.
Playing around with the axioms
Our list of axioms is a little redundant, meaning that we could probably prove some of
the axioms we've listed in terms of the other axioms. In this sense, it doesn't pass the
usual mathematical aesthetic. To see that this is true, you can try to use the other
axioms listed above to prove
Theorem: For any ,
2
In class, we sketched a proof of the following result
Theorem: For any nonzero , if satisfies , then
Proof: We started by noting that we could assume a is either greater than or less than
0; by trichotomy we know that one of (1) (2) or (3) is true, and if
we had then we'd be done with the theorem. By a similar token we know that
or , since is ruled out by assumption. So we broke things into 4
cases based on whether or and whether or .
Case I: and
In this case, one of our axioms on inequality tells us that . This contradicts the
fact that , and so we know this case is impossible.
One could proceed with analyzing the other cases, each time you would find a
contradiction to the given equality . At the end, one concludes that all the
possibilities lead to a contradiction, and hence neither nor are possible.
This leaves only , the desired result.
Though working through axiomatic proofs is good exercise for building your proof-
muscles, in practice we won't be quite so explicit in our use of these familiar axioms
during class. This won't present any real problems since you are more accustomed to
manipulating these axioms then perhaps you realize.
Putting the Elementary in Number Theory
With the basic ground rules set, we had a chance to talk about the most important
property of integers in this whole class: divisibility. It is the study of this property
which makes the number theory we'll study "elementary." One can think of divisibility
as the attempt to carry division into the realm of the integers, made appropriately
cautious to reflect the fact that the integers don't always behave so well under
division.
An integer d is said to divide an integer a if there exists an integer q so that a = dq. If
d divides a we write , and if d does not divide a then we write .
An example
This definition should agree with your own intuitive notion of divisibility in the
integers, so hopefully it isn't too surprising. To see an example in action, notice that
since we can find an integer q to solve the equation
(2)
3
in this case, the integer q is simply 2.
A non-example
Let's try to prove that . For this, we need to show that we cannot find an integer
n
(3)
q satisfying the equatio
For this notice that ; for any integer q satisfying we have (a
n one of ave
(4)
slight modification o our axioms), and hence we h
ikewise we know that L , and for all integers q satisfying we get
(5)
Since all integers fall into one of the categories we have described, we conlude that
for any integer q, and so .
Lecture 2 Introducing Divisibility
Summary
Today we continued our discussion of divisibility and its basic properties. We saw
Divisibility Continued
Last class previous we defined the notion of divisibility in the integers as follows:
some examples of how to put these properties into practice to prove exciting new
results which might otherwise be quite difficult. Today's lecture culminated in the
statement and proof of the division algorithm, one of the foundational results in
number theory.
4
Definition: An integer d is said to divide an integer a if there exists some integer q
satisfying the equation .
of
Some Examples
and We already saw proofs in class on Wednesday. Most divisibility
statements will seem pretty obvious to you just by inspection, but the one exception
ent lvinmight be divisibility statem s invo g 0. Below we provide a few examples.
dsince the equation oesn't have any solution; any value you plug in
for q will still make the right hand side 0.
is true for some integer-value of q (in fact, it's since the equation
true for all q!).
. does h ion, namely ave a solutsince
The c f eve
isibility by using the terms even and odd.
ase o ns and odds
We also single out a special case of div
Specifically, we have
Definition: An integer a is even if , and an integer a is odd if .
of ers which satisfy
a = 2k for some integer k. Indeed, we have the following
You might also be used to thinking even integers as those numb a
Lemma: An integer a is even if and only if there exists so that .
e defi f divisi and
evenness.
Our proof of this result will require us to simply recall th nitions o blity
Proof:
We know that an integer a is even if and only if ; this is just the definition of
that evenness. We also know if and only if there exists an integer k so that
; this is just the definition of divisibility. H e we have enc
(1)
as desired.
You'll extend th is problem in your homework when you show that all odd numbers can
be written in the form .
5
Properties of Divisibility
There are a handful of properties of divisibility which are handy to remember;
ou want to try to divide one integer into
another. You can also think of these lemmas as good exercise for the definitions we've
wn
basically, these are good tools to use when y
encountered in the class: none of the proofs require much more than writing do
definitions, so they are a good chance for you to get used to the new terminology
we've covered.
Lemma: For , if and then .
Proof: We're told that and . By the definition of divisibility, this means we
that
have
an integer d so (using the first divisibility condition), and
an integer e so that (using the second divisibility condition).
(2)
Substituting appropriately, this means that
Since de is an integer, this equation tells us that as desired.
Lemma: For , if and , then .
Proof: Again, we start by just writing down the definitions. In this case, we're told
that and , which m we have eans
an integer d so that (using the first divisibility condition), and
e so that an integer (using the second divisibility condition).
(3)
Hence we have
Since md+ne is an integer, this equation tells us that .
There was another basic property of division we mentioned that allo ed us to
e statement of
this result is
w
compare the size of a divisor to the size of the integer it is dividing. Th
6
Lemma: If for a nonzero integer a, then .
We didn't prove this result, but it might show up on your homework.
rick
rmining when
an integer is divisible by 17. You can think of this as a cousin of the old "casting our
u use to determine whether a given integer is divisible by 9. This
new rule says
A Neat T
One of the examples of divisibility we gave in class was a rule for dete
nines" rule that yo
Theorem: An integer is divisible by 17 if and only if is divisible by 17.
Proof: First, ass . Since ume that is obvious, our result on
integral linear combinations tells us that
(4)
In the other direction, assume that we are told that , and we want to
. Now since we know that prove , our result on integral linear
combinations tells us that
(5)
Example
To see this result in practice, notice that we have 221 = 22(10)+1. Since
, we can conclude that .
A Final Divisibility Result
We finished off with ivisibility proof, when we showed that one last example of a d
. For every positive integer n, we have
Proof: We proved this result by induction, starting with the base case . In this
case it's easy to see that the statement is true: .
7
For the inductive step, we'll assume we know that , and we' to use this
to prove that
ll try
. In order to do thi e'ls, w l try to simplify the
expression into something more user friendly; we decided the bast
way to do this was to just expand the term , which gives us
(6)
by induction and Since due to our clever
factorization, our result on integral that linear combinations tells us
as well.
The Division Algorithm
The following result, though it seems pretty basic, is actually extremely powerful,
g greatest common divisors (Section 1.3) but
f modular arithmetic (Chapter 2).
giving rise not just to a method for findin
also laying the foundation for the notion o
The Division Algorithm: For a positive integer d and an arbitrary integer a, there exist
unique integers q and r with and .
Proof of the division algorithm:
Part 1: Existence
We start by defining the set
(7)
and we claim that S has at least one non-negative element. To back up this claim,
that
if
notice
then we can take and find that ;
otherwise , in which case taking shows that .
