CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(c)
Question 1.
Show that the operation ∗ given by x ∗ y = x + y – xy is a binary operation on Z, Q and R but not on N.
Answer:
The operation ∗ given by
x ∗ y = x + y – xy
Clearly for all x, y ∈ Z
x + y – xy ∈ Z
⇒ x ∗ y ∈ Z
∴ ∗ is a binary operation on Z.
For all x, y ∈ Q
x + y – xy ∈ Q
⇒ x ∗ y ∈ Q
⇒ ∗ is a binary operation on Q for all x, y, ∈ R.
x + y – xy ∈ R
⇒ x ∗ y ∈ R
⇒ ∗ is a binary operation on R
Again 3, 4 ∈ N.
3 + 4 – 3 x 4 = 7 – 12 = – 5 ∉ N
i.e., x, y ∈ N
≠ x ∗ y ∈ N
∴ ∗ is not a binary operation on N.
Question 2.
Determine whether the following operations as defined by ∗ are binary operations on the sets specified in each case. Give reasons if it is not a binary operation.
(i) a ∗ b = 2a + 3b on Z.
(ii) a ∗ b = ma – nb on Q+ where m and n ∈ N.
(iii) a ∗ b = a + b (mod 7) on {0, 1, 2, 3, 4, 5, 6}
(iv) a ∗ b = min {a, b} on N.
(v) a ∗ b = GCD {a, b} on N.
(vi) a ∗ b = LCM {a, b} on N.
(vii) a ∗ b = LCM {a, b} on {0, 1, 2, 3, 4……, 10}
(viii) a ∗ b = \(\sqrt{a^2+b^2}\) on Q+
(ix) a ∗ b =a × b (mod 5) on {0, 1, 2, 3, 4}.
(x) a ∗ b = a2 + b2 on N.
(xi) a ∗ b = a + b – ab on R – {1}.
Answer:
(i) For all a, b ∈ Z
2a + 3b ∈ Z
⇒ a ∗ b ∈ Z
∗ is a binary operation on Z.
(ii) Let a = 1, b = 2
m = 1, n = 3
ma – nb = 1 – 6 = – 5 ∉ Q+
∴ a, b ∈ Q+ ≠ a ∗ b ∈ Q+
⇒ ∗ is not a binary operation on Q+
(iii) a ∗ b = a + b (mod 7) ∈ {0, 1, 2, 3, 4, 5, 6}
for a, b ∈ 7
∗ is a binary operation on the given set.
(iv) a, b ∈ N ⇒ min {a, b} ∈ N
∴ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.
(v) for all a, b ∈ N, GCD [a, b] ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.
(vi) for all a, b ∈ N, LCM {a, b} ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.
(vii) Let A = {0, 1, 2, ….. 10}
4, 5 ∈ A but 4 ∗ 5 = LCM {4, 5}
= 20 ∉ A
⇒ ∗ is not a binary operation on A.
(viii) for all a, b ∈ Q+
a ∗ b = \(\sqrt{a^2+b^2}\) ∉ Q+
⇒ ∗ is not a binary operation on Q+.
(ix) For all a, b ∈ {0, 1, 2, 3, 4}
a ∗ b = a × b (mod 5) ∈ {0, 1, 2, 3, 4}
∴ ∗ is a binary operation on the given set.
(x) for all a, b ∈ N, a * b = a2 + b2 ∈ N
∴ ∗ is a binary operation on N.
(xi) For all a, b ∈ R – {1}
a ∗ b = a + b – ab ∈ R – {1}
∴ ∗ is a binary operation on R – {1}
Question 3.
In case ∗ is a binary operation in Q2 above, test whether it is (i) associative
(ii) commutative, Test further if the identity element exists and the inverse element for any element of the respective set exists.
Answer:
(i) On Z the binary operation is
a ∗ b = 2a + 3b
Commutative:
b ∗ a = 2b + 3a ≠ a ∗ b
∴ ∗ is not commutative.
Associative:
(a ∗ b) ∗ c = (2a + 3b) ∗ c
= 2 (2a + 3b) + 3c
= 4a + 6b + 3c
a ∗ (b ∗ c) = a ∗ (2b + 3c)
= 2a + 3 (2b + 3c)
= 2a + 6b + 9c
As (a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∗ is not associative.
Existance of identity:
Let e is the identity
∴ e ∗ a = a
⇒ 2e + 3a = a
⇒ e = -2a / 2 = -a
which depends on a.
∴ Identity element does not exist.
(iii) A = (0, 1, 2, 3, 4, 5, 6}
Commutative:
a ∗ b = a + b (mod 7)
= The remainder obtained when a + b is divided by 7.
b ∗ a = b + a (mod 7) = a + b (mod 7)
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = {a + b (mod 7)} ∗ c
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
a ∗ (b ∗ c) = a ∗ {b + c (mod 7)}
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
∴ (a ∗ b) ∗ c = a + (b ∗ c)
∴ ∗ is associative.
Existance of identity:
Let e is the identity
⇒ e ∗ a = a ∗ e = a
⇒ e + a mod 7 = a
⇒ e = 0
∴ 0 is the identity.
Existance of inverse:
Let a-1 = the inverse of a
⇒ a ∗ a-1 = a-1 ∗ a = e = 0
⇒ a + a-1 (mod 7) = 0
⇒ a + a-1 is divisible by 7.
1-1 = 6, 6-1 = 1
2-1 = 5, 5-1 = 2
3-1 = 4, 4-1 = 3
(iv) a ∗ b = min {a, b} on N.
