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Ok so welcome to today's talk lecture on number
theory so today we shall be discussing about
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some elementary concepts in number theory
which forms the kind of very a center i mean
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it is very central to the field of the cryptology
so we shall i mean it is a really fascinating
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field and quiet a big vast field but, today
we shall essentially understand some of the
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basic concepts
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So today's objective of this particular talk
shall be on congruences modular arithmetic
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introduction to field theory euler totient
function and fermat's little theorem so we
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shall be talking about these topics so one
of them is congruences which is modular arithmetic
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which is very important to understand field
of cryptology and then we shall try to give
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the define the concepts of what is meant by
mathematical field which is very again very
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central idea to the study and then discuss
about a particular function which is called
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as euler totient function very popular and
very useful kind of parameter and then conclude
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the talk with discussion on fermat's little
theorem so we will be actually discussing
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about fermat's euler theorem
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So starting so essentially we will find that
when we're doing cryptology or when we're
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doing a normal kind of computation i mean
in in related to the ciphers then we essentially
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do not with deal with infinite sequences so
we for example, we do not have a set which
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rather we do not have a set which is which
can be which can take infinitely large number
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of values so which means that there is a finite
number of values and we are supposed do our
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computations on finite values so therefore,
there is i mean we have to define our arithmetic
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in a finite set of values so for that or rather
in order to study such kind of operations
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one basic concept or one very important concept
is something which is called as congruences
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So we say that a is congruent to b modulo
m so a and b are supposed to be integral integer
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numbers and we say that a is congruent to
b modulo m and we denote them as them as follows
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that is a is congruent to b mod m if m divides
a minus b or b minus a so that is the basic
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idea therefore, let us see some examples like
minus two will be is equal to will be congruent
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to nineteen modulo twenty one the reason being
that twenty one divides nineteen minus of
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minus two that is twenty one and similarly,
twenty will be zero mod ten because if you
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take or rather one way of thinking is that
ten will divide twenty minus zero or zero
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minus twenty
So i mean so therefore, i mean the idea is
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that if you just think in terms of numbers
then this number if we divide by the modulus
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of operation and then this particular number
on the right hand side is nothing but, the
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remainder therefore, so that is the way of
how to kind of compute modular computations
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so this is nothing but, the simple modulo
computation now you may note it that congruence
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modulo m is actually an equivalence relation
why now because it is an equivalence relation
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on the set of integers so let us try to reason
it out
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So we know that for in order for a particular
relation to be satisfying their properties
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of an equivalence relation it has to satisfy
three important concepts of reflexivity symmetricity
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and also transitivity so let us try to first
of all argue why it is reflexive so we see
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that reflexive its reflexive because any integer
is congruent to itself modulo m therefore,
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if you take a number a or other integers take
any number then this number is also congruent
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to itself so why it is because of this if
you take a number a and discus that whether
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and i say that a is kind of a is congruent
a mod m right therefore, this is perfectly
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why now because this means that m divides
a minus a which is zero ok
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So this is true therefore, a is congruent
to a modulo m therefore, the reflexivity relationship
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holds therefore, if i define my relation by
congruency which means that number a is related
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to b if a is congruent to b modulo m this
is the basic definition of the of of the relation
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that we are considering of the congruence
relation then this relation satisfies the
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property of reflexivity because of this reason
Now what about symmetricity now you see that
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in terms of symmetricity also it holds because
as we say that if a is congruent to b modulo
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m then it also holds that b is congruent to
a modulo m why now because this says that
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if m divides a minus b then m also divides
b minus a therefore, the symmetric the symmetricity
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relation also holds and this is also true
for all all a and b therefore, it is true
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for all integer values a and b now this is
really important it holds for this relation
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holds the symmetricity properties holds for
all values of a and m b hm
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What about transitivity so you see that in
it is also transitive why because if a is
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related to b mod m and i tell u that b is
congruent to c mod m then this implies that
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a is congruent to c mod m why now because
you see that m divides a minus b and m divides
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b minus c so from there we can conclude that
m also divides a minus c why now because you
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could have written this a minus b mod simply
as some multiple of m similarly, you could
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have written b minus c as some other multiple
of m ok
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So now if you have added these two equations
right if you have simply added these two equations
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then you would have got a minus b plus b minus
c is nothing but, lambda plus mu which is
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again another integer times m now form here
b and b cancels and we have got a minus c
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is equal to some other constant we can call
that to be some epsilon or something so it
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is something some integer multiplied with
m and therefore, this means that m divides
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a minus c and therefore, this this also holds
true so we see that this relation is also
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transitive and therefore, since it is and
this will hold for any such a b and c right
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and therefore, since it satisfies the properties
of reflexivity symmetricity and also of transitivity
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we say that this particular relation of congruence
also satisfies the i mean it is an equivalence
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relation ok
So therefore, it basically induces a partition
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on the set of integers we know that equivalence
relation induces a set of partition- i mean
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induces a induces a partition on the set on
the set right therefore, it holds this property
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holds true
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Now the following are actually some equivalence
statements and you can easily verify them
