UPSC Optional (Maths) Paper-02 Algebra Solution

2014

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UPSC Algebra

\(2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)

1(a) Let \(G\) be the set of all real \(2 \times 2\) matrices \(\left[\begin{array}{ll}x & y \\ 0 & z\end{array}\right]\), where \(x z \neq 0\). Show that \(G\) is a group under matrix multiplication. Let \(N\) denote the subset \(\left\{\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]: a \in \mathbb{R}\right\}\). Is \(N\) a normal subgroup of \(G\)? Justify your answer.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Part 1: Show that G G GGG is a Group under Matrix Multiplication
To show that G G GGG is a group under matrix multiplication, we need to verify the following properties:
  1. Closure: For any two matrices A , B ∈ G A , B ∈ G A,B in GA, B \in GA,B∈G, their product A B A B ABABAB should also be in G G GGG.
  2. Associativity: For any three matrices A , B , C ∈ G A , B , C ∈ G A,B,C in GA, B, C \in GA,B,C∈G, ( A B ) C = A ( B C ) ( A B ) C = A ( B C ) (AB)C=A(BC)(AB)C = A(BC)(AB)C=A(BC).
  3. Identity: There exists an identity element E E EEE such that for any A ∈ G A ∈ G A in GA \in GA∈G, A E = E A = A A E = E A = A AE=EA=AAE = EA = AAE=EA=A.
  4. Inverses: For each A ∈ G A ∈ G A in GA \in GA∈G, there exists an inverse A − 1 ∈ G A − 1 ∈ G A^(-1)in GA^{-1} \in GA−1∈G such that A A − 1 = A − 1 A = E A A − 1 = A − 1 A = E AA^(-1)=A^(-1)A=EAA^{-1} = A^{-1}A = EAA−1=A−1A=E.
Closure
Let A = [ x 1 y 1 0 z 1 ] A = x 1      y 1 0      z 1 A=[[x_(1),y_(1)],[0,z_(1)]]A = \left[\begin{array}{ll}x_1 & y_1 \\ 0 & z_1\end{array}\right]A=[x1y10z1] and B = [ x 2 y 2 0 z 2 ] B = x 2      y 2 0      z 2 B=[[x_(2),y_(2)],[0,z_(2)]]B = \left[\begin{array}{ll}x_2 & y_2 \\ 0 & z_2\end{array}\right]B=[x2y20z2] be two matrices in G G GGG.
The product A B A B ABABAB is:
A B = [ x 1 y 1 0 z 1 ] [ x 2 y 2 0 z 2 ] = [ x 1 x 2 x 1 y 2 + y 1 z 2 0 z 1 z 2 ] A B = x 1      y 1 0      z 1 x 2      y 2 0      z 2 = x 1 x 2      x 1 y 2 + y 1 z 2 0      z 1 z 2 AB=[[x_(1),y_(1)],[0,z_(1)]][[x_(2),y_(2)],[0,z_(2)]]=[[x_(1)x_(2),x_(1)y_(2)+y_(1)z_(2)],[0,z_(1)z_(2)]]AB = \left[\begin{array}{ll}x_1 & y_1 \\ 0 & z_1\end{array}\right] \left[\begin{array}{ll}x_2 & y_2 \\ 0 & z_2\end{array}\right] = \left[\begin{array}{ll}x_1x_2 & x_1y_2 + y_1z_2 \\ 0 & z_1z_2\end{array}\right]AB=[x1y10z1][x2y20z2]=[x1x2x1y2+y1z20z1z2]
Since x 1 z 1 ≠ 0 x 1 z 1 ≠ 0 x_(1)z_(1)!=0x_1z_1 \neq 0x1z1≠0 and x 2 z 2 ≠ 0 x 2 z 2 ≠ 0 x_(2)z_(2)!=0x_2z_2 \neq 0x2z2≠0, we have x 1 x 2 z 1 z 2 ≠ 0 x 1 x 2 z 1 z 2 ≠ 0 x_(1)x_(2)z_(1)z_(2)!=0x_1x_2z_1z_2 \neq 0x1x2z1z2≠0, which implies x 1 x 2 ≠ 0 x 1 x 2 ≠ 0 x_(1)x_(2)!=0x_1x_2 \neq 0x1x2≠0 and z 1 z 2 ≠ 0 z 1 z 2 ≠ 0 z_(1)z_(2)!=0z_1z_2 \neq 0z1z2≠0. Therefore, A B ∈ G A B ∈ G AB in GAB \in GAB∈G.
