Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

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

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

Question 2.
On Z, define ⊗ by (m ⊗ n) =mⁿ + nm: ∀m, n ∈ Z Is ⊗ binary on Z?
Solution:
(m ⊗ n) = mⁿ + nm ∀ 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
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 1

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

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:
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 2
so, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

(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.
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 3
Solution:
Given * is commutative on A = {a, b, c}
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 4
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

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

Question 7.
Consider the binary operation * defined on the set A = {a, b, c, d} by the following table:
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 5
Solution:
* is defined on the set A = {a, b, c, d} Given table
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 6
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
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 7
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
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 8
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 9

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

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:
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 10
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:
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 11
Also, we can show that EA = A
Hence E is the identity element in M
Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1 12

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

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.

Samacheer Kalvi 12th Maths Guide Chapter 12 Discrete Mathematics Ex 12.1

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