Ok so welcome to today's talk lecture on number
theory so today we shall be discussing about
some elementary concepts in number theory
which forms the kind of very a center i mean
it is very central to the field of the cryptology
so we shall i mean it is a really fascinating
field and quiet a big vast field but, today
we shall essentially understand some of the
basic concepts
So today's objective of this particular talk
shall be on congruences modular arithmetic
introduction to field theory euler totient
function and fermat's little theorem so we
shall be talking about these topics so one
of them is congruences which is modular arithmetic
which is very important to understand field
of cryptology and then we shall try to give
the define the concepts of what is meant by
mathematical field which is very again very
central idea to the study and then discuss
about a particular function which is called
as euler totient function very popular and
very useful kind of parameter and then conclude
the talk with discussion on fermat's little
theorem so we will be actually discussing
about fermat's euler theorem
So starting so essentially we will find that
when we're doing cryptology or when we're
doing a normal kind of computation i mean
in in related to the ciphers then we essentially
do not with deal with infinite sequences so
we for example, we do not have a set which
rather we do not have a set which is which
can be which can take infinitely large number
of values so which means that there is a finite
number of values and we are supposed do our
computations on finite values so therefore,
there is i mean we have to define our arithmetic
in a finite set of values so for that or rather
in order to study such kind of operations
one basic concept or one very important concept
is something which is called as congruences
So we say that a is congruent to b modulo
m so a and b are supposed to be integral integer
numbers and we say that a is congruent to
b modulo m and we denote them as them as follows
that is a is congruent to b mod m if m divides
a minus b or b minus a so that is the basic
idea therefore, let us see some examples like
minus two will be is equal to will be congruent
to nineteen modulo twenty one the reason being
that twenty one divides nineteen minus of
minus two that is twenty one and similarly,
twenty will be zero mod ten because if you
take or rather one way of thinking is that
ten will divide twenty minus zero or zero
minus twenty
So i mean so therefore, i mean the idea is
that if you just think in terms of numbers
then this number if we divide by the modulus
of operation and then this particular number
on the right hand side is nothing but, the
remainder therefore, so that is the way of
how to kind of compute modular computations
so this is nothing but, the simple modulo
computation now you may note it that congruence
modulo m is actually an equivalence relation
why now because it is an equivalence relation
on the set of integers so let us try to reason
it out
So we know that for in order for a particular
relation to be satisfying their properties
of an equivalence relation it has to satisfy
three important concepts of reflexivity symmetricity
and also transitivity so let us try to first
of all argue why it is reflexive so we see
that reflexive its reflexive because any integer
is congruent to itself modulo m therefore,
if you take a number a or other integers take
any number then this number is also congruent
to itself so why it is because of this if
you take a number a and discus that whether
and i say that a is kind of a is congruent
a mod m right therefore, this is perfectly
why now because this means that m divides
a minus a which is zero ok
So this is true therefore, a is congruent
to a modulo m therefore, the reflexivity relationship
holds therefore, if i define my relation by
congruency which means that number a is related
to b if a is congruent to b modulo m this
is the basic definition of the of of the relation
that we are considering of the congruence
relation then this relation satisfies the
property of reflexivity because of this reason
Now what about symmetricity now you see that
in terms of symmetricity also it holds because
as we say that if a is congruent to b modulo
m then it also holds that b is congruent to
a modulo m why now because this says that
if m divides a minus b then m also divides
b minus a therefore, the symmetric the symmetricity
relation also holds and this is also true
for all all a and b therefore, it is true
for all integer values a and b now this is
really important it holds for this relation
holds the symmetricity properties holds for
all values of a and m b hm
What about transitivity so you see that in
it is also transitive why because if a is
related to b mod m and i tell u that b is
congruent to c mod m then this implies that
a is congruent to c mod m why now because
you see that m divides a minus b and m divides
b minus c so from there we can conclude that
m also divides a minus c why now because you
could have written this a minus b mod simply
as some multiple of m similarly, you could
have written b minus c as some other multiple
of m ok
So now if you have added these two equations
right if you have simply added these two equations
then you would have got a minus b plus b minus
c is nothing but, lambda plus mu which is
again another integer times m now form here
b and b cancels and we have got a minus c
is equal to some other constant we can call
that to be some epsilon or something so it
is something some integer multiplied with
m and therefore, this means that m divides
a minus c and therefore, this this also holds
true so we see that this relation is also
transitive and therefore, since it is and
this will hold for any such a b and c right
and therefore, since it satisfies the properties
of reflexivity symmetricity and also of transitivity
we say that this particular relation of congruence
also satisfies the i mean it is an equivalence
relation ok
So therefore, it basically induces a partition
on the set of integers we know that equivalence
relation induces a set of partition- i mean
induces a induces a partition on the set on
the set right therefore, it holds this property
holds true
Now the following are actually some equivalence
statements and you can easily verify them
it says that a is congruent to b mod m is
same as saying that there is an integer value
of k which satisfies that a is equal to b
plus k m so this is what i just just wrote
in the previous slide so you see that m divides
a minus b because a minus b is k which is
an integer multiplied with m it is an integral
multiple of m now when divided by m then both
a and b leave the same reminder so these is
also same as saying as a is congruent to b
modulo m now equivalence class of a of a modulo
m that is a modulo m consists of all integer
that are obtained by adding a with integral
multiples of m now this is called or defined
as the residue class of a mod m so which means
that if i say that the class which is called
the residue class of a mod m
Then what we do is that we take a and we add
all possible integral multiples of m therefore,
therefore, by saying you that there is a residue
class of there is a residue class of a mod
m thin it means that this consists of all
integer so it is basically a set of all integers
which are obtained by adding a integer multiples
