Exercise 12.1 - Chapter 12 Discrete Mathematics 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Discrete Mathematics
Ex $12.1$
Question 1.
Determine whether $*$ is a binary operation on the sets given below
(i) $a * b=a .|b|$ on $R$,
(ii) $\mathrm{a} * \mathrm{~b}=\min (\mathrm{a}, \mathrm{b})$ on $\mathrm{A}=\{1,2,3,4,5\}$
(iii) $(\mathrm{a} * \mathrm{~b})=a \sqrt{b}$ is binary on $\mathrm{R}$.
Solution:
(i) Yes.
Reason: $\mathrm{a}, \mathrm{b} \in \mathrm{R}$. So, $|\mathrm{b}| \in \mathrm{R}$ when $\mathrm{b} \in \mathrm{R}$
Now multiplication is binary on $\mathrm{R}$
So $a|b| \in R$ when a,be $R$.
(Le.) $a * b \in R$.
$*$ is a binary operation on $\mathrm{R}$.
(ii) Yes.
Reason: $a, b \in R$ and minimum of $(a, b)$ is either $a$ or $b$ but $a, b \in R$.
So, $\min (a, b) \in R$.
(Le.) $\mathrm{a} * \mathrm{~b} \in \mathrm{R}$.
$*$ is a binary operation on $\mathrm{R}$.
(iii) $\mathrm{a}^* \mathrm{~b}=a \sqrt{b}$ where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$.
No. $*$ is not a binary operation on $\mathrm{R}$.
Reason: $\mathrm{a}, \mathrm{b} \in \mathrm{R}$.
$\Rightarrow \mathrm{b}$ can be -ve number also and square root of a negative number is not real.
So $\sqrt{b} \notin \mathrm{R}$ even when $\mathrm{b} \in \mathrm{R}$.
So $\sqrt{b} \notin \mathrm{R}$. ie., $\mathrm{a} * \mathrm{~b} \notin \mathrm{R}$.
$*$ is not a binary operation on $\mathrm{R}$.
Question 2.
On $\mathbb{Z}$, define $\otimes$ by $(m \otimes n)=m^n+n^m: \forall m, n \in \mathbb{Z}$. Is $\otimes$ binary on $\mathbb{Z}$ ?
Solution:
No. $*$ is not a binary operation on $Z$.
Reason: Since $m, n \in Z$.
So, $\mathrm{m}, \mathrm{n}$ can be negative also.
Now, if $\mathrm{n}$ is negative (Le.) say $\mathrm{n}=-\mathrm{k}$ where $\mathrm{k}$ is +ve.

Then $m^n=m^{-k}=\frac{1}{m^k} \notin \mathbb{Z}$.
Similarly, when $\mathrm{m}$ is negative then $\mathrm{n}^{\mathrm{m}} \notin \mathrm{Z}$. $\therefore \mathrm{m} * \mathrm{n} \notin \mathrm{Z} . \Rightarrow *$ is not a binary operation on $\mathrm{Z}$.
Question 3.
Let $*$ be defined on $\mathrm{R}$ by $(\mathrm{a} * \mathrm{~b})=\mathrm{a}+\mathrm{b}+\mathrm{ab}-1$.Is $*$ binary on $\mathrm{R}$ ? If so, find $3 *\left(\frac{-7}{15}\right)$
Solution:
$
\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+\mathrm{ab}-7
$
Now when $\mathrm{a}, \mathrm{b} \in \mathrm{R}$, then $\mathrm{ab} \in \mathrm{R}$ also $\mathrm{a}+\mathrm{b} \in \mathrm{R}$.
So, $a+b+a b \in R$.
We know $-7 \in \mathrm{R}$.
So, $a+b+a b-7 \in R$.
(ie.) $a * b \in R$.
So, $*$ is a binary operation on $\mathrm{R}$.
Now, to find $3^*\left(-\frac{7}{15}\right)$
$
\begin{aligned}
3 *\left(-\frac{7}{15}\right) & =3+\left(-\frac{7}{15}\right)+(3)\left(-\frac{7}{15}\right)-7 \\
& =3-\frac{7}{15}-\frac{21}{15}-7=\frac{45-7-21-105}{15}=-\frac{88}{15}
\end{aligned}
$
Question 4.
Let $\mathrm{A}=\{\mathrm{a}+\sqrt{5} \mathrm{~b}$ : $\mathrm{a}, \mathrm{b} \in \mathrm{Z}\}$. Check whether the usual multiplication is a binary operation on $\mathrm{A}$.
Solution:
Let $\mathrm{A}=\mathrm{a}+\sqrt{5} \mathrm{~b}$ and $\mathrm{B}=\mathrm{c}+\sqrt{5} \mathrm{~d}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{M}$.
Now $\mathrm{A} * \mathrm{~B}=\{\mathrm{a}+\sqrt{5} \mathrm{~b})(\mathrm{c}+\sqrt{5} \mathrm{~d})$
$=\mathrm{ac}+\sqrt{5} \mathrm{ad}+\sqrt{5} \mathrm{bc}+\sqrt{5} \mathrm{~b} \sqrt{5} \mathrm{~d}$
$=(\mathrm{ac}+5 \mathrm{bd})+\sqrt{5}(\mathrm{ad}+\mathrm{bc}) \in \mathrm{A}$
Where $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{Z}$
So $*$ is a binary operation.