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it says that a is congruent to b mod m is
same as saying that there is an integer value
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of k which satisfies that a is equal to b
plus k m so this is what i just just wrote
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in the previous slide so you see that m divides
a minus b because a minus b is k which is
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an integer multiplied with m it is an integral
multiple of m now when divided by m then both
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a and b leave the same reminder so these is
also same as saying as a is congruent to b
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modulo m now equivalence class of a of a modulo
m that is a modulo m consists of all integer
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that are obtained by adding a with integral
multiples of m now this is called or defined
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as the residue class of a mod m so which means
that if i say that the class which is called
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the residue class of a mod m
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Then what we do is that we take a and we add
all possible integral multiples of m therefore,
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therefore, by saying you that there is a residue
class of there is a residue class of a mod
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m thin it means that this consists of all
integer so it is basically a set of all integers
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which are obtained by adding a integer multiples
of of m therefore, it could be like a plus
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minus m a plus minus two m and so on so the
basic idea is that if we just take any number
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from here and you divide it by m then the
remainder which you obtained is a therefore,
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the remainder defines that particular class
therefore, this class is a residue class of
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a mod m and it is also an equivalence class
because it satisfies i mean it it satisfies
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the properties of equivalence class we are
just seeing that right
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So therefore, so here is an example for residue
class of one modulo four so one modulo four
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will be one and then one plus minus four therefore,
and again one plus minus two multiplied four
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and then one plus minus three multiple multiplied
by four and so on so the set of residue classes
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mod m is denoted by z m z so z this is also
denoted in this in this fashion
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So this particular notation is also,metimes
known as z m it is also denoted as z i mean
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the you can also denote it as like this so
it is also an equivalent way of may be shorter
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way of representing this z so this this just
defines that this is the set of residue classes
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mod m and it is denoted by z slash m z and
that equal to z m therefore, now now this
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particular set we will have all the possible
all the possible values so it'll essentially
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have m values so what are the m values possible
the m values will run from zero one and till
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m minus one so this is the set which essentially
comprises of this values that is from zero
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two m minus one now this is also called an
a complete system or a complete set of incongruent
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residues therefore, you see that if you just
take a number so an example of this would
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like let us take the example of ten and then
define like what is z by ten z so then this
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will comprise of numbers from zero one till
m till nine that is ten minus one is nine
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So that means that if we just take m is equal
to ten then these are the possible values
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of numbers which are which are possible i
mean maximally possible and which also have
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which are also incongruent to each other so
that means that if you take any numbers from
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this from this set for example, let us take
like one example one one random choice could
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be like two and another arbitrary choice could
be say five then two and five are never congruent
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to each other modulo ten so that is why why
is two not congruent to five modulo ten because
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ten will not divide two minus five or five
minus two so that means all the elements in
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this set are actually incongruent to each
other and this is the maximal possible possible
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number of incongruent values that can that
can happen ok
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You can take any number of outside this particular
set for example, if you just take an arbitrary
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choice like say twelve then you'll find one
number in this set which is congruent to this
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number modulo ten which is that number for
example, you can choose two so you see that
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two and twelve are congruent to each other
modulo ten why now because ten will divide
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twelve minus two right therefore, you'll find
that this particular set which is like z defined
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by z slash m z is also called a complete set
of incongruent residues for a complete system
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Now there are example this is an example given
here for complete system for modulo five so
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we know now for mod five it will be like from
zero to five minus one that is zero one till
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four and this an another example you can verify
like that this is minus twelve minus fifteen
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eighty two and minus one and thirty two this
set is also a complete system why now this
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is also a complete system because if you just
take these numbers like you take minus twelve
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so we what we are considering is z slash five
z right so you see that in this set if we
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just take twelve then twelve is actually so
it is minus twelve right so minus twelve if
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you take modulo five so what does it mean
what is this value so if you divide twelve
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by five therefore, you will get that this
is equal to is it correct yes because if you
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just take this number five then five you divide
minus twelve plus two because that is minus
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twelve plus two is nothing but, minus ten
so five divided minus ten ok
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Now what is minus two if you if you would
like to bring in this among this numbers like
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from zero one till/two three and four then
this particular number minus two will be actually
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equal to three therefore, because it is you
can verify this because five will agian divide
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minus two minus three so this is same as minus
two modulo five will be same as three modulo
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five what about the next number the next number
here in this set was given as minus fifteen
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so minus fifteen if you take modulo five what
will be it it will be zero so similarly, you
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can do it for the other once also like eighty
two will be equal to two modulo five because
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five will divide it into minus two and minus
one will be congruent to four mod five and
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thirty one will be congruent one mod four
one one mod five so you see that the numbers
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that i obtain after taking the modulo five
essentially are nothing but, another order
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of the numbers zero one two three and four
therefore, you see that this is also a kind
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of these numbers like this set like minus
twelve minus fifteen eighty two minus one
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and thirty one is also another example of
a complete systems and any two numbers from
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this set are are incongruent to each other
they are not congruent we can check this