of of m therefore, it could be like a plus
minus m a plus minus two m and so on so the
basic idea is that if we just take any number
from here and you divide it by m then the
remainder which you obtained is a therefore,
the remainder defines that particular class
therefore, this class is a residue class of
a mod m and it is also an equivalence class
because it satisfies i mean it it satisfies
the properties of equivalence class we are
just seeing that right
So therefore, so here is an example for residue
class of one modulo four so one modulo four
will be one and then one plus minus four therefore,
and again one plus minus two multiplied four
and then one plus minus three multiple multiplied
by four and so on so the set of residue classes
mod m is denoted by z m z so z this is also
denoted in this in this fashion
So this particular notation is also,metimes
known as z m it is also denoted as z i mean
the you can also denote it as like this so
it is also an equivalent way of may be shorter
way of representing this z so this this just
defines that this is the set of residue classes
mod m and it is denoted by z slash m z and
that equal to z m therefore, now now this
particular set we will have all the possible
all the possible values so it'll essentially
have m values so what are the m values possible
the m values will run from zero one and till
m minus one so this is the set which essentially
comprises of this values that is from zero
two m minus one now this is also called an
a complete system or a complete set of incongruent
residues therefore, you see that if you just
take a number so an example of this would
like let us take the example of ten and then
define like what is z by ten z so then this
will comprise of numbers from zero one till
m till nine that is ten minus one is nine
So that means that if we just take m is equal
to ten then these are the possible values
of numbers which are which are possible i
mean maximally possible and which also have
which are also incongruent to each other so
that means that if you take any numbers from
this from this set for example, let us take
like one example one one random choice could
be like two and another arbitrary choice could
be say five then two and five are never congruent
to each other modulo ten so that is why why
is two not congruent to five modulo ten because
ten will not divide two minus five or five
minus two so that means all the elements in
this set are actually incongruent to each
other and this is the maximal possible possible
number of incongruent values that can that
can happen ok
You can take any number of outside this particular
set for example, if you just take an arbitrary
choice like say twelve then you'll find one
number in this set which is congruent to this
number modulo ten which is that number for
example, you can choose two so you see that
two and twelve are congruent to each other
modulo ten why now because ten will divide
twelve minus two right therefore, you'll find
that this particular set which is like z defined
by z slash m z is also called a complete set
of incongruent residues for a complete system
Now there are example this is an example given
here for complete system for modulo five so
we know now for mod five it will be like from
zero to five minus one that is zero one till
four and this an another example you can verify
like that this is minus twelve minus fifteen
eighty two and minus one and thirty two this
set is also a complete system why now this
is also a complete system because if you just
take these numbers like you take minus twelve
so we what we are considering is z slash five
z right so you see that in this set if we
just take twelve then twelve is actually so
it is minus twelve right so minus twelve if
you take modulo five so what does it mean
what is this value so if you divide twelve
by five therefore, you will get that this
is equal to is it correct yes because if you
just take this number five then five you divide
minus twelve plus two because that is minus
twelve plus two is nothing but, minus ten
so five divided minus ten ok
Now what is minus two if you if you would
like to bring in this among this numbers like
from zero one till/two three and four then
this particular number minus two will be actually
equal to three therefore, because it is you
can verify this because five will agian divide
minus two minus three so this is same as minus
two modulo five will be same as three modulo
five what about the next number the next number
here in this set was given as minus fifteen
so minus fifteen if you take modulo five what
will be it it will be zero so similarly, you
can do it for the other once also like eighty
two will be equal to two modulo five because
five will divide it into minus two and minus
one will be congruent to four mod five and
thirty one will be congruent one mod four
one one mod five so you see that the numbers
that i obtain after taking the modulo five
essentially are nothing but, another order
of the numbers zero one two three and four
therefore, you see that this is also a kind
of these numbers like this set like minus
twelve minus fifteen eighty two minus one
and thirty one is also another example of
a complete systems and any two numbers from
this set are are incongruent to each other
they are not congruent we can check this
So there is a theorem i mean i am i- i really
do not have i- i am not going the proof but,
its very simple you can verify it like if
a is congruent to b modulo m and c is congruent
to d modulo m then this implies that minus
a and minus b are congruent to each other
modulo m we can also add like a plus c is
congruent b plus d modulo m there should be
a modulo m here and a c is congruent to b
d modulo m so you can verify this and its
very simple like what you can do is that you
can start like writing as a is equal to b
plus some constant multiple of b of m and
c is equal to d plus some constant times m
and then try to find out minus a and and this
is very simple so this is this is quite trivial
to prove
Now what is the implication of this result
now there- there is a very is strong implication
of this result so let me reflect on this see
for example, what we understand here is that
this number a may be quite a large value and
this number c maybe also another quite large
value and suppose you're actually interested
in computing doing computation on large values
now if you would like to do computation of
large values but, you always have this modulo
m then the idea is that what it says is that
instead of taking a plus c or rather adding
a plus c and then doing a modulo m is same
as doing a modulo m on a doing a modulo m
on c and reducing this numbers between zero
to m minus one and then adding these two numbers
so let me kind of illustrate this with a very
simple example
So let us take a large value of a so for example,
you consider that a is again let us consider
that our m value is five and let us see that
a is supposed a large value like may be larger
than this so it may be like twenty seven and
let us consider another value which is say
b or rather c and consider that this is say
something like forty nine now let us try to
add these two numbers so we know that if we
add these two numbers then we get seventy
six and therefore, if i do a modulo five here
that means that i divide this by five and
my remainder is one ok