Question 5.
(i) Define an operation * on $\mathrm{Q}$ as follows: $\mathrm{a} * \mathrm{~b}=\left(\frac{a+b}{2}\right)$; $\mathrm{a}, \mathrm{b} \in \mathrm{Q}$. Examine the closure, commutative, and associative properties satisfied by $*$ on $Q$.
(ii) Define an operation * on $\mathrm{Q}$ as follows: $\mathrm{a} * \mathrm{~b}=\left(\frac{a+b}{2}\right) ; \mathrm{a}, \mathrm{b} \in \mathrm{Q}$. Examine the existence of identity and the existence of inverse for the operation * on $\mathrm{Q}$.
Solution:
(i) 1. Closure property:
Let $a, b \in Q$.

Now $a * b=\frac{a+b}{2} \in \mathbb{Q} . \quad(\because a, b, 2 \in \mathbb{Q})$
So, closure property is satisfied.
2. Commutative property:
Let $\mathrm{a}, \mathrm{b} \in \mathrm{Q}$.
Now $a * b=\frac{a+b}{2}$
$b * a=\frac{b+a}{2}$
(1) $=(2) \Rightarrow$ Now $a * b=b * a$
$\Rightarrow$ Commutative property is satisfied.
3. Associative property:
Let a,b,c G Q.^
To prove associative property we have to prove that $\mathrm{a} *(\mathrm{~b} * \mathrm{c})=(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
LHS: $\mathrm{a} *(\mathrm{~b} * \mathrm{c})$
Now $b * c=\frac{b+c}{2}=\mathrm{D}$ (say)
So $a *(b * c)=a * \mathrm{D}=\frac{a+\mathrm{D}}{2}=\frac{a+\frac{b+c}{2}}{2}=\frac{2 a+b+c}{4}$
RHS: $(a * b) * c$
$a * b=\frac{a+b}{2}=\mathrm{K}$ (say)
So $(a * b) * c=\mathrm{K} * c=\frac{\mathrm{K}+c}{2}=\frac{\frac{a+b}{2}+c}{2}=\frac{a+b+2 c}{4}$
$(1) \neq(2)$
So, associative property is not satisfied.

(ii) $a * b=\frac{a+b}{2}$
Let $e \in \mathbb{Q}$ be the identity element. Then $a * e=a$
To find $e: a * e=a$
(i.e.) $\frac{a+e}{2}=a \Rightarrow a+e=2 a \Rightarrow e=2 a-a$
(i.e.) the identity Clement $\mathrm{e}=\mathrm{a}$ which is not possible.
So, the identity element does not exist and so inverse does not exist.
Question 6.
Fill In the following table so that the binary operation * on $A=\{a, b, c\}$ is commutative.