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So there is a theorem i mean i am i- i really
do not have i- i am not going the proof but,
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its very simple you can verify it like if
a is congruent to b modulo m and c is congruent
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to d modulo m then this implies that minus
a and minus b are congruent to each other
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modulo m we can also add like a plus c is
congruent b plus d modulo m there should be
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a modulo m here and a c is congruent to b
d modulo m so you can verify this and its
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very simple like what you can do is that you
can start like writing as a is equal to b
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plus some constant multiple of b of m and
c is equal to d plus some constant times m
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and then try to find out minus a and and this
is very simple so this is this is quite trivial
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to prove
Now what is the implication of this result
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now there- there is a very is strong implication
of this result so let me reflect on this see
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for example, what we understand here is that
this number a may be quite a large value and
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this number c maybe also another quite large
value and suppose you're actually interested
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in computing doing computation on large values
now if you would like to do computation of
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large values but, you always have this modulo
m then the idea is that what it says is that
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instead of taking a plus c or rather adding
a plus c and then doing a modulo m is same
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as doing a modulo m on a doing a modulo m
on c and reducing this numbers between zero
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to m minus one and then adding these two numbers
so let me kind of illustrate this with a very
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simple example
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So let us take a large value of a so for example,
you consider that a is again let us consider
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that our m value is five and let us see that
a is supposed a large value like may be larger
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than this so it may be like twenty seven and
let us consider another value which is say
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b or rather c and consider that this is say
something like forty nine now let us try to
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add these two numbers so we know that if we
add these two numbers then we get seventy
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six and therefore, if i do a modulo five here
that means that i divide this by five and
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my remainder is one ok
So therefore, i write that a plus c modulo
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m is nothing but, equal to one it is congruent
to one so another way of writing this is rather
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the way we are writing this a plus c is equal
to one mod m now let us try to do i mean apply
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the previous result so what we will do is
that we will take twenty seven and perform
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a modulo five operation so if i do that then
i see that twenty five is divisible by five
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so the remainder is two similarly, i take
forty nine and do a modulo five operation
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so we see that forty five is divisible so
this is actually is equal to four now you
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see that four also can be written as minus
one right so therefore, we write the twenty
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seven is equal to two mod five and forty nine
is minus one mod five now if i add these two
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numbers like twenty seven plus forty nine
then i can obtain- add this instead of these
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two numbers and reducing what i can do is
that i can actually add these two small numbers
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and i get one modulo five which matches with
this result
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So which means that i need not do the additions
with large number when i am doing a modular
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operation right when then i really do not
need to add these large numbers but, what
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i can do it is that i can reduce this numbers
modulo five reduce this number modulo five
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and then add these two numbers so this particular
particular technique can actually make our
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computations quite simple and this is actually
applied in the implementation of cryptosystems
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quite regularly to make your implementations
easier ok
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So we will squarely get this idea or rather
find more examples of this as we proceed but,
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this is a very interesting and very important
results which we should understand similarly,
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when we're talking about multiplication we
can also again instead of multiplying two
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large numbers which is even more complicated
we can actually reduce these numbers mod m
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and then just multiply this small numbers
for example, imagine like if you have to multiply
196
00:20:45,539 --> 00:20:50,529
twenty seven and forty nine and then do a
modulo five operation right then this this
197
00:20:50,529 --> 00:20:53,909
multiplication is not i mean it is not proceeding
well you have to do it and you can make mistake
198
00:20:53,909 --> 00:20:58,179
right and most importantly when your computer
does it it may not make mistakes but, what
199
00:20:58,179 --> 00:21:02,570
it may do is it may require large number of
time right so the resource may be useful may
200
00:21:02,570 --> 00:21:06,950
be used up so instead of that what you can
do is that you can reduce it it becomes two
201
00:21:06,950 --> 00:21:10,739
and this becomes minus one and you know that
this result is very simple this resulting
202
00:21:10,739 --> 00:21:17,529
is nothing but, minus two mod five which is
nothing but, three ok
203
00:21:17,529 --> 00:21:22,359
So you can just straight away from these two
results we can obtain this product that actually
204
00:21:22,359 --> 00:21:27,239
makes your computations much easier otherwise
you would have it would have become a little
205
00:21:27,239 --> 00:21:34,239
more difficult right so this result has got
lot of impact therefore, and here is a another
206
00:21:34,690 --> 00:21:39,029
example so you can probably try to understand
the impact of the previous result from this
207
00:21:39,029 --> 00:21:44,729
example so let let us consider so this is
the very important number in number theory
208
00:21:44,729 --> 00:21:48,259
but, let us not go into that say is the two
to the power two to the power of five plus
209
00:21:48,259 --> 00:21:53,129
one is divisible by six forty one and a proof
is required for that so one way of doing it
210
00:21:53,129 --> 00:21:56,479
like it will be two to the power of five is
thirty two therefore, let us compute two to
211
00:21:56,479 --> 00:22:00,649
the power of thirty two and then add one and
then divide by six forty one and then see
212
00:22:00,649 --> 00:22:04,169
that whether it is divisible or not
But this is quite tedious approach instead
213
00:22:04,169 --> 00:22:10,190
of that lets see this we know that six forty
one can be actually written as six forty plus
214
00:22:10,190 --> 00:22:15,359
one and six forty can be factorized as five
multiplied by two to the power of seven plus
215
00:22:15,359 --> 00:22:19,739
one now this factorial is quite easy right
because its divisible by five and the others
216
00:22:19,739 --> 00:22:25,669
are just powers of two therefore, we can write
as five multiplied by two to the power of
217
00:22:25,669 --> 00:22:32,309
seven is nothing but, minus one modulo six
forty one why now because we know that five
218
00:22:32,309 --> 00:22:36,799
two to the power of two to the power of seven
will be equal to six forty one minus one right
219
00:22:36,799 --> 00:22:40,989
so we can rearrange this as five into two
to the power of seven is equal to six forty
220
00:22:40,989 --> 00:22:44,379
one minus one
Now if take modulo six forty one then this
221
00:22:44,379 --> 00:22:49,070
vanishes and therefore, minus one remains
as a remainder now from the previous result
222
00:22:49,070 --> 00:22:54,429
we know that if i raise this to the left hand
side to the power of four it is same as write
223
00:22:54,429 --> 00:22:59,619
i mean raising the right hand side to the
power of minus one hole to the power of four
224
00:22:59,619 --> 00:23:04,039
when you're doing modulo six forty one operation
because this means that we are multiplying
225
00:23:04,039 --> 00:23:08,340
this four times and therefor the right hand
side also needs to be multiplied four times
226
00:23:08,340 --> 00:23:12,720
therefore, straight away from here we get
five to the power of four multiplied by two
227
00:23:12,720 --> 00:23:17,789
power of twenty eight is nothing but, congruent
to one modulo six forty one
228
00:23:17,789 --> 00:23:22,190
