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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.1

Question 1.

Determine whether * is a binary operation on the sets given below

(i) a * b = a.|b| on R ,

(ii) a * b = min (a, b) on A = {1, 2, 3, 4, 5}

(iii) (a * b) = \(a \sqrt{b}\) is binary on R.

Solution:

(i) a * b = a.|b| on R

a, b ∈ R ⇒ a.|b| ∈ R as a ∈ R and |b| ∈ R.

Hence * is a binary operation on R

(ii) a * b = min (a, b) on A = {1, 2, 3, 4, 5}

A = {1, 2, 3, 4, 5}

a * b = min {(a, b)}

Now, 1, 2 ∈ A ⇒1 * 2 = 1 ∈ A

3, 5 ∈ A ⇒ 3 * 5 = 3 ∈ A

Hence * is a binary operation on A

(iii) (a * b) = a√b is binary on R

a√b ∉ R

∴ * is not a binary on R

Question 2.

Solution:

(m ⊗ n) = mⁿ + n^{m} ∀ m, n ∈ Z

Let us take – 2, 2 ∈ Z

(m ⊗ n) = (-2 ⊗ 2) = (-2)² + (2)^{-2}

= 4 + \(\frac { 1 }{ 4 }\) = \(\frac { 17 }{ 4}\) ∉ Z

∴ ⊗ is a binary operation on R

Question 3.

Let * be defined on R by (a * b) = a + b + ab – 1 .Is * binary on R ? If so, find \(3 *\left(\frac{-7}{15}\right)\)

Solution:

a * b = a + b + ab – 7.

Now when a, b ∈ R, then ab ∈ R also a + b ∈ R.

So, a + b + ab ∈ R.

We know – 7 ∈ R.

So, a + b + ab – 7 ∈ R.

(ie.) a * b ∈ R.

So, * is a binary operation on R.

Question 4.

Let A= {a+ \(\sqrt{5}\)b: a, b ∈ Z}. Check whether the usual multiplication is a binary operation on A.

Solution:

A = {a + √5 b: a, b ∈ Z}

Let 1, 2 ∈ Z ⇒ a + √5 b = 1 + 2 √5 ∈ A

and 2, 3 ∈ Z ⇒ a + √5 b = 2 + 3 √5 ∈ A

Taking usual multiplication as binary] operation

(1+2 √5).(2 + 3√5) = (32 + 7 √5) ∈ A

∴ Usual multiplication can be a binary operation on A.

Question 5.

(i) Define an operation * on Q as follows: a * b = \(\left(\frac{a+b}{2}\right)\); a, b ∈ Q. Examine the closure,

commutative, and associative properties satisfied by * on Q.

(ii) Define an operation * on Q as follows: a*b = \(\left(\frac{a+b}{2}\right)\); a, b ∈ Q. Examine the existence of identity and the existence of inverse for the operation * on Q.

Solution:

(i) 1. Closure property:

Let a, b ∈ Q.

So, the closure property is satisfied.

2. Commutative property:

Let a, b ∈ Q.

(1) = (2) ⇒ Now a * b = b * a

⇒ Commutative property is satisfied.

3. Associative property:

Let a, b, c ∈ Q

To prove associative property we have to prove that a * (b * c) = (a * b) * c

LHS: a * (b * c)

(i.e.) the identity Clement e = a which is not possible.

So, the identity element does not exist and so inverse does not exist.

Question 6.

Fill In the following table so that the binary operation * on A = {a, b, c} is commutative.

Solution:

Given that the binary operation * is Commutative.

To find a * b :

a * b = b * a (∵ * is a Commutative)

Here b * a = c. So a * b = c

To find a *c:

a * c = c * a (∵ * is a Commutative)

c * a = a. (Given)

So a * c = a

To find c * b:

c * b = b * c

Here b * c = a.

So c * b = a

Question 7.

Consider the binary operation * defined on the set A = [a, b, c, d] by the following table:

– Is it commutative and associative?

Solution:

From the table

b * c = b

c * b = d

So, the binary operation is not commutative.

To check whether the given operation is associative.

Let a, b, c ∈ A.

To prove the associative property we have to prove that a * (b * c) = (a * b) * c

From the table,

LHS: b * c = b

So, a * (b * c) = a * b = c ……. (1)

RHS: a * b = c

So, (a * b) * c = c * c = a …… (2)

(1) ≠ (2). So, a * (b * c) ≠ (a * b) * c

∴ The binary operation is not associative.

Question 8.

Solution:

Question 9.

and Let * be the matrix multiplication. Determine whether M is closed under * . If so, examine the commutative and associative properties satisfied by * on M.

and let * be the matrix multiplication. Determine whether M is closed under *. If so, examine the existence of an identity, the existence of inverse properties for the operation * on M.

Solution:

So, inverse property is satisfied.

Question 10.

