Numbers history

A decimal place system has been traced back to ca. 500 in India. Before that epoch, the Brahmi numeral system was in use; that system did not encompass the concept of the place-value of numbers. Instead, Brahmi numerals included additional symbols for the tens, as well as separate symbols for hundredand thousand.

The Indian place-system numerals spread to neighboring Persia, where they were picked up by the conquering Arabs. In 662, Severus Sebokht - a Nestorian bishop living in Syria wrote:

I will omit all discussion of the science of the Indians ... of their subtle discoveries in astronomy — discoveries that are more ingenious than those of the Greeks and the Babylonians - and of their valuable methods of calculation which surpass description. I wish only to say that this computation is done by means of nine signs. If those who believe that because they speak Greek they have arrived at the limits of science would read the Indian texts they would be convinced even if a little late in the day that there are others who know something of value.

The addition of zero as a tenth positional digit is documented from the 7th century byBrahmagupta, though the earlier Bakhshali Manuscript, written sometime before the 5th century, also included zero. But it is in Khmer numerals of modern Cambodia where the first extant material evidence of zero as a numerical figure, dating its use back to the seventh century, is found.

As it was from the Arabs that the Europeans learned this system, the Europeans called them Arabic numerals; the Arabs refer to their numerals as Indian numerals. In academic circles they are called the Hindu–Arabic orIndo–Arabic numerals.

The significance of the development of the positional number system is probably best described by the French mathematician Pierre Simon Laplace (1749–1827) who wrote:

It is India that gave us the ingenious method of expressing all numbers by the means of ten symbols, each symbol receiving a value of position, as well as an absolute value; a profound and important idea which appears so simple to us now that we ignore its true merit, but its very simplicity, the great ease which it has lent to all computations, puts our arithmetic in the first rank of useful inventions, and we shall appreciate the grandeur of this achievement when we remember that it escaped the genius of Archimedes and Apollonius, two of the greatest minds produced by antiquity.

If u like this plz follow and give us your query ....

Field theory

Feild ;-

An algebric expression (F,+,.) where F is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

D1. (F, + ) be an abelian group.

1)      closure for addition:-  if a,b ∈F => a+b∈F for all a,b∈F

2)      associate for addition:-if a,b,c ∈F then (a+b)+c=a+(b+c) for all a,b,c∈F

3)      identity for addition:-  if for all a∈F there are exist 0∈F such that a+0=a for all a∈F

4)      inverse for addition:- if for all a∈F then there are exist –a∈F such that a+(-a) ∈F for all a∈F

5)      commutative low    :-  for all a,b∈F such that a+b=b+a for all a,b∈F

D2.  for  (R,.).

      1)   closure for multifi. :-   for all a,b∈F => a.b∈F  for all a,b∈F

      2)  associate for dot    :-   for all a,b,c∈F st    (a.b).c=a.(b.c)  for all a,b,c∈F

      3) Commutative for dot:- for all a,b∈F => a.b =b.a

      4) unit element :- 1 ∈F st a.1 = a =1.a for all a∈F

      5) Inverse for dot:- for all a(≠0) ∈F there are exist  a^-1 ∈F st a a^-1 = 1

D3.  distribution law :- for all a,b,c∈F

       1) left distribution:-             a.(b+c)=a.b+a.c for all a,b,c∈F

       2) Right distribution :-       (b+c).a=b.a+c.a for all a,b,c∈F.

So (F,+,.) is a Field…………..

Subfield:-  A subset F' of F is called subfield if it is satisfies all conditions of a field.i.e. if F' shall be a Feild itself then it is called subfield.
Prime field :- A field ( F,+,.) is called prime field if it have not any proper subfield .
 Theoram:-Prove that Every field always be a integral domain.
proof :- :- let (F,+,.) be a field. Since we know that every field be a commutative ring with a unit element.

To show:- (F,+,.) will a integral domain.

                  For this we will only to show that every field does not contains a zero divisor.

                 Let a,b∈F be orbitary element of F .and let a≠0 such that ab≠0

                Since a≠0 => there are exist a^-1 ∈F .