Solution:
Given that the binary operation $*$ is Commutative. To find $\mathrm{a} * \mathrm{~b}$ : $\mathrm{a} * \mathrm{~b}=\mathrm{b} * \mathrm{a}(\because *$ is a Commutative)
Here $b * a=c$. So $a * b=c$ To find a $*$

$\mathrm{a} * \mathrm{c}=\mathrm{c} * \mathrm{a}(\because *$ is a Commutative)
$\mathrm{c}^* \mathrm{a}=\mathrm{a}$. (Given)
So a $* \mathrm{c}=\mathrm{a}$
To find $\mathrm{c}^* \mathrm{~b}$ :
$\mathrm{c} * \mathrm{~b}=\mathrm{b} * \mathrm{c}$
Here $\mathrm{b} * \mathrm{c}=\mathrm{a}$.
So $\mathrm{c}^* \mathrm{~b}=\mathrm{a}$
Question 7.
Consider the binary operation $*$ defined on the set $\mathrm{A}=[\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}]$ by the following table:

- Is it commutative and associative?
Solution:
From the table
$
\begin{aligned}
& \mathrm{b} * \mathrm{c}=\mathrm{b} \\
& \mathrm{c} * \mathrm{~b}=\mathrm{d}
\end{aligned}
$
So, the binary operation is not commutative.
To check whether the given operation is associative.
Let a, b, c $\in \mathrm{A}$.
To prove the associative property we have to prove that $\mathrm{a} *(\mathrm{~b} * \mathrm{c})=(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
From the table,
LHS: $b * c=b$
So, $a *(b * c)=a * b=c$
RHS: $a * b=c$
So, $(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{c} * \mathrm{c}=\mathrm{a}$
$(1) \neq(2)$. So, $a *(b * c) \neq(a * b) * c$
$\therefore$ The binary operation is not associative.

Question 8.
Let $A=\left(\begin{array}{llll}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1\end{array}\right), B=\left(\begin{array}{llll}0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1\end{array}\right), C=\left(\begin{array}{llll}1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1\end{array}\right)$ be any three boolean matrices of the same type. Find (i) $A \vee B$ (ii) $A \wedge B$ (iii) $(A \vee B) \wedge C$ (iv) $(A \wedge B) \vee C$. Solution:
(i) $\quad \mathrm{A} \vee \mathrm{B}=\left(\begin{array}{llll}1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1\end{array}\right)$
(ii) $\quad \mathrm{A} \wedge \mathrm{B}=\left(\begin{array}{llll}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1\end{array}\right)$
(iii) $\begin{aligned} A \vee B & =\left(\begin{array}{llll}1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1\end{array}\right) \\ (A \vee B) \wedge C & =\left(\begin{array}{llll}1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1\end{array}\right) \wedge\left(\begin{array}{llll}1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1\end{array}\right)=\left(\begin{array}{llll}1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1\end{array}\right)\end{aligned}$
$(i v)$
$
\begin{aligned}
\text { (iv) } & \wedge \mathrm{B}=\left(\begin{array}{llll}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1
\end{array}\right) \\
\text { So, }(\mathrm{A} \wedge \mathrm{B}) \vee \mathrm{C} & =\left(\begin{array}{llll}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1
\end{array}\right) \vee\left(\begin{array}{llll}
1 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 \\
1 & 1 & 1 & 1
\end{array}\right)=\left(\begin{array}{llll}
1 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 \\
1 & 1 & 1 & 1
\end{array}\right)
\end{aligned}
$

Question 9.
(i) Let $M=\left\{\left(\begin{array}{ll}x & x \\ x & x\end{array}\right): x \in \mathbf{R}-\{0\}\right\}$ and Let * be the matrix multiplication. Determine whether M is closed under * . If so, examine the commutative and associative properties satisfied by $*$ on $\mathrm{M}$.
(ii) Let $\mathbf{M}=\left\{\left(\begin{array}{ll}x & x \\ x & x\end{array}\right): x \in \mathbf{R}-\{0\}\right\}$ and let $*$ be the matrix multiplication. Determine whether $\mathrm{M}$ is closed under * If so, examine the existence of identity, existence of inverse properties for the operation
* on $\mathrm{M}$.
Solution:

(i) $\quad \mathrm{M}=\left\{\left(\begin{array}{ll}x & x \\ x & x\end{array}\right), x \in \mathrm{R}-\{0\}\right\}$
$*=$ Matrix multiplication
Let $\mathrm{A}=\left(\begin{array}{ll}x & x \\ x & x\end{array}\right), \mathrm{B}=\left(\begin{array}{ll}y & y \\ y & y\end{array}\right) \in \mathrm{M}$
Where $x, y \in \mathrm{R}-\{0\}$
Now $\mathrm{AB}=\overrightarrow{\left(\begin{array}{ll}x & x \\ x & x\end{array}\right)}\left[\left(\begin{array}{ll}y & y \\ y & y\end{array}\right)=\left(\begin{array}{ll}2 x y & 2 x y \\ 2 x y & 2 x y\end{array}\right) \in \mathrm{M}\right.$
$\because 2 x y \in \mathrm{R}-\{0\}$
So, $\mathrm{M}$ is closed under *.
To verify the existence of identity property:
; Let $\mathrm{E}=\left(\begin{array}{ll}e & e \\ e & e\end{array}\right) \in \mathrm{M}$ such that
$
\mathrm{A} * \mathrm{E}=\mathrm{E} * \mathrm{~A}=\mathrm{A}
$
Where $\mathrm{A}=\left(\begin{array}{ll}x & x \\ x & x\end{array}\right)$
To find $\mathrm{E}: \mathrm{A} * \mathrm{E}=\mathrm{A}$
$
\begin{aligned}
& \Rightarrow\left(\begin{array}{ll}
x & x \\
x & x
\end{array}\right)\left(\begin{array}{ll}
e & e \\
e & e
\end{array}\right)=\left(\begin{array}{ll}
x & x \\
x & x
\end{array}\right) \\
& \text { i.e., }\left(\begin{array}{ll}
2 x e & 2 x e \\
2 x e & 2 x e
\end{array}\right)=\left(\begin{array}{ll}
x & x \\
x & x
\end{array}\right) \\
& \Rightarrow 2 x e=x \\
& \Rightarrow e=x / 2 x=1 / 2 \\
& \therefore \mathrm{E}=\left(\begin{array}{ll}
1 / 2 & 1 / 2 \\
1 / 2 & 1 / 2
\end{array}\right) \in \mathrm{M}
\end{aligned}
$
So, identity property is satisfied.
To verify the inverse property:
Let $\mathrm{A}^{-1}=\left(\begin{array}{ll}y & y \\ y & y\end{array}\right) \in \mathrm{M}$
Such that $\mathrm{A} * \mathrm{~A}^{-1}=\mathrm{A}^{-1} * \mathrm{~A}=\mathrm{E}$

To find $\mathbf{A}^{-1}$ :
$
\begin{aligned}
& \mathrm{A} * \mathrm{~A}^{-1}=\mathrm{E} \\
& \text { i.e., }\left(\begin{array}{ll}
x & x \\
x & x
\end{array}\right)\left(\begin{array}{ll}
y & y \\
y & y
\end{array}\right)=\left(\begin{array}{ll}
1 / 2 & 1 / 2 \\
1 / 2 & 1 / 2
\end{array}\right) \\
& \Rightarrow\left(\begin{array}{ll}
2 x y & 2 x y \\
2 x y & 2 x y
\end{array}\right)=\left(\begin{array}{ll}
1 / 2 & 1 / 2 \\
1 / 2 & 1 / 2
\end{array}\right) \\
& \Rightarrow 2 x y=1 / 2 \Rightarrow 2 y=\frac{1}{2(2 x)}=\frac{1}{4 x} \\
& \therefore \mathrm{A}^{-1}=\left(\begin{array}{cc}
1 / 4 x & 1 / 4 x \\
1 / 4 x & 1 / 4 x
\end{array}\right) \in \mathrm{M}
\end{aligned}
$
So, inverse property is satisfied.

