Maths M.C.Q (1 Marks)

4. Associative and commutative with an identity.

Solution:

Commutativity:

$\text{X}\triangle\text{Y}=(\overline{\text{X}}\cap\text{Y})\cup(\text{X}\cap\overline{\text{Y}})$

$=(\overline{\text{Y}}\cap\text{X})\cup(\text{Y}\cap\overline{\text{X}})$

$=\text{Y}\triangle\text{X}$

Thus,

$\text{X}\triangle\text{Y}=\text{Y}\triangle\text{X}$

Hence, $\triangle$ is commutative on A.

Let $\phi$ be the identity element for $\triangle$ on P.

$\text{A}\triangle\phi=\big(\overline{\text{A}}\cap\phi\big)\cup\big(\text{A}\cap\overline{\phi}\big)$

$=\phi\cup\text{A}$

$=\text{A}$

and,

$\phi\triangle\text{A}=\big(\overline{\phi}\cap\text{A}\big)\cup\big(\phi\cap\overline{\text{A}}\big)$

$=\text{A}\cup\phi$

$=\text{A}$

1. $\frac{9}8$

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

Then,

$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$

e = 3, $\forall\text{ a}\in\text{Q}^+$

Thus, 3 is the identity element in Q⁺ with respect to *.

Let $\text{a}\in\text{Q}^+$ and $\text{b}\in\text{Q}^+$ be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$

$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$

Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.

Given: $\text{a}*\text{b}=\frac{\text{ab}}3$

$4*6=\frac{4\times6}3=8$

Now,

$\text{a}^{-1}=\frac{9}{\text{a}}$

$(4*6)^{-1}=8^{-1}$

$=\frac{9}8$

4. $0$

Solution:

Let e be the identity element in Q - {1} with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}-\{-1\}$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}-\{-1\}$

a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{Q}-\{-1\}$

e(1 - a) = 0, $\forall\text{ a}\in\text{Q}-\{-1\}$ $[\because \text{a}\neq1]$

Thus, 0 is the identity element in Q - {1} with respect to *.

4. 2

Solution:

The number of commutative binary operations on a set of n elements is $\text{n}\frac{\text{n}(\text{n}-1)}{2}$.

Therefore,

Number of commutative binary operations an a set of 2 elements $=2\frac{2(2-1)}{2}=2^1$

$=2$

2. $\frac{1}2$

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

Then,

$\frac{\text{ae}}{2}=\text{a}\text{ and }\frac{\text{ea}}{2}=\text{a},\forall\text{ a}\in\text{Q}^+$

e = 2, $\forall\text{ a}\in\text{Q}^+$

Thus, 2 is the identity element in Q⁺ with respect to *.

Let $\text{b}\in\text{Q}^+$ be the inverse of 8. Then,

8 * b = e = b * 8

8 * b = e and b * 8 = e

$\frac{(8)\text{b}}2=2\text{ and }\frac{\text{b}(8)}2=2$ $[\because\ \text{e}=2]$

b = 12

Thus, $\frac{1}2$ is the inverse of 8.

3. $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$

Solution:

Let the identity of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ be $\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}$, then

$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}=\begin{bmatrix}2&3\\-3&3\end{bmatrix}$

⇒ 2e - 3f = 2 →(1)

2f + 3e = 3 →(2)

Solving (1) and (2) we get e = 1 and f = 0

So, the identity is $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.

Let the inverse be $\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}$, then

$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

⇒ 2a - 3b = 1 →(1)

2b + 3a = 0 →(2)

Solving (1) and (2), we get $\text{a}=\frac{2}{13}$ and $\text{b}=\frac{-3}{13}$

So, the inverse of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ is $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$

3. $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$

Solution:

Let $\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\in\text{G}$ and $\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}\in\text{G}$ such that

$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$

$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$

$\begin{bmatrix}2\text{ex}&2\text{ex}\\2\text{ex}&2\text{ex}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$

$2\text{ex}=\text{x}$

$\text{e}=\frac{1}2\in\text{R}-\{0\}$

Thus, $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}\in\text{G}$, is the identity element in G.

2. $-\frac{\text{a}}{\text{a}-1}$

Solution:

Let e be the identity element in R - {1} with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$

a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$

Then,

a + e + ae = a and e + a + ea = a, $\forall\text{ a}\in\text{R}-\{1\}$

e(1 + a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$

$\text{e}=0\in\text{R}-\{1\}$

Thus, 0 is the identity element in R - {1} with respect to *.

Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

⇒ a + b + ab = 0 and b + a + ba = 0

$\Rightarrow\text{b}(1+\text{a})=-\text{a}\in\text{R}-\{1\}$

$\Rightarrow\text{b}=\frac{-\text{a}}{\text{a}-1}\in\text{R}-\{1\}$

Thus, $\frac{-\text{a}}{\text{a}-1}$ is the inverse of $\text{a}\in\text{R}-\{1\}$.

1. Commutative but not associative.

Solution:

Commutativity:

Let $\text{a, b}\in\text{R}$

a * b = ab + 1

= ba + 1

= b * a

Therefore,

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

Therefore, * is commutative on R.

Associativity:

Let $\text{ a, b, c}\in\text{R}$

a * (b * c) = a * (bc + 1)

= a(bc + 1) + 1

= abc + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1)c + 1

= abc + c + 1

$\therefore$ a * (b * c) $\neq$ (a * b) * c

For example: a = 1, b = 2 and c = 3 [which belong to R]

Now,

1 * (2 * 3) = 1 * (6 + 1)

= 1 * 7

= 7 + 1

= 8

(1 * 2) * 3 = (2 + 1) * 3

= 3 * 3

= 9 + 1

= 10

⇒ 1 * (2 * 3) $\neq$ (1 * 2) * 3

Therefore, $\exists$ a = 1, b = 2 and c = 3 which belong to R such that

a * (b * c) $\neq$ (a * b) * c

Hence, * is not associative on R.