                So, ab=0 => a^-1 (ab) = a^-1 .0    [a.0=0]

                                  =>( a^-1 a) b = 0      [associate law]

                                  =>1.b = 0                              [a^-1 a = 1]

                                  => b=0                  [1.b=b]

                 Similarly, suppose  ab = 0 and b≠0

                b≠0 => b^-1 ∈F.

therefore ab = 0 => (ab) b^-1 = b^-1 .0

                                => a(b b^-1 ) = 0

                                => a.1 = 0

                                => a =0

Therefore in a field ab=0 => a=0 or b=0

Therefore there are no zero divisor in a field and hence every field is a integral domain.

                                                                                                Hence proved………………

{note : - but its conversaly is not true. We can show it like below}

Proof :- let (F,+,.) be a field and let it contains distinct unit and zero elements i.e. 1≠0 .

                Let a be a non-zero element of F then

a^-1 = 0 => a a^-1 = a.0                [left multiple with a]

                => 1  = 0 [a a^-1 =1 and a.0 =0]

                => a.1 = a.0             [left multiplel with a]

                => 1 = 0  [ cencellation law]

This is a contradiction therefore integral domain is not a field.

                                                                Hence proved………

Integral Domain

Integral Domain à An algebric expression (D,+,.) where D is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

D1. (R, + ) be an abelian group.

1)      closure for addition:-  if a,b ∈D => a+b∈D for all a,b∈D

2)      associate for addition:-if a,b,c ∈D then (a+b)+c=a+(b+c) for all a,b,c∈D

3)      identity for addition:-  if for all a∈D there are exist 0∈D such that a+0=a for all a∈D

4)      inverse for addition:- if for all a∈D then there are exist –a∈D such that a+(-a) ∈D for all a∈D

5)      commutative low    :-  for all a,b∈D such that a+b=b+a for all a,b∈D

D2.  (R,.) be a semi group.

      1)   closure for multifi. :-   for all a,b∈D => a.b∈D  for all a,b∈D

      2)  associate for dot    :-   for all a,b,c∈D st    (a.b).c=a.(b.c)  for all a,b,c∈D

      3) Commutative for dot:- for all a,b∈D => a.b =b.a

      4) unit element :- 1 ∈ D st a.1 = a =1.a for all a∈D

D3.  distribution law :- for all a,b,c∈D

       1) left distribution:-             a.(b+c)=a.b+a.c for all a,b,c∈D

       2) Right distribution :-       (b+c).a=b.a+c.a for all a,b,c∈D

D4. D is without zero divisor :- i.e. a.b = 0 => a=0 or b=0 or both zero.

So (D,+,.) is a Integral Domain…………..

Integral Domain:- A ring (R,+,.) is called integral domain if it satisfies three more exioms

(1)   Commutative.

(2)   It have a unit element.

(3)   Without zero divisor.

But it have atleast two element.

Examples :- set of rational numbers is a integral domain which have a zero element and unit element.

2) set of real number is a integral domain .

Theoram:- Ring with modulo p of integers be a integral domain iff p be a prime number.

Proof :- suppose  (Ip,+p , .p) be a ring of integers with modulo p.

To show:- p be a prime number.

Assume contrary , p is not a prime number and let p =m . n where 1<m<p ,1<n<p.

Therefore         [n] .p [m] = [n.m] 

     =[p] = [0] (mod p)

                        i.e.     [n] .p [m] =[0] (mod p)

            but       [m] …[0] and [n]≠ [0].

So, (Ip,+p , .p) is not a integral domain.

Therefore  p is a prime number.

 Conversaly :-  let p be a prime number and [m],[n]∈Ip such that [m].p[n]=[0]
 =>  r.s = 0 (modp)
 =>    r.s is divided by p
   =>either r divided by p or s divided s divided by p, since p is prime number.
 =>Either r=0(mod p) or s = 0(mod P)
    =>[r] = [0] or [s] = [0]
Therefore (Ip,+p , .p) is a integral domain.

