2014

Estimated reading: 50 minutes 21 views

UPSC Algebra

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

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.

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

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

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Part 1: Show that ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is a Field
To show that ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ 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: ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is a commutative ring under addition and multiplication modulo 7.
2. Unity: The element $\left[1\right]$$\left[1\right]$[1][1]$\left[1\right]$ serves as the multiplicative identity.
3. Multiplicative Inverses: We need to show that every non-zero element $\left[a\right]$$\left[a\right]$[a][a]$\left[a\right]$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ has a multiplicative inverse.
To find the multiplicative inverses, we can check that $gcd\left(7,a\right)=1$$gcd\left(7,a\right)=1$gcd(7,a)=1\gcd(7, a) = 1$gcd\left(7,a\right)=1$ for $a=1,2,3,4,5,6$$a=1,2,3,4,5,6$a=1,2,3,4,5,6a = 1, 2, 3, 4, 5, 6$a=1,2,3,4,5,6$.
After calculating, we find that $gcd\left(7,a\right)=1$$gcd\left(7,a\right)=1$gcd(7,a)=1\gcd(7, a) = 1$gcd\left(7,a\right)=1$ for all $a$$a$aa$a$ in $\left\{1,2,3,4,5,6\right\}$$\left\{1,2,3,4,5,6\right\}${1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}$\left\{1,2,3,4,5,6\right\}$.
This means every non-zero element in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ has a multiplicative inverse, confirming that ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is a field.
The GCD of 7 with each of the numbers $\left\{1,2,3,4,5,6\right\}$$\left\{1,2,3,4,5,6\right\}${1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}$\left\{1,2,3,4,5,6\right\}$ is 1. This confirms that every non-zero element in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ has a multiplicative inverse, making ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ a field.
Part 2: Calculations in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$
$\left(\left[5\right]+\left[6\right]{\right)}^{-1}$$\left(\left[5\right]+\left[6\right]{\right)}^{-1}$([5]+[6])^(-1)([5]+[6])^{-1}$\left(\left[5\right]+\left[6\right]{\right)}^{-1}$
First, we find $\left[5\right]+\left[6\right]$$\left[5\right]+\left[6\right]$[5]+[6][5] + [6]$\left[5\right]+\left[6\right]$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$:
$\left[5\right]+\left[6\right]=\left[11\right]=\left[4\right]$$\left[5\right]+\left[6\right]=\left[11\right]=\left[4\right]$[5]+[6]=[11]=[4][5] + [6] = [11] = [4]$\left[5\right]+\left[6\right]=\left[11\right]=\left[4\right]$
After calculating, we find that $\left[4\right]$$\left[4\right]$[4][4]$\left[4\right]$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is $\left[4\right]$$\left[4\right]$[4][4]$\left[4\right]$.
Now, we find $\left[4{\right]}^{-1}$$\left[4{\right]}^{-1}$[4]^(-1)[4]^{-1}$\left[4{\right]}^{-1}$:
$\left[4{\right]}^{-1}=\left[2\right]$$\left[4{\right]}^{-1}=\left[2\right]$[4]^(-1)=[2][4]^{-1} = [2]$\left[4{\right]}^{-1}=\left[2\right]$
After calculating, we find that $\left[4{\right]}^{-1}$$\left[4{\right]}^{-1}$[4]^(-1)[4]^{-1}$\left[4{\right]}^{-1}$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is $\left[2\right]$$\left[2\right]$[2][2]$\left[2\right]$.
$\left(-\left[4\right]{\right)}^{-1}$$\left(-\left[4\right]{\right)}^{-1}$(-[4])^(-1)(-[4])^{-1}$\left(-\left[4\right]{\right)}^{-1}$
First, we find $-\left[4\right]$$-\left[4\right]$-[4]-[4]$-\left[4\right]$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$:
$-\left[4\right]=\left[3\right]$$-\left[4\right]=\left[3\right]$-[4]=[3]-[4] = [3]$-\left[4\right]=\left[3\right]$
After calculating, we find that $-\left[4\right]$$-\left[4\right]$-[4]-[4]$-\left[4\right]$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is $\left[3\right]$$\left[3\right]$[3][3]$\left[3\right]$.
Now, we find $\left[3{\right]}^{-1}$$\left[3{\right]}^{-1}$[3]^(-1)[3]^{-1}$\left[3{\right]}^{-1}$:
$\left[3{\right]}^{-1}=\left[5\right]$$\left[3{\right]}^{-1}=\left[5\right]$[3]^(-1)=[5][3]^{-1} = [5]$\left[3{\right]}^{-1}=\left[5\right]$
After calculating, we find that $\left[3{\right]}^{-1}$$\left[3{\right]}^{-1}$[3]^(-1)[3]^{-1}$\left[3{\right]}^{-1}$ in ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is $\left[5\right]$$\left[5\right]$[5][5]$\left[5\right]$.
Summary
1. ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$ is a field.
2. $\left(\left[5\right]+\left[6\right]{\right)}^{-1}=\left[2\right]$$\left(\left[5\right]+\left[6\right]{\right)}^{-1}=\left[2\right]$([5]+[6])^(-1)=[2]([5]+[6])^{-1} = [2]$\left(\left[5\right]+\left[6\right]{\right)}^{-1}=\left[2\right]$
3. $\left(-\left[4\right]{\right)}^{-1}=\left[5\right]$$\left(-\left[4\right]{\right)}^{-1}=\left[5\right]$(-[4])^(-1)=[5](-[4])^{-1} = [5]$\left(-\left[4\right]{\right)}^{-1}=\left[5\right]$
5/5
$$a=b\:cos\:C+c\:cos\:B$$
$$2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$

