Exercise 12.1-Additional Problems - Chapter 12 - Discrete Mathematics - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Show that the set $\mathrm{G}=\{\mathrm{a}+\mathrm{b} \sqrt{2} / \mathrm{a}, \mathrm{b} \in \mathrm{Q}\}$ is an infinite abelian group with respect to Binary operation addition. Satisfies closure, associative, identity and inverse properties.
Solution:
(i) Closure axiom:
Let $\mathrm{x}, \mathrm{y} \in \mathrm{G}$. Then $\mathrm{x}=\mathrm{a}+\mathrm{b} \sqrt{2}, \mathrm{y}=\mathrm{c}+\mathrm{d} \sqrt{2} ; \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{Q}$. $\mathrm{x}+\mathrm{y}=(\mathrm{a}+\mathrm{b} \sqrt{2})+(\mathrm{c}+\mathrm{d} \sqrt{2})=(\mathrm{a}+\mathrm{c})+(\mathrm{b}+\mathrm{d}) \sqrt{2} \in \mathrm{G}$,
Since $(a+c)$ and $(b+d)$ are rational numbers.
$\therefore \mathrm{G}$ is closed with respect to addition.
(ii) Associative axiom : Since the elements of $\mathrm{G}$ are all real numbers, addition is associative.
(iii) Identity axiom : There exists $0=0+0 \sqrt{2} \in G$ such that for all $\mathrm{x}=\mathrm{a}+\mathrm{b} \sqrt{2} \in \mathrm{G}$.
$
\begin{aligned}
& \mathrm{x}+0=(\mathrm{a}+\mathrm{b} \sqrt{2})+(0+0 \sqrt{2}) \\
& =\mathrm{a}+\mathrm{b} \sqrt{2}=\mathrm{x}
\end{aligned}
$
Similarly, we have $0+\mathrm{x}=\mathrm{x} . \therefore 0$ is the identity element of $\mathrm{G}$ and satisfies the identity axiom.
(iv) Inverse axiom: For each $\mathrm{x}=\mathrm{a}+\mathrm{b} \sqrt{2} \in \mathrm{G}$,
there exists $-\mathrm{x}=(-\mathrm{a})+(-\mathrm{b}) \sqrt{2} \in \mathrm{G}$
such that $\mathrm{x}+(-\mathrm{x})=(\mathrm{a}+\mathrm{b} \sqrt{2})+((-\mathrm{a})+(-\mathrm{b}) \sqrt{2})$
$
=(\mathrm{a}+(-\mathrm{a}))+(\mathrm{b}+(-\mathrm{b})) \sqrt{2}=0
$
Similarly, we have $(-\mathrm{x})+\mathrm{x}=0$.
$\therefore(-\mathrm{a})+(-\mathrm{b}) \sqrt{2}$ is the inverse of $\mathrm{a}+\mathrm{b} \sqrt{2}$ and satisfies the inverse axiom.
(v) Commutative axiom:
$
\begin{aligned}
& \mathrm{x}+\mathrm{y}=(\mathrm{a}+\mathrm{c})+(\mathrm{b}+\mathrm{d}) \sqrt{2}=(\mathrm{c}+\mathrm{a})+(\mathrm{d}+\mathrm{b}) \sqrt{2} \\
& =(\mathrm{c}+\mathrm{d} \sqrt{2})+(\mathrm{a}+\mathrm{b} \sqrt{2}) \\
& =\mathrm{y}+\mathrm{x}, \text { for all } \mathrm{x}, \mathrm{y} \in \mathrm{G}
\end{aligned}
$
$\therefore$ The commutative property is true.
$\therefore(\mathrm{G},+)$ is an abelian group. Since $\mathrm{G}$ is infinite, we see that $(\mathrm{G},+)$ is an infinite abelian group.

Question 2.
Show that $\left(\mathrm{Z}_7-\{[0]\}, .7\right)$ write to the binary operation multiplication modul07 satisfies closure, associative, identity and inverse properties.
Solution:
Let $\mathrm{G}=[[1],[2], \ldots[6]]$
The Cayley's table is

From the table:
(i) all the elements of the composition table are the elements of $\mathrm{G}$.
$\therefore$ The closure axiom is true.
(ii) multiplication modulo 7 is always associative.
(iii) the identity element is $[1] \in \mathrm{G}$ and satisfies the identity axiom.
(iv) the inverse of [1] is [1]; [2] is [4]; [3] is [5]; [4] is [2]; [5] is [3] and [6] is [6] and it satisfies the inverse axiom.