Ordered Integral Domain:-  A Integral domain (D,+,.) is called ordered integral domain if a subset D+ of D such that

1)      D+ ,will be closure corresponding to D with respect to ‘+’ and ‘.’ i.e. if a,b∈D+ => a+b∈D+ and a.b∈D+.

2)      For all a∈D+ its true for one and only one, a=0 , a∈D+,  -a∈D+ .

Theoram:- prove that set of complex numbers is not a ordered integral domain.

Proof:- let C be a set of complex numbers.we know that (C,+,.) be a integral domain. suppose C+ be a set of all positive complex numbers of C. where 0+i0 is a zero element of C+.

since i≠0 therefore for trichotomy law it will either i∈C+ or -i∈C+.

now C+ is always closure for '.' and '+'. 

let i∈C+ => i.i ∈C+=> -1∈C+

so i∈C+ , -1∈C+ => i(-1)∈C+ => -i∈C+

so i∈C+ => -i∈C+ this is contradiction of trichonotomy law. 

similarly -i∈C+ => i∈C+  ,this is again contradiction .

therefore (C+,+,.) is not a ordered integral domain.

                                                                    Hence proved.....

Subring

 Subring-

Definition-  suppose S be a non-empty set of (R, +, .) and if S be stable in R for addition and multification then it is called subring if

                                a ∈S , b∈S => a+b∈S

                             a ∈S , b∈S => ab ∈S for all a,b∈S.

                in other word, A non-empty set S is called subring if it is satisfies all axioms of a Ring.

Theoram:- suppose (R,+,.) be a Ring and S be a non-empty set of R.then (S,+,.)  will be subring of (R,+,.) iff

a) a ∈S , b∈S => a-b∈S

b) a ∈S ,b∈S => a.b ∈S for all a,b∈S.

Proof:- suppose (S,+,.) be a subring of (R,+,.) i.e. S is a ring itself.

To show:-  a) a ∈S , b∈S => a-b∈S

                  b) a ∈S ,b∈S => a.b ∈S for all a,b∈S

 since          a ∈S , b∈S    => a∈S, -b∈S      [inverse law]

                        =>a+(-b) ∈S              [closure low]

                        =>a-b∈S     for all a,b∈S

Now ,                 a ∈S ,b∈S => a.b ∈S for all a,b∈S.  [ closure for dot]

                                                                                    Proved first part.

Conversaly :-

                        Let S is a non empty set of R which is satisfies two axioms.

                        a) a ∈S , b∈S => a-b∈S     ………(1)

                         b) a ∈S ,b∈S => a.b ∈S for all a,b∈S…..(2)

to show:-          S is a subring of (R,+,.).

                        from (1)

a)       a ∈S , a∈S => a-a∈S

                           => 0   i.e. identity exist in R.

Now 0 ∈S , b∈S => 0-b∈S          [from 1]

                         => -b∈S.  i.e. inverse of every element exist.

                        Now a ∈S , b∈S    => a∈S, -b∈S      [inverse law]

                        =>a-(-b) ∈S                         [from 1]

                        =>a+b∈S     for all a,b∈S

i.e. closure for addition axist.

Now take a,b,c ∈S and S⊆R then a,b,c ∈R.

Therefore a+(b+c) = (a+b)+c for all a,b,c ∈S

i.e. associate law satisfies.

a,b∈S and S⊆R then a,b ∈R.

Therefore a+b = b+a  for all a,b ∈S

i.e. commutative law satisfies.

Therefore        (S,+) is an abelian group.

 Now                 (S,.) is an semi group.

                        Form (2)

                        a ∈S ,b∈S => a.b ∈S for all a,b∈S

                        closure for dot is satisfies.

Associate satisfies because a,b,c ∈S and S⊆R then a,b,c ∈R.

Therefore a.(b.c) = (a.b).c for all a,b,c ∈S

Therefore (S,.) is a semi group.