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.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the set $\left\{a+b\omega :{\omega }^{3}=1\right\}$$\left\{a+b\omega :{\omega }^{3}=1\right\}${a+b omega:omega^(3)=1}\left\{ a + b\omega : \omega^3 = 1 \right\}$\left\{a+b\omega :{\omega }^{3}=1\right\}$ is a field, we need to verify the following properties:
1. Closure under Addition and Multiplication: For any two elements $x,y$$x,y$x,yx, y$x,y$ in the set, $x+y$$x+y$x+yx+y$x+y$ and $x×y$$x×y$x xx yx \times y$x×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
Let $x={a}_{1}+{b}_{1}\omega$$x={a}_{1}+{b}_{1}\omega$x=a_(1)+b_(1)omegax = a_1 + b_1\omega$x={a}_{1}+{b}_{1}\omega$ and $y={a}_{2}+{b}_{2}\omega$$y={a}_{2}+{b}_{2}\omega$y=a_(2)+b_(2)omegay = a_2 + b_2\omega$y={a}_{2}+{b}_{2}\omega$ be two elements in the set. Then,
$x+y=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)\omega$$x+y=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)\omega$x+y=(a_(1)+a_(2))+(b_(1)+b_(2))omegax + y = (a_1 + a_2) + (b_1 + b_2)\omega$x+y=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)\omega$
Since ${a}_{1}+{a}_{2}$${a}_{1}+{a}_{2}$a_(1)+a_(2)a_1 + a_2${a}_{1}+{a}_{2}$ and ${b}_{1}+{b}_{2}$${b}_{1}+{b}_{2}$b_(1)+b_(2)b_1 + b_2${b}_{1}+{b}_{2}$ are real numbers, $x+y$$x+y$x+yx + y$x+y$ is also in the set.
Closure under Multiplication
Let $x={a}_{1}+{b}_{1}\omega$$x={a}_{1}+{b}_{1}\omega$x=a_(1)+b_(1)omegax = a_1 + b_1\omega$x={a}_{1}+{b}_{1}\omega$ and $y={a}_{2}+{b}_{2}\omega$$y={a}_{2}+{b}_{2}\omega$y=a_(2)+b_(2)omegay = a_2 + b_2\omega$y={a}_{2}+{b}_{2}\omega$ be two elements in the set. Then,
$x×y=\left({a}_{1}+{b}_{1}\omega \right)\left({a}_{2}+{b}_{2}\omega \right)={a}_{1}{a}_{2}+{a}_{1}{b}_{2}\omega +{b}_{1}{a}_{2}\omega +{b}_{1}{b}_{2}{\omega }^{2}$$x×y=\left({a}_{1}+{b}_{1}\omega \right)\left({a}_{2}+{b}_{2}\omega \right)={a}_{1}{a}_{2}+{a}_{1}{b}_{2}\omega +{b}_{1}{a}_{2}\omega +{b}_{1}{b}_{2}{\omega }^{2}$x xx y=(a_(1)+b_(1)omega)(a_(2)+b_(2)omega)=a_(1)a_(2)+a_(1)b_(2)omega+b_(1)a_(2)omega+b_(1)b_(2)omega^(2)x \times y = (a_1 + b_1\omega)(a_2 + b_2\omega) = a_1a_2 + a_1b_2\omega + b_1a_2\omega + b_1b_2\omega^2$x×y=\left({a}_{1}+{b}_{1}\omega \right)\left({a}_{2}+{b}_{2}\omega \right)={a}_{1}{a}_{2}+{a}_{1}{b}_{2}\omega +{b}_{1}{a}_{2}\omega +{b}_{1}{b}_{2}{\omega }^{2}$
Using ${\omega }^{3}=1$${\omega }^{3}=1$omega^(3)=1\omega^3 = 1${\omega }^{3}=1$, we can simplify ${\omega }^{2}={\omega }^{-1}$${\omega }^{2}={\omega }^{-1}$omega^(2)=omega^(-1)\omega^2 = \omega^{-1}${\omega }^{2}={\omega }^{-1}$ and rewrite $x×y$$x×y$x xx yx \times y$x×y$ as:
