5 Marks Questions

Question 15 Marks

Let S be the set of all real numbers except -1 and let '*' be an operation defined by a * b = a + b + ab for all a, b ∈ S. Determine whether '*' is a binary operation on S. If yes, check its commutativity and associativity. Also, solve the equation (2 * x) * 3 = 7.

Answer

Checking for binary operation: Let $\text{a, b}\in\text{S.}$ Then, $\text{a, b}\in\text{R}$ and $\text{a}\neq-1,\text{b}\neq-1$ a * b = a + b + ab We need to prove that $\text{a}+\text{b}+\text{ab}\in\text{S.}$ $\big[$For this we have to prove that $\text{a}+\text{b}+\text{ab}\in\text{R}$ and $\text{a}+\text{b}+\text{ab}\neq-1\big]$ Since, $\text{a, b}\in\text{R},\ \text{a}+\text{b}+\text{ab}\in\text{R},$ let us assume that a + b + ab = -1. a + b + ab + 1 = 0 a + ab + b + 1 = 0 a(1 + b) + 1(1 + b) = 0 (a + 1)(b+ 1) = 0 a = -1, b = -1 [which is false] Hence, $\text{a}+\text{b}+\text{ab}\neq-1$ Therefore, $\text{a}+\text{b}+\text{ab}\in\text{S}$ Thus, * is a binary operation on S. Commutativity: Let $\text{a, b}\in\text{S.}$ Then, a * b = a + b + ab = b + a + ba = b * aTherefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{S}$ Thus, * is commutative on N. Associativity: Let $\text{a, b, c}\in\text{S.}$ a * (b * c) = a * (b + c + bc) = a + b + c + bc + a(b + ac + bc) = a + b + c + bc + ab + ac + abc (a * b) * c = (a + b + ab) * c = a + b + ab + c + (a + b + ab)c = a + b + ab + c + ac + bc + abc Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{S}$ Thus, * is associative on S. Now, Given: (2 * x) * 3 = 7 Implies that (2 + x + 2x) * 3 = 7 Implies that (2 + 3x) * 3 = 7

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Question 25 Marks

Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b - ab, for all a, b ∈ S.
Prove that:

* is a binary operation on S.
* is commutative as well as associative.

Answer

We have,
S = R - {1} and * is defined on S as a * b = a + b - ab, for all a, b ∈ S.

It is seen that for each a, b ∈ S, there is a unique element a + b - ab in S.

This means that * carries each pair (a, b) to a unique element a * b = a + b - ab in S.

Therefore, * is a binary operation on S.

Commutativity: Let a, b ∈ S. Then,

a * b = a + b - ab

= b + a - ba

= b * a

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

Thus, * is commutative on S.

Associativity: Let a, b, c ∈ S. Then,

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

= a + b + c - bc - a(b + c - bc)

= a + b + c - bc - ab - ac + abc

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

= a + b - ab + c = (a + b - ab)c

= a + b + c - ab - ac - bc + abc

Therefore,

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

Thus, * is associative on S.

Therefore, * is commutative as well as associative.

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Question 35 Marks

On R − {1}, a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.

Answer

Commutativity:Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a},\text{b}\in\text{R}-\{1\}$
Thus, * is commutative on R - {1}.
Associativity:
Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * (b * c) = a * (b + c - bc)
= a + b + c - bc - a(b + c - bc)
= a + b + c - bc - ab - ac - abc
(a * b) * c = (a + b - ab) * c
= a + b - ab + c - (a + b - ab)c
= a + b + c - ab - ac - bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\text{ a},\text{b},\text{c}\in\text{R}-\{1\}$
Thus, * is associative on R - {1}.
Finding identity element:
Let e be the 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, $\forall\text{ a}\in\text{R}-\{1\}$
⇒ a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{R}-\{1\}$
e(1 - a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$
$\text{e}=0\in\forall\text{ a}\in\text{R}-\{1\},\forall\text{ a}\in\text{R}-\{1\}$ $[\because\ \text{a}\neq1]$
Thus, 0 is the identity element in R - {1} with respect to *.
Finding inverse:
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
⇒ a = ab - b
⇒ a = b(a - 1)
$\Rightarrow\text{b}=\frac{\text{a}}{\text{a}-1}$
Thus, $\frac{\text{a}}{\text{a}-1}$ is inverse of $\text{a}\in\text{R}-\{1\}.$

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Question 45 Marks

Let A =R×R and * be a binary operation on A defined by,
(a, b) * (c, d) = (a + c, b + d).
Show that * is commutative and associative. Find the binary element for * on A, if any.

Answer

We have,
A = R × R and * is a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d).
Now,
(a, b) * (c, d) = (a + c, b + d) = (c + a, d + b)
⇒ (a, b) * (c, d) = (c, d) * (a, b)
So, * is commutative.
Also,
(a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f)
= (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
= (a + c, b + d) * (e, f)
= [(a, b) * (c, d)] * (e, f)
⇒ (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e, f)
So, * is associative.
Let (x, y) be the binary element for * on A.
(a, b) * (x, y) = (a, b) = (x, y) * (a, b)
⇒ (a + x, b + y) = (a, b)
⇒ a + x = a and b + y = b
⇒ x = 0 and y = 0
Hence, (0, 0) is the binary element for * on A.