Since a,b,c ∈S and S⊆R then a,b,c ∈R.

 a(b+c) = a.b+a.c for all a,b,c ∈S

(b+c).a = b.a+c.a for all a,b,c ∈S

                   Therefore (S,+,.) is a subring … Hence Proved

Theoram :- prove that Intersection of two subrings is also a subring.

Proof:-      suppose (S,+,.) and (K,+,.) are two subrings. Since S ≠ i, K ≠ i=> S∩K ≠ i

To prove:- S∩K is also a subgroup.

For this we shall prove :-  a) a ∈ S∩K , b∈ S∩K => a-b∈ S∩K

b)       a ∈ S∩K ,b∈ S∩K => a.b ∈ S∩K for all a,b∈ S∩K.

now

                                a ∈ S∩K => a∈S and a∈K

                                b ∈ S∩K => b∈S and b∈K

Since S and K are subring then

                                a ∈S , b∈S => a-b∈S ,ab∈S ………………A

                        a ∈K , b∈K => a-b∈K ,ab∈K……………..B

now                  a ∈ S∩K ,b∈ S∩K =>a,b∈S and a,b∈K

                                      =>a-b∈S ,ab∈S ,and a-b∈K ,ab∈K

                                       => a-b∈S and a-b∈K, ab∈S and ab∈K

                                       =>a-b S∩K , a.b ∈ S∩K

Therefore S∩K is a subring..

Definations:-Left Ideals: suppose (R,+,.) be a ring then a non-empty set of S of R is called left ideals if

1)       S is a additive subgroup of R.

2)       For all s∈S, r∈R => sr∈S.

Right Ideals: suppose (R,+,.) be a ring then a non-empty set of S of R is called right ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => rs∈S.

Ideals :- suppose (R,+,.) be a ring then a non-empty set of S of R is called ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => sr∈S , rs∈S

Ring

Ringà An algebric expression (R,+,.) where R is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

R1. (R, + ) be an abelian group.

1)      closure for addition:-  if a,b ∈R => a+b∈R for all a,b∈R

2)      associate for addition:-if a,b,c ∈R then (a+b)+c=a+(b+c) for all a,b,c∈R

3)      identity for addition:-  if for all a∈R there are exist 0∈R such that a+0=a for all a∈R

4)      inverse for addition:- if for all a∈R then there are exist –a∈R such that a+(-a) ∈R for all a∈R

5)      commutative low    :-  for all a,b∈R such that a+b=b+a for all a,b∈R

R2.  (R,.) be a semi group.

      1)   closure for multifi. :-   for all a,b∈R => a.b∈R   for all a,b∈R

      2)  associate for dot    :-   for all a,b,c∈R st    (a.b).c=a.(b.c)  for all a,b,c∈R

R3.  distribution law n:- for all a,b,c∈R

       1) left distribution:-             a.(b+c)=a.b+a.c for all a,b,c∈R

       2) Right distribution :-       (b+c).a=b.a+c.a for all a,b,c∈R

So (R,+,.) is a Ring…………..

Types of Ring:à
·         Ring with unity:à A ring with multification identity 1 which is define as a.1=1.a=a for all a∈R is called Ring with unity…
·         Commutative Ring:à A ring with multification identity is called commutative ring i.e. a.b=b.a for all a,b∈R ….
Examples. Set of intergers is a Ring Set of real number is a ringSet of rational is a ring…..
·         Boolean ring:à A set (R,+,.) is called Boolean Ring if its all element are idempotent. i.e. a²=a for all a∈R
·         Zero devisors:àA ring is called with zero divisor if it is difined as ab=0 but a…0 , b…0..
·         Ring without zero divisor:à A ring is called ring without zero divisor if it is defined as for all a,b∈R  ,   ab=0 => a=0 or b=0 or a=b=0 both are zero…

Theoram:à A ring R is a without zero divisor iff R is satisfies cencellation law..

