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Pigeon Hole Principle

If

A field, on the other hand, is a commutative ring with unit element in which every non-zero element has a multiplicative inverse in the ring.

Let

Let

The elements

We claim that these elements are all distinct.

Let if possible

Since

Thus

Now we shall show that

Thus

**n**objects are distributed over m places and if**n > m**, then some places receives at least two objects.**Theorem:**Prove that finite integral domain is a field.**Proof:**We know that an Integral domain is a commutative ring**R**without zero division, i.e. if**a, b R**then**ab = 0**either**a = 0**or**b = 0**.A field, on the other hand, is a commutative ring with unit element in which every non-zero element has a multiplicative inverse in the ring.

Let

**D**be a finite integral domain. In order to prove that**D**is a field we must show that**1.**There exists an element**1 D**such that**a . 1 = a**for every**a D**i.e.**2.**For every element**a ≠ 0, a D**, there exists an element**b D**such that**ab = 1**, i.e. the multiplicative inverse of every elements in**D**exists.Let

**O(D) = n**. let**x**_{1}, x_{2}, x_{3}, …. , x_{n}be all the elements of**D**and suppose that**a ≠ 0, a D**.The elements

**x**(closure property in_{1}a, x_{2}a, x_{3}a, … , x_{n}a D**D**).We claim that these elements are all distinct.

Let if possible

**x**for_{i}a = x_{j}a**i ≠ j**.**x**for_{j}a – x_{j}a = 0**i ≠ j**.**(x**_{i}– x_{j})a = 0 for i ≠ j.Since

**D**is an integral domain and**a ≠ 0**, so**(x**_{i}– x_{j})a = 0 x_{i}– x_{j}= 0**x**contradicting_{i}= x_{j}**i ≠ j**.Thus

**x**are_{1}a, x_{2}a, x_{3}a, …. , x_{n}a**n**distinct elements belonging to**D**, which has exactly n elements. By the pigeon-hole principle, all these elements must account for all the elements of**D**. Thus every element**y D**must be of the form**x**for some_{i}a**x**_{i}. In particular**a D**, so**a = x**for some_{i0}a**x**._{i0}D**∴ a = x**_{10}a = ax_{10}. (**∵ D**is a commutative ring)Now we shall show that

**x**_{10}acts as a unit element for every element of**D**. Let**y**be an arbitrary element of**D**.**∴ y D y = x**for some_{1}a**x**and so._{i}D**yx**(associative law holds in_{i0}= (x_{i}a)x_{i}0 = x_{i}(ax_{i0})**D**).**= x**._{i}a = y (∵ ax_{10}= a)Thus

**x**_{10}is a unit element for**D**and we write it as**1**. Now**1 D**, so it must be multiple of**a**there exists**a, b D**such that**1 = ab b**is the inverse of**a ∀ a ≠ 0 D**.**Services: -**Pigeon Hole Principle Homework | Pigeon Hole Principle Homework Help | Pigeon Hole Principle Homework Help Services | Live Pigeon Hole Principle Homework Help | Pigeon Hole Principle Homework Tutors | Online Pigeon Hole Principle Homework Help | Pigeon Hole Principle Tutors | Online Pigeon Hole Principle Tutors | Pigeon Hole Principle Homework Services | Pigeon Hole PrincipleSubmit Your Query ???

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