So therefore, i write that a plus c modulo
m is nothing but, equal to one it is congruent
to one so another way of writing this is rather
the way we are writing this a plus c is equal
to one mod m now let us try to do i mean apply
the previous result so what we will do is
that we will take twenty seven and perform
a modulo five operation so if i do that then
i see that twenty five is divisible by five
so the remainder is two similarly, i take
forty nine and do a modulo five operation
so we see that forty five is divisible so
this is actually is equal to four now you
see that four also can be written as minus
one right so therefore, we write the twenty
seven is equal to two mod five and forty nine
is minus one mod five now if i add these two
numbers like twenty seven plus forty nine
then i can obtain- add this instead of these
two numbers and reducing what i can do is
that i can actually add these two small numbers
and i get one modulo five which matches with
this result
So which means that i need not do the additions
with large number when i am doing a modular
operation right when then i really do not
need to add these large numbers but, what
i can do it is that i can reduce this numbers
modulo five reduce this number modulo five
and then add these two numbers so this particular
particular technique can actually make our
computations quite simple and this is actually
applied in the implementation of cryptosystems
quite regularly to make your implementations
easier ok
So we will squarely get this idea or rather
find more examples of this as we proceed but,
this is a very interesting and very important
results which we should understand similarly,
when we're talking about multiplication we
can also again instead of multiplying two
large numbers which is even more complicated
we can actually reduce these numbers mod m
and then just multiply this small numbers
for example, imagine like if you have to multiply
twenty seven and forty nine and then do a
modulo five operation right then this this
multiplication is not i mean it is not proceeding
well you have to do it and you can make mistake
right and most importantly when your computer
does it it may not make mistakes but, what
it may do is it may require large number of
time right so the resource may be useful may
be used up so instead of that what you can
do is that you can reduce it it becomes two
and this becomes minus one and you know that
this result is very simple this resulting
is nothing but, minus two mod five which is
nothing but, three ok
So you can just straight away from these two
results we can obtain this product that actually
makes your computations much easier otherwise
you would have it would have become a little
more difficult right so this result has got
lot of impact therefore, and here is a another
example so you can probably try to understand
the impact of the previous result from this
example so let let us consider so this is
the very important number in number theory
but, let us not go into that say is the two
to the power two to the power of five plus
one is divisible by six forty one and a proof
is required for that so one way of doing it
like it will be two to the power of five is
thirty two therefore, let us compute two to
the power of thirty two and then add one and
then divide by six forty one and then see
that whether it is divisible or not
But this is quite tedious approach instead
of that lets see this we know that six forty
one can be actually written as six forty plus
one and six forty can be factorized as five
multiplied by two to the power of seven plus
one now this factorial is quite easy right
because its divisible by five and the others
are just powers of two therefore, we can write
as five multiplied by two to the power of
seven is nothing but, minus one modulo six
forty one why now because we know that five
two to the power of two to the power of seven
will be equal to six forty one minus one right
so we can rearrange this as five into two
to the power of seven is equal to six forty
one minus one
Now if take modulo six forty one then this
vanishes and therefore, minus one remains
as a remainder now from the previous result
we know that if i raise this to the left hand
side to the power of four it is same as write
i mean raising the right hand side to the
power of minus one hole to the power of four
when you're doing modulo six forty one operation
because this means that we are multiplying
this four times and therefor the right hand
side also needs to be multiplied four times
therefore, straight away from here we get
five to the power of four multiplied by two
power of twenty eight is nothing but, congruent
to one modulo six forty one
So now we see that five to the power of four
is nothing but, six twenty one six twenty
five and we know we know that need not actually
require to multiply these two numbers but,
we can actually deduce this number modulo
six forty one so that should be make our computations
easy right and then multiplied by two power
of twenty eight and that is congruent one
modulo six forty one so this equation is not
disturbed by this now we see that six twenty
five modulo six forty one is nothing but,
equal to minus sixteen therefore, this result
is nothing but, minus two power of four multiplied
by two power of twenty eight is congruent
to one modulo six forty one and that means
that two power of thirty two is nothing but,
equal to minus one modulo six forty one so
that means that six forty one divides from
the definition of congruence we know that
six forty one divides two power of thirty
two minus of minus one that is six forty one
divides two power of thirty two plus one so
this particular thing is proved
But you see that throughout this we are actually
not done a single bit large computation and
that is the advantage of previous theorem
therefore, idea that when you need do multiplications
or additions or computations on large numbers
and when you're doing a modular operation
then it is better to reduce each individual
numbers and perform the operations that'll
you're your computations easier
So now we actually i mean so this is we actually
shift our focus and go into a little bit of
field theory now we fields are actually very
central to the understanding of crypto that
is for example, the common that is the recent
standard which is known as the advanced encryption
standard relies heavily on finite fields therefore,
we will try to understand some of the basic
of fields that is what are what are the fields
and what are the properties of a mathematical
field ok
So let us start with something which is called
semi groups so first of all let us define
something which is called a transformation
or an operation so here is an x is a set and
let us define a map which is defined by this
circle has nothing but, a transformation which
takes two numbers from this set x let let
it be x one and the other one is x two and
perform an operation therefore, this example
of this could be an addition that is just
a simple plus in the integer domain or multiplication
so, the so you can generalize it means that
i just take two numbers from the set x and
i perform these computations and therefore,
this element therefore, this transforms an
element x one coma x two to the element x
one operation x two and this is vocalize-noise
so this is just this is a simple transformation
ok.