·         Proof:à  firstly let R is a Ring without zero divisor i.e. a,b∈R are orbitary element  s.t. ,   ab=0 => a=0 or b=0 or a=b=0 both are zero…
T show:-                       ab=ac (a…0) => b=c
                                     ba=ca (b…0) => b=c
for this , take                ab=ac (a…0) => a(b-c) = 0
                                                       =>b-c = 0 since a…0
                                                        => b=c
                                  Similarly we can prove   ba=ca (b…0) => b=c
Conversaly:- now let R be a ring which satisfies cancellation law.
To prove:- R is a without zero divisor ring.
Assume contrary, suppose R is a ring with zero divisor then there are exist a,b∈R such that ab=0 but a…0 , b…0.

Now ,                ab= 0, a is not 0 => ab = a.0                       [a.0=0

                                           => b = 0                 [ by cancellation law]
                             This is  a contradiction because b…≠0

Similarly           ab= 0, b is not 0… => ab = 0.b                      [0.b=0]
                                       => a = 0                [ by cancellation law]
   Again this is a contradiction because a is not. therefore R is a Ring without zero divisor......
Hence proved...........
Theoram:- if R is a ring with additive identity 0 then for all a,∈R a.0=0=0.a    for all  a,∈R
Proof:-       for all a,∈R
  0+0 = 0  
 a.(0+0)=a.0     (left multification of a)
  a.0+a.0=a.0+0  (distribution and ideal)
 a.0 = 0               (cencellation law)
  similarly we can prove 0.a=0                                                  proved…..
theorem:- :- if R is a ring with additive identity 0 then for all a,b∈R   a(-b) = -(ab) = (-a)b
proof :-     for all a,b∈R
  b + (-b) = 0           (inverse law)
  a.(b + (-b) ) = 0                       ( left multification of a)
 a.b + a.(-b) =0                         (distribution)
 ab + a (-b) =0                          (inverse law)
 a (-b) = -(ab)    for all a,b∈R  
 similarly we can prove (-a) b = -(ab)   for all a,b∈R.
Theoram:- :- if R is a ring with additive identity 0 then for all a,b∈R   (-a)(-b)=ab
Proof:-     (-a)(-b) = -[(-a)b]        
                 [a (-b) = -(ab)]
                            =>-[-(ab)]       [a (-b) = -(ab)]
                            => ab  for all a,b∈R.   [ -(-a) = a ]
                                                                             proved

Theoram :- if R is a ring with additive identity 0 then for all a,b∈R  , -(a+b) = (-a) + (-b)Proof :-     now we can take
   [(-a) + (-b) ]+ (a+b) =[(-b) + (-a)] + (a+b)        (commutative)
=>(-b) + [(-a) + (a+b)]      (associate 
 =>(-b) +[ {(-a) + (a)}+b]    (associate)
 =>(-b) +[ 0+b]                 (inverse)  
=>(-b) +( b )                   (identity)
=> 0                                             (inverse)
                                           Proved…………

Theoram:- if R is a ring with additive identity 0 then for all a,b∈R  a.(b-c)= ac-bc   for all a,b,c∈R 
Proof |-  for all a,b,c∈R 
 a.(b-c)  = a.[b(+-c)]
=>a.b + a.(-c)      [distribution law]
 =>ab – ac           {a(-b)=-(ab)}
  Similarly we can prove (b-c)a = ba-ca  for all a,b,c∈R 
Theoram :- if R is Boolean ring st a² =a for all a∈R     then prove that

(1)    a+a =0

(2)    a + b = 0 => a = b for all a,b,∈R

(3)     R be a commutative ring.

proof- 

     since R is a ring such that a² =a for all a∈R .

 a∈R => a,a ∈R 

   =>a+a∈R 

 now,

 (a+a)² = (a+a)              [since a² =a]

  (a+a) (a+a) = (a+a)

ð       (a+a)a+(a+a)a= (a+a) [left distribution]

ð       a.a+a.a+a.a+a.a =a+a [right distribution]

ð       a² + a² + a² + a² =a+a

ð       (a+a) + (a+a) = (a+a) + 0 [ since a² =a]

ð       a+a =0

proved (1)

II proof-

    Now since  a+a =0

(1)          a+b = 0 => a+b = a+a

 =>  b=a          [ cancellation law]

  proved

III) proof-

   (a+b)² = (a+b)         [since a² =a]