$x×y={a}_{1}{a}_{2}+\left({a}_{1}{b}_{2}+{b}_{1}{a}_{2}\right)\omega +{b}_{1}{b}_{2}{\omega }^{-1}$$x×y={a}_{1}{a}_{2}+\left({a}_{1}{b}_{2}+{b}_{1}{a}_{2}\right)\omega +{b}_{1}{b}_{2}{\omega }^{-1}$x xx y=a_(1)a_(2)+(a_(1)b_(2)+b_(1)a_(2))omega+b_(1)b_(2)omega^(-1)x \times y = a_1a_2 + (a_1b_2 + b_1a_2)\omega + b_1b_2\omega^{-1}$x×y={a}_{1}{a}_{2}+\left({a}_{1}{b}_{2}+{b}_{1}{a}_{2}\right)\omega +{b}_{1}{b}_{2}{\omega }^{-1}$
Since all the coefficients are real numbers, $x×y$$x×y$x xx yx \times y$x×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
The additive identity is $0$$0$00$0$ and the additive inverse of $a+b\omega$$a+b\omega$a+b omegaa + b\omega$a+b\omega$ is $-a-b\omega$$-a-b\omega$-a-b omega-a – b\omega$-a-b\omega$.
Multiplicative Identity and Inverse
The multiplicative identity is $1$$1$11$1$.
To find the multiplicative inverse of $a+b\omega$$a+b\omega$a+b omegaa + b\omega$a+b\omega$, let’s assume $\left(a+b\omega \right)\left(c+d\omega \right)=1$$\left(a+b\omega \right)\left(c+d\omega \right)=1$(a+b omega)(c+d omega)=1(a + b\omega)(c + d\omega) = 1$\left(a+b\omega \right)\left(c+d\omega \right)=1$.
After calculating, we find:
$\left(a+b\omega \right)\left(c+d\omega \right)=ac+ad\omega +bc\omega +bd{\omega }^{2}=ac+\left(ad+bc\right)\omega +bd{\omega }^{-1}=1$$\left(a+b\omega \right)\left(c+d\omega \right)=ac+ad\omega +bc\omega +bd{\omega }^{2}=ac+\left(ad+bc\right)\omega +bd{\omega }^{-1}=1$(a+b omega)(c+d omega)=ac+ad omega+bc omega+bdomega^(2)=ac+(ad+bc)omega+bdomega^(-1)=1(a + b\omega)(c + d\omega) = ac + ad\omega + bc\omega + bd\omega^2 = ac + (ad + bc)\omega + bd\omega^{-1} = 1$\left(a+b\omega \right)\left(c+d\omega \right)=ac+ad\omega +bc\omega +bd{\omega }^{2}=ac+\left(ad+bc\right)\omega +bd{\omega }^{-1}=1$
Solving these equations gives us the multiplicative inverse $c+d\omega$$c+d\omega$c+d omegac + d\omega$c+d\omega$.
Summary
All the properties required for a field are satisfied. Therefore, the set $\left\{a+b\omega :{\omega }^{3}=1\right\}$$\left\{a+b\omega :{\omega }^{3}=1\right\}${a+b omega:omega^(3)=1}\left\{ a + b\omega : \omega^3 = 1 \right\}$\left\{a+b\omega :{\omega }^{3}=1\right\}$ is a field with respect to usual addition and multiplication.
5/5
$$sin^2\left(\frac{\theta }{2}\right)=\frac{1-cos\:\theta }{2}$$
$$cos\left(\theta +\phi \right)=cos\:\theta \:cos\:\phi -sin\:\theta \:sin\:\phi$$

