Mathquery

Theorem Related To Rings :     Theorem 1 :                    If  R is a commutative ring with unit element and M is an ideal of R , then M is maximal ideal of R iff R / M is a field . Proof :          Since R is a commutative ring with unity , therefore R / M is also a commutative ring with unity . The zero element of the ring R / M is M and the unit element of the coset M +1 where 1 is the unit element of R .              Let the ideal M be maximal . Then to prove that R / M is a field .        Let M + b be any non zero element of    R  / M  . Then M + b ≠ M i.e b doesn't belongs to M . To prove that M + b is inversible .          If (b) is the principal ideal of R generated by b, then M(b) is also an ideal of R . Since b doesn't belong to M , therefore the ideal M is pro...

Definition and Examples Of Rings :  Definition of Rings :                  A non empty set R is said to be a ring (or an associative ring) if in R there are defined two binary operations called addition and multiplication denoted by  + and * such that for all a,b,c ∈ R .  1. a+b ∈ R (closure axiom for addition)  2. a+b = b+a (commutative axiom for                                                                       addition)  3. a+(b+c) = (a+b)+c   (associative axiom for                                                        addition)  4. There exists an element 0 in...

Newton's Forward Difference Interpolation Formula:      Let y = f(x) be a function of x and let us suppose that yᵢ = f(xᵢ) ...(1) for i = 1,2,3,.....,n satisfying the condition xᵢ = x₀+ih where  'h' is the interval of difference .        Now our aim is to constuct a function Φ(x) of degree not higher than n such that   Φ(xᵢ) = yᵢ   ............(2)           Since Φ(x) is a polynomial of degree n  then we can write Φ(x) = a₀ + a₁(x-x₀) + a₂(x-x₀)(x-x₁)     +                                 a₃(x-x₀)(x-x₁)(x-x₂)+......                   + aₙ(x-x₀)(x-x₁)(x-x₂).....(x-xₙ₋₁)...(3) Let us find the value of a₀,a₁,a₂.....aₙ satisfying the equations (2) and (3) From equation (2) , we get Φ(x₀) = y₀  From equation (3) , we get Φ(x₀) = a₀ ...

Derivation of Newton's Fundamental Interpolation Formula:     Let y = f (x) be a function with given values yᵢ = f(xᵢ) for (n+1) points x₀,x₁,x₂,.....,xₙ . Our aim is to construct a polynomial Φ(x) of degree not higher than n satisfying the following conditions        Φ(xᵢ) =yᵢ=f(xᵢ) ........(1)                              for i = 0,1,2...,n Let us take the polynomial Φ(x) in the following form  Φ(x) =a₀+a₁(x-x₀)+a₂(x-x₀)(x-x₁)+a₃(x-x₀)(x-x₁)(x-x₂) + ......+aₙ(x-x₀)(x-x₁)(x-x₂)....(x-xₙ)                                                         ........(2) where a₀,a₁,....aₙ i.e aᵢ's are constants to be determined . Putting  i=0 in equation (1) ,we get  Φ(x₀) = y₀ = f(x₀) i.e f(x₀) = Φ(x₀) Again ,...

Lagrangian Interpolation Formula :        Let y = f(x) be a real valued function which is defined in an interval [a,b] . Let  x₀ , x₁ ,x₂ ,............xₙ be n+1 distinct points in that interval at which the respective values y₀ ,y₁,y₂ ............yₙ are tabulated.      Now our aim is to construct a polynomial Φ(x) of degree ≤ n , which interpolates f(x) such that       Φ(xᵢ) = y(xᵢ) , i = 1,2,3,.......,n .........(1)    Let us suppose that the polynomial Φ(x)                                   n be given by Φ(x) = Σ  lᵢ(x) y(xᵢ)  .........(2)                                  i= 0      where each lᵢ(x) is a polynomial of degree ≤n in xᵢ , called  Lagrangian function.      The function ...

Three Theorems of Isomorphism: First Theorem of Isomorphism:              If f: G →G' be an onto homomorphism with kernel K = ker f , then G/K ≈ G'  In other words , every homomorphic image of a group G is isomorphic to a quotient group of G.  Proof :       Define a map Φ : G/K →G' such that ,                                       Φ (Ka) = f(a) , a∈G We Show Φ  is an isomorphism :        That Φ is well defined follows by                            Ka = Kb                       ⇒ab⁻¹ ∈K = Ker f                      ⇒f(ab⁻¹) = e'                ...

Fundamental Theorem Of Homomorphism Of Group :           Let Φ be a homomorphism of G onto G̅ with kernel K .Then G/K ≈ G̅ . Proof:               Let G̅ be the homomorphie image of a group G and Φ be the corresponding homomorphie . Then K is normal subgroup of G .       To prove that G/K ≈ G̅ . If   a ∈ G , the Ka ∈ G /K and Φ(a) ∈G̅  Let ψ : G/K →G̅ such that ψ(Ka) = Φ(a) ∀ a∈G  Where Ka is called right coset and Kb is called left coset. To Show The Mapping ψ is well defined :       i.e if a,b ∈G and Ka = Kb ,then  ψ(Ka) = ψ(Kb)  We have Ka = Kb => ab⁻¹ ∈ K                   => Φ(ab⁻¹) = e̅     (identity of G̅)                =>Φ(a) Φ(b⁻¹) = e̅              => Φ(a)...