Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 12 Discrete Mathematics Ex 12.1 Textbook Questions and Answers, Notes.

## 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√b is binary on R

Solution:

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.

On Z, define ⊗ by (m ⊗ n) =mⁿ + n^{m}: ∀m, n ∈ Z Is ⊗ binary on Z?

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 – 7. Is * binary on R? If so, find 3 * (\(\frac { -7 }{ 15}\))

Solution:

(a * b) = a + b + ab – 7 ∀ a, b ∈ R

If a ∈ R, b ∈ R then ab ∈ R

∴ (a * b) = a + b + ab – 7 ∈ R

For example, let 1, 2 ∈ R

(1 * 2) = 1 + 2 + (1)(2) – 7

= -2 ∈ R

∴ * is a binary operation on R

Question 4.

Let A = {a + √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 = (\(\frac { a+b }{ 2}\)); a, b ∈ Q. Examine the closure, commutative and associate properties satisfied by * on Q.

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

Solution:

so, a * (b * c) ≠ (a * b) * c

Hence, the binary operation * is not associative.

(ii) a * b = (\(\frac { a+b }{ 2}\)); a, b ∈ Q

For identity, a * e = e * a = a

Now; a * e = a

\(\frac { a+e }{ 2}\) = a

a + e = 2a

e = 2a – a = a

Which is not possible

∴ Identity does not exist and hence the 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 * is commutative on A = {a, b, c}

From the table, it is given that b * a = c

⇒ a * b = c

as * is commutative

c * a = a = a * c and

b * c = a = c * b

Question 7.

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

Solution:

* is defined on the set A = {a, b, c, d} Given table

From the table a * b = c and b * a = d

⇒ a * b ≠ b * a

∴ * is not commutative.

Now(a * b) * c = c * c = a

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

⇒ (a * b) * c ≠ a * (b * c)

∴ * is not associative

Question 8.

Let

be any three boolean matrices of the same type. Find

(i) A v B

(ii) A ∧ B

(iii) (A v B) ∧ C

(iv) (A ∧ B) v C.

Solution:

Given boolean matrices

Question 9.

(i) Let M = { \(\left(\begin{array}{ll}

x & x \\

x & x

\end{array}\right)\) : x ∈ R – {0} } and let * be the matrix multiplication. Determine whether M is closed under *. If so, examine the commutative and associative properties satisfied by * on M.

(ii) Let M = { \(\left(\begin{array}{ll}

x & x \\

x & x

\end{array}\right)\) : x ∈ R – {0} } and let * be the matrix multiplication. Determine whether M is closed under *. If so, examine the existence of identity, existence of inverse properties for the operation * on M

Solution:

Since x ≠ 0, y ≠ 0, we see that 2xy ≠ 0 and so AB ∈ M. This shows that M is closed under matrix multiplication.

Commutative axiom:

AB = \(\left(\begin{array}{ll}

2xy & 2xy \\

2xy & 2xy

\end{array}\right)\)

= BA for all A, B ∈ M

Here, Matrix multiplication is commutative (though in general, matrix multiplication is not commutative)

Associative axiom:

Since matrix multiplication is associative, this axiom holds goods for M.

(ii) Identity axiom:

Also, we can show that EA = A

Hence E is the identity element in M

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 identity, 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.