## Question 1:Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.(i) On Z+, define * by a * b = a − b(ii) On Z+, define * by a * b = ab(iii) On R, define * by a * b = ab2(iv) On Z+, define * by a * b = |a − b|(v) On Z+, define * by a * b = a

(i) On Z+, * is defined by * b = a − b.
It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2
= −1 ∉ Z+.

(ii) On Z+, * is defined by a * b = ab.
It is seen that for each ab ∈ Z+, there is a unique element ab in Z+.
This means that * carries each pair (ab) to a unique element * b ab in Z+.
Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab2.
It is seen that for each ab ∈ R, there is a unique element ab2 in R.
This means that * carries each pair (ab) to a unique element * b abin R.
Therefore, * is a binary operation.

(iv) On Z+, * is defined by * b = |a − b|.
It is seen that for each ab ∈ Z+, there is a unique element |a − b| in Z+
This means that * carries each pair (ab) to a unique element * b
|a − b| in Z+.
Therefore, * is a binary operation.

(v) On Z+, * is defined by a * b = a.
* carries each pair (ab) to a unique element * b a in Z+.
Therefore, * is a binary operation.

## Question 2:For each binary operation * defined below, determine whether * is commutative or associative.(i) On Z, define a * b = a − b(ii) On Q, define a * b = ab + 1(iii) On Q, define a * b (iv) On Z+, define a * b = 2ab(v) On Z+, define a * b = ab(vi) On R − {−1}, define

(i) On Z, * is defined by a * b = a − b.
It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.
∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z
Hence, the operation * is not commutative.
Also we have:
(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4
1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z
Hence, the operation * is not associative.

(ii) On Q, * is defined by * b = ab + 1.
It is known that:
ab = ba &mnForE; a, b ∈ Q
⇒ ab + 1 = ba + 1 &mnForE; a, b ∈ Q
⇒ * b = * b &mnForE; a, b ∈ Q
Therefore, the operation * is commutative.
It can be observed that:
(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10
1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q
Therefore, the operation * is not associative.

(iii) On Q, * is defined by * b
It is known that:
ab = ba &mnForE; a, b ∈ Q
⇒ &mnForE; a, b ∈ Q
⇒ * b = * a &mnForE; a, b ∈ Q
Therefore, the operation * is commutative.
For all a, b, c ∈ Q, we have:
Therefore, the operation * is associative.

(iv) On Z+, * is defined by * b = 2ab.
It is known that:
ab = ba &mnForE; a, b ∈ Z+
⇒ 2ab = 2ba &mnForE; a, b ∈ Z+
⇒ * b = * a &mnForE; a, b ∈ Z+
Therefore, the operation * is commutative.
It can be observed that:
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+
Therefore, the operation * is not associative.

(v) On Z+, * is defined by * b = ab.
It can be observed that:
and
∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z+
Therefore, the operation * is not commutative.
It can also be observed that:
∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+
Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by
It can be observed that and
∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ − {−1}
Therefore, the operation * is not commutative.
It can also be observed that:
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ − {−1}
Therefore, the operation * is not associative.

## Question 3:Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by a ∨b = min {a, b}. Write the operation table of the operation∨.

The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as  b = min {ab}
&mnForE; ab ∈ {1, 2, 3, 4, 5}.
Thus, the operation table for the given operation ∨ can be given as:

 ∨ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

## Question 4:Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table. (i) Compute (2 * 3) * 4 and 2 * (3 * 4)(ii) Is * commutative?(iii) Compute (2 * 3) * (4 * 5).(Hint: use the following table)

 * 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5

(i) (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have * b = b * a. Therefore, the operation * is commutative.

(iii) (2 * 3) = 1 and (4 * 5) = 1
∴(2 * 3) * (4 * 5) = 1 * 1 = 1

## Question 5:Let*′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as *′ b = H.C.F of a and b.
The operation table for the operation *′ can be given as:

 *′ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5

We observe that the operation tables for the operations * and *′ are the same.
Thus, the operation *′ is same as the operation*.

## Question 6:Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find(i) 5 * 7, 20 * 16 (ii) Is * commutative?(iii) Is * associative? (iv) Find the identity of * in N(v) Which elements of N are invertible for the operation *?

The binary operation * on N is defined as * b = L.C.M. of a and b.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35
20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:
L.C.M of a and b = L.C.M of b and a &mnForE; a, b ∈ N
a * b = * a
Thus, the operation * is commutative.

(iii) For a, b∈ N, we have:
(* b) * c = (L.C.M of a and b) * c = LCM of ab, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of ab, and c
∴(* b) * c = a * (* c
Thus, the operation * is associative.

(iv) It is known that:
L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N
⇒ a * 1 = a = 1 * a &mnForE; a ∈ N
Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that * b = e = b * a.
Here, e = 1
This means that:
L.C.M of a and b = 1 = L.C.M of b and a

This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.

## Question 13:Consider a binary operation * on N defined as a * b = a3 + b3. Choose the correct answer.(A) Is * both associative and commutative?(B) Is * commutative but not associative?(C) Is * associative but not commutative?(D) Is * neither commutative nor associative?

#### On N, the operation * is defined as a * b = a3 + b3. For, a, b, ∈ N, we have: a * b = a3 + b3 = b3 + a3 = b * a[Addition is commutative in N] Therefore, the operation * is commutative. It can be observed that: ∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N Therefore, the operation * is not associative. Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.

PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

# 2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

# HSCMaharashtraBoardPapers2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

# SSCMaharashtraBoardPapers2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

Geography Model Set 1 2020-2021

MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates

NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!