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Question 55 Marks

Define a binary operation * on the set ${0, 1, 2, 3, 4, 5}$ as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
Show that 0 is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.

Answer

Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
An element $\text{e}\in\text{X}$ is the identity element for the operation *, if
a * e = a = e * a $\forall\text{ a}\in\text{X}$.
For $\text{a}\in\text{X}$, we observed that:
a * 0 = a + 0 = a $[\text{a}\in\text{X}\Rightarrow\text{a}+0<6]$
0 * a = a = 0 * a $[\text{a}\in\text{X}\Rightarrow0+\text{a}<6]$
$\therefore$ a * 0 = a = 0 * a $\forall\text{ a}\in\text{X}$
Thus, 0 is the identity element for the given operation *.
An element $\text{a}\in\text{X}$ is invertible if there exists be X such that a * b = 0 = b * a.
i.e.,$\begin{cases}\text{a + b}=0=\text{b + a},&\text{if }\text{a + b}<6\\\text{a + b}-6=0=\text{b + a}-6,&\text{if }\text{a + b}\geq6\end{cases}$
i.e.,
a = -b or b = 6 - a
But, X = {0, 1, 2, 3, 4, 5} and $\text{a, b}\in\text{X}$. Then, $\text{a}\neq-\text{b}$.
Therefore, b = 6 - a is the inverse of $\text{a}\in\text{X}$.
Hence, the inverse of an element $\text{a}\in\text{X},\text{a}\neq0$ is $6 - a i.e., a^{-1} = 6 - a.$

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Question 65 Marks

Check the commutativity and associativity of the following binary operations:
'*' on N defined by a * b = gcd(a, b) for all a, b ∈ N.

Answer

Commutativity: Let $\text{a, b}\in\text{N}.$ Then,a * b = gcd(a, b)
= gcd(b, a)
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{N}$
Thus '*' is commutative on N.
Associativity: Let $\text{a, b, c}\in\text{N}.$ Then,
a * (b * c) = a * [gcd(a, b)]
= gcd(a, b, c)
(a * b) * c = [gcd(a, b)] * c
= gcd(a, b, c)
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{N}$
Thus, '*' is not associative on N.

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Question 75 Marks

On Z, the set of all integers, a binary operation * is defined by a * b = a + 3b - 4. Prove that * is neither commutative nor associative on Z.

Answer

Commutativity: Let $\text{a, b}\in\text{Z.}$ Then,
a * b = a + 3b - 4
b * a = b + 3a - 4
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Let a = 1, b = 2
1 * 2 = 1 + 6 - 4
= 3
2 * 1 = 2 + 3 - 4
= 1
Therefore, $\exists\ \text{a}=1,\text{b}=2\in\text{Z}$ such that $\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z.}$ Then,
a * (b * c) = a * (b + 3c - 4)
= a + 3(b + 3c - 4) - 4
= a + 3b + 9c - 12 - 4
= a + 3b + 9c - 16
(a * b) * c = (a + 3b - 4) * c
= a + 3b - 4 + 3c - 4
= a + 3b + 3c - 8
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
If a = 1, b = 2, c = 3
1 * (2 * 3) = 1 * (2 + 9 - 4)
= 1 * 7
= 1 + 21 - 4
= 18
(1 * 2) * 3 = (1 + 6 - 4) * 3
= 3 * 3
= 3 + 9 - 4
= 8
Therefore, $\exists\ \text{a}=1,\text{b}=2,\text{c}=3\in\text{Z}$ such that $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Z.

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Question 85 Marks

Consider the binary operation * and o defined by the following tables on set S = {a, b, c, d}.

*	a	b	c	d
a	a	b	c	d
b	b	a	d	c
c	c	d	a	b
d	d	c	b	a

Answer

Commutativity:The table is symmetrical about the leading element. It means * is commutative on S.
Associativity:
$a * (b * c) = a * d$
$= d$
$(a * b) * c = b * c$
$=d$
Therefore,
a * (b * c) = (a * b) * c $\forall\text{ a, b, c}\in\text{S}$
So, * is Associative on S.
Finding identity element:
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at a.
$\Rightarrow x * a = a * x = x,$ $\forall\text{ x}\in\text{S}$
So, a is the identity element:
$a * a = a$
$\Rightarrow a^{-1} = a$
$b * b = a$
$\Rightarrow b^{-1} = b$
$c * c = a$
$\Rightarrow c^{-1} = c$
$d * d = a$
$\Rightarrow d^{-1} = d$

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Question 95 Marks

On Q, the set of all rational numbers a binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}.$ Show that * is not associative on Q.