Now since d is positive by assumption we know that , and so the
product is either a positive number (if ) or (if ).
In either case we see that is a non-negative element of
ordering principle tells us that S contains a least non-ne e element. We'll call this
element r, and notice that r takes the form
S.
In either case we see that S contains a non-negative element, and hence the well
gativ
8
(8)
Hence we get . To show this satisfies the conditions of the division
ithm, we simply need to show that . The condition algor is satisfied
since r is chosen to be non-negative, so we only need to ve y rif .
To see that , assume to the contrary that , and we'll derive a
contradiction. In this case we have that
(9)
by assumption we have Since is a non-negative element of S which is smaller
r. This is a contradiction to the selection of r as the smallest non-negative
element of S, so we must conclude that
than
as desired.
rt of
uppo
Part 2: Uniqueness
To finish the proof we need to show that the q and r we found in the previous pa
the theorem are, indeed, unique. Hence s se we have
(10)
This tells us that
, and therefore that . But since we also have
by our conditions , we are in the
e divisor d has larger absolute value than the number it is dividing
into — namely,
scenario where th
. This tells us that we must have , and hence
.
With this in hand, we see that the equation then becomes
. Usin cancellation law of multiplication, w e have . g the e therefor
Lectur
st class period we talked at length about divisibility and the division algorithm.
day we moved on to discuss the concept of greatest common divisors. We finished
e 3 Greatest Common Divisors
Recap and Summary
La
To
9
by describing some properties that gr
(surprisingly powerful) result that the gc
eatest common divisors enjoy, including the
d of two integers a and b can be expressed as
Along with the language we already established — namely that an integer d divides a,
For
r q so that
an integral linear combination of a and b.
A few comments on the divisibility and the division algorithm
or that d is a divisor of a — there are plenty of other equivalent expressions.
instance, if there exists an intege
(1)
then one can say that "d divides a," that " d is a divisor of a," that ''dq is a
factorization of a," or that "a is a multiple of d." All of these expressions capture the
equation above, and they should be pretty familiar vocabulary to all of you.
e the
same
It is also worth pointing out the division algorithm gives us a way to measur
success or failure of one integer dividing another. What we mean by this is the
following. In the case that , we have an equation which is satisfied. In
the case that , however, the definition of divisibility doesn't give us an equation
the
form
we can write down. The division algorithm, however, let's use write an equation of
(with ) regardless of the divisibility of d and a. In fact, the
remainder term tells us whether we're in the caseprecisely or the case .
Being able to w down such an equation winds up being critically important in many
circumstances, as you'll find in this week's assignment.
Finally, we provide an example of the division algorithm in action.
Example: The Division Algorithm
rite
Suppose you wanted to run the division algorithm on and . Playing
around with various multiples of 11, we see that is the smallest multiple of d
se q as 10, then we get the equation which is less than or equal to a. Hence if we choo
(2)
In this case, we see that the division algorithm gives us .
10
Greatest Common Divisors
For two integers a and b, it is often useful to know if there are any numbers d so that
and . For obvious reasons, such a nu
Certainly common divisors exist for any pair of i
mber is called a common divisor.
ntegers a and b, since we know that 1
always divides any integer. We also know that common divisors can't get too big since
diviso f a a
divisors can't be any larger than the number they are dividing; hence a common
r d o nd b must have and , so that . With all
as motivation, we have the following
this
Definition: The greatest common divisor of two integers a and b, written , is the
largest integer d so that and . More nerally, if ction of ge you have a colle
integers , then the greatest common divisor of the collection ,
written , is the largest integer d so that for every i.
Example: Non-trivial G
that
CD
Suppose we'd like to know the greatest common divisor of 12 and 15. We can see
the divisors of 12 are
the divisors of 15 are
The largest number which is a member of both of these sets — and hence the
and 9 — is therefore 3 e have shown (12,9) = greatest common divisor of 12 . So w
3.
Example: Trivial GCD
If we want to know the greatest common divisor of 21 and 10, then we write down
the r i divisors:
the divisors of 21 are
the divisors of 10 are .
Hence we know that (21, 10) = 1.
Example: GCD of a collection
Looking at the lists of divisors we've already written out, we can see that the greatest
1, so that (12, 15, 10) = 1. common divisor of 12, 15 and 10 is
11
Example: GCDs with 0
We finish by noting that for any . This follows since 0 has the
r divides it. Since property that every intege is the largest divisor of n, this means
that .
Of particular interest in number theory are in ers which do not share a common
diviso u
teg
r, and beca se of their importance they get their own special name.
Definition: Two integers a and b are said to be relatively prime if = 1; i.e., if a
and b share no common non-trivial divisors. A collection is relatively prime
. A collection if is said to be pairwise relatively prime if
whenever .
In the examples above, we see t 10 are relatively prime, and that the hat 21 and
collection is rel vely prime. Notice in this last example that the
collection is relatively prime even though each pair of integers from the collection is
a
pairwise re
Having met and played around with greatest common divisors a bit, we'll now
oy.
First, we'll see what we get when we remove the gcd of two integers.
ati
not relatively prime. (As a general rule of thumb, you'll care more about whether
collection is latively prime than whether it's relatively prime).
Properties of the GCD
introduce a few properties that they enj
Removing the GCD
Lemma: For any pair of integers a and b, we have .
Proof: and , there exist integers Since so that
(3)
Our goal, then, is to show that . For this, suppose that there were some
common divisor of both and . This would imply that there exist integers
satisfying
(4)
12
Putting this togethe ith the p us equation, we'd then have r w revio
(5)
Hence the integer would be a common divisor of a and b which is larger than
(a,b). This is a contradiction to the definition of greatest common divisor, and hence
e left to conclude that
(6)
we ar
You might be tempted to think that the integers a and are relatively prime; resist
the temptation! In general, it is not true that .
GCD as a linear combonation
foll
expressed as a linear combination of
the a and b.
Another surprisingly useful result to have around is the owing Proposition, which
says that the gcd of two integers a and b can be
Proposition: For any two integers a and b, we have
Proof: To prove this result, we'll define , and we'll
show that it is in fact the greatest common divisor. Toward this end, we'll start by
then we'll show that all other
common divisors divide d (and so all other divisors are no bigger than d).
rt by using the Division Algorithm t
showing that d is a common divisor of both a and b,
To show that d is a divisor of a, we'll sta o find
integers q and r with that satisfy . Using the fact that
for appropriately chosen integers m and n, this means we have
(7)
But since d is chosen as v ear combination ofthe minimum positi e integral lin a and b,
we therefore have , and so . Hence , and a similar proof shows that
. So d is a common divisor.