(i) Let A be Q\{1). Define * on A by x * y = x + y – xy. Is * binary on A ? If so, examine the commutative and associative properties satisfied by * on A.

(ii) Let A be Q\{1}. Define *on A by x * y = x + y – xy. Is * binary on A ? If so, examine the existence of an identity, the existence of inverse properties for the operation * on A.

Solution:

Let A = Q\{1}

Let x, y ∈ A. Then x and y are rational numbers, and x ≠ 1, y ≠ 1

Closure axiom:

Clearly x * y = x + y – xy is a rational number.

But to prove x * y ∈ A we have to prove that

x * y ≠ 1

Suppose x * y = 1

⇒ x + y – xy = 1

i.e., y – xy = 1 – x

y(1 – x) = 1 – x

y = 1 (as x ≠ 1, 1 – x ≠ 0)

This is impossible as y ≠ 1.

Hence our assumption is wrong. Thus x * y ≠ 1 and hence x, y ∈ A.

This shows that * is binary on A.

Commutative axiom:

For any x, y ∈ A, x * y = x + y – xy

and y * x = y + x – yx

⇒ x * y = y * x for all x, y ∈ A

This shows that * is commutative

Associative axiom:

Now

x * (y * z) = x * (y + z – yz)

= x + y + z – yz – x(y + z – yz)

= x + y + z – xy – yz – xz + xyz

Also (x * y) * z = (x + y – xy) * z

= x + y – xy + z – (x + y – xy)z

= x + y + z – xy – yz – xz + xyz

Thus x * (y * z) = (x * y) * z ∀ x, y, z ∈ A

Hence, the associative axiom is true.

(ii) Identity axiom:

Let ‘e’ be the identity element.

By definition of e, x * e = x

By definition of *, we have x * e = x + e – xe

Thus x + e – xe = x ⇒ e (1 – x) = 0

⇒ e = 0, since x ≠ 1

Since e = 0 ∈ A

The identity element is e = 0 ∈ A

Inverse axiom:

Let x^{-1} denote the inverse of x ∈ A

By definition of inverse x * x^{-1} = e = 0

Also, by definition of *, x * x^{-1} = x + x^{-1} – x x^{-1}

Thus x + x^{-1} – x x^{-1} = 0

x^{-1} (1 – x) = -x

∴ x^{-1} = \(\frac { -x }{ 1-x }\) = \(\frac { x }{ x-1 }\)

since x ≠ 1, x – 1 ≠ 0 and

so x^{-1} = \(\frac { x }{ x-1 }\) ∈ A

as \(\frac { x }{ x-1 }\) ≠ 1

∴ Inverse axiom holds good for A.

### Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.1 Additional Problems

Question 1.

Show that the set G = {a + b\(\sqrt{2}\)/ a, b ∈ Q} is an infinite abelian group with respect to Binary operation addition. Satisfies closure, associative, identity and inverse properties.

Solution:

(i) Closure axiom :

Let x, y ∈ G. Then x = a + b\(\sqrt{2}\), y = c + d\(\sqrt{2}\); a, b, c, d ∈ Q.

x + y = (a + b\(\sqrt{2}\)) + (c + d\(\sqrt{2}\)) = (a + c) + (b + d) \(\sqrt{2}\) ∈ G,

Since (a + c) and (b + d) are rational numbers.

∴ G is closed with respect to addition.

(ii) Associative axiom : Since the elements of G are all real numbers, addition is associative.

(iii) Identity axiom : There exists 0 = 0 + 0 \(\sqrt{2}\) ∈ G

such that for all x = a + b\(\sqrt{2}\) ∈ G.

x + 0 = (a + b\(\sqrt{2}\) ) + (0 + 0\(\sqrt{2}\))

= a + b\(\sqrt{2}\) = x

Similarly, we have 0 + x = x. ∴ 0 is the identity element of G and satisfies the identity axiom.

(iv) Inverse axiom: For each x = a + b\(\sqrt{2}\) ∈ G,

there exists -x = (-a) + (-b) \(\sqrt{2}\) ∈ G

such that x + (-x) = (a + b\(\sqrt{2}\)) + ((-a) + (- b)\(\sqrt{2}\))

= (a + (-a)) + (b + (-b)) \(\sqrt{2}\) = 0

Similarly, we have (- x) + x = 0 .

∴ (- a) + (-b)\(\sqrt{2}\) is the inverse of a + b \(\sqrt{2}\) and satisfies the inverse axiom.

(v) Commutative axiom:

x + y = (a + c) + (b + d) \(\sqrt{2}\) = (c + a) + (d + b) \(\sqrt{2}\)

= (c + d\(\sqrt{2}\)) + (a + b\(\sqrt{2}\))

= y + x, for all x, y ∈ G.

∴ The commutative property is true.

∴ (G, +) is an abelian group. Since G is infinite, we see that (G, +) is an infinite abelian group.