  (a+b) (a+b) = (a+b)

ð       (a+b)a+(a+b)b= (a+b) [left distribution]

ð       a.a+b.a+a.b+b.b =a+b [right distribution]

ð       a² + ba + ab+ b² =a+b

ð       a+ba+ab+b =a+b

=>(a+b)+ (ab+ba) = (a+b) + 0 [ since a² =a]

=> ab+ba = 0

(2)    => ab =ba    [a + b = 0 => a = b ]

Therefore R is commutative

Hence proved…

Right Ideals: suppose (R Definations:-Left Ideals: suppose (R,+,.) be a ring then a non-empty set of S of R is called left ideals if

1)       S is a additive subgroup of R.

2)       For all s∈S, r∈R => sr∈S.

(R,+,.) be a ring then a non-empty set of S of R is called right ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => rs∈S. 

Ideals :- suppose (R,+,.) be a ring then a non-empty set of S of R is called ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => sr∈S , rs∈S

Improper and Proper Ideals:- suppose (R,+,.) be a ring . Ideal R and {0} are called Improper or trivals ideals and different from it are all called Proper or non-trival ideals.

Unit and zero ideals:- (R,+,.) be a ring then R and {0} are called Unit and zero ideals.

Simple ideals:- A ring (R, +, .) is called a simple ring if it does not contains any proper ideals.

Theorem:- Intersection of two ideals is also an ideal.

Proof:- suppose M and N are two ideals where M and N are both non-empty.then we can say both have 0 element ie

0∈M ,0∈N => 0∈M∩N therefore M∩N also a non-empty.

Since M is an ideal therefore it satisfies two condition ie

 M is a subgroup and s∈M , r∈R => sr,rs∈M …………….(1)

N is also an ideal therefore it also satisfies two contion ie

N is a subgroup and s∈N ,r∈R => sr,rs∈N………………..(2)

We have to prove :- M∩N is an ideal.

Since we know that intersection of two subgroup is also a subgroup. Therefore M∩N is also a subgroupFor this, take s∈M∩N , and r∈R

Now

s∈M∩N , and r∈R => s∈M and s∈N ,r∈R

=>s∈M ,r∈R and s∈N ,r∈R

  => rs,sr∈M and rs,sr∈ N       from (1) and (2)

   => rs,sr ∈ M∩N

It satisfies both conditions .

and hence M∩N is an ideal, but union of two ideals is not necessary be an ideal . let us proof this by an example. Now we take a set M ={2n : n∈Z} and N = {5n : n∈I} are ideals for Ring R. but if you try to solve , it does not satisfies first low . for example 2∈M ,5∈ N => 2,5 ∈ M∪N but 2-5 = -3 ∉MUN because -3 does not exist both ideals , so MUN is not necessary a ideal.

Note:- similarly we can prove this forcollection of  family of intersecton of all ideals.

Theoram:- if (R,+,.) be a ring and M and N are two ideals then MUN is also an ideal iff they contains each others.

Proof:-let M and N are two ideals

To prove:-M ∪N is a Ideal of R.

  let  M and N are two ideal.and let they contains each others. i.e. M⊆N and N⊆M.

then  M∪N=M or N . and M and N are Ideal .therefore MUN  is a ideal of R.

conversaly:-  now let MUN is an ideal. Then we will show that they contains each other.assume contrary- MéN and NéM

now                  MéN=> there exist  a∈M then a∉N.               ……..(1)

 NéM => there exist b∈N then b∉M                        ……..(2)

From (1) and (2)

a∈M then a∉N=>a∈ M∪N

 b∈Nthen b∉M=>b∈ M∪N.

so,    a∈ M∪N , b∈ M∪N=>a-b ∈ M∪N   (since M∪N is a subgroup)

 a-b∈ MUN then a-b ∈ M or a-b∈ N

now

a ∈ M, a-b∈ M=> a-(a-b)= b∈ M but b∉M this is a contradiction.  {by 1}

Similarly a-b∈ N , b∈ N => a-b+b=a∈ N but a∉N  this a contradiction. {by 2}

Therefore M⊆N and N⊆M.