13
Now we show that any other common divisor k of a and b is also a divisor of d, from
which we conclude that ; this ensures that d is the greatest common divisor,
as claimed. To show that , we note that since and , then we have k
linea tion of In p
(8)
divides any integral r combina a and b. articular, we have
Corollary: Two integers e relatively prime if and only if 1 can be written as a and b ar
an integral linear combination of a and b.
Another interesting property which the gcd of two integers has is that all other
on divisors of a and b will divide (a,b). We actually proved this in the midst of
the proof of the last theorem, so we can write is as a
comm
Corollary: For any pair of integers a and b, a common divisor d has .
.
As a final corollary, we note that since the gcd of two numbers is their smallest
positive integral linear combination, any positive number smaller than their gcd
cannot be expressed as an integral linear combination
Corollary: If k is a positive integer which is smaller than , then there are no
integers x and y so that the equation holds.
One of the real benefits of using relatively prime integers is that they let you conclude
ua
Divisibility and Relatively Prime Integers
certain statements about divisibility which you might not us lly get to make. For
instance, if you are told that , you mi pted to ght be tem conclude that or
. In general, though, this is false (can you find a counterexample?). When you
gers, however, you can call
on a result such as this.
have some ''nice'' property involving relatively prime inte
Lemma: If and , then .
I won't prove this for you now, since this is one of your homework exercises.
14
Lecture 4 The Euclidean Algorithm;
Prime Numbers
divisors in class last time,
dn't come up with a very effective way of computing GCDs. We remedy this with
e Euclidean Algorithm, and we show how this algorithm can also be used to express
mbers a and b as an explicit linear combination of a and b.
oduced prime numbers and started proving some results about
u
aw GCDs when you were in high school. The kinds of GCDs you were after
then — between pairs of numbers which are relatively small and easily factored — are
care about in practice — namely
o large to easily factor. For this reason,
Summary
Although we introduced the concept of greatest common
we di
th
the GCD of two nu
Afterwards we intr
them.
The Euclidean Algorithm
Greatest Common Divisors might seem like a boring subject to you, since likely yo
already s
quite different from the kinds of GCDs people
between pairs of numbers which are much to
we need a method for computing GCDs effeciently. The idea for this method comes
from the remarkably simple
Lemma: Suppose that a and b are integers and that as per the Division
Algorithm. Then .
Proof: This is really not a hard proof. Suppose that d is a common divisor of a and b.
Then since , we have that d is a divisor of r as well (using our "divisibility
Likewise, if d is a common divisor of b and r, then since
of integral linear combinations" result).
we have that d is
mmon diviso
gst these common divisors are the
same.
also a divisor of a.
Hence all co rs of a and b are also common divisors of b and r, and vice
versa, proving that the greatest elements amon
We can apply this idea iteratively to give a computationally effective way to compute
GCDs.
Example: Computing GCDs with the Euclidean Algorithm
15
Suppose you're interested in computing (1921,493). Using the previous result, since
(1)
we know that this GCD is the same as (493,442). In fact, we can keep using this
result over and over again, provided we keep using the division algorithm every time
rithm are: we get a non-zero remainder. These applications of the division algo
(2)
Hence we have
(3)
Example: Expressing GCD as a linear combo
This same procedure can be used to write the GCD be en a and b as an integral
n of the two. In the previous computation, for instance, the second-
to-last division algorithm application gives
(4)
twe
linear combinatio
The previous application of the division algorithm, though, showed us that
, and hence we can substitute this into our expression for 17:
(5)
Using the equality and substituting again gives
(6)
Continuing this kind of back substitution gives
(7)
16
Prime Numbers
tars of number theory are the prime numbers. The s
A number is said to be prime if the only positive divisors of p are 1 and itself. A
number is said to be composite if it is not prime; i.e., n is composite if there
so that . exist
The reason
in the multiplicat
they are foundati
that prime numbers are so exciting is that, despite their foundational role
ive structure of the integers, they are very elusive. When I say that
onal in the multiplicative structure of the integers, I mean that any
factorization of an integer n involves prime numbers as the atomic pieces — in the
s built out of elements from the
periodic table. This is made precise by the Fundamental Theorem of Arithmetic, a
tart d
s a
ideas — as me co applications of primality — throughout the
remainder of the course.
same way that any physical substance we encounter i
topic we'll s iscussing tomorrow. And when I say elusive, I mean just that: the
damn thing re hard to pin down and understand. We'll talk more about both of
these well as so ol
For now, though, we'll take a step in the first direction: showing that prime numbers
are the building blocks of integers. For this, we begin with a nice lemma that says
that any number is divisible by at least 1 prime number.
Lemma: For any integer , there exists some prime number p which divides n.
Proof: We'll prove this result by contradiction: assuming the opposite of what we w
to prove, manipulating this assumption until it reaches a contradiction, and the
concluding that our assum
ant
n
ption must be false — and hence our desired conclusion is
true.
(8)
So suppose that not every integer n has a prime factor. This means that the set
is non-empty. As a non-empty set of positive integers, S must have a least element.
We'll call this least element N.
Now N is an element of S, and hence has no prime divisor. Since N is a divisor of itself
— , after all — this means in particular that N cannot be prime. Therefore N is
composite, meaning there exist integers so that N = ab. Being positive
integers less than N, both a and b must live outside of S, and hence each has a prime
17
and . But then and , so that factor: say , contrary to the
defining property of N.
we conclude that S must, indeed, be empty, and so
every integer greater than 1 has a prime factor.
Having reached a contradiction,
Thi t gives us a method for finding prime numbers using a sieve technique.
Before we get there, we need to first mak ng
s resul
e the followi
Observation: If is a composite number, then there exists a prime divisor p of n
such that .
Proof: If n is composite, then there exist integers a and b so that n = ab. Now one of
a or b must be less than , since otherwise th ir e product would be greater than n.
Without loss of generality, we can assume that . Now a has a prime factor p
from the previous lemma, and so . Since and , we further have
, giving the desired result.
A Sieve Example
The idea behind a sieve is to find prime numbers by eliminating multiples of known
Suppose, for instance, that you wanted to find all prime numbers less than 60. The
previous observation says that any composite number
prime numbers. The magic, though, is that one has to use relatively few primes to
sieve out larger ones.
must have a prime
factor which is smaller than . Hence any composite number smaller
le by one of the primes which is smaller than 8 — namely one
f we listed all the numbers between 2 and 60 and crossed out
2 3 4
than 60 must be divisib
of 2, 3, 5, or 7. Hence i
the multiples of the primes listed above, the remaining numbers would all have to be
prime. Let's try it out:
5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
18
51 52 53 54 55 56 57 58 59 60
After sin t u les 2, , see that the primes less than 60
are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 21, 37, 41, 43, 47, 53, and 59.
