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:

(i) Yes.

Reason: a, b ∈ R. So, |b| ∈ R when b ∈ R

Now multiplication is binary on R

So a|b| ∈ R when a, b ∈ R.

(Le.) a * b ∈ R.

* is a binary operation on R.

(ii) Yes.

Reason: a, b ∈ R and minimum of (a, b) is either a or b but a, b ∈ R.

So, min (a, b) ∈ R.

(Le.) a * b ∈ R.

* is a binary operation on R.

(iii) a* b = \(a \sqrt{b}\) where a, b ∈ R.

No. * is not a binary operation on R.

Reason: a, b ∈ R.

⇒ b can be -ve number also and the square root of a negative number is not real.

So \(\sqrt{b}\) ∉ R even when b ∈ R.

So \(\sqrt{b}\) ∉ R. ie., a * b ∉ R.

* is not a binary operation on R.

Question 2.

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

Solution:

No. * is not a binary operation on Z.

Reason: Since m, n ∈ Z.

So, m, n can be negative also.

Now, if n is negative (Le.) say n = -k where k is +ve.

Similarly, when m is negative then n^{m} ∉ Z.

∴ m * n ∉ Z. ⇒ * is not a binary operation on Z.

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:

Let A = a + \(\sqrt{5}\) b and B = c + \(\sqrt{5}\)d, where a, b, c, d ∈ M.

Now A * B ={a + \(\sqrt{5}\)b)(c + \(\sqrt{5}\)d)

= ac + \(\sqrt{5}\)ad + \(\sqrt{5}\)bc + \(\sqrt{5}\)b\(\sqrt{5}\)d

= (ac + 5bd) + \(\sqrt{5}\)(ad+ bc) ∈ A

Where a, b, c, d ∈ Z

So * is a binary operation.

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

x & x \\

x & x

\end{array}\right): x \in R-\{0\}\right\}\) 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\{\left(\begin{array}{ll}

x & x \\

x & x

\end{array}\right): x \in R-\{0\}\right\}\) 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 an identity, the existence of inverse properties for the operation * on A.

Solution:

(i) Let a, b ∈ A (i.e.) a ≠ ±1 , b ≠ 1

Now a * b = a + b – ab

If a + b – ab = 1 ⇒ a + b – ab – 1 = 0

(i.e.) a(1 – b) – 1(1 – b) = 0

(a – 1)(1 – b) = 0 ⇒ a = 1, b = 1

But a ≠ 1 , b ≠ 1

So (a – 1) (1 – 6) ≠ 1

(i.e.) a * b ∈ A. So * is a binary on A.

To verify the commutative property:

Let a, b ∈ A (i.e.) a ≠ 1 , b ≠ 1

Now a * b = a + b – ab

and b * a = b + a – ba

So a * b = b * a ⇒ * is commutative on A.

To verify the associative property:

Let a, b, c ∈ A (i.e.) a, b, c ≠ 1

To prove the associative property we have to prove that

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

LHS: b * c = b + c – bc = D(say)

So 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 = K(say)

So (a * b) * c = K * c = K + c – Kc

= (a + b – ab) + c – (a + b – ab) c

= a + b – ab + c – ac – bc + abc

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

(ii) To verify the identity property:

Let a ∈ A (a ≠ 1)

If possible let e ∈ A such that

a * e = e * a = a

To find e:

a * e = a

(i.e.) a + e – ae = a

So, e = (≠ 1) ∈ A

(i.e.) Identity property is verified.

To verify the inverse property:

Let a ∈ A (i.e. a ≠ 1)

If possible let a’ ∈ A such that

To find a’:

a * a’ = e

(i.e.) a + a’ – aa’ = 0

⇒ a'(1 – a) = – a

⇒ For every ∈ A there is an inverse a’ ∈ A such that

a* a’ = a’ * a = e

⇒ Inverse property is verified.