UPSC Optional (Maths) Paper-02 Algebra Solution

2014

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UPSC Algebra

\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)

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

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
\(c=a\:cos\:B+b\:cos\:A\)
\(a^2=b^2+c^2-2bc\:Cos\left(A\right)\)

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
\(cos\:2\theta =cos^2\theta -sin^2\theta\)
\(c=a\:cos\:B+b\:cos\:A\)

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}