In the same way that knowing the primes less than
cros g ou all m ltip of 3, 5 and 7, we
gave us a method for finding
r h 0, owing th t of ss than n gives us a way to
er h o im es an
the p imes less t an 6 kn e lis primes le
gen ate t e list f pr es l s th . W t s a great way to conclusively
generate prime numbers, the downside is that this technique takes a LONG times to
le t. ce a fe b m tic e for finding really big prime
numbers.
Having actually gone through and found a handful of small primes, we now begin to
ow many primes there are of a given magnitude?
do we know how "spread out" the prime numbers are? do they come in
ys far apart from each other?
is there a formula which allows us to quickly generate prime numbers?
ome of these questions are exceedingly difficult to investigate. We'll
cover a sampling in class tomorrow.
Today we spent the first half of the class exploring questions about prime numbers.
ong the way we proved that there are infinitely many prime numbers and that there
are arbitrarily large gaps between prime numbers. We also saw a formula which gives
a rough count for the number of integers up to a given number x, and we saw some
hile his i
imp men Hen it is n ef ctive ut i prac al m thod
Asking Questions About Primes
wonder what can be said about primes. Here are a few basic questions you might
want to know
how many primes are there? for instance, is the number of primes finite?
if the number of primes isn't finite, do we at least have a reasonable guess as
to h
clusters, or should we expect that they are alwa
These are all good questions, and some of them have nice, easy answers.
Alternatively, s
Lecture 5 Prime Numbers; The Fundamental Theorem of Arithmetic
Summary
Al
19
conjectures about other behaviors about prime numbers. In the last half of the cl
we started a proof of the Fundamental Theorem of Arithmetic.
ass
sking Questions About Primes
ne through and talked about the basics regarding prime numbers,
onder what can be said about primes. Here are a few basic
hey come in
clusters, or should we expect that they are always far apart from each other?
erate prime numbers?
for instance, are there more
primes behave under addition?
A
Having actually go
we now begin to w
questions you might want to know
how many primes are there? for instance, is the number of primes finite?
if the number of primes isn't finite, do we at least have a reasonable guess as
to how many primes there are of a given magnitude?
do we know how "spread out" the prime numbers are? do t
is there a formula which allows us to quickly gen
do prime numbers obey any special properties?
primes which leave remainder 1 after division by 4 than there are primes
which leave remainder 3 after division by 1?
what can you say about how the
These are all good questions, and some of them have nice, easy answers.
Alternatively, some of these questions are exceedingly difficult to investigate. We'll
cover a sampling now.
The Infinitude of Primes
The question on the number of primes has been around for a long time, and the
answer was known at least two thousand years ago. Here's the proof that Euclid gave
in his Elements.
Theorem: There are infinitely many primes.
Proof: Again, we'll proceed by contradiction: assuming there are finitely ma
massaging this condition into a contradiction, and then concluding that a finite
number of primes is imp
ny primes,
ossible.
e prime numbers, and call them . So suppose you have a list of all th
. Notice that for any Then we'll form the integer in our list of
primes, we cannot have
; if we did, then we'd also know that
(1)
20
But we know that N has to have at least 1 prime factor p. Since this prime number
isn't one of the primes in our list, we conclude that the list of primes we started
with was incomplete.
off
Gaps and Clusters in Primes
reasonable idea of how the primes are spaced out amongst the intege
Now that we know there are infinitely many primes, we might want to have a
rs. Displaying
l quirkiness, the answer to this question seems to be on both extremes:
imes have wide gaps to their next neighbor, while — conjecturally, at least —
others are as close as can be.
their typica
some pr
On the one extreme, we have a theorem which tells us that large gaps between
primes numbers are known to exist.
For any positive integer M, there is a string of at least M consecutive composite
integers.
Proof: The M integers between
(2)
are all composite, since the first is divisible by 2, the second by 3, etc.
suggests that there are also lots of primes
which are quite close to each other. The most famous result in this vein is
On the other hand, empirical evidence
The Twin Prime Conjecture: There are infinitely many primes p such that p+2 is also
prime.
For those who are interested, the record largest twin primes to date can be found at
argest Known Primes Page; as of this morning, the largest twin primes were
(3)
The L
two numbers which have something like 60,000 digits.
The Prime Number Theorem
the spreading out and bunching up between the prime numbers, one might
think that it would be hard to give an estimate for the number of primes of a given
With all
magnitude. However, one of the biggest results in number theory — and one which is
almost always proved using techniques from complex analysis (!) — tells us exactly
21
this information. It uses a function , which is defined as the number of primes
less than or equal to a given numbe . (So, for instance, we have r x since the
primes less than or equal to 11 are .
The Prime Number Theorem: as .
This says that for large values of x, the number of primes less than or equal to x is
about .
Primes of a particular form
Now that we know a little bit about primes, it is natural to ask: how can we go about
finding them? The answer to this question, sadly, is that there's not really a general
method for finding all primes aside from ''brute force'' techniques like our sieve
in finding pri is one of the hard problems which
helps keep our world afloat right now: encryption online is dependent on the fact that
oug it's hard to come up with an exhaustive list of all primes, there are some
. Although finding large primes
ematicians a hundred years or so
ago, today it is big business: the aforementioned internet security applications of
appen to take a particular form: they can be
expressed as
method. Indeed, the difficulty mes
it's really hard to factor large numbers.
Even th h
places where prime hunters go to search for big game
was a kind of pleasant amusement amongst math
primality require large primes to work. Hopefully we'll be able to talk about all this
more at the end of the term.
Mersenne Primes
The largest primes found these days all h
for a prime number p. These are the so-called Mersenne Primes.
There was a recent development (i.e., early last semester), when the Great Internet
Mersenne Prime Search (GIMPS) came across the new largest prime number. This
number is
and has around 13 million digits. If you want, you can use your computer to help
e your computer which finds the next largest prime!
numbers? Since primes are defined based upon a multiplicative property, one might
GIMPS out; maybe it will b
The Primes Under Addition
Finally, we consider the question: what happens when you add together prime
22
not expect that they really have a lot of interesting additive structure. It seems,
however, that they have a very rich additive structure. For instance, here's a long-
standing conjecture about how the primes behave under addition:
Goldbach's Conjecture: Every even integer at least 4 can be expressed as the sum of
two prime numbers.
Though plenty of smart people have been thinking about this problem for a couple of
en verified for "lots" of even numbers (can
someone post to the Wiki how many even integers have been verified to satisfy this
wn the
Fundamental Theorem of Arithmetic. This theorem is something which you all have
ase
hundred years, and although it has be
condition?), no one has yet been able to prove that it is true.
The Fundamental Theorem of Arithmetic
Having covered many of the basics, it's now time for us to knock do
seen many times before — whether explicitly or not — and is an incredibly useful tool
in number theory.
We need a preliminary lemma before we can knock down the Fundamental Theorem.
This preliminary result is known as Euclid's Lemma, and it is essentially a special c
of one of your homework problems for the week (44a).