Associativity
Matrix multiplication is associative, so this property is automatically satisfied.
Identity
The identity matrix E = [ 1 0 0 1 ] E = 1      0 0      1 E=[[1,0],[0,1]]E = \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]E=[1001] is in G G GGG and serves as the identity element because A E = E A = A A E = E A = A AE=EA=AAE = EA = AAE=EA=A for any A ∈ G A ∈ G A in GA \in GA∈G.
Inverses
For any A = [ x y 0 z ] ∈ G A = x      y 0      z ∈ G A=[[x,y],[0,z]]in GA = \left[\begin{array}{ll}x & y \\ 0 & z\end{array}\right] \in GA=[xy0z]∈G, the inverse A − 1 A − 1 A^(-1)A^{-1}A−1 can be found as:
A − 1 = [ x − 1 − x − 1 y z − 1 0 z − 1 ] A − 1 = x − 1      − x − 1 y z − 1 0      z − 1 A^(-1)=[[x^(-1),-x^(-1)yz^(-1)],[0,z^(-1)]]A^{-1} = \left[\begin{array}{ll}x^{-1} & -x^{-1}y z^{-1} \\ 0 & z^{-1}\end{array}\right]A−1=[x−1−x−1yz−10z−1]
It’s easy to verify that A A − 1 = A − 1 A = E A A − 1 = A − 1 A = E AA^(-1)=A^(-1)A=EAA^{-1} = A^{-1}A = EAA−1=A−1A=E, and A − 1 A − 1 A^(-1)A^{-1}A−1 is also in G G GGG.
Summary for Part 1
All the group properties are satisfied, so G G GGG is a group under matrix multiplication.
Part 2: Is N N NNN a Normal Subgroup of G G GGG?
A subgroup N N NNN of G G GGG is normal if for every n ∈ N n ∈ N n in Nn \in Nn∈N and g ∈ G g ∈ G g in Gg \in Gg∈G, g n g − 1 ∈ N g n g − 1 ∈ N gng^(-1)in Ngng^{-1} \in Ngng−1∈N.
Let n = [ 1 a 0 1 ] n = 1      a 0      1 n=[[1,a],[0,1]]n = \left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]n=[1a01] be any element in N N NNN and g = [ x y 0 z ] g = x      y 0      z g=[[x,y],[0,z]]g = \left[\begin{array}{ll}x & y \\ 0 & z\end{array}\right]g=[xy0z] be any element in G G GGG.
We find g n g − 1 g n g − 1 gng^(-1)gng^{-1}gng−1:
g n g − 1 = [ x y 0 z ] [ 1 a 0 1 ] [ x − 1 − x − 1 y z − 1 0 z − 1 ] g n g − 1 = x      y 0      z 1      a 0      1 x − 1      − x − 1 y z − 1 0      z − 1 gng^(-1)=[[x,y],[0,z]][[1,a],[0,1]][[x^(-1),-x^(-1)yz^(-1)],[0,z^(-1)]]gng^{-1} = \left[\begin{array}{ll}x & y \\ 0 & z\end{array}\right] \left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}x^{-1} & -x^{-1}y z^{-1} \\ 0 & z^{-1}\end{array}\right]gng−1=[xy0z][1a01][x−1−x−1yz−10z−1]
After calculating, we get:
g n g − 1 = [ 1 a z 0 1 ] g n g − 1 = 1      a z 0      1 gng^(-1)=[[1,az],[0,1]]gng^{-1} = \left[\begin{array}{ll}1 & az \\ 0 & 1\end{array}\right]gng−1=[1az01]
Since a z a z azazaz is a real number, g n g − 1 g n g − 1 gng^(-1)gng^{-1}gng−1 is in N N NNN.