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.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To prove that the set $\mathbb{Q}\left(\sqrt{5}\right)=\left\{a+b\sqrt{5}:a,b\in \mathbb{Q}\right\}$$\mathbb{Q}\left(\sqrt{5}\right)=\left\{a+b\sqrt{5}:a,b\in \mathbb{Q}\right\}$Q(sqrt5)={a+bsqrt5:a,b inQ}\mathbb{Q}(\sqrt{5}) = \{a + b\sqrt{5} : a, b \in \mathbb{Q}\}$\mathbb{Q}\left(\sqrt{5}\right)=\left\{a+b\sqrt{5}:a,b\in \mathbb{Q}\right\}$ is a commutative ring with identity, we need to show the following properties:
1. Closure under Addition and Multiplication: The set is closed under addition and multiplication.
2. Associativity of Addition and Multiplication: $\left(a+b\right)+c=a+\left(b+c\right)$$\left(a+b\right)+c=a+\left(b+c\right)$(a+b)+c=a+(b+c)(a + b) + c = a + (b + c)$\left(a+b\right)+c=a+\left(b+c\right)$ and $\left(a×b\right)×c=a×\left(b×c\right)$$\left(a×b\right)×c=a×\left(b×c\right)$(a xx b)xx c=a xx(b xx c)(a \times b) \times c = a \times (b \times c)$\left(a×b\right)×c=a×\left(b×c\right)$.
3. Commutativity of Addition and Multiplication: $a+b=b+a$$a+b=b+a$a+b=b+aa + b = b + a$a+b=b+a$ and $a×b=b×a$$a×b=b×a$a xx b=b xx aa \times b = b \times a$a×b=b×a$.
4. Existence of Additive and Multiplicative Identity: There exists an element $0$$0$00$0$ such that $a+0=a$$a+0=a$a+0=aa + 0 = a$a+0=a$ for all $a$$a$aa$a$ in the set, and an element $1$$1$11$1$ such that $a×1=a$$a×1=a$a xx1=aa \times 1 = a$a×1=a$ for all $a$$a$aa$a$ in the set.
5. Existence of Additive Inverse: For each $a$$a$aa$a$ in the set, there exists an element $-a$$-a$-a-a$-a$ such that $a+\left(-a\right)=0$$a+\left(-a\right)=0$a+(-a)=0a + (-a) = 0$a+\left(-a\right)=0$.
6. Distributive Law: $a×\left(b+c\right)=\left(a×b\right)+\left(a×c\right)$$a×\left(b+c\right)=\left(a×b\right)+\left(a×c\right)$a xx(b+c)=(a xx b)+(a xx c)a \times (b + c) = (a \times b) + (a \times c)$a×\left(b+c\right)=\left(a×b\right)+\left(a×c\right)$.
7. Closure under Addition and Multiplication