. Euclid's Lemma: If p is a prime number and , then either or
Proof: Suppose that , and we'll argue that . For this, notice that forces
— the only divisors of p are 1 and itself, and we already know that p isn't
divisor of a. Applying 44a from your homework gives the desired result.
a
The Proof of the Fundamental Theorem
We're now ready to prove the Fundamental Theorem of Arithmetic. Recall that it says
The Fundamental Theorem of Arithmetic - Every positive integer at least 2 can be
uniquely expressed as a product of prime numbers.
We'll break our proof into two parts
Existence1. : that every can be written as for some prime
numbers
2. Uniqueness: there is only one such way to factor a given integer
23
We only had time in class to cover the first statement; we'll prove the second in class
on Monday.
Existence: Suppose that there were integers greater than 1 which couldn't be factored
into a product of primes. This would mean that there is a smallest such integer (by
l call this smallest element n. Now n can't be
prime since otherwise n is already an expression of itself as a product of primes.
the well-ordering principle), and we'l
Hence for some . Since both a and ller than n, this
means that they m s
b are sma
u t be elements which do have prime factorizations (since n was
Therefore selected as the smallest positive integer which didn't have this property).
and for appropriate primes . But then we have
(4)
a prime factorization of n. Since this contradicts the selection of n as the least
element without a prime factorization, we must conclude that every integer greate
than 1 can be factored as a product of primes.
undament Theorem
r
Lecture 6 The F al
and its Applications
Today we began by finishing off the proof of the fundamental theorem of arithmetic.
ter we completed the proof, we saw how the fundamental theorem could be used to
cilitate the computation of GCDs and LCMs, and we also used it to prove that there
got
Finishing off the FTA
e were in the midst of proving the Fundamental Theorem of
says
Summary
Af
fa
are infinitely many primes which leave remainder 3 after division by 4. Finally, we
a sneak peak at the fundamental concept in chapter 2: congruence of integers.
In class last time w
Arithmetic, which
The Fundamental Theorem of Arithmetic: Any integer can be uniquely
expressed as a product of prime numbers.
24
We had already shown that every integer has a factorization into primes, but we had
left to show that this factorization was unique. That's what we'll do now.
To show that any integer n has only o
factorizations of an integer n:
ne prime factorization, suppose we're given two
(1)
This expression just means that each of the are prime, and that the
corresponding exponents are positive.
We aim to show that the list of primes and are indeed the
same, and moreover that the corresponding exponents match up as well.
For this, we start by noting that for each , the term clearly divides
the first expression of n as a product of primes. For this reason we must also have
, and the supped up version of Euclid's Lemma says that for
some j. But since is prime, this means that in fact . Hence the list of primes
for the first factorization is a subset of the list of prim the second factorization.
e
es for
Running the same argument for a given prim in the second factorizatio
that the list of primes for the second factor
the first. Hence the list of primes are, in al. An entical as sets,
but — by virtue of our increasing ordering of the
n, we have
ization is a subset of the list of primes for
fact, identic d not just id
and — we must in fact have
and .
Now that the lists of primes are identical, we just need to show that . For this,
suppose that . Then we have
(2)
Now clearly divides the first expression (since ), whereas it cannot
divide the second expression (since doesn't sho up in the factorization). This is a
iction so
w
contrad , and we must have . A similar argument shows that is
impossible, and so we have .
GCDs, LCMs and FTA
To see that the Fundamental Theorem can be used to make our lives easier, we're
going to show how it relates to a concept already discussed (GCDs) as well as a close
25
cousin (LCMs or least common multiples). This se has yet to be mentio
in this class, so we give a definition
, cond term ned
Definition: For two positive integers a and b, the least common multiple of a and b —
written either or sometimes — is the smallest number m so that
and .
Example: Computing
In order to compute the least common multiple of 10 and 6, we should write down all
common multiples of these two numbers:
(3)
Notice that the first number common to both lists is 30, and so that means that 30 is
le of and 6. Notice that in this case the least common
multiple of a and b was *not* simply ab; there was a smaller common multiple than
the least common multip 10
the "obvious" common multiple.
Instead of writing out a list of common multiples, it would be nice if we had a uniform
o computer LCMs. In fact, there is a connection between GCDs and LCMs that
makes one easy to compute whenever you have the other. This is given by the
following
way t
. Theorem: For any integers a and b, one has
One can prove this result from the definition of GCDs and LCMs, but it because quite
cumbersome. Instead, one can prove it by taking advantage of the following result
that's borne from the Fundamental Theorem.
Theorem: For integers a and b with prime factorizations and
, one has
and
.
26
We won't prove this result in class, but we'll use it to prov relae the theorem ting the
GCD and LCM of two numbers above.
//Proof that // :
We'll assume that a and b have factorizations given by and
as in the theorem above. This then lets us substitute in the values of
and in the product :
(4)
Now we just notice that for any integers x and y we have
. Hence the product above becomes
es of a Particular form
We asked a while back how primes spread out amongst different classes of integers.
can be written as
(5)
Prim
For instance, we asked: how many primes are there which for
some integer k? We saw pretty t took this form.
We also asked: how many primes can be written in the form
quickly that there were no primes tha
for some integer
is time we argued that there was only 1 such prime (namely 2). This left primes
of the form
k? Th
or , and we asked: are there "more" primes of one form
than the other? We'll start to give an answer to this question today in class with a
roof of the following p
Theorem: There exist infinitely many primes p for which there exists with
.
In order to do this, we first note the following
Lemma: The product of two integers of the form and is another integer
of the form .
Proof: It isn't hard to see that
(6)
27
gives the desired result.Taking .
Now we're ready to prove our theorem above
Proof of Theorem: Suppose, to the contrary, that there are only finitely many such
primes. We'll list these primes out in order: , with the largest such
prime denoted . We claim that the integer contains a prime
divisor not on our list.
To see this, note first that N is an odd number, so its prime factorization contains o y
odd primes. If all these primes were of the form
nl
, then so too would N be of
(using induction H ce there exists at least one this form on our previous lemma). en
prime divisor p of N for which for some integer n.
We claim that p is not included in our list of primes. Suppose first that . By ou
result on divisibility of integral linear combina that
r
tions, this implies
. Hence Euclid's Lemma implies that either (which it
doesn't) or for some i (also impossible). Hence we're led to a contradiction, and
so we must have .
ve remSince we have a complete list of primes which ha ainder 3 after division by 4,
this means that for some . But then we have
— another clear contradiction. o conclude that for any of the We're left t in
our supposed complete list of primes of the form , and hence our list must
. have been incomplete.