Summary for Part 2
For every n ∈ N n ∈ N n in Nn \in Nn∈N and g ∈ G g ∈ G g in Gg \in Gg∈G, g n g − 1 g n g − 1 gng^(-1)gng^{-1}gng−1 is in N N NNN. Therefore, N N NNN is a normal subgroup of G G GGG.
Verified Answer
5/5
\(tan\:\theta =\frac{sin\:\theta }{cos\:\theta }\)
\(c=a\:cos\:B+b\:cos\:A\)

2(a) Show that \(\mathbb{Z}_7\) is a field. Then find \(([5]+[6])^{-1}\) and \((-[4])^{-1}\) in \(\mathbb{Z}_7\).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Part 1: Show that Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is a Field
To show that Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is a field, we need to verify that it is a commutative ring with unity and that every non-zero element has a multiplicative inverse.
  1. Commutative Ring: Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is a commutative ring under addition and multiplication modulo 7.
  2. Unity: The element [ 1 ] [ 1 ] [1][1][1] serves as the multiplicative identity.
  3. Multiplicative Inverses: We need to show that every non-zero element [ a ] [ a ] [a][a][a] in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 has a multiplicative inverse.
To find the multiplicative inverses, we can check that gcd ( 7 , a ) = 1 gcd ( 7 , a ) = 1 gcd(7,a)=1\gcd(7, a) = 1gcd(7,a)=1 for a = 1 , 2 , 3 , 4 , 5 , 6 a = 1 , 2 , 3 , 4 , 5 , 6 a=1,2,3,4,5,6a = 1, 2, 3, 4, 5, 6a=1,2,3,4,5,6.
After calculating, we find that gcd ( 7 , a ) = 1 gcd ( 7 , a ) = 1 gcd(7,a)=1\gcd(7, a) = 1gcd(7,a)=1 for all a a aaa in { 1 , 2 , 3 , 4 , 5 , 6 } { 1 , 2 , 3 , 4 , 5 , 6 } {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}{1,2,3,4,5,6}.
This means every non-zero element in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 has a multiplicative inverse, confirming that Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is a field.
The GCD of 7 with each of the numbers { 1 , 2 , 3 , 4 , 5 , 6 } { 1 , 2 , 3 , 4 , 5 , 6 } {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}{1,2,3,4,5,6} is 1. This confirms that every non-zero element in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 has a multiplicative inverse, making Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 a field.
Part 2: Calculations in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7
( [ 5 ] + [ 6 ] ) − 1 ( [ 5 ] + [ 6 ] ) − 1 ([5]+[6])^(-1)([5]+[6])^{-1}([5]+[6])−1
First, we find [ 5 ] + [ 6 ] [ 5 ] + [ 6 ] [5]+[6][5] + [6][5]+[6] in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7:
[ 5 ] + [ 6 ] = [ 11 ] = [ 4 ] [ 5 ] + [ 6 ] = [ 11 ] = [ 4 ] [5]+[6]=[11]=[4][5] + [6] = [11] = [4][5]+[6]=[11]=[4]
After calculating, we find that [ 4 ] [ 4 ] [4][4][4] in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is [ 4 ] [ 4 ] [4][4][4].
Now, we find [ 4 ] − 1 [ 4 ] − 1 [4]^(-1)[4]^{-1}[4]−1:
[ 4 ] − 1 = [ 2 ] [ 4 ] − 1 = [ 2 ] [4]^(-1)=[2][4]^{-1} = [2][4]−1=[2]
After calculating, we find that [ 4 ] − 1 [ 4 ] − 1 [4]^(-1)[4]^{-1}[4]−1 in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is [ 2 ] [ 2 ] [2][2][2].