Let’s consider two arbitrary elements $x={a}_{1}+{b}_{1}\sqrt{5}$$x={a}_{1}+{b}_{1}\sqrt{5}$x=a_(1)+b_(1)sqrt5x = a_1 + b_1\sqrt{5}$x={a}_{1}+{b}_{1}\sqrt{5}$ and $y={a}_{2}+{b}_{2}\sqrt{5}$$y={a}_{2}+{b}_{2}\sqrt{5}$y=a_(2)+b_(2)sqrt5y = a_2 + b_2\sqrt{5}$y={a}_{2}+{b}_{2}\sqrt{5}$ in $\mathbb{Q}\left(\sqrt{5}\right)$$\mathbb{Q}\left(\sqrt{5}\right)$Q(sqrt5)\mathbb{Q}(\sqrt{5})$\mathbb{Q}\left(\sqrt{5}\right)$.
$x+y=\left({a}_{1}+{b}_{1}\sqrt{5}\right)+\left({a}_{2}+{b}_{2}\sqrt{5}\right)$$x+y=\left({a}_{1}+{b}_{1}\sqrt{5}\right)+\left({a}_{2}+{b}_{2}\sqrt{5}\right)$x+y=(a_(1)+b_(1)sqrt5)+(a_(2)+b_(2)sqrt5)x + y = (a_1 + b_1\sqrt{5}) + (a_2 + b_2\sqrt{5})$x+y=\left({a}_{1}+{b}_{1}\sqrt{5}\right)+\left({a}_{2}+{b}_{2}\sqrt{5}\right)$
$x+y=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)\sqrt{5}$$x+y=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)\sqrt{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=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)\sqrt{5}$
Since ${a}_{1},{a}_{2},{b}_{1},{b}_{2}$${a}_{1},{a}_{2},{b}_{1},{b}_{2}$a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2${a}_{1},{a}_{2},{b}_{1},{b}_{2}$ are rational numbers, ${a}_{1}+{a}_{2}$${a}_{1}+{a}_{2}$a_(1)+a_(2)a_1 + a_2${a}_{1}+{a}_{2}$ and ${b}_{1}+{b}_{2}$${b}_{1}+{b}_{2}$b_(1)+b_(2)b_1 + b_2${b}_{1}+{b}_{2}$ are also rational numbers. Therefore, $x+y$$x+y$x+yx + y$x+y$ is also in $\mathbb{Q}\left(\sqrt{5}\right)$$\mathbb{Q}\left(\sqrt{5}\right)$Q(sqrt5)\mathbb{Q}(\sqrt{5})$\mathbb{Q}\left(\sqrt{5}\right)$.
$x×y=\left({a}_{1}+{b}_{1}\sqrt{5}\right)×\left({a}_{2}+{b}_{2}\sqrt{5}\right)$$x×y=\left({a}_{1}+{b}_{1}\sqrt{5}\right)×\left({a}_{2}+{b}_{2}\sqrt{5}\right)$x xx y=(a_(1)+b_(1)sqrt5)xx(a_(2)+b_(2)sqrt5)x \times y = (a_1 + b_1\sqrt{5}) \times (a_2 + b_2\sqrt{5})$x×y=\left({a}_{1}+{b}_{1}\sqrt{5}\right)×\left({a}_{2}+{b}_{2}\sqrt{5}\right)$
$x×y={a}_{1}{a}_{2}+5{b}_{1}{b}_{2}+\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\sqrt{5}$$x×y={a}_{1}{a}_{2}+5{b}_{1}{b}_{2}+\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\sqrt{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}