So now the sum of the free residue classes
that is so here is an example like examples
of this transformation as i told you that
integer is one example similarly, you can
also define in the residue classes like that
we just take two residue classes a plus m
z and b plus m z so this was so we know that
a plus m z and b plus m z and if i just need
to find out the some of these residue classes
then this is nothing but, a plus b plus m
z similarly, the product of the residue classes
a plus m z and b plus m z is equal to a multiplied
b plus m z so this is the same thing which
i just told you before this that is if i need
to kind of multiply or add two large numbers
in the modulo m domain then i just find the
remainders right and just add the remainder
or multiply the remainders this is perfectly
defined ok
So now an operation this circle on x is associative
so we now we would like to define the associativity
of this it means that a operation b operation
c and if i do this it can also be associated
as a this operation can also be performed
as this that is a operation b operation c
being given a priority and this holds for
all a b c in x so this is how associativity
is defined ok
And we know what is commutativity which means
that if a operation b is same as b operation
a for all a and b in x then it is said to
be commutative so examples we know that integer
addition is an associative operation right
and it is is it commutative also it is also
commutative but, what about an example which
is not commutative for example, matrix multiplication
is not commutative we know that if i take
two matrix a and another matrix as b and if
i multiply a and b then it may not be the
same as that b and a ok
So now we define what is meant by a semi group
so it says that there is a pair of h so it
is a set so we define a transformation as
just as we told right now so it consists of
a set h and an associative operation so that
means if the operation is also associative
on h and then it is called a semi group so
of semi-groups means simply that there is
a set h with an operation circle and this
operation is also associative so if the operation
is also associative then if the operation
is associative then we say it to be a semi-group
now the semi-group is also called abelian
or commutative if the operation is also commutative
and examples of this could be like integer
set the addition defined over it integer set
multiplication defined over it residue class
set and addition defined over this and the
residue class set and multiplication defined
over this ok
So we know that these are examples of abelian
or commutative semi-groups because this is
also associative because it is a semi-group
and also further its abelian or commutative
so which means you can do it either as a operation
b or as b operation a for all a and b which
belongs in the set so, implications are as
follows so there are some interesting implications
So let h and this operation be defined as
a semi-group and let us define like a let
us define some powers of a as follows so a
power one is nothing but, a and a power n
plus one is defined as a operated with a power
n so this is the recursive definition of a
power n plus one for a which is in h and natural
value of n so this is a natural number n now
if i need to compute a power n and operate
it with a power m then it is same as doing
as a power n plus m or if i do as a power
n and take a whole power m then this is same
as doing a power n m and where a is in h and
n and m are nothing but, natural values or
natural numbers
Now you see that it is a further interest
result which says that a and b are in h and
we perform and if a and if the operation is
commutative that is if a operation b is same
as b operation a then a power the a operation
b and if i raise it to power of n is same
as a raise to power of n operated with b raise
to power of n so we can actually reflect upon
these are kind of definitions and we can actually
i mean we can actually reflect upon this definition
by by for example, choosing the value of n
to be two ok
So let us take n is equal two and consider
the operation as a transformation b raise
the power of two so we now that we can actually
write this as a power of b by our recursive
definition as a power b now we apply we know
that this is this operation is so the first
thing that we start with is this so that is
the first definition but, since this is a
semi-group so we know that it is associative
so we can actually do like a b operation a
operated with b and we know that because if
i apply commutativeity then we actually write
this as a with b and we know that again we
can actually do this that is a apply associativity
and perform this operation like this and this
is nothing but, a square operated with b square
therefore, this we can actually inductively
apply and obtain the results that is a operated
operation b if i raise it to the power of
n is same as a raise to the power of n operation
operated with b power n
So now we let us the define some something
which is called monoids so first of all one
of the definition the central definition of
the monoids is something which is called a
neutral element a neutral element of the semi-group
h operation is an element e in h which satisfies
e operation a as same as a operation e and
that that means that this is equal to a for
all a in h therefore, we see that an easy
way of understanding the neutral element would
be to consider an example so let us consider
the example of let us consider the example
of for example, a field of z and so this is
an integer set this is an integer set and
there is a plus operation defined so we so
we have just defined that this is a this is
an example of if so this was an example of
semi-group so we have defined what is meant
by a semi-group
And let us consider this particular example
and see that how or rather what is the neutral
element in this set so i think all of us know
what is the neutral element in this set so
if i so the by the definition it means that
if i take a and if i apply the operation-
operation plus and there should be a neutral
element a so that i get back a or so this
is called a right neutral element and similarly,
there is something which is called a left
neutral element which says that if i add e
with a and if i obtain a so this is the left
neutral element so this is the right right
neutral element this is the left neutral element
that i get back here now you see that this
defines that so this actually gives us an
idea that this e is nothing but, zero right
so for example, if i add a with zero then
