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Maximum Marks: 25
Due Date: May 31, 2020
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INSTRUCTIONS

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these questions.

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Note: Attempt all the following questions.

Question: 1 Marks: 5

Determine whether the binary operation * defined by *:R×R→R and given for all a,b∈R as : a*b=〖(a+b)〗^2 is associative or not? Explain your answer.

Question: 2 Marks: 5

Show that C, the set of all non-zero complex numbers is a multiplicative group.

Question: 3 Marks: 5

Show that the following function f:Z_2→Z_2 is a ring homomorphism:

f(x)=x^2

Question: 4 Marks: 5

Show that the following function g:Z→Z is not a ring homomorphism:

f(x)=2x

Question: 5 Marks: 5

Show that in a principal ideal domain, every nonzero prime ideal is maximal.

]]>Question: 5 Marks: 5

Show that in a principal ideal domain, every nonzero prime ideal is maximal.

]]>Question: 4 Marks: 5

Show that the following function g:Z→Z is not a ring homomorphism:

f(x)=2x

]]>Question: 3 Marks: 5

Show that the following function f:Z_2→Z_2 is a ring homomorphism:

f(x)=x^2

Question: 2 Marks: 5

Show that C, the set of all non-zero complex numbers is a multiplicative group.

Answer:

Let C={z:z=x+iy, x,y∈R}C={z:z=x+iy, x,y∈R}. Here R is the set of all real numbers and i=√(-1).

(G1) Closure Axiom: If a+ib∈C and c+id∈C, then by the definition of multiplication of complex numbers

(a+ib)(c+id)=(ac–bd)+i(ad+bc)∈C

Since ac–bd,ad+bc∈R, for a,b,c,d∈R. Therefore,C is closed under multiplication.

(G2) Associative Axiom:

(a+ib){(c+id)(e+if)}=(ace–adf–bcf–bde)+i(acf+ade+bce–bdf)

={(a+ib)(c+id)}(e+if) for a,b,c,d∈R .

(G3) Identity Axiom: e=1(=1+i0) is the identity in C.

(G4) Inverse Axiom: Let (a+ib)(≠0)∈C, then

(a+ib)^(-1)=1/(a+ib)=(a-ib)/(a^2+b^2 )

=a/(a^2+b^2 )-i b/(a^2+b^2 )=m+in∈∁

Hence C is a multiplicative group.

Question: 1 Marks: 5

Determine whether the binary operation * defined by :R×R→R and given for all a,b∈R as : ab=〖(a+b)〗^2 is associative or not? Explain your answer.

Answer:

Consider the elements 1,3,6∈R. Then we have that:

1∗(2∗3)=1∗(2+3)2=1∗25=(1+25)2=676

We also have that:

(1∗2)∗3=(1+2)2∗3=9∗3=(9+3)2=122=144

Clearly 676≠144 and so ∗ is nonassociative on R since a∗(b∗c)≠(a∗b)∗c for 1,3,6∈R.