Hence proved.

 Definition :-suppose R is a commutative ring with unit element and a∈R then the ideal of it all multiples of a ie (a) = {ra : r∈R } is an ideal generate from a and its denoted by (a). then an Ideal M of R is called principal ideal if for a∈R st M=(a)..

*Result:- if M and N are two ideals of ring R then the set M+N = { x∈R ; x=a+b, a∈M ,b∈N }, is an ideal generate from MUN .

Proof:- since M and N are two ideals and we have to show M+N  is an ideal generate from MUN . for this we have to firstly show that M+N is an ideal then its an ideal generate from MUN.

Let x,y ∈M+N =>there are exist  a₁,a₂ ∈M and b₁,b₂∈N   where x=a₁+b₁ and y=a₂+b₂.

  *         Now, since x,y ∈M+N => x-y=(a₁+b₁)-(a₂+b₂)

                                                        =(a₁-a₂)+(b₁-b₂) ∈ M+N 

[because M and N are ideals therefore  a₁-a₂∈M and b₁-b₂∈N ]

So we can say that if x,y ∈M+N => x-y∈M+N for all x,y is in M+N.

·         let r∈R and x∈M+N => xr=(a₁+b₁)r = a₁r +b₁r∈M+N

=>xr∈M+N

Similarly r∈R and x∈M+N then rx∈M+N. this shows that M+N is an ideal of R.

After this we will prove that M+N is an ideal generate from MUN. for this we show that M+N=(M+N)

For this we take element a∈M so a= a+0 ∈M+N [0∈N] therefore M⊆M+N and similarly N⊆M+N…….

Since M⊆M+N , N⊆M+N => MUN⊆M+N………….(a)

We suppose that X ve an ideal of R such that MUN⊆X .

If z∈M+N then z=a+b where a∈M,b∈N.

Now , a∈MUN and b∈ MUN => a,b ∈X because MUN⊆X

                                         => a+b∈X

                                         => z∈X .

ie M+N ⊆X. then we can say M+N =(MUN) ie M+N is an ideal generate from MUN.

                                      hence proved.

*Note :- multiple of two ideals is also an ideal.

*Result :- prove that the ring of all integer be an principal ideal ring.

Proof:-

           Let (Z,+, .) be a ring of integer .let S be and Ideal of Z .

Condition 1:- if S=(0) then nothing to prove because S be a trival principal ideal generate from 0∈Z.

Condition2:- if S≠(0). Ie S is a non-trival ideal. Then we can say S contains only positive elements.then we can say there are exist n be a least positive integer such that n∈S .

then (n) ⊆ S……(1)

Now only to show that S⊆ (n). and then we can say S will be principal ideal of Z.

 Let a∈Z such that a∈S. then by the definition of division function there are exist q,r in Z such that a= qn+r where 0≤r<n.  or a-qn=r..

Since S be an ideal therefore n∈S , q∈Z=> nq∈S. [definition of ideal] .

Again a∈S ,qn∈ S => a-qn∈S [because S is a addition subgroup of Z]

ð       r∈S. since 0≤r<n where n is a least positive integer therefore n is in S and then the value of r will be zero ie r=0 .

therefore a=qn , where a is the multiple of n and then a∈(n).[definition of principal ideal]

ie

a∈S => a∈(n) so S⊆(n)……..(2)

by (1) and (2)

(n) ⊆ S , S⊆(n) => S=(n)

And hence S is a principal ideal generate form Z.

                                                            Hence proved.

Euclidean Ring:- A commutative ring is called  an Euclidean ring if for all x∈R there are exist a non-negative integer d(x)  define as

1)      d(x) = 0 iff  x=0 ,where 0∈R

2)       d(x,y) ≥ d(x) when x,y ≠ 0

3)       for all x∈R and y≠z∈R there are exist q,r∈R st x=yq+r   , 0≤d(r)<d(y).

 if u like this then give us feedback and follow ..........