So now we see that five to the power of four
is nothing but, six twenty one six twenty
229
00:23:22,190 --> 00:23:27,239
five and we know we know that need not actually
require to multiply these two numbers but,
230
00:23:27,239 --> 00:23:32,690
we can actually deduce this number modulo
six forty one so that should be make our computations
231
00:23:32,690 --> 00:23:37,100
easy right and then multiplied by two power
of twenty eight and that is congruent one
232
00:23:37,100 --> 00:23:42,609
modulo six forty one so this equation is not
disturbed by this now we see that six twenty
233
00:23:42,609 --> 00:23:49,119
five modulo six forty one is nothing but,
equal to minus sixteen therefore, this result
234
00:23:49,119 --> 00:23:54,139
is nothing but, minus two power of four multiplied
by two power of twenty eight is congruent
235
00:23:54,139 --> 00:23:59,129
to one modulo six forty one and that means
that two power of thirty two is nothing but,
236
00:23:59,129 --> 00:24:04,489
equal to minus one modulo six forty one so
that means that six forty one divides from
237
00:24:04,489 --> 00:24:08,649
the definition of congruence we know that
six forty one divides two power of thirty
238
00:24:08,649 --> 00:24:14,629
two minus of minus one that is six forty one
divides two power of thirty two plus one so
239
00:24:14,629 --> 00:24:18,570
this particular thing is proved
But you see that throughout this we are actually
240
00:24:18,570 --> 00:24:23,149
not done a single bit large computation and
that is the advantage of previous theorem
241
00:24:23,149 --> 00:24:29,409
therefore, idea that when you need do multiplications
or additions or computations on large numbers
242
00:24:29,409 --> 00:24:33,450
and when you're doing a modular operation
then it is better to reduce each individual
243
00:24:33,450 --> 00:24:40,450
numbers and perform the operations that'll
you're your computations easier
244
00:24:41,429 --> 00:24:47,649
So now we actually i mean so this is we actually
shift our focus and go into a little bit of
245
00:24:47,649 --> 00:24:53,539
field theory now we fields are actually very
central to the understanding of crypto that
246
00:24:53,539 --> 00:24:58,190
is for example, the common that is the recent
standard which is known as the advanced encryption
247
00:24:58,190 --> 00:25:03,799
standard relies heavily on finite fields therefore,
we will try to understand some of the basic
248
00:25:03,799 --> 00:25:09,419
of fields that is what are what are the fields
and what are the properties of a mathematical
249
00:25:09,419 --> 00:25:12,499
field ok
So let us start with something which is called
250
00:25:12,499 --> 00:25:16,259
semi groups so first of all let us define
something which is called a transformation
251
00:25:16,259 --> 00:25:22,460
or an operation so here is an x is a set and
let us define a map which is defined by this
252
00:25:22,460 --> 00:25:29,409
circle has nothing but, a transformation which
takes two numbers from this set x let let
253
00:25:29,409 --> 00:25:34,769
it be x one and the other one is x two and
perform an operation therefore, this example
254
00:25:34,769 --> 00:25:41,629
of this could be an addition that is just
a simple plus in the integer domain or multiplication
255
00:25:41,629 --> 00:25:47,450
so, the so you can generalize it means that
i just take two numbers from the set x and
256
00:25:47,450 --> 00:25:52,279
i perform these computations and therefore,
this element therefore, this transforms an
257
00:25:52,279 --> 00:25:57,619
element x one coma x two to the element x
one operation x two and this is vocalize-noise
258
00:25:57,619 --> 00:26:00,979
so this is just this is a simple transformation
ok.
259
00:26:00,979 --> 00:26:07,979
So now the sum of the free residue classes
that is so here is an example like examples
260
00:26:09,570 --> 00:26:13,330
of this transformation as i told you that
integer is one example similarly, you can
261
00:26:13,330 --> 00:26:18,440
also define in the residue classes like that
we just take two residue classes a plus m
262
00:26:18,440 --> 00:26:25,440
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
263
00:26:26,799 --> 00:26:31,210
to find out the some of these residue classes
then this is nothing but, a plus b plus m
264
00:26:31,210 --> 00:26:38,210
z similarly, the product of the residue classes
a plus m z and b plus m z is equal to a multiplied
265
00:26:38,210 --> 00:26:43,259
b plus m z so this is the same thing which
i just told you before this that is if i need
266
00:26:43,259 --> 00:26:48,849
to kind of multiply or add two large numbers
in the modulo m domain then i just find the
267
00:26:48,849 --> 00:26:54,159
remainders right and just add the remainder
or multiply the remainders this is perfectly
268
00:26:54,159 --> 00:26:55,909
defined ok
269
00:26:55,909 --> 00:27:02,779
So now an operation this circle on x is associative
so we now we would like to define the associativity
270
00:27:02,779 --> 00:27:08,489
of this it means that a operation b operation
c and if i do this it can also be associated
271
00:27:08,489 --> 00:27:13,999
as a this operation can also be performed
as this that is a operation b operation c
272
00:27:13,999 --> 00:27:20,269
being given a priority and this holds for
all a b c in x so this is how associativity
273
00:27:20,269 --> 00:27:23,080
is defined ok
And we know what is commutativity which means
274
00:27:23,080 --> 00:27:28,559
that if a operation b is same as b operation
a for all a and b in x then it is said to
275
00:27:28,559 --> 00:27:34,460
be commutative so examples we know that integer
addition is an associative operation right
276
00:27:34,460 --> 00:27:41,460
and it is is it commutative also it is also
commutative but, what about an example which
277
00:27:41,719 --> 00:27:47,059
is not commutative for example, matrix multiplication
is not commutative we know that if i take
278
00:27:47,059 --> 00:27:54,059
two matrix a and another matrix as b and if
i multiply a and b then it may not be the
279
00:27:55,299 --> 00:27:59,559
same as that b and a ok
So now we define what is meant by a semi group
280
00:27:59,559 --> 00:28:03,559
so it says that there is a pair of h so it
is a set so we define a transformation as
281
00:28:03,559 --> 00:28:08,389
just as we told right now so it consists of
a set h and an associative operation so that
282
00:28:08,389 --> 00:28:13,869
means if the operation is also associative
on h and then it is called a semi group so
283
00:28:13,869 --> 00:28:18,899
of semi-groups means simply that there is
a set h with an operation circle and this
284
00:28:18,899 --> 00:28:24,769
operation is also associative so if the operation
is also associative then if the operation
285
00:28:24,769 --> 00:28:29,649
is associative then we say it to be a semi-group
now the semi-group is also called abelian
286
00:28:29,649 --> 00:28:34,499
or commutative if the operation is also commutative
and examples of this could be like integer
287
00:28:34,499 --> 00:28:40,609
set the addition defined over it integer set
multiplication defined over it residue class
288
00:28:40,609 --> 00:28:46,249
set and addition defined over this and the
residue class set and multiplication defined
289
00:28:46,249 --> 00:28:47,149
over this ok
290
00:28:47,149 --> 00:28:52,989
So we know that these are examples of abelian
or commutative semi-groups because this is
291
00:28:52,989 --> 00:28:56,309
also associative because it is a semi-group
and also further its abelian or commutative
292
00:28:56,309 --> 00:29:03,090
so which means you can do it either as a operation
b or as b operation a for all a and b which
293
00:29:03,090 --> 00:29:08,379
belongs in the set so, implications are as
follows so there are some interesting implications
294
00:29:08,379 --> 00:29:14,570
So let h and this operation be defined as
a semi-group and let us define like a let
295
00:29:14,570 --> 00:29:20,899
us define some powers of a as follows so a
power one is nothing but, a and a power n
296
00:29:20,899 --> 00:29:26,529
plus one is defined as a operated with a power
n so this is the recursive definition of a
297
00:29:26,529 --> 00:29:33,419
power n plus one for a which is in h and natural
value of n so this is a natural number n now
298
00:29:33,419 --> 00:29:38,710
if i need to compute a power n and operate
it with a power m then it is same as doing
299
00:29:38,710 --> 00:29:44,679
as a power n plus m or if i do as a power
n and take a whole power m then this is same
300
00:29:44,679 --> 00:29:50,639
as doing a power n m and where a is in h and
n and m are nothing but, natural values or
301
00:29:50,639 --> 00:29:54,299
natural numbers
Now you see that it is a further interest
302
00:29:54,299 --> 00:29:59,899
result which says that a and b are in h and
we perform and if a and if the operation is
303
00:29:59,899 --> 00:30:05,729
commutative that is if a operation b is same
as b operation a then a power the a operation
304
00:30:05,729 --> 00:30:11,389
b and if i raise it to power of n is same
as a raise to power of n operated with b raise
305
00:30:11,389 --> 00:30:17,789
to power of n so we can actually reflect upon
these are kind of definitions and we can actually
306
00:30:17,789 --> 00:30:23,429
i mean we can actually reflect upon this definition
by by for example, choosing the value of n
307
00:30:23,429 --> 00:30:24,379
to be two ok
308
00:30:24,379 --> 00:30:30,799
So let us take n is equal two and consider
the operation as a transformation b raise
309
00:30:30,799 --> 00:30:37,149
the power of two so we now that we can actually
write this as a power of b by our recursive
310
00:30:37,149 --> 00:30:44,149
definition as a power b now we apply we know
that this is this operation is so the first
311
00:30:45,059 --> 00:30:49,710
thing that we start with is this so that is
the first definition but, since this is a
312
00:30:49,710 --> 00:30:55,590
semi-group so we know that it is associative
so we can actually do like a b operation a
313
00:30:55,590 --> 00:31:01,279
operated with b and we know that because if
i apply commutativeity then we actually write
314
00:31:01,279 --> 00:31:08,279
this as a with b and we know that again we
can actually do this that is a apply associativity
315
00:31:10,479 --> 00:31:16,529
