Let $A=\{1,2,3, \ldots, 9\}$ and $R$ be the relation in $A \times A$ defined by $(a, b) R(c, d)$ if $a+d=b+c$ for $(a, b),(c, d)$ in $A \times A$. Prove that $R$ is an equivalence relation and also obtain the equivalent class $[(2,5)]$.
Given that, $A=\{1,2,3, \ldots, 9\}$ and $(a, b) R(c, d)$ if $a+d=b+c$ for $(a, b) \in A \times A$ and $(c, d) \in A \times A$.
Let $(a, b) R(a, b)$
$$\Rightarrow \quad a+b=b+a, \forall a, b \in A$$
which is true for any $a, b \in A$.
Hence, $R$ is reflexive.
Let $(a, b) R(c, d)$
$$\begin{aligned} & a+d=b+c \\ & c+b=d+a \quad \Rightarrow \quad(c, d) R(a, b) \end{aligned}$$
So, $R$ is symmetric.
$$\begin{aligned} &\text { Let }\\ &\begin{aligned} (a, b) R(c, d) \text { and } & (c, d) R(e, f) \\ a+d & =b+c \text { and } c+f=d+e \\ a+d & =b+c \text { and } d+e=c+f \\ (a+d)-(d+e) & =(b+c)-(c+f) \\ (a-e) & =b-f \\ a+f & =b+e \\ (a, b) & R(e, f) \end{aligned} \end{aligned}$$
So, $R$ is transitive.
Hence, $R$ is an equivalence relation.
Now, equivalence class containing $[(2,5)]$ is $\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$.
Using the definition, prove that the function $f: A \rightarrow B$ is invertible if and only if $f$ is both one-one and onto.
A function $f: X \rightarrow Y$ is defined to be invertible, if there exist a function $g=Y \rightarrow X$ such that gof $=I_X$ and fog $=I_Y$. The function is called the inverse of $f$ and is denoted by $f^{-1}$. A function $f=X \rightarrow Y$ is invertible iff $f$ is a bijective function.
Functions $f, g: R \rightarrow R$ are defined, respectively, by $f(x)=x^2+3 x+1$, $g(x)=2 x-3$, find
(i) $f \circ g$ (ii) $g \circ f$ (iii) $f \circ f$ (iv) $g \circ g$
Given that,
$$f(x)=x^2+3 x+1, g(x)=2 x-3$$
(i)
$$\begin{aligned} f \circ g & =f\{g(x)\}=f(2 x-3) \\ & =(2 x-3)^2+3(2 x-3)+1 \\ & =4 x^2+9-12 x+6 x-9+1=4 x^2-6 x+1 \end{aligned}$$
(ii)
$$\begin{aligned} g \circ f & =g\{f(x)\}=g\left(x^2+3 x+1\right) \\ & =2\left(x^2+3 x+1\right)-3 \\ & =2 x^2+6 x+2-3=2 x^2+6 x-1 \end{aligned}$$
(iii)
$$\begin{aligned} f \circ f & =f\{f(x)\}=f\left(x^2+3 x+1\right) \\ & =\left(x^2+3 x+1\right)^2+3\left(x^2+3 x+1\right)+1 \\ & =x^4+9 x^2+1+6 x^3+6 x+2 x^2+3 x^2+9 x+3+1 \\ & =x^4+6 x^3+14 x^2+15 x+5 \end{aligned}$$
(iv)
$$\begin{aligned} g \circ g & =g\{g(x)\}=g(2 x-3) \\ & =2(2 x-3)-3 \\ & =4 x-6-3=4 x-9 \end{aligned}$$
Let * be the binary operation defined on $Q$. Find which of the following binary operations are commutative
(i) $a * b=a-b, \forall a, b \in Q$
(ii) $a * b=a^2+b^2, \forall a, b \in Q$
(iii) $a * b=a+a b, \forall a, b \in Q$
(iv) $a * b=(a-b)^2, \forall a, b \in Q$
Given that * be the binary operation defined on $Q$.
(i) $a * b=a-b, \forall a, b \in Q$ and $b * a=b-a$
So, $$a * b \neq b * a\quad$$ $$[\because b-a \neq a-b]$$
Hence, * is not commutative.
(ii)
$$\begin{aligned} & a * b=a^2+b^2 \\ & b * a=b^2+a^2 \end{aligned}$$
So, * is commutative.
[since, ' + ' is on rational is commutative]
$$\begin{aligned} &\text { (iii) }\\ &\begin{aligned} a * b & =a+a b \\ b * a & =b+a b \\ \text { Clearly, } \quad a+a b & \neq b+a b \end{aligned}\\ &\text { So, * is not commutative. } \end{aligned}$$
(iv)
$$\begin{aligned} a * b & =(a-b)^2, \forall a, b \in Q \\ b * a & =(b-a)^2 \\ \because \quad(a-b)^2 & =(b-a)^2 \end{aligned}$$
Hence, * is commutative.
If * be binary operation defined on $R$ by $a * b=1+a b, \forall a, b \in R$. Then, the operation $*$ is
(i) commutative but not associative.
(ii) associative but not commutative.
(iii) neither commutative nor associative.
(iv) both commutative and associative.
$$\begin{aligned} &\text { (i) Given that, }\\ &\begin{aligned} & a * b=1+a b, \forall a, b \in R \\ & a * b=a b+1=b * a \end{aligned} \end{aligned}$$
So, * is a commutative binary operation.
Also,
$$\begin{aligned} a *(b * c) & =a *(1+b c)=1+a(1+b c) \\ a *(b * c) & =1+a+a b c \quad\text{.... (i)}\\ (a * b) * c & =(1+a b) * c \\ & =1+(1+a b) c=1+c+a b c\quad\text{.... (ii)} \end{aligned}$$
So, * is not associative.
Hence, * is commutative but not associative.