( − [ 4 ] ) − 1 ( − [ 4 ] ) − 1 (-[4])^(-1)(-[4])^{-1}(−[4])−1
First, we find − [ 4 ] − [ 4 ] -[4]-[4]−[4] in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7:
− [ 4 ] = [ 3 ] − [ 4 ] = [ 3 ] -[4]=[3]-[4] = [3]−[4]=[3]
After calculating, we find that − [ 4 ] − [ 4 ] -[4]-[4]−[4] in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is [ 3 ] [ 3 ] [3][3][3].
Now, we find [ 3 ] − 1 [ 3 ] − 1 [3]^(-1)[3]^{-1}[3]−1:
[ 3 ] − 1 = [ 5 ] [ 3 ] − 1 = [ 5 ] [3]^(-1)=[5][3]^{-1} = [5][3]−1=[5]
After calculating, we find that [ 3 ] − 1 [ 3 ] − 1 [3]^(-1)[3]^{-1}[3]−1 in Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is [ 5 ] [ 5 ] [5][5][5].
Summary
  1. Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7 is a field.
  2. ( [ 5 ] + [ 6 ] ) − 1 = [ 2 ] ( [ 5 ] + [ 6 ] ) − 1 = [ 2 ] ([5]+[6])^(-1)=[2]([5]+[6])^{-1} = [2]([5]+[6])−1=[2]
  3. ( − [ 4 ] ) − 1 = [ 5 ] ( − [ 4 ] ) − 1 = [ 5 ] (-[4])^(-1)=[5](-[4])^{-1} = [5](−[4])−1=[5]
Verified Answer
5/5
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)
\(cos\:3\theta =4\:cos^3\:\theta -3\:cos\:\theta \)

3(a) Show that the set \(\left\{a+b \omega: \omega^3=1\right\}\), where \(a\) and \(b\) are real numbers, is a field with respect to usual addition and multiplication.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the set { a + b ω : ω 3 = 1 } a + b ω : ω 3 = 1 {a+b omega:omega^(3)=1}\left\{ a + b\omega : \omega^3 = 1 \right\}{a+bω:ω3=1} is a field, we need to verify the following properties:
  1. Closure under Addition and Multiplication: For any two elements x , y x , y x,yx, yx,y in the set, x + y x + y x+yx+yx+y and x × y x × y x xx yx \times yx×y should also be in the set.
  2. Associativity and Commutativity: The set should be associative and commutative under both addition and multiplication.
  3. Existence of Identity and Inverse Elements: There should exist additive and multiplicative identity elements, as well as additive and multiplicative inverses for each element in the set.
Step 1: Closure
Closure under Addition
Let x = a 1 + b 1 ω x = a 1 + b 1 ω x=a_(1)+b_(1)omegax = a_1 + b_1\omegax=a1+b1ω and y = a 2 + b 2 ω y = a 2 + b 2 ω y=a_(2)+b_(2)omegay = a_2 + b_2\omegay=a2+b2ω be two elements in the set. Then,
x + y = ( a 1 + a 2 ) + ( b 1 + b 2 ) ω x + y = ( a 1 + a 2 ) + ( b 1 + b 2 ) ω x+y=(a_(1)+a_(2))+(b_(1)+b_(2))omegax + y = (a_1 + a_2) + (b_1 + b_2)\omegax+y=(a1+a2)+(b1+b2)ω
Since a 1 + a 2 a 1 + a 2 a_(1)+a_(2)a_1 + a_2a1+a2 and b 1 + b 2 b 1 + b 2 b_(1)+b_(2)b_1 + b_2b1+b2 are real numbers, x + y x + y x+yx + yx+y is also in the set.