4. $\frac{4}{\text{a}}$

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

$\frac{\text{ae}}2=\text{a}$ and $\frac{\text{ea}}2=\text{a}$, $\forall\text{ a}\in\text{Q}^+$

$\text{e}=2\in\text{Q}^+, \forall\text{ a}\in\text{Q}^+$

Thus, 2 is the identity element in Q⁺ with respect to *.

Let $\text{ a}\in\text{Q}^+$ and $\text{ b}\in\text{Q}^+$ be the inverse of a.

Then,

a * e = a = e * a

a * b = e and b * a = e

$\frac{\text{ab}}2=2$ and $\frac{\text{ba}}2=2$

$\text{b}=\frac{4}{\text{a}}\in\text{Q}^+$

Thus, $\frac{4}{\text{a}}$ is the inverse of $\text{ a}\in\text{Q}^+$.

4. None of these.

Solution:

* is not clouser because when a = 1 and b = 2,

$\text{a}*\text{b}=\frac{\text{a}}{\text{b}}=\frac{1}{2}\in\text{Z}$

* is not commutative because when a = 1, b = 2 and c = 3,

$1*(2*3)=1*\Big(\frac{2}3\Big)$

$=\frac{1}{\big(\frac{2}{3}\big)}$

$=\frac{3}2$

$(1*2)*3=\frac{1}2*3$

$=\frac{\big(\frac{1}2\big)}{3}$

$=\frac{1}6$

Thus,

$1*(2*3)\neq(1*2)*3$

3. Not commutative.

Solution:

Let $\text{a, b}\in\text{Z}$

a * b = 3a + b

b * a = 3b + a

Thus, a * b $\neq$ b * a

If a = 1 and b = 2,

1 * 2 = 3(1) + 2

= 5

2 * 1 = 3(2) + 1

= 7

1 * 2 $\neq$ 2 * 1

Thus, * is not commutative on Z.

1. $\frac{3}{160}$

Solution:

Given $\text{a}\odot\text{b}=\frac{\text{ab}}4$

$\Rightarrow\Big(\frac{1}5\odot\frac{1}2\Big)$

$=\frac{\frac{1}5.\frac{1}2}{4}$

$=\frac{1}{40}$

$3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$

$=3\odot\frac{1}{40}$

$=\frac{\frac{1}{40}.3}{4}$

$=\frac{3}{160}$

3. 16

Solution:

Total number of binary operations on a set containing n elements is

$\text{(n)}^{\text{n}^2}$ so for n = 2 we have $(2)^{2^2}=2^4=16$

1. 10⁵

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

$\frac{\text{ae}}{100}=\text{a}\text{ and }\frac{\text{ea}}{100}=\text{a},\forall\text{ a}\in\text{Q}^+$

$\text{e}=100,\forall\text{ a}\in\text{Q}^+$

Thus, 100 is the identity element in Q⁺ with repect to *.

0.1 * b = e = b * 0.1

0.1 * b = e and b * 0.1 = e

$\frac{(0.1)\text{b}}{100}=100\text{ and }\frac{\text{b}(0.1)}{100}=100$

$\text{b}=\frac{100\times100}{0.1}$

$=10^5\in\text{Q}^+$

Thus, 10⁵ is the inverse of 0.1.

1. $\frac{4}3$

Solution:

Let us first find the identity element.

We know that if e is the identity element then,

$\text{a}\odot\text{e}=\text{e}$

Given $\text{a}\odot\text{e}=\frac{\text{ae}}2$

$\Rightarrow\text{a}=\frac{\text{ae}}2$

$\Rightarrow\text{e}=2$

Let b be the inverse of 3, then

$3\odot\text{b}=\text{e}$ 

$\Rightarrow\frac{3\text{b}}2=2$

$\Rightarrow\text{b}=\frac{4}3$

2. * defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on Q.

Solution:

For option a, if we take 3 and 2 then

$3*2=\frac{5}2\in\text{Z}$. So, option a is not true.

For option b, if we take any two numbers a and b

then $\frac{\text{a + b}}2$ belongs to Q for $\text{a, b}\in\text{Q}$.

So, option b is correct.

For option d, if we take 2, 3 then $2-3=-1\in\text{N}$.

So, option d is not true.

Option c is not true.

1. Neither commutative nor associative.

Solution:

Let $\text{a, b}\in\text{Z}$, then

a * b = a - b

b * a = b - a

⇒ a * b $\neq$ b * a

Substraction is not commutative.

(a * b) * c

= (a - b) * c

= a - b - c

a * (b * c)

= a * (b - c)

= a - b + c

⇒ (a * b) * c $\neq$ a * (b * c)

Substraction is not associative.

3. Not associative.

Solution:

a * b = a² + b²

b * a = b² + a²

⇒ a * b = b * a

So * is commutative.

Now

(a * b) * c

= (a² + b²) * c

= (a² + b²)² + c²

a * (b * c)

= a * (b² + c²)

= a² + (b² +c²)²

⇒ (a * b) * c $\neq$ a * (b * c)

So * is not associative.

1. 0

Solution:

Let e be the identity element in Q - {1} with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}-\{-1\}$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}-\{-1\}$

Then,

a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{Q}-\{-1\}$

e(1 - a) = 0, $\forall\text{ a}\in\text{Q}-\{-1\}$

$\text{e}=0\in\text{Q}-\{-1\}$ $[\because\text{ a}\neq1]$

Thus, 0 is the identity element in Q - {1} with respect to *.