Though an awfully nice w t re
ny pr
result, we can't adapt this technique to sho hat there a
infinitely ma imes of the form — we would need a result that says the
product of two prim hich toes w ok the form again takes that form, but this is
NOT true. Hence we have to be more clever want to prove such a result. Indeed,
if we
studying problems such as these makes us wonder how many primes there are of the
form or — of other . Though the proof
goes beyond the means
primes of that form
or plenty possible prime types
we have in this class, there is a big result which tells us about
Dirichlet's Theorem on Primes in an Arithmetic P grero ssion: For any integers a and b
with , the sequence
contains infinitely many prime numbers.
28
The proof of this result uses complex analysis to show that
(7)
diverges. Crazy!
The topics we've covered so far — basic ide
A Preview of Chapter 2: Congruence
as which are born from the concept of
ory as of a few hundred
years ago. Our next concept — the notion of modular congruence — was developed
uss and was a key result for moving forward in number theory. The basic idea
centers around the following
divisibility — cover most of the basic tools used in number the
by Ga
Definition: Two integers a and b are said to be congruence (or equivalent) module an
integer m — written — if .
We'll see that this definition provides a relation that has a lot of the properties of our
zes that this new
ultiplication) just
odular Arithmetic
ut
o addition and multiplication. We started to explore
can use these arithmetic properties of congruences to prove results about
tegers.
e key idea in this chapter centers around the following
"usual" equality. What is truly powerful, though, is when one reali
version of "equality" admits arithmetic operations (addition and m
like our usual notion of equality.
Lecture 7 M
Summary
Today in class we introduced the notion of modular congruence and saw that it can be
used to give an equivalence relation to the integers. Not only does this provide us
with a way to split the integers up into distinct subgroups (a so-called "partition"), b
it also behaves well with respect t
how we
in
Congruence
Th
29
Definition: Two integers a and b are said to be congruence (or equivalent) module an
integer m — written — if .
flexive: for any integer a and any modulus m, we have
One of the benefits of modular congruence is that it behaves an awful lot like the
regular "equals" you're used to playing with. In fact, modular congruence is an
equivalence relation, which means it has the following properties
1. Re .
2. Symmetric: for any integers a and b and any modulus m, if
then .
3. Transitive: for any integers a,b and c, and any modulus m, if
and , then .
Proof: To prove the reflexive property, note that just means that we
want to verify . We s hile back, though, that any integer m
divides 0, so this statement is valid.
aw a w
To prove symmetry, we need to show that implies . If
, though, the definition of modular congruence tells us that ,
so that . But then we have , and so
. By the definition of modular congruence, we therefore have
.
ansiti that Finally, for tr vity we are supposed to assume and
, and somehow conclude that . To prove this result, we
us that note that the first two congruence conditions tells and .
Our result on d tegral linear combinations, then, tells us that ivisibility of in
. Hence the definition of modular congruence tells us
that .
The benefit of t thi
tells us that congruence class partition the integers into distinct sets. For instance,
e modulus is ery integer fits into one of t
collections
(1)
showing that modular congruence is an equivalence relation is tha s
when th 3, we know that ev he three
30
We know this has to be true b r
emainder eithe 0 2 after trying to divide by 3.
ecause the division algorithm tells us that any numbe
has r r ,1 or
Example: Negative Numbers and Congruences
Suppose you want to know what the integer -2 is congruent to mod 3. The definition
tells us that is the same as saying is divisible by 3. Note that
choosing makes something which isn't divisible by 3, whereas
ing . Since leaves us with choos , we have
.
Notice, however, that the way we've written these subsets isn't unique. For instance,
since , the transitive property of congruence hows that s
(2)
With this observation in mind, one might be curious to know all the different ways of
writing representatives for the congruence classes of a given modulus. This leads to
the following
Definition: A ction of s called a complete residue system for modul colle integers i us m
if every in is congruent m from the cteger odulo m to exactly one element ollection.
Example: Complete residue systems for
The di m tells us that vision algorith is a complete residue system for .
But notice that so too are and . On the other hand, note that
is not a complete residue system, since it has a repeated congruence
cally, class; specifi . On the other hand, the set fails to be a
complete residue system because not every integer is congruence to either 1 or 2. In
particular, and .
The fact that is a complete residue system for comes from the
following more general result
Lemma: For any integer m, the set is a complete residue system
modulo m.
This complete residue system is so important that it gets its own name: it is called the
. least non-negative residue system for m
31
Congruence ithmetic
The reason
and Ar
that congruences are so important in number theory is that the notion of
congruence plays well with addition and multiplication. By this we mean
and For integers with , we have
and
This result is important because it tells us that when we're doing arithmetic
m. We'll see an example of this after we
Proof:
computations module m, we can do our computations by choosing any integers which
sit in the given congruence classes module
prove the theorem.
We're given that and , and these statements
translate into the divisibility statements and . By our result on
divisibility of integral linear combinations, we have that
(3)
This d tatem nt,ivisibility s e in turn, tells us that . To verify the
second statement, we'll choose a different integral linear combination:
(4)
and by the definition of congruence we have .
ion work modulo six
Example: Arithmetic Modulo 6
The following tables tells us how addition and multiplicat
+ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
32
4 4 5 0 1 2 3
5 5 0 1 2 3 4
x 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 1
An important thing to notice about this table is that it gives us examples where
d .oes not imply For example, notice that we can find
a and b so that , and yet . Canceling coefficients is
e in ou p ab e used to, so you need to be wary when doing
modular arithmetic that you aren't carelessly "dividing" by constants. The following
m te us ac w t r tionship such a,b have to each other.
som th g y 're rob ly r ally
lem a lls ex tly ha ela
Lemma : is e ivalent toqu .
am le anceling" ef ients in modular equations
tic ha n e p ab e, anytime we have a and b such that
Ex p : "C co fic
No e t t i the xam le ov
, we also have . For instance, we could choose
and , in which case we'd have . Likewise if we have a and
//b/ such that , t that hen it follows ; for example, if we
choose and , then we get .
Though we didn't get to prove this lemma in class, I'll give a sketch of part of the
proof below.
Proof: We 'll only prove the direction, le direaving the other ction for the
enthusiastic student. Now if we're told that , then this translates to
the divisibility statement . Hence there is some integer e so that
33
. If we write , then we can divide each of m an
of integers
d c by d and
get an equation
(5)
Now since this is an equation of integers, we can cancel out the d on both sides, an
we're left with
d
the div
(6)
from which we have . We know that by an old result, and we
also know that this relative primality result together with our divisibility condition
implies that
(7)
from which we find as desired.
ple: A divisibility criterion for 11
People have been talking about divisibility criteria for integers on the forum, so I
thought we might talk about how one goes about proving such a result. We'll prove
Exam
the following
Divisibility Criterion: Suppose that a number n has digits , meaning that
. Then n is divisible by 11 if and only if the alternating sum of its
digits — — is divisible by 11.