Closure under Multiplication
Let x = a 1 + b 1 ω x = a 1 + b 1 ω x=a_(1)+b_(1)omegax = a_1 + b_1\omegax=a1+b1ω and y = a 2 + b 2 ω y = a 2 + b 2 ω y=a_(2)+b_(2)omegay = a_2 + b_2\omegay=a2+b2ω be two elements in the set. Then,
x × y = ( a 1 + b 1 ω ) ( a 2 + b 2 ω ) = a 1 a 2 + a 1 b 2 ω + b 1 a 2 ω + b 1 b 2 ω 2 x × y = ( a 1 + b 1 ω ) ( a 2 + b 2 ω ) = a 1 a 2 + a 1 b 2 ω + b 1 a 2 ω + b 1 b 2 ω 2 x xx y=(a_(1)+b_(1)omega)(a_(2)+b_(2)omega)=a_(1)a_(2)+a_(1)b_(2)omega+b_(1)a_(2)omega+b_(1)b_(2)omega^(2)x \times y = (a_1 + b_1\omega)(a_2 + b_2\omega) = a_1a_2 + a_1b_2\omega + b_1a_2\omega + b_1b_2\omega^2x×y=(a1+b1ω)(a2+b2ω)=a1a2+a1b2ω+b1a2ω+b1b2ω2
Using ω 3 = 1 ω 3 = 1 omega^(3)=1\omega^3 = 1ω3=1, we can simplify ω 2 = ω − 1 ω 2 = ω − 1 omega^(2)=omega^(-1)\omega^2 = \omega^{-1}ω2=ω−1 and rewrite x × y x × y x xx yx \times yx×y as:
x × y = a 1 a 2 + ( a 1 b 2 + b 1 a 2 ) ω + b 1 b 2 ω − 1 x × y = a 1 a 2 + ( a 1 b 2 + b 1 a 2 ) ω + b 1 b 2 ω − 1 x xx y=a_(1)a_(2)+(a_(1)b_(2)+b_(1)a_(2))omega+b_(1)b_(2)omega^(-1)x \times y = a_1a_2 + (a_1b_2 + b_1a_2)\omega + b_1b_2\omega^{-1}x×y=a1a2+(a1b2+b1a2)ω+b1b2ω−1
Since all the coefficients are real numbers, x × y x × y x xx yx \times yx×y is also in the set.
Step 2: Associativity and Commutativity
The set is associative and commutative under both addition and multiplication because the real numbers are associative and commutative under these operations.
Step 3: Existence of Identity and Inverse Elements
Additive Identity and Inverse
The additive identity is 0 0 000 and the additive inverse of a + b ω a + b ω a+b omegaa + b\omegaa+bω is − a − b ω − a − b ω -a-b omega-a – b\omega−a−bω.
Multiplicative Identity and Inverse
The multiplicative identity is 1 1 111.
To find the multiplicative inverse of a + b ω a + b ω a+b omegaa + b\omegaa+bω, let’s assume ( a + b ω ) ( c + d ω ) = 1 ( a + b ω ) ( c + d ω ) = 1 (a+b omega)(c+d omega)=1(a + b\omega)(c + d\omega) = 1(a+bω)(c+dω)=1.
After calculating, we find:
( a + b ω ) ( c + d ω ) = a c + a d ω + b c ω + b d ω 2 = a c + ( a d + b c ) ω + b d ω − 1 = 1 ( a + b ω ) ( c + d ω ) = a c + a d ω + b c ω + b d ω 2 = a c + ( a d + b c ) ω + b d ω − 1 = 1 (a+b omega)(c+d omega)=ac+ad omega+bc omega+bdomega^(2)=ac+(ad+bc)omega+bdomega^(-1)=1(a + b\omega)(c + d\omega) = ac + ad\omega + bc\omega + bd\omega^2 = ac + (ad + bc)\omega + bd\omega^{-1} = 1(a+bω)(c+dω)=ac+adω+bcω+bdω2=ac+(ad+bc)ω+bdω−1=1
Solving these equations gives us the multiplicative inverse c + d ω c + d ω c+d omegac + d\omegac+dω.
Summary
All the properties required for a field are satisfied. Therefore, the set { a + b ω : ω 3 = 1 } a + b ω : ω 3 = 1 {a+b omega:omega^(3)=1}\left\{ a + b\omega : \omega^3 = 1 \right\}{a+bω:ω3=1} is a field with respect to usual addition and multiplication.
