Elementary Group Theory - Exercise 3.1 : Class 12 Math

$Elementary Group Theory - Exercise 3.1 : Class 12 Math$

Chapter 1: Elementary Group Theory.

Exercise: 3.1

Complete Exercise of Elementary Group Theory - Exercise 3.1 : Class 12 Mathematics 2080 NEB.

Exercise 3.1 is about: Binary Operation on Sets of Integers which includes, Binary Operation (Addition, Multiplication), Properties of Binary Operations ( Addition, Multiplication, Additive Inverse (or negative), The associative properties of Addition and Multiplication, Commutative properties of addition and multiplication, Distributive Properties), and Composition table or Operation table (Binary Arithmetic, Set Relation, Congruent Modulo).

Read: NEB Class 12 Mathematics All Chapter Exercise

**1. A binary operation * is defined on the set of integers by**

a) m*n=m+n
b) m*n=m-n
c) m*n=mn+m+n
Find m*n if
i) m=3 and n=5
ii) m=2 and n=-5

a. Soln:

(i) m = 3, n = 5

Given, m * n = m + n

So, 3 * 5 = 3 + 5 = 8

(ii) m = 2, n = - 5

So, 2 * (-5) = 2 + (-5) = 2 – 5 = - 3.

b. Soln:

(i) m = 3, n = 5

Given, m * n = m – n

So, 2 * (-5) = 2 + (-5) = 2 – 5 = - 3.

(ii) m = 2, n = - 5

Given, m + n = m – n.

So, 2 * (-5) = 2 – (-5) = 2 + 5 = 7.

c. Soln:

(i) m = 3, n = 5

Given, m * n = nm + m + n

So, 3 * 5 = 3 * 5 + 3 + 5 = 23.

(ii) m = 2, n = - 5

Given, m * n = mn + m + n

So, 2 * (-5) = 2 * (-5) + 2 + (-5) = - 10 + 2 – 5 = - 13.

**2. S is a given set and a, b ∊ S. Prove that the operation * defined by**

a. a*b=a+b on S = {-1, 0, 1} is not a binary operation.

Soln:

S = {-1,0,1}

Consider 1,1 ԑ S.

Here, a * b = a + b

So, 1 * 1 = 1 + 1 = 2 ∉ S.

Thus, a * b = a + b Is not a binary operation on S = {-1,0,1}

b. a*b=ab on S = {1, 2, 4} is not a binary operation.

Soln:

S = {1,2,4}

Consider 2,4 ԑ S.

Here, a * b = a.b

So, 2 * 4 = 2 * 4 = 8 ∉ S.

Thus, a * b = ab is not a binary operation on S = {1,2,4}

c. a*b=a+b on S = {2, 4, 6, 8, 10, ...} is a binary operation.

Soln:

S = {2,4,6,8,10,…}

Here, a * b = a + b

We know that the addition of two even numbers is always an even number which belong to the set S.

So, a * b = a + b is a binary operation on the set S = {2,4,6,8,10,….}

d. a*b=a-b on S = set of integers is a binary operation.

Soln:

S = Set of integers.

Given, a * b = a – b

Since, the difference of two integers always yields an integer.

Thus, for all,a,b ԑ S, a * b = a – b ԑ S.

So, a * b = a – b is a binary operation on the set ‘S’ of integers.

3. Let S = {-1, 1} and x denotes the usual operation of multiplication. Represent it by Cayley's table. Show that the multiplication is a binary operation on S.

Soln:

Cayley’s table.

X	-1	1
-1	1	-1
1	-1	1

From the above table,

(-1) * (-1) = 1 ԑ S, (-1) * 1 = - 1 ԑ S.

1 * (-1) = - 1 ԑ S, 1 * 1 = 1 ԑ S.

So, multiplication is a binary operation on set a S.

4. Examine the set of positive integers for

Soln:

The set of positive integers is denoted by Z

a. closure property under addition.

Any m, n ԑ Z+

m + n ԑ Z+

So, Z+

is closed under addition.

b. commutative property under multiplication.

Any m,n ԑ Z+

mn = nm

So, Z+

is commutative under multiplication.

c. associative property under subtraction.

Consider, 2,3,4 ԑ Z+.

2 – (3 – 4) = 2 – (-1) = 2 + 1 = 3.

And (2 – 3) – 4 = - 1 – 4 = - 5

Thus, 2 – (3 – 4) ≠ (2 – 3) – 4.

SO, Z+

is not associative under subtraction.

5. Test the closure, associative and commutative properties for each of the following cases.

Soln:

a. the operation defined by m*n = 1/2 (m-n) on Z, m, n є Z.

For closure,

Consider 1,2 ԑ Z.

Then , m * n = 1/2(m – n)

So, 1 * 2 = $\frac{1}{2}$ (1 – 2) = $- \frac{1}{2}$ ∉ Z.

So, Z is not closed.

For Associative,

Consider 2,4,8 ԑ Z.

= 2 * (4 * 8) = 2 * {12(4−8)}{12(4−8)}

= 2 * (-2) = 1212 {2 – (-2)} = 2

And (2 * 4) * 8 = {12(2−4)}{12(2−4)} * 8

= (-1) * 8 = 1212 {(-1) – 8} = 1212 (-9) = −92−92∉ Z.

Hence, the operation * defined on Z is now associative,

For Commutative,

For, m , n ԑ Z, m * n = 1212(m – n)

= −12−12(n – m) = - (n * m).

So, the operation * is not commutative on Z.

b. the operation defined by m*n = n on Z, m, n є Z.

For closure,

For, m,n ԑ Z, m * n = n ԑ Z.

So, the operation ‘*’ is closed on Z.

For associative,

Since, for m,n and p ԑ Z.

(m * n) * p = n * p = p ԑ Z.

And m * (n * p) = m * p = p ԑ Z.

So, (m * n) * p = m * (n * p)

So, the operation ‘*’ is associative.

For commutative.

Since, for m,n ԑ Z, m * n = n ԑ Z.

And n * m = m ԑ Z.

But m ≠ n; so the operation * is not commutative on Z.

c. the operation defined by m*n = m + n + 1, Z, m, n є Z.

For closure

Since, for m,n ԑ Z, m * n = m + n + 1 ԑ Z.

So, * is closed on Z,

For associative

Since, for n,m,p ԑ Z.

(m * n) * p = (m + n + 1) * p

= (m + n + 1) + p + 1 = m + n + p + 2.

And m * (n * p) = m * (m + p + 1)

= m + n (n + p + 1) + 1 = m + n + p + 2.

So, (m * n) * p = m * (n * p), for all m,n,p ԑ Z.

So, * is associative on Z.

For commutative.

For m,n ԑ Z.

m * n = m + n + 1.

And n * m = n + m + 1 = m + n + 1

So, m * n = n * m, for all m,n ԑ Z.

So, * is commutative on Z.