Proof: To see that th l take the equis is true, we'l ation and consider
what it gives us modulo 11. On the left hand side we just get , but on the
right hand side we get
. Notice that . Now since "modding by 11"
plays nicely with addition and multiplication, this means that
.
Hence this me ans that our equation become
(8)
s
34
and only if In particular, if , meaning that n
gits is divisible is divisible by 11 if and only if the alternating sum of its di by 11.
Example: Computing
large powers of a given integer can be computed. As an exam
large powers (modularly)
One of the benefits of modular arithmetic is that it provides a context in which really
ple, we'll compute
. (This is more than the ms in the u number of ato niverse!). To do this,
we start by computing successive square powers of 10. To make this as efficient as
ble, I'll often use the trick of substituting a given integer with another small
integer which is equivalent modulo 13 (for instance, using the facts that
possi
and that . Witness:
(9)
Now when it comes to computing , we just notice that
(10)
Lecture 8 Linear Congruence Equations
Summary
Today we started by reviewing how one can go about "canceling" common factors in
congruence equations. Afterwards we introduced the notion of a linear congruence
an equation has solutions
(and, indeed, how many solutions exist).
equation, giving a theorem which told us exactly when such
35
Cancellation in Modular Equations
st class period we stated the following La
Proposition: is equivalent to .
This result will play a central role in today's lecture, so we'll start by giving a full
proof.
Proof: We'll use the notation to make our exposition easier to read.
Suppose first that . This means that , so that there exists an
integer e with . Multiplying this equation by c then gives .
Now notice that since we can rewrite the left hand side as , where each of
these terms are bona fide integers. Hence we have
(1)
Since the left hand side is a multiple of m, we conclude that .
'll prove the direction. Since we're told that Now we , this
translates to the divisibility statement . Hence there is some integer e so
that . We can divide each of m and c by d (since ) and get an
equation of integers
(2)
Now since this is an equation of integers, we can cancel out the d on both sides, and
left with the div
(3)
we're
from which we have . We know that by an old result, and we
sult togethe
implies that
also know that this relative primality re r with our divisibility condition
(4)
36
from which we find as desired.
Linear Congruences
Now that we've played around a bit with modular arithmetic, it's time that we take
one of our favorite problems in mathematics and give t a modular spin: solving
equations. We'll start off at the beginning, dealing with linear equations.
i
Definition: For integers a,b and m, the equation is called a linear
congruence.
The goal, of course, is to find all integers x which solve this equation. Given that the
equation is really a statement about modular congruence, though, you won't be
surprised to hear that we're actually most interes n knowing solutions to the
system modulo m; that is to say, we want to know which congruence classes modulo
uation.
e equations
.
ted i
m solve the given eq
Example: Some linear congruenc
From our multiplication table, we can read off solutions to some equations modulo 6
(5)
So we see that our linear congruences can have either no solutions, 1 solution, or
several solutions (where by "solutions" we mean more precisely "distinct solutions
modulo m"). The question, th
en, is how to distinguish when an equation does have a
solution from when it doesn't. And if it does have a solution, how can we produce all
solutions? How many solutions will there be?
Big Theorem on Linear Congruences: The congruence has integer
solutions if and only if . If is such a solution, then all other integral
, where solutions take the form . A complete list of the distinct
when solutions modulo m is given by .
37
Proof: For notational convenience, we'll write d for the gcd of a and m. Now we'll
proceed with the proof in steps: (1) show that solutions exist if and only if ; (2)
t
show that other solutions can be expressed in terms of one fixed solution; (3)
determine when two integer solutions are congruent modulo m. (Note: we didn't ge
to prove this last part in class today.)
Step 1: First, suppose that a solution exists to the equation . This
implies that , so that there exists an integer e with .
Rearranging, we therefore have
(6)
Now since d is the gcd of a and m we have and , and therefor divides
any integral linear combination of a and m. In particular,
e d
(7)
Hence if our congruence equation has a solution, then .
en Now we'll prove the converse, showing that a solution exists wh . We st
ere exists
(8)
art by
noting that th integers r and s such that
this follows because the gcd of two integers can be expressed as an integral
the fact that combination of the two integers. Now using , we find an integer e s
that
o
. Multiplying the displayed equation by e then gives
(9)
Taking this equation modulo m, we therefore have , and hence
is an integer solution to the equation .
Step 2: Now suppose we are given two solutions to the equation, and , and we
wish to show that . In order to do this, note that we have
(10)
38
This tells us that , and so it follows that —
we turn this divisibility statement into an equation, there ts some integer
or, if
exis k so
that . We'll divide this equation on both sides by d — a legal move
d is a common divisor of a and m — and we find that since . This
is equivalent to the divisibility condition . Notice, however, that
, and hence homework problem 44a tells us that . This is the
same as saying that , so that as desired.
Step 3: (Note: we didn't get to discuss this proof in class, but I'm including it in the
e we pick up tw
notes for people interested in seeing the full proof.) To find the distinct solutions
(modulo m), suppos o solutions and which are the same modulo
m. Since by the previous step, this means that we have
(11)
Getting rid of the that is common to both sides, we turn this divisibility condition
into an equation: . Hence we have
(12)
and after canceling th s of the e al move since m's on both side quation (a leg e this is
an equation in integers, not a congruence equation) we're left with —
. i.e., that
This tells us that two solutions and are distinct if and only if
. Hence the distinct solutions to are given as
when .
Example: Solving
Suppose you want to solve the equation . Notice that the gcd of 6
and 15 is 3, and that . Our big theorem tells us that this congruence equation
solutions. has no
Example: Solving
Let's put these ideas in practice to try to solve . To
decide wh en
start, we need to
ether this congru ce will have solutions or not. For this, we just notice that
39
, and that . Hen lutions, and we're expectin
be 2 distinct solutions modulo 14
h soluti
ce we know there are so g
that there should .
To find one suc on, we need to do two things:
xpress 2 as a linear combination of 4 and 4, and
2. we need to express 6 as a multiple of 2.
we k
e so:
1. we need to e
Toward the first goal, now that we can to use the Euclidean Algorithm. The
algorithm runs lik
(13)
and from this we see that
(14)
Now for the second goal, it isn't too hard to see that . Finding a solution,
ation by 3:
(15)
then means we should multiply our expression of 2 as a linear combin
Taking this equation modulo 14 leaves us with
(16)
and hence is one integer solution.
at we have one solution , we can find all solutions by taking Now th for
. Doing so shows that the distinct solutio lo 14 are given by ns modu
.
Notice that if we had been interested in least non-negative solutions, we would write
5 in place of -9 (since ) and 12 in place of -2 (since
).