Question 10.
(i) Let $A$ be $Q \backslash\{1)$. Define $*$ on $A$ by $x * y=x+y-x y$. Is * binary on $A$ ? If so, examine the commutative and associative properties satisfied by $*$ on $\mathrm{A}$.
(ii) Let $A$ be $Q \backslash\{1\}$. Define *on A by $x * y=x+y-x y$. Is * binary on $A$ ? If so, examine the existence of identity, existence of inverse properties for the operation * on A.
Solution:
(i) Let $a, b \in A$ (i.e.) $a \neq \pm 1, b \neq 1$
Now $a * b=a+b-a b$
If $\mathrm{a}+\mathrm{b}-\mathrm{ab}=1 \Rightarrow \mathrm{a}+\mathrm{b}-\mathrm{ab}-1=0$
(i.e.) $a(1-b)-1(1-b)=0$
$(a-1)(1-b)=0 \Rightarrow a=1, b=1$
But $\mathrm{a} \neq 1, \mathrm{~b} \neq 1$
So $(a-1)(1-6) \neq 1$
(i.e.) $\mathrm{a} * \mathrm{~b} \in \mathrm{A}$. So $*$ is a binary on $\mathrm{A}$.
To verify the commutative property:
Let $a, b \in A$ (i.e.) $a \neq 1, b \neq 1$
Now $a * b=a+b-a b$
and $\mathrm{b} * \mathrm{a}=\mathrm{b}+\mathrm{a}-\mathrm{ba}$
So $\mathrm{a} * \mathrm{~b}=\mathrm{b} * \mathrm{a} \Rightarrow *$ is commutative on $\mathrm{A}$.
To verify the associative property:
Let a, b, c $\in \mathrm{A}$ (i.e.) a, b, $\mathrm{c} \neq 1$
To prove the associative property we have to prove that
$\mathrm{a} *(\mathrm{~b} * \mathrm{c})=(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
LHS: $\mathrm{b} * \mathrm{c}=\mathrm{b}+\mathrm{c}-\mathrm{bc}=\mathrm{D}$ (say)
So $a *(b * c)=a * D=a+D-a D$
$=a+(b+c-b c)-a(b+c-b c)$

$
\begin{aligned}
& =a+b+c-b c-a b-a c+a b c \\
& =a+b+c-a b-b c-a c+a b c \\
& \text { RHS: }(a * b)=a+b-a b=K(\text { say) } \\
& \text { So }(a * b) * c=K * c=K+c-K c \\
& =(a+b-a b)+c-(a+b-a b) c \\
& =a+b-a b+c-a c-b c+a b c \\
& =a+b+c-a b-b c-a c+a b c
\end{aligned}
$
(ii) To verify the identity property:
Let $a \in A(a \neq 1)$
If possible let e $\in \mathrm{A}$ such that
$
\mathrm{a} * \mathrm{e}=\mathrm{e} * \mathrm{a}=\mathrm{a}
$
To find e:
$
\begin{aligned}
& \mathrm{a} * \mathrm{e}=\mathrm{a} \\
& \text { (i.e.) } \mathrm{a}+\mathrm{e}-\mathrm{ae}=\mathrm{a} \\
& \Rightarrow \quad e(1-a)=0 \Rightarrow e=\frac{0}{1-a}=0 \quad(\because a \neq 1)
\end{aligned}
$
So, e $=(\neq 1) \in \mathrm{A}$
(i.e.) Identity property is verified.
To verify the inverse property:
Let $a \in A$ (i.e. $a \neq 1$ )
If possible let a' $\in \mathrm{A}$ such that
To find a':
$
\begin{aligned}
& \mathrm{a}^* \mathrm{a}^{\prime}=\mathrm{e} \\
& \text { (i.e.) } \mathrm{a}+\mathrm{a}^{\prime}-\mathrm{aa}^{\prime}=0 \\
& \Rightarrow \mathrm{a}^{\prime}(1-\mathrm{a})=-\mathrm{a} \\
& \Rightarrow \quad a^{\prime}=\frac{-a}{1-a}=\frac{a}{a-1} \in \mathrm{A} \quad(\because a \neq 1) \\
& \text { So, } a^{\prime} \in \mathrm{A}
\end{aligned}
$
$\Rightarrow$ For every a $\in \mathrm{A}$ there is an inverse a' $\in \mathrm{A}$ such that $a^* a^{\prime}=a^{\prime} * a=e$
$\Rightarrow$ Inverse property is verified.