Question 3.
Show that the set $\mathrm{G}$ of all positive rationals with respect to composition $*$ defined by ab $\mathrm{a}^* \mathrm{~b}=\frac{\mathrm{ab}}{3}$ for all $\mathrm{a}, \mathrm{b} \in \mathrm{G}$ satisfies closure, associative, identity and inverse properties.
Solution:
Let $\mathrm{G}=$ Set of all positive rational number and $*$ is defined by,
$
a * b=\frac{a b}{3}
$
(i) Closure axiom: Let a, b $\in \mathrm{G}$
$
a * b=\frac{a b}{3} \in \mathrm{G} \quad[\because a, b, 3 \in \mathrm{G}]
$
$\therefore$ closure axiom is satisfied.
(ii) Associative axiom: Let a, b, $c \in \mathrm{G}$.
To prove the associative property, we have to prove that
$
\begin{aligned}
& (a * b) * c=a *(b * c) \\
& \text { LHS: } a * b=\frac{a b}{3}=\mathrm{D} \text { (say) } \\
& \therefore(a * b) * c=\mathrm{D} * c=\frac{\mathrm{D} c}{3}=\frac{\frac{a b}{3}(c)}{3}=\frac{a b c}{9} \\
& \text { RHS: } b * c=\frac{b c}{3}=\mathrm{DA}(\mathrm{say}) \\
& \therefore a *(b * c)=a * \mathrm{~A}=\frac{a \mathrm{~A}}{3}=\frac{a\left(\frac{b c}{3}\right)}{3}=\frac{a b c}{9}
\end{aligned}
$
$(1)=(2) \Rightarrow \mathrm{LHS}=\mathrm{RHS}$ i.e., associative axiom is satisfied.
(iii) Identity axiom: Let a $\in \mathrm{G}$.
Let e be the identity element.
By the definition, $\mathrm{a} * \mathrm{e}=\mathrm{a}$
i.e., $\frac{a e}{3}=a \Rightarrow e=\frac{3 a}{a}=3$
$\mathrm{e}=3 \in \mathrm{G} \Rightarrow$ identity axiom is satisfied.
(iv) Inverse axiom: Let $\mathrm{a} \in \mathrm{G}$ and a' be the inverse of $\mathrm{a}^* \mathrm{a}^{\prime}=\mathrm{e}=3$.
$
\text { i.e., } \frac{a a^{\prime}}{3}=3 ; \quad \therefore a^{\prime}=\frac{a}{a} \in \mathrm{G}[\because a, a \in \mathrm{G}]
$

Question 4.
Show that the set $\mathrm{G}$ of all rational numbers except $-1$ satisfies closure, associative, identity and inverse property with respect to the operation * given by $\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+\mathrm{ab}$ for all $\mathrm{a}, \mathrm{b} \in \mathrm{G}$
Solution:
$
\mathrm{G}=[\mathrm{Q},-\{-1\}]
$
$*$ is defined by $\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+\mathrm{ab}$
To prove $\mathrm{G}$ is an abelian group.
$\mathrm{G}_1$ : Closure axiom: Let $\mathrm{a}, \mathrm{b} \in \mathrm{G}$.
i.e., $a$ and $b$ are rational numbers and $a \neq-1, b \neq-1$.
So, $a * b=a+b+a b$
If $\mathrm{a}+\mathrm{b}+\mathrm{ab}=-1$
$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{ab}+1=0$
i.e., $(a+a b)+(b+1)=0$
$a(1+b)+(b+1)=0$
i.e., $(a+1)(1+b)=0$
$\Rightarrow \mathrm{a}=-1, \mathrm{~b}=-1$
But $\mathrm{a} \neq-1, \mathrm{~b} \neq-1$
$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{ab} \neq-1$
i.e., $a+b+a b \in G \forall a, b \in G$
$\Rightarrow$ Closure axiom is verified.
$\mathrm{G}_2$ : Associative axiom: Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{G}$.
To prove $\mathrm{G}_2$, we have to prove that,
$
\mathrm{a} *\{\mathrm{~b} * \mathrm{c})=(\mathrm{a} * \mathrm{~b}) * \mathrm{c}
$
LHS:
$
\begin{aligned}
& b * c=b+c+b c=D \text { (say) } \\
& a *(b * c)=a * D=a+D+a D \\
& =a+(b+c+b c)+a(b+c+b c) \\
& =a+b+c+b c+a b+a c+a b c \\
& =a+b+c+a b+b c+a c+a b c
\end{aligned}
$
RHS:
$
\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+\mathrm{ab}=\mathrm{E} \text { (say) }
$
$\therefore(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{E} * \mathrm{c}=\mathrm{E}+\mathrm{c}+\mathrm{Ec}$
$=\mathrm{a}+\mathrm{b}+\mathrm{ab}+\mathrm{c}+(\mathrm{a}+\mathrm{b}+\mathrm{ab}) \mathrm{c}$
$=\mathrm{a}+\mathrm{b}+\mathrm{ab}+\mathrm{c}+\mathrm{ac}+\mathrm{bc}+\mathrm{abc}$
$=\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{ab}+\mathrm{be}+\mathrm{ac}+\mathrm{abc}$
$(1)=(2) \Rightarrow$ Associative axiom is verified.