40
Lecture 9 Multiplicative Inverses; the Chinese Remainder Theorem
Summary
talking about a spec ce equations,
namely those of the form
We started off today by ial class of linear congruen
. These led to multiplicative inverses, which
we saw were useful in solving certain congruence equations. We drove this point
me when we used multiplicative inverses to prove the Chinese Remainder Theorem,
tool that is used to solve simultaneous linear congruence equations.
ime we talked about solving linear congruence equations. Let's do another
ample of this kind of problem.
g the linear congruence
ho
a
Multiplicative Inverses
Last t
ex
Example: Solvin
Suppose we want to solve the equation . We first check to see if
solutions exist. In this case, we know that , and since , we
know there are solutions. In fact, we know that there is exactly 1 solution mod 67
compute it, we first need to write
. To
as a linear combination. We'll use the
Euclidean Algorithm. This gives
(1)
Now we can use these equations to express 1 n as a combinatio of 5 and 67:
(2)
Taking this equation modulo 67 shows that , and so 27 is the
multiplicative inverse of 5 modulo 6
7.
This example leads to the following
Definition: A solution to the linear congruence is called a
multiplicative inverse for a modulo m.
41
Example: The Inverse of 5 mod 67
The previous example can be translated to say "27 is the multiplicative inverse of 5
modulo 67."
Notice that we already have machinery that tells us when multiplicative inverses exist.
Theorem: An integer a has a multiplicative inverse modulo m if and only if
. When a and m are relatively prime, the multiplicative inverse of a mod m is unique
mod m.
Proof: Recall that has a solution if and only if . Of course
there aren't a lot of choices for what can be if this divisibility is going to hold;
in fact, is the only way this divisibility can hold. Hence must be
ultiplicative inverse mod m.
on solving linear congruences says that the
en by the gcd of a and m. We've already
a and m
relatively prime if a is going to have a m
When a solution does exist, our theorem
number of distinct solutions modulo m is giv
seen that a solution exists if and only if , and so in this case there is only
one solution modulo . m
Solving Congruences Using Inverses
Multiplicative inverses can be quite useful in solving other linear congruence
they allow one to solve a congruence by a simple multiplication.
s, since
Example: Solving
Suppose we wish to solve . We could proceed as e before —
finding a gcd, writing that gcd as a lin mbination, etc. Alternatively, we can use
the fa 've already computed the multiplicative inverse of 5 as 27. To take
this latter route, notice that we have
we hav
ear co
ct that we
(3)
(Notice: we're allowed to multiply by 27 des of the expression without
disturbing the solution s t b
on both si
e ecause , and you'll recall our theorem which
if and only if ). says that
Using the fact that by our previous example, this means that our
solution is .
42
The Chinese e Remaind r Theorem
tic on congruenc
Example: Simultaneous Congruence equations
Suppose you want to find an integer x which satisfies both of the congruen s
(4)
We've now defined arithme e classes mod m, and we've also managed
to solve linear equations mod m. Now we're going to try to solve simultaneous linear
congruences mod m.
ce
We don't have a really g this systematically right now, but you can
that we can
good way for doin
try out some small numbers to see if you can find a solution. For instance, we know
't have since this fails the second congruence; we also can't have
and , since this fails the first congruence. We can similarly rule out
but notice that does satisfy both of these equations. A little more
experimentation shows that works too, and the particularly diligent student
might also come across the solution .
This example shows us that we "experimentally" solve these simultaneous
ic (or efficient) way of computing
solutions. For this, we turn to
congruences, but they don't provide a very systemat
The Chinese Remainder Theorem: If are pairwise relatively prime
integers, then the congruence equations for each have a
unique solution modulo .
Proof: We'll break the proof into two pieces: first we'll construct a simultaneous
solution to the given congruences, and then we'll show this solution is unique in the
given modulus.
To start, we'll de for . fine , and for each we'll write
Now since the are pairwis ively prime, you showed in your homework (in th
course of #43(c) in chapter 1) that
e relat e
. Hence for every i, there exists an
integer which satisfies .
With the so constructed, we claim that
(5)
43
is a solution to all the congruences . To see this is true, fix an integer
i, and we'll show that . Notice that for every we have ,
is the product of all the moduli except for — in particular, since shows
the product which defines N_j$]]. Hence we have
up in
(6)
But now recall t t ha , and so the previous equation becomes
as desired.
Hence we've constructed a solution. To show that all solutions are equivalent modulo
, notice that if are two solutions to the congruence equations,
then we have for every i/. It follows that for
i, and so . By homework 43(c) in Chapter 1, since the every are
relatively prime we can conclude that .
Example: CRT in Action
se nc
(7)
Suppo that we're given the simultaneous congrue es
Our proof of the t we need to start by computing CRT says tha , which in
this case are given as , and . With
these numbers in hand, we now need to solve the congruence equations
for each i.
To solve , notice that . Hence we're rea
solve
lly trying to
. Now we could ing the Eusolve this equation by us clidean
Algorithm to express the gcd of 10 and 3 as a linear combination of the two, but since
can just use "guess and check" to find this inverse. For
this, notice that
the modulus is so small, we
, and that . Hence we have .
, we'll do a similar trick: since , we're
ng to solve
To solve
really tryi . But this makes it clear that we can take
.
44
Finally, we need to solve . Since , we're trying to
solve . Using "guess and check", we see that is the solution
we're after.
tion Now that we've computed all the appropriate terms, our desired solu modulo
is therefore
(8)
You can e d see what its tive residue is, but you don't
need t ant.
xpand that out an least non-nega
o if you don't w
The Chinese Remainder Theorem is a really powerful tool for solving simultaneous
congruences, but lls us how s where the given li are it only te to solve problem modu
pairwise relatively prime. There are plenty of "real life" scenarios in which the moduli
ystem of congrue in which case it'
rsion of the CRT
for your s nces won't be so nice, though, s handy to
know this stronger ve :
Strengthened Chinese Remainder Theorem: For arbitrary integers and
congruence equations , th s a simultaneo ere exist us solution if and
only if for every . When a solution exis s unique modulots, it i
the least common multiple of the .
Example: The Strengthened CRT
Suppose someone asks you to solve the simultaneous equations
We won't worry about proving this for now, but it is good to have in mind.
(9)
Since you know that and since , you know that this
simultaneous system has no solutions.
Example: The Strengthened CRT
Suppose someone asks you to solve the simultaneous equations
(10)
45
Note that since , there will be a solution to this equation (and the
solution will be unique modulo ). To find this solution, we note that the
first equation translates to the exi eger e so that stence of an int . This
means that , and we can plug this value of x into the second equation:
(11)
Of course this equation is equivalent to , which is just a linear
ble e. W the methodo veloped
w that
congruence in the varia e can use logy we've already de
to solve this equation, and doing so will sho are the solutions mod
15. Plugging these values back into our initial expr , we see that ession for x
46
. Notice that all these values of x are congruence mod , so they
all give the same solution to our equation (mod ).