i get back a and if i add zero with a i also
get back a so zero is the neutral element
in the semi-group z plus
Similarly all so therefore, if the semi-group
contains a neutral element like this then
it is actually defined to be something which
is called a monoid so, a semi-group has there
is a result which says that a semi-group has
at most one neutral element and this is actually
not very difficult to prove so the first result
that we can actually kind of reflect is that
the left neutral element and the right neutral
element are the same so if i just consider
the left neutral element so the left neutral
element and the right neutral element are
the same why
Now you can just say that the left the left
neutral element be e one and let the right
neutral element be e two so what about e one
operation e two now you see that if i just
think that this is the left neutral element
then by the definition i know that e one operation
with e two should be e two because this is
the left neutral element now what what if
i think this to be the right neutral element
like if i just think that e two if i think
that e two is the right neutral element then
i know that e one operation e two will be
e one therefore, from here i understand that
e one and e two are the same right
So therefore, the left neutral element and
the right neutral element are the same now
the other thing that holds is that if there
is one neutral element then there can be at
most one right neutral element so this proof
i i mean it is quite easy actually you can
follow the similar principle but, i leave
this as an exercise that is if there is a
left neutral element that is for a left left
neutral element there can be at most one right
neutral element so therefore, so from here
we know that if there can be at most one right
neutral element then this right neutral element
is again same as the left neutral element
by this definition so there can be at most
so from these two results we can conclude
that there can be at most one neutral element
ok
So therefore, if the semi-group which essentially
has a neutral element which is called a monoid
has can can have at most one neutral element
therefore, this result holds true that is
semi-group has got at most one neutral element
now if e which belongs to h is a neutral element
of the semi-group and and of the semi-group
and then b also belongs to h then there is
a definition of inverse like this that is
if i take a and if i operation operate it
with b then this is same as and it is same
as b being operated on a and i obtain e so
for example, when i am doing when i consider
the semi-group z coma plus then what will
be the inverse of a it will be minus a so
because if i know that if i add with minus
a that i obtain back the neutral element which
is zero
So if a has an inverse then a is called invertible
in the semi-group h and in a monoid each element
has at most one inverse so this also result
can be proved that is if there is a monoid
so in the monoid then each element has at
most one inverse so examples are like this
that is if i consider z coma plus the neutral
element is zero and inverse is minus a if
you consider z under multiplication the neutral
element is one and the only invertible elements
are plus one and minus one because if i just
consider any other integer let's- says five
then there is no integer say integer number
if you multiply with which you will obtain
back the neutral element except therefore,
the neutral element exists only for plus one
and minus one
What about the residue class z slash m z plus
and operation is plus then neutral element
is m z itself and the inverse is minus a so
this should be actually minus a plus m z so
it means that if i take if if i just consider
this particular set that is z z m z which
is a semi-group and then the operation is
defined as plus then a neutral element is
m z so that is the neutral element and the
inverse is so this is the neutral element
and the inverse is minus a plus m z so that
is the inverse and you can verify this it
is quite simple so, what about z m z and product
z m z coma product in this case the neutral
element is one plus m z and the inverse and
this result we will see actually is are those
elements t which are actually which are actually
co-plained to m that is every element does
not have an inverse but, only those elements
for only those elements inverse exist which
are actually co-plained to m so which means
that the g c d or the greatest common devisor
of that number and m is actually equal to
one so we will see this as we go ahead
So then we define what is called a group a
group is a monoid in which every element is
invertible so a group is a monoid in which
every element is invertible and the group
is commutative or abelian if the monoid is
also commutative so examples of this will
be like z coma plus which is an abelian group
and z product which is not a group so so this
is actually not an example of a group and
then we have got z coma m m z and plus z slash
m z and plus which is an abelian group so
why is this not a group this is not a group
because from the definition of a group a group
is a monoid in which every element is invertible
and we have just defined just discussed in
the previous slide that is in this particular
set only invertible elements are plus one
and minus one so every element does not i
mean rather every element does not have does
not have an inverse therefore, z coma multiplication
is actually not a group because every element
does not is not invertible in this is not
invertible in these semi-group ok
So therefore, that is that is the definition
of a group and there is higher concept to
this which is called ring and ring is nothing
but, r and they are actually triplets so there
is not only one operation as plus and there
is another operation also along with it such
that r coma plus is an abelian group the way
we have defined and r coma product is r coma
the second operation is actually a monoid
in addition it satisfies the properties of
distributivity which says that x if we multiply-
i mean product if we take product with y plus
z then which is same as x dot y plus x dot
z for x y z which belongs to this ring the
ring is also called commutative if the semi-group
r coma dot that is this one is also commutative
and a unit element of the ring is the neutral
element of the semi-group r coma dot so a
unit element of the ring is a neutral element
of the semi-group r coma dot therefore, a
neutral unit element of this ring will be
as defined as the neutral element of this
of the monoid r coma dot.