Commutative:
a ∗ b = min {a, b}
b ∗ a – min {b, a} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = min {a, b} ∗ c
= min {a, b, c}
a ∗ (b ∗ c)= a ∗ min {b, c}
= min {a, b, c}
⇒ a ∗ (b ∗ c) = (a ∗ b) ∗ c
∴ ∗ is associative.
Existance of Identity:
Let e is the identity
∴ For all a ∈ N
e ∗ a = a ∗ e = a
⇒ min {e, a} = a
No such element exists in N.
∴ ∗ has no identity element on N.
(v) a ∗ b = GCD {a, b} on N.
b ∗ a = GCD {b, a} = GCD {a, b} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = GCD {a, b} ∗ c
= GCD {a, b, c}
a ∗ (b ∗ c) = a ∗ GCD {b, c}
= GCD {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.
Existance of Identity:
Let e is the identity
∴ a ∗ e = e ∗ a = a
⇒ GCD {e, a} = a
No such element exists in N
⇒ ∗ has no indentity element.
(vi) a ∗ b = LCM {a, b} on N
Commutative:
a ∗ b = LCM {a, b}
= LCM {b, a}
= b ∗ a
∴ ∗ is commutative.
Associative
(a ∗ b) ∗ c = LCM {a, b} ∗ c
= LCM {a, b, c}
a ∗ (b ∗ c) =» a ∗ LCM {b, c}
= LCM {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
∴ ∗ is associative.
Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ LCM {e, a} = a
⇒ e – 1
∴ 1 is the identity element.
Existance of inverse:
Let a-1 is the inverse of a
⇒ a * a-1 = e = 1
⇒ LCM [a, a-1} = 1
a = a-1 = 1
Only 1 is invertible with 1-1 = 1.
(ix) a ∗ b = a × (mod 5) on {0, 1, 2, 3, 4}
Commutative:
a ∗ b = a x b (mod 5)
= Remainder on dividing a x b by 5
= Remainder on dividing b x a by 5
= b x a (mod 5)
= b x a
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c=a x b (mod 5) ∗ c
= a x b x c (mod 5)
a and a ∗ {b ∗ c} = a ∗ {b x c (mod 5)}
= a x b x c (mod 5)
∴ (a ∗ b) ∗ c -=a ∗ (b ∗ c)
⇒ ∗ is associative.
Existance of identity:
Let e is the identity
∴ For all a ∈ {0, 1, 2, 3, 4}
a ∗ a = e ∗ a = a
a × e (mod 5) = a
⇒ e = 1
∴ 1 is the identity element.
Existance of inverse:
Let a-1 is the inverse of a
∴ a ∗ a-1= a-1 ∗ a = e = 1
⇒ a x a-1 (mod 5) = 1
⇒ 1-1 = 1
2-1 = 3, 3-1 = 2, 4-1 = 4
0 has no inverse.
(x) a ∗ b = a2 + b2 on N.
Commutative:
a ∗ b = a2 + b2
b ∗ a = b2 + a2 = a2 + b2 = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = (a2 + b2) ∗ c
= (a2 + b2)2 + c2
a ∗ (b ∗ c) = a ∗ (b2 + c2)
= a2 + (a2 + b2)2
(a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∴ ∗ is not associative.
Existance of Identity:
Let e is the identity
a ∗ e = e ∗ a = a
⇒ a2 + e2 = a
⇒ e = \( \sqrt{a-a^2}\) which depends on a
∴ Identity does not exist.
(xi) a ∗ b = a + b – ab on R – {1}
Commutative:
a ∗ b = a + b – ab
b ∗ a = b + a – ba
a ∗ b = b ∗ a
∴ ∗ is commutative.
Associative:
a ∗ (b ∗ c) = a ∗ (b + c – bc)
= a + (b + c – bc) – a (b + c – bc)
= a + b + c – bc – ab – ac + abc
(a ∗ b) ∗ c = (a + b – ab) ∗ c
= a + b – ab + c – (a + b – ab) c
= a + b + c – ab – bc – ca + abc
∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.
Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ a + e – ae = a
⇒ e (1 – a) = 0
⇒ e = 0 (∵ a ≠ 1)
∴ 0 is the identity.
Existance of inverse:
Let a-1 is the inverse of a
⇒ a ∗ a-1 = a-1 ∗ a = e
⇒ a + a-1 – aa-1 = 0
⇒ a-1 (1 – a) = – a
⇒ a-1 = \(\frac{a}{a-1}\) for a ∈ R – {1}
Question 4.
Construct the composition table/multiplication table for the binary operation ∗ defined on {0, 1, 2, 3, 4} by a ∗ b = a × b {mod 5). Find the identity element if any. Also find the inverse elements of 2 and 4.
[This operation is called multiplication moduls 5 and denoted by x5. In general, on a finite subset of N, xm denotes the operation of multiplication modulo m where m is a fixed positive integer].
Answer:
A = {0. 1, 2, 3, 4}
a ∗ b = a × b mod 5
| ∗ | 0 | 1 | 2 | 3 | 4 |
| 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 |
| 2 | 0 | 2 | 4 | 1 | 3 |
| 3 | 0 | 3 | 1 | 4 | 2 |
| 4 | 0 | 4 | 3 | 2 | 1 |
As 3rd row is identical to the first row we have 1 is the identity clearly 2-1 = 3 and 4-1 = 4.