and perform this operation like this and this
is nothing but, a square operated with b square
316
00:31:16,529 --> 00:31:23,229
therefore, this we can actually inductively
apply and obtain the results that is a operated
317
00:31:23,229 --> 00:31:29,619
operation b if i raise it to the power of
n is same as a raise to the power of n operation
318
00:31:29,619 --> 00:31:32,729
operated with b power n
319
00:31:32,729 --> 00:31:37,749
So now we let us the define some something
which is called monoids so first of all one
320
00:31:37,749 --> 00:31:42,009
of the definition the central definition of
the monoids is something which is called a
321
00:31:42,009 --> 00:31:49,009
neutral element a neutral element of the semi-group
h operation is an element e in h which satisfies
322
00:31:49,429 --> 00:31:55,639
e operation a as same as a operation e and
that that means that this is equal to a for
323
00:31:55,639 --> 00:32:02,639
all a in h therefore, we see that an easy
way of understanding the neutral element would
324
00:32:03,389 --> 00:32:10,389
be to consider an example so let us consider
the example of let us consider the example
325
00:32:11,519 --> 00:32:17,289
of for example, a field of z and so this is
an integer set this is an integer set and
326
00:32:17,289 --> 00:32:20,679
there is a plus operation defined so we so
we have just defined that this is a this is
327
00:32:20,679 --> 00:32:26,239
an example of if so this was an example of
semi-group so we have defined what is meant
328
00:32:26,239 --> 00:32:31,429
by a semi-group
And let us consider this particular example
329
00:32:31,429 --> 00:32:37,269
and see that how or rather what is the neutral
element in this set so i think all of us know
330
00:32:37,269 --> 00:32:41,279
what is the neutral element in this set so
if i so the by the definition it means that
331
00:32:41,279 --> 00:32:47,059
if i take a and if i apply the operation-
operation plus and there should be a neutral
332
00:32:47,059 --> 00:32:54,059
element a so that i get back a or so this
is called a right neutral element and similarly,
333
00:32:58,820 --> 00:33:02,139
there is something which is called a left
neutral element which says that if i add e
334
00:33:02,139 --> 00:33:09,139
with a and if i obtain a so this is the left
neutral element so this is the right right
335
00:33:11,070 --> 00:33:16,089
neutral element this is the left neutral element
that i get back here now you see that this
336
00:33:16,089 --> 00:33:22,089
defines that so this actually gives us an
idea that this e is nothing but, zero right
337
00:33:22,089 --> 00:33:27,769
so for example, if i add a with zero then
i get back a and if i add zero with a i also
338
00:33:27,769 --> 00:33:32,389
get back a so zero is the neutral element
in the semi-group z plus
339
00:33:32,389 --> 00:33:37,919
Similarly all so therefore, if the semi-group
contains a neutral element like this then
340
00:33:37,919 --> 00:33:43,249
it is actually defined to be something which
is called a monoid so, a semi-group has there
341
00:33:43,249 --> 00:33:48,299
is a result which says that a semi-group has
at most one neutral element and this is actually
342
00:33:48,299 --> 00:33:53,979
not very difficult to prove so the first result
that we can actually kind of reflect is that
343
00:33:53,979 --> 00:33:58,499
the left neutral element and the right neutral
element are the same so if i just consider
344
00:33:58,499 --> 00:34:05,499
the left neutral element so the left neutral
element and the right neutral element are
345
00:34:06,039 --> 00:34:13,039
the same why
Now you can just say that the left the left
346
00:34:13,109 --> 00:34:18,869
neutral element be e one and let the right
neutral element be e two so what about e one
347
00:34:18,869 --> 00:34:23,770
operation e two now you see that if i just
think that this is the left neutral element
348
00:34:23,770 --> 00:34:30,679
then by the definition i know that e one operation
with e two should be e two because this is
349
00:34:30,679 --> 00:34:34,760
the left neutral element now what what if
i think this to be the right neutral element
350
00:34:34,760 --> 00:34:39,210
like if i just think that e two if i think
that e two is the right neutral element then
351
00:34:39,210 --> 00:34:46,210
i know that e one operation e two will be
e one therefore, from here i understand that
352
00:34:46,290 --> 00:34:53,169
e one and e two are the same right
So therefore, the left neutral element and
353
00:34:53,169 --> 00:34:57,950
the right neutral element are the same now
the other thing that holds is that if there
354
00:34:57,950 --> 00:35:04,950
is one neutral element then there can be at
most one right neutral element so this proof
355
00:35:06,410 --> 00:35:11,510
i i mean it is quite easy actually you can
follow the similar principle but, i leave
356
00:35:11,510 --> 00:35:18,150
this as an exercise that is if there is a
left neutral element that is for a left left
357
00:35:18,150 --> 00:35:25,150
neutral element there can be at most one right
neutral element so therefore, so from here
358
00:35:44,570 --> 00:35:48,330
we know that if there can be at most one right
neutral element then this right neutral element
359
00:35:48,330 --> 00:35:53,530
is again same as the left neutral element
by this definition so there can be at most
360
00:35:53,530 --> 00:35:57,930
so from these two results we can conclude
that there can be at most one neutral element
361
00:35:57,930 --> 00:35:58,230
ok
362
00:35:58,230 --> 00:36:04,540
So therefore, if the semi-group which essentially
has a neutral element which is called a monoid
363
00:36:04,540 --> 00:36:09,790
has can can have at most one neutral element
therefore, this result holds true that is
364
00:36:09,790 --> 00:36:16,790
semi-group has got at most one neutral element
now if e which belongs to h is a neutral element
365
00:36:19,580 --> 00:36:25,460
of the semi-group and and of the semi-group
and then b also belongs to h then there is
366
00:36:25,460 --> 00:36:29,520
a definition of inverse like this that is
if i take a and if i operation operate it
367
00:36:29,520 --> 00:36:35,840
with b then this is same as and it is same
as b being operated on a and i obtain e so
368
00:36:35,840 --> 00:36:41,160
for example, when i am doing when i consider
the semi-group z coma plus then what will
369
00:36:41,160 --> 00:36:46,480
be the inverse of a it will be minus a so
because if i know that if i add with minus
370
00:36:46,480 --> 00:36:50,160
a that i obtain back the neutral element which
is zero
371
00:36:50,160 --> 00:36:55,830
So if a has an inverse then a is called invertible
in the semi-group h and in a monoid each element
372
00:36:55,830 --> 00:37:00,800
has at most one inverse so this also result
can be proved that is if there is a monoid
373
00:37:00,800 --> 00:37:06,780
so in the monoid then each element has at
most one inverse so examples are like this
374
00:37:06,780 --> 00:37:11,660
that is if i consider z coma plus the neutral
element is zero and inverse is minus a if
375
00:37:11,660 --> 00:37:17,670
you consider z under multiplication the neutral
element is one and the only invertible elements
376
00:37:17,670 --> 00:37:22,120
are plus one and minus one because if i just
consider any other integer let's- says five
377
00:37:22,120 --> 00:37:27,670
then there is no integer say integer number
if you multiply with which you will obtain
378
00:37:27,670 --> 00:37:32,100
back the neutral element except therefore,
the neutral element exists only for plus one
379
00:37:32,100 --> 00:37:33,320
and minus one
380
00:37:33,320 --> 00:37:39,040
What about the residue class z slash m z plus
and operation is plus then neutral element
381
00:37:39,040 --> 00:37:46,040
is m z itself and the inverse is minus a so
this should be actually minus a plus m z so
382
00:37:46,760 --> 00:37:53,760
it means that if i take if if i just consider
this particular set that is z z m z which
383
00:37:54,250 --> 00:37:58,940
is a semi-group and then the operation is
defined as plus then a neutral element is
384
00:37:58,940 --> 00:38:05,940
m z so that is the neutral element and the
inverse is so this is the neutral element
385
00:38:08,920 --> 00:38:15,920
and the inverse is minus a plus m z so that
is the inverse and you can verify this it
386
00:38:19,150 --> 00:38:26,150
is quite simple so, what about z m z and product
z m z coma product in this case the neutral
387
00:38:27,300 --> 00:38:32,780
element is one plus m z and the inverse and
this result we will see actually is are those
388
00:38:32,780 --> 00:38:37,530
elements t which are actually which are actually
co-plained to m that is every element does
389
00:38:37,530 --> 00:38:42,120
not have an inverse but, only those elements
for only those elements inverse exist which
390
00:38:42,120 --> 00:38:46,380
are actually co-plained to m so which means
that the g c d or the greatest common devisor
391
00:38:46,380 --> 00:38:51,690
of that number and m is actually equal to
one so we will see this as we go ahead
392
00:38:51,690 --> 00:38:55,700
So then we define what is called a group a
group is a monoid in which every element is
393
00:38:55,700 --> 00:39:00,250
invertible so a group is a monoid in which
every element is invertible and the group
394
00:39:00,250 --> 00:39:04,490
is commutative or abelian if the monoid is
also commutative so examples of this will
395
00:39:04,490 --> 00:39:10,850
be like z coma plus which is an abelian group
and z product which is not a group so so this
396
00:39:10,850 --> 00:39:17,850
is actually not an example of a group and
then we have got z coma m m z and plus z slash
397
00:39:19,400 --> 00:39:25,580
m z and plus which is an abelian group so
why is this not a group this is not a group
398
00:39:25,580 --> 00:39:30,580
because from the definition of a group a group
is a monoid in which every element is invertible
399
00:39:30,580 --> 00:39:35,020
and we have just defined just discussed in
the previous slide that is in this particular
400
00:39:35,020 --> 00:39:39,920
set only invertible elements are plus one
and minus one so every element does not i
401
00:39:39,920 --> 00:39:46,920
mean rather every element does not have does
not have an inverse therefore, z coma multiplication