So then we define what is called a unit group
which means that let r be a ring with unit
element an element a of r is called invertible
or a unit if it is invertible in the multiplicative
semi-group of r so an element a i repeat this
definition a element a of r is called invertible
or a unit if it is invertible in the multiplicative
semi-group of r that means multiplicative
semi-group of r will be this if it is invertible
in this particular semi-group then it is said
to be invertible and the element a is called
a zero divisor if it is non-zero that is if
the element itself is non-zero and there is
a non-zero b in r such that if i take the
product of a and b or the product of b and
a then we get back zero now units of a commutative
ring actually from a group and this is called
the unit group of the ring and it is often
denoted by r star ok
So that is the definition of a unit group
and we can actually have one very i mean some
example because i think this is little abstract
so in order to understand this let us take
some examples so let us consider the set or
rather let us consider the ring z m z and
let us define let us define rather let us
take the value of m to be something like ten
so we know that the elements here will be
like zero one two and so on till nine now
you consider like whether all the elements
so let us consider like some some let us consider
some interesting facts let us consider the
numbers for example, two and let us consider
like let us multiply two with all possible
elements which are there all possible nonzero
elements which are there in the set
So let us consider two into one two into two
two into three two into four two into five
and so on if you consider these multiply-
these products and if you take the modulo
ten operation that is if you take mod ten
for all of them then you'll find that this
is actually equal to two this is actually
equal to four this is six this is eight but,
this one is zero so this shows us an example
where there are two elements which are nonzero
like two is nonzero and five is also nonzero
but, if i multiply this and if we take i mean
in these particular if i take a modulo ten
then what i obtain back is zero therefore,
from this definition we say that two is actually
a zero divisor the element two is actually
a zero divisor therefore, this particular
thing i mean therefore, i mean this this is
an example which shows that zero divisor exists
therefore, let us come back to this and see
whether element a is called a zero divisor
if it is a nonzero and there is a nonzero
b in r such that a multiplied with b or b
multiplied with a is zero so this is an example
to understand this particular aspect
Now what about this an element a of r is called
invertible or unit if it is invertible in
the multiplicative semi-group of r so let
us see that among these particular i mean
numbers like zero one two and- nine what are
the elements which are invertible so let us
leave out zero because zero is obviously not
invertible let us consider one so we know
that if i take one and multiply it with one
i get back one itself what about two so if
i take two and if i multiply with with any
of the numbers will i get back one so actually
we will see that no you will not get back
two is not invertible ok
So the numbers which are invertible in this
particular set from zero to nine can actually
be found out like this there we see one similarly,
three will invertible four will not be invertible
five could not invertible six will not be
invertible seven will be invertible nine will
be invertible eight will also be invertible
sorry eight not will not be invertible nine
will be invertible and the reason rather the
one you see we have checking is that if i
take the g c d of any of these numbers a and
with ten if a is invertible then this g c
d of and ten should be equal to one so if
the g c d of a coma ten is equal to one then
we we say that a is invertible so a is invertible
if and only if the g c d of a coma ten will
be equal to one so that means you'll find
that this particular set that is one three
seven and nine will actually form something
which is called as unit-group therefore, this
will form a unit-group so i mean it is often
defined as r star in this ring so we will
see more examples of this but, as we proceed
we will see more examples of this
So let us consider zero divisor therefore,
the zero divisors of the residue class z slash
m z is actually a plus m z and this is the
generalization of what we said that is zero
divisors of the residue class z slash m z
is actually a plus m z such that if you take
the g c d of a and with m then it is neither
one nor m but, it is somewhere in between
so that is nontrivial g c d of a and m and
the proof is very simple it says that if a
plus m z if the zero divisor of z slash m
z then there is an integer b this follows
take down the definition such that a b is
congruent to zero modulo m but, neither a
nor b is zero modulo m so which means that
m divides a b so a b is congruent to zero
means m divides a b but, neither a nor b is
actually divisible by m
So this atomically implies that if i take
the g c d of a and m then this should be some
significant value that is it means that if
you take m so, say that a b is congruent to
zero modulo m right so that means that a b
will be equal to i mean rather m divides a
b right so if a m divides a b so which means
that a b divided by rather a b divided by
m this should be a number right this should
be an integer number this is an integer value
but, we know that neither a nor b are actually
divisible by m therefore, a by m is not an
integer b by m is not an integer so that means
that a and b definitely i mean i mean there
should be a cancellation between a and m and
that means that the g c d of a and m should
be actually something which is between one
and m so it is neither one nor m see for example,
if you just consider the example just now
what we considered with m is equal to ten
and a being equal to two and b being equal
to five ok
So now let us consider two into five divided
by ten this ratio so you see that two into
five by ten therefore, it means that the g
c d of this particular thing is actually not
equal to its not it is neither one nor ten
but, it is in between therefore, this shows
that there should be a i mean there should
be a number d which divides both a and d also
divides m there should be a number d and that
number is neither d is not neither equal to
one nor nor is equal to m so this proves that
this shows that m divides a b but, neither
a not nor nor b and therefore, this implies
that the g c d of a and m lies between one
and m.