$\mathrm{G}_3$ : Identity axiom: Let a $\in \mathrm{G}$. To prove $\mathrm{G}_3$ we have to prove that there exists an element e $\in \mathrm{G}$ such that a $* \mathrm{e}=\mathrm{e} * \mathrm{a}=\mathrm{a}$
To find $\mathrm{e}: \mathrm{a} * \mathrm{e}=\mathrm{a}$ i.e., $a+e+a e=a$
$\Rightarrow \mathrm{e}(1+\mathrm{a})=\mathrm{a}-\mathrm{a}=0$
$\therefore e=\frac{0}{1+a}=0 \quad\{\because a \neq-1 ; 1+a \neq 0\}$
So, $\mathrm{e}=0 \in \mathrm{G} \Rightarrow$ Identity axiom is verified.
$\mathrm{G}_4$ : Inverse axiom: Let a $\in \mathrm{G}$. To prove $\mathrm{G}_4$, we have to prove that there exists an element a' $\in \mathrm{G}$ such that $a^* a^{\prime}=a^{\prime} * a=e$.
To find $a^{\prime}: a^* a^{\prime}=e$
i.e., $a+a^{\prime}+a a^{\prime}=: 0\{\because \mathrm{e}=0\}$
$
\begin{aligned}
& \Rightarrow \mathrm{a}^{\prime}(1+\mathrm{a})=-\mathrm{a} \\
& \therefore a^{\prime}=\frac{-a}{1+a} \in \mathrm{G} \quad\{\because a \neq-1\}
\end{aligned}
$
Thus, inverse axiom is verified.

Question 5.
Show that the set $\{[1],[3],[4],[5],[9]\}$ under multiplication modulo 11 satisfies closure, associative, identity and inverse properties.
Solution:
$\mathrm{G}=\{[1],[3],[4],[5],[9]\}$
$*$ is defined by multiplication modulo 11.
To prove $\mathrm{G}$ is an abelian group with respect to *
Since we are given a finite number of elements i.e., since the given set is finite, we can frame the multiplication table called Cayley's table.
The Cayle's table is as follows:

$\mathrm{G}_1$ : The elements in the above table are [1], [3], [4], [5] and [9] which are elements of $\mathrm{G}$.
$\therefore$ closure axiom is verified.
$\mathrm{G}_2$ : Consider [3], [4], [5] which are elements of $\mathrm{G}$.
$\{[3] *[4]\} *[5]=[1] *[5]=[5]$
$[3] *\{[4] *[5]\}=[3] *[9]=[5]$
(2)
$(1)=(2) \Rightarrow(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c})$ i.e., associative axiom is verified.
$\mathrm{G}_3$ : The first row elements are the same as that of the given elements in the same order. ie., from the table, the identity element is $[1] \in G$. So identity axiom is verified.
$\left.\begin{array}{r}
\mathbf{G}_4:[1] *[1]=[1] \Rightarrow \text { inverse of }[1] \text { is }[1] . \\
{[3] *[4]=[1] \Rightarrow \text { inverse of }[3] \text { is }[4]} \\
{[4] *[3]=[1] \Rightarrow \text { inverse of }[4] \text { is }[3]} \\
{[5] *[9]=[1] \Rightarrow \text { inverse of }[5] \text { is }[9]} \\
{[9] *[5]=[1] \Rightarrow \text { inverse of }[9] \text { is }[5]}
\end{array}\right\} \quad \in \mathrm{G}$

$\therefore \text { inverse axiom is verified. }$
$\mathrm{G}_5$ : From the table $*$ is commutative i.e., the entries equidistant from the leading diagonal on either sides are equal $\Rightarrow \mathrm{a} * \mathrm{~b}=\mathrm{b} * \mathrm{a}$