Question 2.

Show that (Z_{7} – { [0]}, .7) write to the binary operation multiplication modul07 satisfies closure, associative, identity and inverse properties.

Solution:

Let G = [[1], [2],… [6]]

The Cayley’s table is

From the table:

(i) all the elements of the composition table are the elements of G.

∴ The closure axiom is true.

(ii) multiplication modulo 7 is always associative.

(iii) the identity element is [1] ∈ G and satisfies the identity axiom.

(iv) the inverse of [1] is [1]; [2] is [4]; [3] is [5]; [4] is [2]; [5] is [3] and [6] is [6] and it satisfies the inverse axiom.

Question 3.

Show that the set G of all positive rationals with respect to composition * defined by ab

a* b = \(\frac{a b}{3}\) for all a, b ∈ G satisfies closure, associative, identity and inverse properties.

Solution:

Let G = Set of all positive rational number and * is defined by,

(i) Closure axiom: Let a, b ∈ G

∴ closure axiom is satisfied.

(ii) Associative axiom: Let a, b, c ∈ G.

To prove the associative property, we have to prove that

(1) = (2) ⇒ LHS = RHS i.e., associative axiom is satisfied.

(iii) Identity axiom: Let a ∈ G.

Let e be the identity element.

By the definition, a * e = a

e = 3 ∈ G ⇒ identity axiom is satisfied.

(iv) Inverse axiom: Let a ∈ G and a’ be the inverse of a * a’ = e = 3.

Question 4.

Show that the set G of all rational numbers except – 1 satisfies closure, associative, identity, and inverse property with respect to the operation * given by a * b = a + b + ab for all a, b ∈ G

Solution:

G = [Q, -{-l}]

* is defined by a * b = a + b + ab

To prove G is an abelian group.

G_{1}: Closure axiom: Let a, b ∈ G.

i.e., a and b are rational numbers and a ≠ -1, b ≠ -1.

So, a * b = a + b + ab

If a + b + ab = – 1

⇒ a + b + ab + 1 = 0

i.e., (a + ab) + (b + 1) = 0

a (1 + b) + (b + 1) = 0

i.e., (a + 1)(1 + b) = 0

⇒ a = -1, b = -1

But a ≠ -1, b ≠ -1

⇒ a + b + ab ≠ -1

i.e., a + b + ab ∈ G ∀ a, b ∈ G

⇒ The closure axiom is verified.

G_{2}: Associative axiom: Let a, b, c ∈ G.

To prove G_{2}, we have to prove that,

a * {b * c) = (a * b) * c

LHS:

b * c = b + c + bc = D (say)

a * (b * c) = a * D = a + D + aD

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

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

= a + b + c + ab + bc + ac + abc ……. (1)

RHS:

a * b = a + b + ab = E (say)

∴ (a * b) * c = E * c = E + c + Ec

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

= a + b + ab + c + ac + bc + abc ……. (2)

= a + b + c + ab +be+ ac + abc

(1) = (2) ⇒ Associative axiom is verified.

G_{3}: Identity axiom: Let a ∈ G. To prove G_{3} we have to prove that there exists an element e ∈ G such that a * e = e * a = a.

To find e: a * e = a

i.e., a + e + ae = a

⇒ e(1 + a) = a – a = 0

So, e = 0 ∈ G ⇒ Identity axiom is verified.

G_{4} : Inverse axiom: Let a ∈ G. To prove G_{4}, we have to prove that there exists an element a’ ∈ G such that a * a’ = a’ * a = e.

To find a’: a * a’ = e

i.e., a + a’ + aa’=: 0 {∵ e = 0}

⇒ a'(1 + a) = -a

Thus, inverse axiom is verified.

Question 5.

Show that the set { [1], [3], [4], [5], [9]} under multiplication modulo 11 satisfies closure, associative, identity and inverse properties.

Solution:

G = {[1], [3], [4], [5], [9]}

* is defined by multiplication modulo 11.

To prove G is an abelian group with respect to *

Since we are given a finite number of elements i.e., since the given set is finite, we can frame the multiplication table called Cayley’s table.

The Cayle’s table is as follows:

G_{1}: The elements in the above table are [1], [3], [4], [5] and [9] which are elements of G.

∴ closure axiom is verified.

G_{2}: Consider [3], [4], [5] which are elements of G.

{[3] * [4]} * [5] = [1] * [5] = [5] ……. (1)

[3] * {[4] * [5]} = [3] * [9] = [5] …… (2)

(1) = (2) ⇒ (a * b) * c = a * (b * c) i.e., associative axiom is verified.

G_{3}: The first row elements are the same as that of the given elements in the same order. ie., from the table, the identity element is [1] ∈ G. So identity axiom is verified.

G_{5}: From the table * is commutative i.e., the entries equidistant from the leading diagonal on either side are equal ⇒ a * b = b * a