Now conversely if one this lesser than equal
to g c d is less than g c d is a coma m less
than m then define then if i define a number
b as m divided by g c d a coma m then both
a and b are nonzero modulo m so then both
a and b are nonzero modulo m that is true
from the definition itself right but, if what
what if i multiply a and b so if i multiply
a and b then what i obtain is a so if i if
i take this definition of m divided by g c
d of a coma m and if i multiply this with
a then i obtain zero modulo m the reason is
if i if i multiply this with a and this is
the g c d of a coma m then this divides a
and therefore, what i obtain is an integer
multiple of m and therefore, if i take a congruence
or rather if i divide it by a and take the
remainder then the remainder is zero so this
proves that a b is congruent to zero modulo
m thus a plus m z is a zero divisor of z slash
m z
So therefore, we will see that i mean zero
zero divisors i mean it is very easy to detect
zero divisors how i mean because we just this
simple test reveals the zero divisor if i
take the g c d of number a and along with
m and if the g c d lies between one and m
then actually it forms the zero divisors so
there is a natural corollary to this that
if p is a prime then set z plus z slash p
z will have no zero divisors because because
if i take the g c d of a number with p then
it is always equal to one because that is
the definition of primelity right so that
means that it automatically implies that there
are no zero devisors if p is a prime number
in this particular z slash p z
So then we come to the definition of field
which says that a field is a commutative ring
r plus multiplication in which every element
in the semi-group r multiplication is invertible
therefore, if every element is invertible
then it is a field so examples of this would
be like the set of integers is not a field
because you know why i mean the set of integers
cannot form a field because every number is
not invertible the set of real and complex
numbers from a field and the residue class
modulo prime number except zero is a field
why because we are seeing that because of
this particular result if p is a prime then
z slash p z will have no zero divisors so
that automatically implies that this is not
a field
So then we come to the concept of euler's
totient function which says that if a is greater
than equal to one and m is greater than equal
to two are integers and if g c d of a coma
m is equal to one then we say that a and m
are relatively prime now the number of integers
in z m where m is greater than one that are
relatively prime to m and does not exceed
m is denoted by phi m or which is also called
as euler's totient function or phi function
therefore, we will we will start with this
define it in a recursive fashion and it says
that phi of one is equal to one ok
Now what about let us consider phi of twenty
six so twenty six is a number of english letters
which are there and we let us consider the
value of phi of twenty six we can see that
phi of twenty six is equal to thirteen if
p is a prime then phi of p is p minus one
and if n is equal to one two twenty four then
the values of phi n are as follows so this
is just a enumeration so, you'll see that
you can verify these results but, the fact
is that this function is very irregular that
is you'll not find any distinct even like
it is not like as n increases phi n always
increases you can see that there is defect
time also ok
So now the let us see the properties of phi
n so there is a very interesting result which
says that if m and n are relatively prime
numbers then phi of m n is nothing but, the
product of phi m and phi n now this result
is very interesting and helpful to compute
the phi of higher numbers or larger numbers
for example, phi of seventy seven is same
as phi of seven multiplied by eleven which
is same as phi of seven multiplied by phi
of eleven and we know that if seven is a prime
number then phi of seven is nothing but, six
and phi of eleven is nothing but, ten so the
product is sixty similarly, phi of one eight
nine six divide i mean if i obtain the prime
factorizations i can obtain the result of
six twenty four so this result can be extended
to more than two arguments comprising of pair
wise co-prime integers
So let us see some interesting results that
is it says that if the first results says
that if there are m terms of an arithmetic
progression a p and has got common differences
which are prime to m then the remainders form
z m so if there are m terms of an arithmetic
progression and has got common difference
that is if the common difference is prime
to m then the remainders form z m so this
is a quite simple result which you can check
the other one says that an integer a is relatively
prime to m if and only if its remainder is
relatively prime to m so if an integer a is
relatively prime to m then it automatically
it is an bidirectional implication is that
same as saying as that its remainders is also
relatively prime to m
And the other interesting results is that
if there are m terms of an a p and i've got
a common difference prime to m then if it
is actually a combination of these two results
which says that then there are phi m elements
in the arithmetic progression which are relatively
prime to m because so you can follow like
because it says that the reminders form z
m and you know that if the reminders form
z m and and because i mean an integer is relatively
prime to m only when the remainder is relatively
prime to m so we actually need to find out
the number of remainders which are relatively
prime to m and that follows from the definition
as phi m right therefore, if there are n terms
of an arithmetic progression and the common
difference is also prime to m then there are
actually phi m elements in the arithmetic
progression which are relatively prime to
m
So we will apply this result to obtain this
nice uh result the nice observation so let
us consider phi of m n so we know from the
definition phi of m n means those numbers
or rather means the number of values inside
from one to m minus n which are actually co-prime
to m n so let us now arrange this numbers
from one to m n minus so this is nothing but,
m minus one i mean so this is m n minus n
plus n so this is m n therefore, from one
to m n all this numbers we have arranged them
in this fashion so it is like one two and
so on till n we arrange them in this fashion
So now you consider that among these numbers
if you need to find out so if since m and
n are relatively prime to both m and both
n now you see that there are i mean among
this numbers if you consider like this number
like for example, is so, you'll find that
there are phi n columns that is there are
phi n columns in which all the elements are
co-prime to n so if you see that for example,
this particular column that is the last column
that is n n plus one this this column all
the elements are not co-prime to n because
this is actually you will find that n multiply
this n plus n this m minus one into n plus