402
00:39:48,760 --> 00:39:53,850
is actually not a group because every element
does not is not invertible in this is not
403
00:39:53,850 --> 00:39:56,040
invertible in these semi-group ok
404
00:39:56,040 --> 00:40:00,980
So therefore, that is that is the definition
of a group and there is higher concept to
405
00:40:00,980 --> 00:40:05,450
this which is called ring and ring is nothing
but, r and they are actually triplets so there
406
00:40:05,450 --> 00:40:09,920
is not only one operation as plus and there
is another operation also along with it such
407
00:40:09,920 --> 00:40:16,270
that r coma plus is an abelian group the way
we have defined and r coma product is r coma
408
00:40:16,270 --> 00:40:21,300
the second operation is actually a monoid
in addition it satisfies the properties of
409
00:40:21,300 --> 00:40:26,080
distributivity which says that x if we multiply-
i mean product if we take product with y plus
410
00:40:26,080 --> 00:40:32,330
z then which is same as x dot y plus x dot
z for x y z which belongs to this ring the
411
00:40:32,330 --> 00:40:39,330
ring is also called commutative if the semi-group
r coma dot that is this one is also commutative
412
00:40:41,320 --> 00:40:47,950
and a unit element of the ring is the neutral
element of the semi-group r coma dot so a
413
00:40:47,950 --> 00:40:52,560
unit element of the ring is a neutral element
of the semi-group r coma dot therefore, a
414
00:40:52,560 --> 00:40:57,820
neutral unit element of this ring will be
as defined as the neutral element of this
415
00:40:57,820 --> 00:41:02,570
of the monoid r coma dot.
416
00:41:02,570 --> 00:41:06,590
So then we define what is called a unit group
which means that let r be a ring with unit
417
00:41:06,590 --> 00:41:13,590
element an element a of r is called invertible
or a unit if it is invertible in the multiplicative
418
00:41:14,100 --> 00:41:21,100
semi-group of r so an element a i repeat this
definition a element a of r is called invertible
419
00:41:21,670 --> 00:41:28,670
or a unit if it is invertible in the multiplicative
semi-group of r that means multiplicative
420
00:41:29,660 --> 00:41:34,720
semi-group of r will be this if it is invertible
in this particular semi-group then it is said
421
00:41:34,720 --> 00:41:41,060
to be invertible and the element a is called
a zero divisor if it is non-zero that is if
422
00:41:41,060 --> 00:41:45,480
the element itself is non-zero and there is
a non-zero b in r such that if i take the
423
00:41:45,480 --> 00:41:51,880
product of a and b or the product of b and
a then we get back zero now units of a commutative
424
00:41:51,880 --> 00:41:56,480
ring actually from a group and this is called
the unit group of the ring and it is often
425
00:41:56,480 --> 00:41:58,750
denoted by r star ok
426
00:41:58,750 --> 00:42:05,290
So that is the definition of a unit group
and we can actually have one very i mean some
427
00:42:05,290 --> 00:42:10,670
example because i think this is little abstract
so in order to understand this let us take
428
00:42:10,670 --> 00:42:17,670
some examples so let us consider the set or
rather let us consider the ring z m z and
429
00:42:18,960 --> 00:42:25,960
let us define let us define rather let us
take the value of m to be something like ten
430
00:42:26,320 --> 00:42:33,320
so we know that the elements here will be
like zero one two and so on till nine now
431
00:42:33,370 --> 00:42:40,370
you consider like whether all the elements
so let us consider like some some let us consider
432
00:42:40,560 --> 00:42:47,560
some interesting facts let us consider the
numbers for example, two and let us consider
433
00:42:47,690 --> 00:42:53,620
like let us multiply two with all possible
elements which are there all possible nonzero
434
00:42:53,620 --> 00:42:55,920
elements which are there in the set
435
00:42:55,920 --> 00:43:02,000
So let us consider two into one two into two
two into three two into four two into five
436
00:43:02,000 --> 00:43:07,030
and so on if you consider these multiply-
these products and if you take the modulo
437
00:43:07,030 --> 00:43:11,240
ten operation that is if you take mod ten
for all of them then you'll find that this
438
00:43:11,240 --> 00:43:18,240
is actually equal to two this is actually
equal to four this is six this is eight but,
439
00:43:19,080 --> 00:43:26,030
this one is zero so this shows us an example
where there are two elements which are nonzero
440
00:43:26,030 --> 00:43:32,310
like two is nonzero and five is also nonzero
but, if i multiply this and if we take i mean
441
00:43:32,310 --> 00:43:37,160
in these particular if i take a modulo ten
then what i obtain back is zero therefore,
442
00:43:37,160 --> 00:43:42,470
from this definition we say that two is actually
a zero divisor the element two is actually
443
00:43:42,470 --> 00:43:47,840
a zero divisor therefore, this particular
thing i mean therefore, i mean this this is
444
00:43:47,840 --> 00:43:54,230
an example which shows that zero divisor exists
therefore, let us come back to this and see
445
00:43:54,230 --> 00:43:59,090
whether element a is called a zero divisor
if it is a nonzero and there is a nonzero
446
00:43:59,090 --> 00:44:04,980
b in r such that a multiplied with b or b
multiplied with a is zero so this is an example
447
00:44:04,980 --> 00:44:07,520
to understand this particular aspect
448
00:44:07,520 --> 00:44:12,170
Now what about this an element a of r is called
invertible or unit if it is invertible in
449
00:44:12,170 --> 00:44:18,620
the multiplicative semi-group of r so let
us see that among these particular i mean
450
00:44:18,620 --> 00:44:24,460
numbers like zero one two and- nine what are
the elements which are invertible so let us
451
00:44:24,460 --> 00:44:28,950
leave out zero because zero is obviously not
invertible let us consider one so we know
452
00:44:28,950 --> 00:44:34,260
that if i take one and multiply it with one
i get back one itself what about two so if
453
00:44:34,260 --> 00:44:41,180
i take two and if i multiply with with any
of the numbers will i get back one so actually
454
00:44:41,180 --> 00:44:45,140
we will see that no you will not get back
two is not invertible ok
455
00:44:45,140 --> 00:44:50,760
So the numbers which are invertible in this
particular set from zero to nine can actually
456
00:44:50,760 --> 00:44:57,760
be found out like this there we see one similarly,
three will invertible four will not be invertible
457
00:44:58,270 --> 00:45:05,270
five could not invertible six will not be
invertible seven will be invertible nine will
458
00:45:07,240 --> 00:45:13,740
be invertible eight will also be invertible
sorry eight not will not be invertible nine
459
00:45:13,740 --> 00:45:17,670
will be invertible and the reason rather the
one you see we have checking is that if i
460
00:45:17,670 --> 00:45:24,670
take the g c d of any of these numbers a and
with ten if a is invertible then this g c
461
00:45:25,580 --> 00:45:31,800
d of and ten should be equal to one so if
the g c d of a coma ten is equal to one then
462
00:45:31,800 --> 00:45:36,480
we we say that a is invertible so a is invertible
if and only if the g c d of a coma ten will
463
00:45:36,480 --> 00:45:41,830
be equal to one so that means you'll find
that this particular set that is one three
464
00:45:41,830 --> 00:45:46,200
seven and nine will actually form something
which is called as unit-group therefore, this
465
00:45:46,200 --> 00:45:52,350
will form a unit-group so i mean it is often
defined as r star in this ring so we will
466
00:45:52,350 --> 00:45:57,580
see more examples of this but, as we proceed
we will see more examples of this
467
00:45:57,580 --> 00:46:03,290
So let us consider zero divisor therefore,
the zero divisors of the residue class z slash
468
00:46:03,290 --> 00:46:10,290
m z is actually a plus m z and this is the
generalization of what we said that is zero
469
00:46:10,300 --> 00:46:16,300
divisors of the residue class z slash m z
is actually a plus m z such that if you take
470
00:46:16,300 --> 00:46:21,790
the g c d of a and with m then it is neither
one nor m but, it is somewhere in between
471
00:46:21,790 --> 00:46:27,920
so that is nontrivial g c d of a and m and
the proof is very simple it says that if a
472
00:46:27,920 --> 00:46:33,100
plus m z if the zero divisor of z slash m
z then there is an integer b this follows
473
00:46:33,100 --> 00:46:38,830
take down the definition such that a b is
congruent to zero modulo m but, neither a
474
00:46:38,830 --> 00:46:45,360
nor b is zero modulo m so which means that
m divides a b so a b is congruent to zero
475
00:46:45,360 --> 00:46:52,360
means m divides a b but, neither a nor b is
actually divisible by m
476
00:46:52,580 --> 00:46:59,290
So this atomically implies that if i take
the g c d of a and m then this should be some
477
00:46:59,290 --> 00:47:06,290
significant value that is it means that if
you take m so, say that a b is congruent to
478
00:47:09,300 --> 00:47:16,300
zero modulo m right so that means that a b
will be equal to i mean rather m divides a
479
00:47:16,570 --> 00:47:22,760
b right so if a m divides a b so which means
that a b divided by rather a b divided by
480
00:47:22,760 --> 00:47:28,900
m this should be a number right this should
be an integer number this is an integer value
481
00:47:28,900 --> 00:47:35,900
but, we know that neither a nor b are actually
divisible by m therefore, a by m is not an
482
00:47:40,300 --> 00:47:47,300
integer b by m is not an integer so that means
that a and b definitely i mean i mean there
483
00:47:48,260 --> 00:47:53,530
should be a cancellation between a and m and
that means that the g c d of a and m should
484
00:47:53,530 --> 00:47:58,780
be actually something which is between one
and m so it is neither one nor m see for example,
485
00:47:58,780 --> 00:48:03,930
if you just consider the example just now
what we considered with m is equal to ten
486
00:48:03,930 --> 00:48:07,400
and a being equal to two and b being equal
to five ok
487
00:48:07,400 --> 00:48:14,400
So now let us consider two into five divided
by ten this ratio so you see that two into
488
00:48:15,350 --> 00:48:21,610
five by ten therefore, it means that the g
c d of this particular thing is actually not
489