Verified Answer
5/5
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)
\(a=b\:cos\:C+c\:cos\:B\)

4(a) Prove that the set \(\mathbb{Q}(\sqrt{5})=\{a+b \sqrt{5}: a, b \in \mathbb{Q}\}\) is a commutative ring with identity.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To prove that the set Q ( 5 ) = { a + b 5 : a , b ∈ Q } Q ( 5 ) = { a + b 5 : a , b ∈ Q } Q(sqrt5)={a+bsqrt5:a,b inQ}\mathbb{Q}(\sqrt{5}) = \{a + b\sqrt{5} : a, b \in \mathbb{Q}\}Q(5)={a+b5:a,b∈Q} is a commutative ring with identity, we need to show the following properties:
  1. Closure under Addition and Multiplication: The set is closed under addition and multiplication.
  2. Associativity of Addition and Multiplication: ( a + b ) + c = a + ( b + c ) ( a + b ) + c = a + ( b + c ) (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)(a+b)+c=a+(b+c) and ( a × b ) × c = a × ( b × c ) ( a × b ) × c = a × ( b × c ) (a xx b)xx c=a xx(b xx c)(a \times b) \times c = a \times (b \times c)(a×b)×c=a×(b×c).
  3. Commutativity of Addition and Multiplication: a + b = b + a a + b = b + a a+b=b+aa + b = b + aa+b=b+a and a × b = b × a a × b = b × a a xx b=b xx aa \times b = b \times aa×b=b×a.
  4. Existence of Additive and Multiplicative Identity: There exists an element 0 0 000 such that a + 0 = a a + 0 = a a+0=aa + 0 = aa+0=a for all a a aaa in the set, and an element 1 1 111 such that a × 1 = a a × 1 = a a xx1=aa \times 1 = aa×1=a for all a a aaa in the set.
  5. Existence of Additive Inverse: For each a a aaa in the set, there exists an element − a − a -a-a−a such that a + ( − a ) = 0 a + ( − a ) = 0 a+(-a)=0a + (-a) = 0a+(−a)=0.
  6. Distributive Law: a × ( b + c ) = ( a × b ) + ( a × c ) a × ( b + c ) = ( a × b ) + ( a × c ) a xx(b+c)=(a xx b)+(a xx c)a \times (b + c) = (a \times b) + (a \times c)a×(b+c)=(a×b)+(a×c).
  7. Closure under Addition and Multiplication
Addition
Let’s consider two arbitrary elements x = a 1 + b 1 5 x = a 1 + b 1 5 x=a_(1)+b_(1)sqrt5x = a_1 + b_1\sqrt{5}x=a1+b15 and y = a 2 + b 2 5 y = a 2 + b 2 5 y=a_(2)+b_(2)sqrt5y = a_2 + b_2\sqrt{5}y=a2+b25 in Q ( 5 ) Q ( 5 ) Q(sqrt5)\mathbb{Q}(\sqrt{5})Q(5).
Let’s substitute the values into the formula for addition:
x + y = ( a 1 + b 1 5 ) + ( a 2 + b 2 5 ) x + y = ( a 1 + b 1 5 ) + ( a 2 + b 2 5 ) x+y=(a_(1)+b_(1)sqrt5)+(a_(2)+b_(2)sqrt5)x + y = (a_1 + b_1\sqrt{5}) + (a_2 + b_2\sqrt{5})x+y=(a1+b15)+(a2+b25)
After simplifying, we get:
x + y = ( a 1 + a 2 ) + ( b 1 + b 2 ) 5 x + y = ( a 1 + a 2 ) + ( b 1 + b 2 ) 5 x+y=(a_(1)+a_(2))+(b_(1)+b_(2))sqrt5x + y = (a_1 + a_2) + (b_1 + b_2)\sqrt{5}x+y=(a1+a2)+(b1+b2)5
Since a 1 , a 2 , b 1 , b 2 a 1 , a 2 , b 1 , b 2 a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2a1,a2,b1,b2 are rational numbers, a 1 + a 2 a 1 + a 2 a_(1)+a_(2)a_1 + a_2a1+a2 and b 1 + b 2 b 1 + b 2 b_(1)+b_(2)b_1 + b_2b1+b2 are also rational numbers. Therefore, x + y x + y x+yx + yx+y is also in Q ( 5 ) Q ( 5 ) Q(sqrt5)\mathbb{Q}(\sqrt{5})Q(5).