n and so on so that means that you need to
find out the number of columns in which all
the elements are co-prime to n and we know
that there're there are actually phi n columns
in which all the elements are co-prime to
n
Now among this columns which are actually
co-prime to n we will like to find out what
are the number which are actually co-prime
to m so we see that we apply the previous
result and see that for example, let us consider
this so in this particular column so assume
that k is actually co-prime to n and consider
this column and the result says the previous
result says that if so what is that what is
the what is the difference here what is the
common difference the common difference is
k now if this k is actually therefore, co-prime
to m then the the previous result says that
if there are m terms of an arithmetic progression
and has a common difference which is prime
to m then there are phi phi m elements in
the arithmetic progression which are relatively
prime to m so that means that these elements
i mean there are phi m elements which are
actually co-prime to m
So if there are phi m elements here which
are co-prime to n and there are phi n five
elements which are actually i mean which all
the elements are co-prime to n then combining
this we obtain that the phi m into phi n numbers
which are actually co-prime to both m and
both n and that actually forms as phi gives
us the number of phi m n ok
So therefore, we conclude like thus there
are phi n columns with phi m elements in each
which are co-prime to both m and n and thus
there are phi m phi n elements which are co-prime
to m n this proves the result
We can actually apply this result and we can
obtain some interesting observations like
phi p ot the power of a will be equal to p
to the power of a minus p to the power of
a minus one and these is quite evident for
a equal to one but, for a greater than one
out of the elements one two p p to the power
of a the elements p p square p to the power
of a minus one are multiplied with p are not
co-prime to p to the power of a the rest will
be co-prime therefore, we can obtain like
phi p to power of a will be p to the power
of a minus p to the power of a minus one and
that is actually equal to p power of a into
one minus one by p
We can actually extend this result and apply
this like follows and compute the phi of any
value in this case we know from the fundamental
theory of arithmetic that we can actually
take n and obtain phi if we and we can obtain
this prime factorizations then it will be
easy to compute the value of phi n so that
means the if factorization of n is available
then computation of phi n can be obtained
using this this this formula so for example,
if i need to the compute phi of sixty then
we can and we know prime factorization of
sixty as four multiplied by three multiplied
by five and phi of sixty will be sixty into
one minus one by two into one minus one by
three into one minus one by five so that is
this is equal to two square therefore, this
is this is straight away application of this
result
So then we see that just let us conclude our
todays talk with theorem of fermat which says
that it is called fermat's little theorem
and its very useful so it is so you see that
if g c d of a coma m is equal to one then
a power phi m is actually congruent to one
modulo m actually this result is euler's fermat's
theorem and variation of this is same as fermat's
theorem it says that if in fermat's little
theorem this this m is actually a prime number
we know we know that if phi m is prime then
phi of p will be equal to p minus one and
therefore, a to the power of p minus one will
be congruent to one modulo m so that is fermat's
little theorem but, let us consider the generalized
theorem and let us consider a set r which
is formed of r one to r phi m we know that
there are phi m elements which forms a reduced
set modulo m
Now if g c d of a coma m is equal to one we
see that a multiplied with r one so we just
take this numbers and you multiply each of
them with a now this if g c d of a coma m
is one then this results also reduced systems
modulo m now this is just to be a permutation
of the set r so this is we have considered
na we have considered one example previously
where we have seen that if you've taken some
numbers and those numbers essentially where
nothing but, rearrangement right like if you
see that in the previous example of this this
this particular set so this particular set
was just a rearrangement of the original of
the original remainder ok
So therefore, similarly, i mean applying this
if you just take this set of remainders and
if you multiply them with a number a which
is co-prime to m then you obtain another order
of the of the remainder now if we just take
therefore, therefore, therefore, it is basically
the same set of the remainders but, in some
other order therefore, the product of these
particular elements will of these elements
will also be the same so if i just take this
numbers and i multiply them this should be
the same as the product of these numbers so
that means writing them i mean in one this
will this will work out to be a to the power
of phi m because there are phi m numbers and
on the left hand side you'll have r one to
r phi m multiplied and the right hand side
you'll also have r one to r phi m multiplied
now note that since these numbers are actually
co-prime with m therefore, they can be cancelled
out and therefore, what remains is a power
phi m is congruent to one modulo m and that
is the euler's fermat's formula
So example of this can be applied to find
out very interesting results like suppose
seventy two to the power of one thousand one
is divided by thirty one and we need to go
argue that so you know that seventy two is
nothing but, equal to ten modulo thirty one
hence seventy two to the power of one thousand
one will be equal to ten to the power of thousand
one mod modulo thirty one now if i apply fermat's
theorem then since p is a prime number then
ten to the power of thirty should be equal
to or congruent to one modulo thirty one note
that thirty one is prime therefore, raising
both sides to the power thirty three will
be ten to the power of nine hundred ninety
is congruent to one modulo thirty one and
therefore, we find that it it is this can
be worked out like this ten to the power of
thousand one will be ten to the power of nine
hundred ninety which is the nearest number
and these are the subsequent remaining numbers
So similar number these small numbers you
can reduce quite easily using and apply the
previous results to these computations then
you can see that this will work out to nineteen
modulo thirteen one similarly, you can work
out this example take this as an example exercise
that is find the least residue of seven to
the power of nine seventy three modulo seventy
two note seventy two is not a prime number
so i conclude here and we have followed these
texts from telang and from buchmann for the
i man for this part and next day's topic will
be probability and information theory