Answer

Let $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{b}+\text{c}}{2}\Big)$
$=\frac{\text{a}+\big(\frac{\text{b}+\text{c}}{2}\big)}{2}$
$=\frac{2\text{a}+\text{b}+\text{c}}{4}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{a}+\text{b}}{2}\big)+\text{c}}{2}$
$=\frac{\text{a}+\text{b}+2\text{c}}{4}$
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
If a = 1, b = 2, c = 3
$1\ ^*\ (2\ ^*\ 3)=1\ ^*\ \Big(\frac{2+3}{2}\Big)$
$=1\ ^*\ \frac{5}{2}$
$=\frac{1+\frac{5}{2}}{2}$
$=\frac{7}{4}$
$(1\ ^*\ 2)\ ^*3=\Big(\frac{1+2}{2}\Big)\ ^*\ 3$
$=\frac{3}{2}\ ^*\ 3$
$=\frac{\frac{3}{2}+3}{2}$
$=\frac{9}{4}$
Therefore, $\exists\text{ a}=1,\text{b}=2,\text{c}=3\in\text{Q}$ such that $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.

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Question 105 Marks

Check the commutativity and associativity of the following binary operations:
'*' on Z defined by a * b = a + b + ab for all a, b ∈ Z.

Answer

Commutativity: Let $\text{a, b}\in\text{Z}.$ Then,
a * b = a + b + ab
= b + a + ba
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Z}$
Thus, * is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z}.$ Then,
a * (b * c) = a * (b + c + bc)
= a + (b + c + bc) + a(b + c + bc)
= a + b + c + bc + ab + ac + abc
(a * b) * c = (a + b + ab) * c
= a + b + ab + c + (a + b + ab)c
= a + b + ab + c + ac + bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Z}$
Thus, * is associative on Z.

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Question 115 Marks

Check the commutativity and associativity of the following binary operations:
'*' on Q defined by a * b = a - b for all a, b ∈ Q.

Answer

Commutativity: Let $\text{a, b}\in\text{Q.}$ Then,
a * b = a - b
b * a = b - a
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q.}$ Then,
a * (b * c) = a * (b - c)
= a - (b - c)
= a - b + c
(a * c) * c = (a - b) * c
= a - b - c
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.

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Question 125 Marks

Consider the binary operation * and o defined by the following tables on set S = {a, b, c, d}.

o	a	b	c	d
a	a	a	a	a
b	a	b	c	d
c	a	c	d	b
d	a	d	b	c

Answer

Commutativity:The table is symmetrical about the leading element. It means that o is commutative on S.
a o (b o c) = a o c
= a
(a o b) o c = a o c
= a
thus,
a o (b o c) = (a o b) o $\text{c }\forall\text{ a, b, c}\in\text{S}$
Associativity:
Therefore, o is associative on S.
Finding identity element:
We observe that the second row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at b.
Implies that x o b = b o x
$=\text{x, }\forall\text{ x}\in\text{S}$
Therefore,
b is the identity element.
Finding inverse elements:
In the first row, we don't have b,
i.e. there does not exist an element x such that a o x = x o a = b.
Therefore,
$a^{-1}$ does not exists.
b o b = b
Implies that $b^{-1}= b$
c o d = b
Implies that $c^{-1} = d$
d o c = b
Implies that $d^{-1} = c$

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Question 135 Marks

Check the commutativity and associativity of the following binary operations:
'*' on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{4}$ for all a, b ∈ Q.

Answer

Commutativity: Let $\text{a, b}\in\text{Q}.$ Then,$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{4}$
$=\frac{\text{ba}}{4}$
$=\text{b}\ ^*\ \text{a}$
Therefore,
$\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\ \forall\ \text{a, b}\in\text{Q}$
Thus '*' is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{4}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{4}\big)}{4}$
$=\frac{\text{abc}}{16}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{ab}}{4}\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{4}\big)\text{c}}{4}$
$=\frac{\text{abc}}{16}$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=(\text{a}\ ^*\ \text{b})\ ^*\ \text{c},\ \forall\ \text{a, b, c}\in\text{Q}$
Thus, '*' is associative on Q.

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Question 145 Marks

Let * be the binary operation on N defined by,
a * b = H.C.F. of a and b.
Does there exist identity for this binary operation one N?

Answer

The binary operation * on N is defined as:
a * b = H.C.F. of a and b.
it is known that:
H.C.F. of a and b = H.C.F. of b and a. $\text{a, b}\in\text{N}$.
Therefore, a * b = b * a
Thus, the operation * is commutative.
For $\text{a, b, c}\in\text{N}$, we have:
(a * b) * c = (H.C.F of a and b) * c = H.C.F. of a, b and c
a * (b * c) = a * (H.C.F. of b and c) = H.C.F. of a, b and c
Therefore, (a * b) * c = a * (b * c)
Thus, the operation * is associative.
Now, an element $\text{e}\in\text{N}$ will be the identity for the operation
if a * e = a = e * a, $\forall\text{ a}\in\text{N}$.
But this relation is not true for any $\text{a}\in\text{N}$.
Thus, the operation * does not have any identity in N.

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