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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 + 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 – 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 (Z7 – { [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.
G1: 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.
G2: Associative axiom: Let a, b, c ∈ G.
To prove G2, 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.
G3: Identity axiom: Let a ∈ G. To prove G3 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.
G4 : Inverse axiom: Let a ∈ G. To prove G4, 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:
G1: The elements in the above table are [1], [3], [4], [5] and [9] which are elements of G.
∴ closure axiom is verified.
G2: 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.
G3: 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.
G5: From the table * is commutative i.e., the entries equidistant from the leading diagonal on either side are equal ⇒ a * b = b * a