00:48:21,610 --> 00:48:27,520
equal to its not it is neither one nor ten
but, it is in between therefore, this shows
490
00:48:27,520 --> 00:48:34,520
that there should be a i mean there should
be a number d which divides both a and d also
491
00:48:35,480 --> 00:48:42,200
divides m there should be a number d and that
number is neither d is not neither equal to
492
00:48:42,200 --> 00:48:49,200
one nor nor is equal to m so this proves that
this shows that m divides a b but, neither
493
00:48:54,450 --> 00:48:59,940
a not nor nor b and therefore, this implies
that the g c d of a and m lies between one
494
00:48:59,940 --> 00:49:01,330
and m.
495
00:49:01,330 --> 00:49:07,190
Now conversely if one this lesser than equal
to g c d is less than g c d is a coma m less
496
00:49:07,190 --> 00:49:14,190
than m then define then if i define a number
b as m divided by g c d a coma m then both
497
00:49:15,020 --> 00:49:22,020
a and b are nonzero modulo m so then both
a and b are nonzero modulo m that is true
498
00:49:22,320 --> 00:49:26,970
from the definition itself right but, if what
what if i multiply a and b so if i multiply
499
00:49:26,970 --> 00:49:33,570
a and b then what i obtain is a so if i if
i take this definition of m divided by g c
500
00:49:33,570 --> 00:49:40,570
d of a coma m and if i multiply this with
a then i obtain zero modulo m the reason is
501
00:49:41,370 --> 00:49:48,270
if i if i multiply this with a and this is
the g c d of a coma m then this divides a
502
00:49:48,270 --> 00:49:52,730
and therefore, what i obtain is an integer
multiple of m and therefore, if i take a congruence
503
00:49:52,730 --> 00:49:57,480
or rather if i divide it by a and take the
remainder then the remainder is zero so this
504
00:49:57,480 --> 00:50:03,950
proves that a b is congruent to zero modulo
m thus a plus m z is a zero divisor of z slash
505
00:50:03,950 --> 00:50:07,980
m z
So therefore, we will see that i mean zero
506
00:50:07,980 --> 00:50:12,740
zero divisors i mean it is very easy to detect
zero divisors how i mean because we just this
507
00:50:12,740 --> 00:50:17,520
simple test reveals the zero divisor if i
take the g c d of number a and along with
508
00:50:17,520 --> 00:50:23,420
m and if the g c d lies between one and m
then actually it forms the zero divisors so
509
00:50:23,420 --> 00:50:28,000
there is a natural corollary to this that
if p is a prime then set z plus z slash p
510
00:50:28,000 --> 00:50:33,650
z will have no zero divisors because because
if i take the g c d of a number with p then
511
00:50:33,650 --> 00:50:37,830
it is always equal to one because that is
the definition of primelity right so that
512
00:50:37,830 --> 00:50:43,500
means that it automatically implies that there
are no zero devisors if p is a prime number
513
00:50:43,500 --> 00:50:48,120
in this particular z slash p z
514
00:50:48,120 --> 00:50:52,010
So then we come to the definition of field
which says that a field is a commutative ring
515
00:50:52,010 --> 00:50:59,010
r plus multiplication in which every element
in the semi-group r multiplication is invertible
516
00:50:59,290 --> 00:51:04,120
therefore, if every element is invertible
then it is a field so examples of this would
517
00:51:04,120 --> 00:51:09,030
be like the set of integers is not a field
because you know why i mean the set of integers
518
00:51:09,030 --> 00:51:14,060
cannot form a field because every number is
not invertible the set of real and complex
519
00:51:14,060 --> 00:51:18,290
numbers from a field and the residue class
modulo prime number except zero is a field
520
00:51:18,290 --> 00:51:23,680
why because we are seeing that because of
this particular result if p is a prime then
521
00:51:23,680 --> 00:51:30,610
z slash p z will have no zero divisors so
that automatically implies that this is not
522
00:51:30,610 --> 00:51:31,850
a field
523
00:51:31,850 --> 00:51:36,300
So then we come to the concept of euler's
totient function which says that if a is greater
524
00:51:36,300 --> 00:51:40,970
than equal to one and m is greater than equal
to two are integers and if g c d of a coma
525
00:51:40,970 --> 00:51:46,990
m is equal to one then we say that a and m
are relatively prime now the number of integers
526
00:51:46,990 --> 00:51:51,870
in z m where m is greater than one that are
relatively prime to m and does not exceed
527
00:51:51,870 --> 00:51:57,330
m is denoted by phi m or which is also called
as euler's totient function or phi function
528
00:51:57,330 --> 00:52:02,860
therefore, we will we will start with this
define it in a recursive fashion and it says
529
00:52:02,860 --> 00:52:05,960
that phi of one is equal to one ok
530
00:52:05,960 --> 00:52:10,650
Now what about let us consider phi of twenty
six so twenty six is a number of english letters
531
00:52:10,650 --> 00:52:14,410
which are there and we let us consider the
value of phi of twenty six we can see that
532
00:52:14,410 --> 00:52:20,580
phi of twenty six is equal to thirteen if
p is a prime then phi of p is p minus one
533
00:52:20,580 --> 00:52:25,950
and if n is equal to one two twenty four then
the values of phi n are as follows so this
534
00:52:25,950 --> 00:52:32,020
is just a enumeration so, you'll see that
you can verify these results but, the fact
535
00:52:32,020 --> 00:52:36,720
is that this function is very irregular that
is you'll not find any distinct even like
536
00:52:36,720 --> 00:52:42,750
it is not like as n increases phi n always
increases you can see that there is defect
537
00:52:42,750 --> 00:52:44,850
time also ok
538
00:52:44,850 --> 00:52:48,760
So now the let us see the properties of phi
n so there is a very interesting result which
539
00:52:48,760 --> 00:52:53,680
says that if m and n are relatively prime
numbers then phi of m n is nothing but, the
540
00:52:53,680 --> 00:52:57,750
product of phi m and phi n now this result
is very interesting and helpful to compute
541
00:52:57,750 --> 00:53:02,690
the phi of higher numbers or larger numbers
for example, phi of seventy seven is same
542
00:53:02,690 --> 00:53:07,690
as phi of seven multiplied by eleven which
is same as phi of seven multiplied by phi
543
00:53:07,690 --> 00:53:12,170
of eleven and we know that if seven is a prime
number then phi of seven is nothing but, six
544
00:53:12,170 --> 00:53:16,860
and phi of eleven is nothing but, ten so the
product is sixty similarly, phi of one eight
545
00:53:16,860 --> 00:53:21,960
nine six divide i mean if i obtain the prime
factorizations i can obtain the result of
546
00:53:21,960 --> 00:53:26,520
six twenty four so this result can be extended
to more than two arguments comprising of pair
547
00:53:26,520 --> 00:53:28,400
wise co-prime integers
548
00:53:28,400 --> 00:53:34,570
So let us see some interesting results that
is it says that if the first results says
549
00:53:34,570 --> 00:53:39,910
that if there are m terms of an arithmetic
progression a p and has got common differences
550
00:53:39,910 --> 00:53:46,390
which are prime to m then the remainders form
z m so if there are m terms of an arithmetic
551
00:53:46,390 --> 00:53:51,660
progression and has got common difference
that is if the common difference is prime
552
00:53:51,660 --> 00:53:58,040
to m then the remainders form z m so this
is a quite simple result which you can check
553
00:53:58,040 --> 00:54:05,040
the other one says that an integer a is relatively
prime to m if and only if its remainder is
554
00:54:05,430 --> 00:54:10,780
relatively prime to m so if an integer a is
relatively prime to m then it automatically
555
00:54:10,780 --> 00:54:14,950
it is an bidirectional implication is that
same as saying as that its remainders is also
556
00:54:14,950 --> 00:54:19,120
relatively prime to m
And the other interesting results is that
557
00:54:19,120 --> 00:54:25,320
if there are m terms of an a p and i've got
a common difference prime to m then if it
558
00:54:25,320 --> 00:54:30,240
is actually a combination of these two results
which says that then there are phi m elements
559
00:54:30,240 --> 00:54:35,060
in the arithmetic progression which are relatively
prime to m because so you can follow like
560
00:54:35,060 --> 00:54:38,970
because it says that the reminders form z
m and you know that if the reminders form
561
00:54:38,970 --> 00:54:45,740
z m and and because i mean an integer is relatively
prime to m only when the remainder is relatively
562
00:54:45,740 --> 00:54:50,030
prime to m so we actually need to find out
the number of remainders which are relatively
563
00:54:50,030 --> 00:54:55,280
prime to m and that follows from the definition
as phi m right therefore, if there are n terms
564
00:54:55,280 --> 00:54:59,550
of an arithmetic progression and the common
difference is also prime to m then there are
565
00:54:59,550 --> 00:55:04,170
actually phi m elements in the arithmetic
progression which are relatively prime to
566
00:55:04,170 --> 00:55:05,300
m
567
00:55:05,300 --> 00:55:10,810
So we will apply this result to obtain this
nice uh result the nice observation so let
568
00:55:10,810 --> 00:55:15,820
us consider phi of m n so we know from the
definition phi of m n means those numbers
569
00:55:15,820 --> 00:55:22,110
or rather means the number of values inside
from one to m minus n which are actually co-prime
570
00:55:22,110 --> 00:55:29,110
to m n so let us now arrange this numbers
from one to m n minus so this is nothing but,
571
00:55:29,420 --> 00:55:34,430
m minus one i mean so this is m n minus n
plus n so this is m n therefore, from one
572
00:55:34,430 --> 00:55:39,660
to m n all this numbers we have arranged them
in this fashion so it is like one two and
573
00:55:39,660 --> 00:55:44,090
so on till n we arrange them in this fashion
So now you consider that among these numbers
574
00:55:44,090 --> 00:55:51,090
if you need to find out so if since m and
n are relatively prime to both m and both
575
00:55:54,760 --> 00:56:00,810
n now you see that there are i mean among
this numbers if you consider like this number
576
00:56:00,810 --> 00:56:05,930
like for example, is so, you'll find that
there are phi n columns that is there are
577
00:56:05,930 --> 00:56:11,020
phi n columns in which all the elements are
co-prime to n so if you see that for example,
578
