Maths 3 Marks

$\forall\ \text{x}\in\text{R}_0\ \&\text{ y}\in\text{R}$ such that

x ⊙ E = X = E ⊙ X, $\forall\text{ x}\in\text{A}$

⇒ X ⊙ E = X and E ⊙ X = X

⇒ (ax, bx + y) = (a, b) and (xa, ya + b) = (a, b)

Considering (ax, bx + y) = (a, b)

⇒ ax = a

⇒ x = 1

& bx + y = b

⇒ y = 0 $[\because\text{ x}=1]$

Considering (xa, ya + b) = (a, b)

⇒ xa = a

⇒ x = 1

& ya + b = b

⇒ y = 0 $[\because\text{ x}=1]$

$[\because\ 1,0]$ is the identity element in A with respect to ⊙.

eg. 2 + 4 = 6 ⇒ 2 +₅ 4 = 1 $\because$ [we get 1 as remainder when 6 is divided by 5]

2 + 4 = 7 ⇒ 3 +₅ 4 = 2 $\because$ [we get 2 as remainder when 7 is divided by 5]

The composition table for +₅ on set S = {0, 1, 2, 3, 4}.

Let $\text{a}=\frac{1}{3},\ \text{b}=\frac{5}{3}\in\text{S}$

$\text{a}\ ^*\ \text{b}=\text{ab}$

$=\frac{1}{3}\times\frac{5}{3}$

$=\frac{5}{9}\notin\text{S}\ \big[\because\ 9\notin\{1,2,3\}\big] $

Therefore, $\exists\text{ a, b}\in\text{S},$ such that $\text{a}\ ^*\ \text{b}\notin\text{S}$

Thus, * is not a binary operation.

a * b = 1.c.m. a, b

= 1.c.m. b, a

= b * a

Therefore,

$\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a}\ \forall\ \text{a, b}\in\text{N}$

Thus, * is Commutative on N.

Associativity: Let $\text{a, b, c}\in\text{N}$

a * b * c = a * 1.c.m. b, c

= 1.c.m. a, b, c

a * b * c = 1.c.m. a, b * c

= 1.c.m. a, b, c

Therefore,

$\text{a}\ ^*\ \text{b}\ ^*\ \text{c}=\text{a}\ ^*\ \text{b}\ ^*\ \text{c}\ \forall\ \text{a, b, c}\in\text{N}$

Thus, * is associative on N.

Let $\text{x}=(\text{a},\text{b})$ and $\text{y}=(\text{c},\text{d})\in\text{A},\forall\text{ a},\text{c}\in\text{R}_0\ \&\text{ b},\text{d}\in\text{R}.$ Then,

X ⊙ Y = (ac, bc + d ) & Y ⊙ X = (ca, da + b)

Therefore, x ⊙ Y = Y ⊙ X, $\forall\ \text{X},\text{Y}\in\text{A}$

Thus, ⊙ is commutative on A.

Associativity:

Let $\text{X}=(\text{a},\text{b}),\text{Y}=(\text{c},\text{d})$ and $\text{Z}=(\text{e},\text{f}),\forall\ \text{a},\text{c},\text{e}\in\text{R}_0\ \&\text{ b},\text{d},\text{f}\in\text{R}$

x ⊙ Y ⊙ Z = (a, b) ⊙ (ce, de + f)

= (ace, bce + de + f)

x ⊙ Y ⊙ Z = (ac, bc + d) ⊙ (e, f)

= (ace, (bc + d)e + f)

= (ace, bce + de + f)

$\therefore$ x ⊙ Y ⊙ Z = x ⊙ Y ⊙ Z, $\forall\ \text{X},\text{Y},\text{Z}\in\text{A}$

Thus, ⊙ is accosiative on A.

Let $\text{a}, \text{b}\ \&\ \text{c}, \text{d}\in\text{A}\forall\text{ a, b, c, d}\in\text{R}_0$. Then,

(a, b) * (c, d) = (ac, bd)

= (ca, db)

= (c, d) * (a, b)

$\therefore$ (a, b) * (c, d) = (c, d) * (a, b)

Thus, * is commutative on A.

Associativity:

Let (a, b), (c, d) & (e, f) $\in\text{A}\forall\text{ a, b, c, d, e, f}\in\text{R}_0$. Then,

(a, b) * ((c, d) * (e, f)) = (a, b) * (ce, df)

= (ace, bdf)

((a, b) * (c, d)) * (e, f) = (ac, bd) * (e, f)

= (ace, bdf)

$\therefore$ (a, b) * ((c, d) * (e, f)) = ((a, b) * (c, d) * (e, f))

Thus, * is associative on A.

a * e = a = e * a, $\forall\text{ a}\in\text{R}$

a * e = a and e * a = a, $\forall\text{ a}\in\text{R}$

Then,

$\sqrt{\text{a}^2+\text{e}^2}=\text{a}$ and $\sqrt{\text{e}^2+\text{a}^2}=\text{a},\forall\text{ a}\in\text{R}$

Implies that $\sqrt{\text{a}^2+\text{e}}=\text{a}$ and $\sqrt{\text{e}+\text{a}^2}=\text{a},\forall\text{ a}\in\text{R}$ [$\because$ e² = e]

Implies that a² + e = a² and e + a² = a², $\forall\text{ a}\in\text{R}$

Implies that $\text{e}=0\in\text{R},\forall\text{ a}\in\text{R}$

Thus, 0 is the identity element in R with respect to *.

The composition table for ×₁₀ on set S = {1, 3, 7, 9}

We know that an element $\text{b}\in\text{S}$ will be the inverse of $\text{a}\in\text{S}$

if a×₁₀ b = 1 [$\because$ 1 is the identity element with respect to multiplication.]

⇒ 3×₁₀ b = 1

From the above table b = 7

$\therefore$ Inverse of 3 is 7.

x ⊙ F = X = E and F ⊙ X = E

Implies that (am, bm + n) = (1, 0) and (ma, na + b) = (1,0)

Considering (am, bm + n) = (1, 0)

Implies that am = 1

Implies that $\text{m}=\frac{1}{\text{a}}$

& bm + n = 0

Implies that $\text{n}=\frac{-\text{b}}{\text{a}}$ $\Big[\because\ \text{m}=\frac{1}{\text{a}}\Big]$

Considering (ma, na + b) = (1,0)

Implies that ma = 1

Implies that $\text{m}=\frac{1}{\text{a}}$

& na + b = 0

Implies that $\text{n}=\frac{-\text{b}}{\text{a}}$ 

$\therefore$ The inverse of $(\text{a},\text{b})\in\text{A}$ with respect to ⊙ is $\Big(\frac{1}{\text{a}},\frac{-\text{b}}{\text{a}}\Big).$

$(\text{a}\ ^*\ \text{b}) *\ \text{c}=2^{\text{ab}}\ ^*\ \text{c}=2^{2{\text{ab}}.\text{c}}\ ...(\text{i})$

$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$

$\text{a}\odot\text{b}\odot\text{c}\neq\text{a}\odot\text{b}\odot\text{c}$

$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=(\text{a}\ ^*\ \text{b})\ ^*\ \text{c},\ \forall\ \text{a, b, c}\in\text{A}$