Question

Q12.A solution of NH4OH of density 2.04 g/ml is taken into consideration,
whose weight percentage is 20% .Calculate molarity, molality and mole fraction
for the following solution.​

Answer 1

Congruence Classes of Polynomials Modulo p(x) over a Field

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Congruence Classes of Polynomials Modulo p(x) over a Field

Congruence Classes of Polynomials Modulo p(x) over a Field

Recall that if a,b∈Z and m∈Z then we say that a is congruent to b modulo m written a≡b(modm) if m|(a−b). In other words, two integers are congruent modulo m if upon division by m we have that a and b have the same remainder.

We now extend this notion to polynomials over a field F.

Definition: Let (F,+,⋅) be a field and let p∈F[x]. If a,b∈F[x] then we say that a is Congruent to b modulo p denoted a(x)≡b(x)(modp(x)) if p(x)|(a(x)−b(x)).

In the following proposition we show that congruence modulo p is an equivalence relation.

Proposition 1: Let (F,+,⋅) be a field and let p∈F[x]. Then congruence modulo p is an equivalence relation.

Proof: Let a∈F[x]. Then clearly p(x)|(a(x)−a(x))=0. So a(x)≡a(x)(modp(x)) so congruence modulo p is reflexive.

Let a,b∈F[x] and suppose that a(x)≡b(x)(modp(x)). Then p|(a(x)−b(x)). So there exists a polynomial q∈F[x] such that p(x)q(x)=a(x)−b(x). Hence p(x)[−q(x)]=b(x)−a(x). So p(x)|(b(x)−a(x)) and hence b(x)≡a(x)(modp(x)). So congruence modulo p is symmetric.

Lastly, let a,b,c∈F[x] and suppose that a(x)≡b(x)(modp(x)) and b(x)≡c(x)(modp(x)). Then p(x)|(a(x)−b(x)) which implies there exists a q1(x)∈F[x] such that p(x)q1(x)=a(x)−b(x) and p(x)|(b(x)−c(x)) which implies there exists a q2(x)∈F[x] such that p(x)q2(x)=b(x)−c(x). So by substituting the first equation into the second we get:

(1)

p(x)q2(x)=b(x)−c(x)=[a(x)−p(x)q1(x)]−c(x)

Hence p(x)[q1(x)+q2(x)]=a(x)−c(x) so a(x)≡c(x)(modp(x)). So congruence modulo p is transitive.

Therefore equivalence modulo p is an equivalence relation. ■

From the previous proposition we can define congruence classes and the set of all congruence classes modulo p.

Definition: Let (F,+,⋅) be a field and let p∈F[x]. For any polynomial a∈F[x] the Congruence Class of a modulo p denoted [a(x)]p(x) is defined as [a(x)]p(x)={b(x)∈F[x]:b(x)≡a(x)(modp(x)). The set of all congruence classes modulo p is denoted by F[x]/<p(x)>.

When no ambiguity arises we may use the notation "[a(x)]" in place of "[a(x)]p(x)".

Note that a polynomial b∈[a(x)]p(x) is always of the form:

(2)

b(x)=a(x)+q(x)p(x)

Theorem 1: Let (F,+,⋅) be a field and let p∈F[x] with p(x)≠0. Let a∈F[x]. If p is not a divisor of a then there exists exactly one polynomial r(x)∈[a(x)]p(x) such that deg(r)<deg(p).

Proof: Let a∈F[x]. Since p∈F[x] is such that p(x)≠0, by The Division Algorithm for Polynomials over a Field there exists unique polynomials q,r∈F[x] such that:

(3)

a(x)=p(x)q(x)+r(x)

Where r(x)=0 or deg(r)<deg(p). Since p is not a divisor of a we have that r(x)≠0. We rewrite the equation above as:

(4)

r(x)=a(x)−p(x)q(x)=a(x)+[−q(x)]p(x)

So r(x)∈[a(x)]p(x) is such that deg(r)<deg(p).

We now show that only one such r(x) exists. Suppose that r′(x)∈[a(x)]p(x) is such that deg(r′)<deg(p). Then since r′(x)∈[a(x)]p(x) we have that r′(x)≡a(x)(modp(x)).

Since r(x)≡a(x)(modp(x)) and r′(x)≡a(x)(modp(x)) we have that r(x)≡r′(x)(modp(x)) so p(x)|(r(x)−r′(x)). But since deg(p)>deg(r),deg(r′) can only happen if r(x)−r′(x)=0, so r(x)=r′(x). ■