00:56:11,020 --> 00:56:16,130
this particular column that is the last column
that is n n plus one this this column all
579
00:56:16,130 --> 00:56:21,630
the elements are not co-prime to n because
this is actually you will find that n multiply
580
00:56:21,630 --> 00:56:27,960
this n plus n this m minus one into n plus
n and so on so that means that you need to
581
00:56:27,960 --> 00:56:33,240
find out the number of columns in which all
the elements are co-prime to n and we know
582
00:56:33,240 --> 00:56:37,680
that there're there are actually phi n columns
in which all the elements are co-prime to
583
00:56:37,680 --> 00:56:38,460
n
584
00:56:38,460 --> 00:56:42,600
Now among this columns which are actually
co-prime to n we will like to find out what
585
00:56:42,600 --> 00:56:48,170
are the number which are actually co-prime
to m so we see that we apply the previous
586
00:56:48,170 --> 00:56:54,320
result and see that for example, let us consider
this so in this particular column so assume
587
00:56:54,320 --> 00:57:00,140
that k is actually co-prime to n and consider
this column and the result says the previous
588
00:57:00,140 --> 00:57:06,830
result says that if so what is that what is
the what is the difference here what is the
589
00:57:06,830 --> 00:57:13,420
common difference the common difference is
k now if this k is actually therefore, co-prime
590
00:57:13,420 --> 00:57:18,470
to m then the the previous result says that
if there are m terms of an arithmetic progression
591
00:57:18,470 --> 00:57:23,369
and has a common difference which is prime
to m then there are phi phi m elements in
592
00:57:23,369 --> 00:57:28,520
the arithmetic progression which are relatively
prime to m so that means that these elements
593
00:57:28,520 --> 00:57:31,910
i mean there are phi m elements which are
actually co-prime to m
594
00:57:31,910 --> 00:57:36,680
So if there are phi m elements here which
are co-prime to n and there are phi n five
595
00:57:36,680 --> 00:57:40,630
elements which are actually i mean which all
the elements are co-prime to n then combining
596
00:57:40,630 --> 00:57:46,070
this we obtain that the phi m into phi n numbers
which are actually co-prime to both m and
597
00:57:46,070 --> 00:57:53,070
both n and that actually forms as phi gives
us the number of phi m n ok
598
00:57:53,340 --> 00:57:57,780
So therefore, we conclude like thus there
are phi n columns with phi m elements in each
599
00:57:57,780 --> 00:58:02,550
which are co-prime to both m and n and thus
there are phi m phi n elements which are co-prime
600
00:58:02,550 --> 00:58:05,350
to m n this proves the result
601
00:58:05,350 --> 00:58:09,310
We can actually apply this result and we can
obtain some interesting observations like
602
00:58:09,310 --> 00:58:13,190
phi p ot the power of a will be equal to p
to the power of a minus p to the power of
603
00:58:13,190 --> 00:58:17,900
a minus one and these is quite evident for
a equal to one but, for a greater than one
604
00:58:17,900 --> 00:58:23,410
out of the elements one two p p to the power
of a the elements p p square p to the power
605
00:58:23,410 --> 00:58:27,869
of a minus one are multiplied with p are not
co-prime to p to the power of a the rest will
606
00:58:27,869 --> 00:58:31,940
be co-prime therefore, we can obtain like
phi p to power of a will be p to the power
607
00:58:31,940 --> 00:58:37,380
of a minus p to the power of a minus one and
that is actually equal to p power of a into
608
00:58:37,380 --> 00:58:39,980
one minus one by p
609
00:58:39,980 --> 00:58:44,020
We can actually extend this result and apply
this like follows and compute the phi of any
610
00:58:44,020 --> 00:58:48,580
value in this case we know from the fundamental
theory of arithmetic that we can actually
611
00:58:48,580 --> 00:58:53,540
take n and obtain phi if we and we can obtain
this prime factorizations then it will be
612
00:58:53,540 --> 00:58:58,150
easy to compute the value of phi n so that
means the if factorization of n is available
613
00:58:58,150 --> 00:59:02,700
then computation of phi n can be obtained
using this this this formula so for example,
614
00:59:02,700 --> 00:59:08,430
if i need to the compute phi of sixty then
we can and we know prime factorization of
615
00:59:08,430 --> 00:59:13,140
sixty as four multiplied by three multiplied
by five and phi of sixty will be sixty into
616
00:59:13,140 --> 00:59:19,350
one minus one by two into one minus one by
three into one minus one by five so that is
617
00:59:19,350 --> 00:59:26,350
this is equal to two square therefore, this
is this is straight away application of this
618
00:59:27,360 --> 00:59:28,200
result
619
00:59:28,200 --> 00:59:32,750
So then we see that just let us conclude our
todays talk with theorem of fermat which says
620
00:59:32,750 --> 00:59:38,250
that it is called fermat's little theorem
and its very useful so it is so you see that
621
00:59:38,250 --> 00:59:44,520
if g c d of a coma m is equal to one then
a power phi m is actually congruent to one
622
00:59:44,520 --> 00:59:50,119
modulo m actually this result is euler's fermat's
theorem and variation of this is same as fermat's
623
00:59:50,119 --> 00:59:55,570
theorem it says that if in fermat's little
theorem this this m is actually a prime number
624
00:59:55,570 --> 01:00:01,280
we know we know that if phi m is prime then
phi of p will be equal to p minus one and
625
01:00:01,280 --> 01:00:06,400
therefore, a to the power of p minus one will
be congruent to one modulo m so that is fermat's
626
01:00:06,400 --> 01:00:12,250
little theorem but, let us consider the generalized
theorem and let us consider a set r which
627
01:00:12,250 --> 01:00:17,160
is formed of r one to r phi m we know that
there are phi m elements which forms a reduced
628
01:00:17,160 --> 01:00:18,270
set modulo m
629
01:00:18,270 --> 01:00:24,380
Now if g c d of a coma m is equal to one we
see that a multiplied with r one so we just
630
01:00:24,380 --> 01:00:29,500
take this numbers and you multiply each of
them with a now this if g c d of a coma m
631
01:00:29,500 --> 01:00:35,950
is one then this results also reduced systems
modulo m now this is just to be a permutation
632
01:00:35,950 --> 01:00:40,230
of the set r so this is we have considered
na we have considered one example previously
633
01:00:40,230 --> 01:00:46,640
where we have seen that if you've taken some
numbers and those numbers essentially where
634
01:00:46,640 --> 01:00:51,800
nothing but, rearrangement right like if you
see that in the previous example of this this
635
01:00:51,800 --> 01:00:57,170
this particular set so this particular set
was just a rearrangement of the original of
636
01:00:57,170 --> 01:00:58,580
the original remainder ok
637
01:00:58,580 --> 01:01:03,630
So therefore, similarly, i mean applying this
if you just take this set of remainders and
638
01:01:03,630 --> 01:01:09,700
if you multiply them with a number a which
is co-prime to m then you obtain another order
639
01:01:09,700 --> 01:01:14,050
of the of the remainder now if we just take
therefore, therefore, therefore, it is basically
640
01:01:14,050 --> 01:01:18,740
the same set of the remainders but, in some
other order therefore, the product of these
641
01:01:18,740 --> 01:01:23,230
particular elements will of these elements
will also be the same so if i just take this
642
01:01:23,230 --> 01:01:27,830
numbers and i multiply them this should be
the same as the product of these numbers so
643
01:01:27,830 --> 01:01:32,040
that means writing them i mean in one this
will this will work out to be a to the power
644
01:01:32,040 --> 01:01:36,160
of phi m because there are phi m numbers and
on the left hand side you'll have r one to
645
01:01:36,160 --> 01:01:39,990
r phi m multiplied and the right hand side
you'll also have r one to r phi m multiplied
646
01:01:39,990 --> 01:01:45,160
now note that since these numbers are actually
co-prime with m therefore, they can be cancelled
647
01:01:45,160 --> 01:01:50,180
out and therefore, what remains is a power
phi m is congruent to one modulo m and that
648
01:01:50,180 --> 01:01:51,640
is the euler's fermat's formula
649
01:01:51,640 --> 01:01:56,880
So example of this can be applied to find
out very interesting results like suppose
650
01:01:56,880 --> 01:02:00,619
seventy two to the power of one thousand one
is divided by thirty one and we need to go
651
01:02:00,619 --> 01:02:05,530
argue that so you know that seventy two is
nothing but, equal to ten modulo thirty one
652
01:02:05,530 --> 01:02:10,530
hence seventy two to the power of one thousand
one will be equal to ten to the power of thousand
653
01:02:10,530 --> 01:02:15,890
one mod modulo thirty one now if i apply fermat's
theorem then since p is a prime number then
654
01:02:15,890 --> 01:02:20,080
ten to the power of thirty should be equal
to or congruent to one modulo thirty one note
655
01:02:20,080 --> 01:02:25,609
that thirty one is prime therefore, raising
both sides to the power thirty three will
656
01:02:25,609 --> 01:02:30,820
be ten to the power of nine hundred ninety
is congruent to one modulo thirty one and
657
01:02:30,820 --> 01:02:34,340
therefore, we find that it it is this can
be worked out like this ten to the power of
658
01:02:34,340 --> 01:02:38,369
thousand one will be ten to the power of nine
hundred ninety which is the nearest number
659
01:02:38,369 --> 01:02:40,609
and these are the subsequent remaining numbers
660
01:02:40,609 --> 01:02:44,359
So similar number these small numbers you
can reduce quite easily using and apply the
661
01:02:44,359 --> 01:02:49,170
previous results to these computations then
you can see that this will work out to nineteen
662
01:02:49,170 --> 01:02:53,650
modulo thirteen one similarly, you can work
out this example take this as an example exercise
663
01:02:53,650 --> 01:02:58,400
that is find the least residue of seven to
the power of nine seventy three modulo seventy
664
01:02:58,400 --> 01:03:03,140
two note seventy two is not a prime number
so i conclude here and we have followed these
665
01:03:03,140 --> 01:03:08,690
texts from telang and from buchmann for the
i man for this part and next day's topic will
666
01:03:08,690 --> 01:03:09,650
be probability and information theory