Multiplication
Similarly, for multiplication, we have:
x × y = ( a 1 + b 1 5 ) × ( a 2 + b 2 5 ) x × y = ( a 1 + b 1 5 ) × ( a 2 + b 2 5 ) x xx y=(a_(1)+b_(1)sqrt5)xx(a_(2)+b_(2)sqrt5)x \times y = (a_1 + b_1\sqrt{5}) \times (a_2 + b_2\sqrt{5})x×y=(a1+b15)×(a2+b25)
After simplifying, we get:
x × y = a 1 a 2 + 5 b 1 b 2 + ( a 1 b 2 + a 2 b 1 ) 5 x × y = a 1 a 2 + 5 b 1 b 2 + ( a 1 b 2 + a 2 b 1 ) 5 x xx y=a_(1)a_(2)+5b_(1)b_(2)+(a_(1)b_(2)+a_(2)b_(1))sqrt5x \times y = a_1a_2 + 5b_1b_2 + (a_1b_2 + a_2b_1)\sqrt{5}x×y=a1a2+5b1b2+(a1b2+a2b1)5
Since a 1 , a 2 , b 1 , b 2 a 1 , a 2 , b 1 , b 2 a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2a1,a2,b1,b2 are rational numbers, a 1 a 2 + 5 b 1 b 2 a 1 a 2 + 5 b 1 b 2 a_(1)a_(2)+5b_(1)b_(2)a_1a_2 + 5b_1b_2a1a2+5b1b2 and a 1 b 2 + a 2 b 1 a 1 b 2 + a 2 b 1 a_(1)b_(2)+a_(2)b_(1)a_1b_2 + a_2b_1a1b2+a2b1 are also rational numbers. Therefore, x × y x × y x xx yx \times yx×y is also in Q ( 5 ) Q ( 5 ) Q(sqrt5)\mathbb{Q}(\sqrt{5})Q(5).
  1. Associativity of Addition and Multiplication
    Associativity of addition and multiplication follows directly from the associativity of addition and multiplication of rational numbers.
  2. Commutativity of Addition and Multiplication
    Commutativity of addition and multiplication also follows directly from the commutativity of addition and multiplication of rational numbers.
  3. Existence of Additive and Multiplicative Identity
    The additive identity is 0 = 0 + 0 5 0 = 0 + 0 5 0=0+0sqrt50 = 0 + 0\sqrt{5}0=0+05 and the multiplicative identity is 1 = 1 + 0 5 1 = 1 + 0 5 1=1+0sqrt51 = 1 + 0\sqrt{5}1=1+05.
  4. Existence of Additive Inverse
    For each x = a + b 5 x = a + b 5 x=a+bsqrt5x = a + b\sqrt{5}x=a+b5, the additive inverse is − x = − a − b 5 − x = − a − b 5 -x=-a-bsqrt5-x = -a – b\sqrt{5}−x=−a−b5.
  5. Distributive Law
    The distributive law a × ( b + c ) = ( a × b ) + ( a × c ) a × ( b + c ) = ( a × b ) + ( a × c ) a xx(b+c)=(a xx b)+(a xx c)a \times (b + c) = (a \times b) + (a \times c)a×(b+c)=(a×b)+(a×c) also holds, as it does for rational numbers.
In summary, Q ( 5 ) Q ( 5 ) Q(sqrt5)\mathbb{Q}(\sqrt{5})Q(5) satisfies all the required properties to be a commutative ring with identity.
Verified Answer
5/5
\(cos\:2\theta =1-2\:sin^2\theta \)

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