TOPICS
Exercise - 1.4

Question-1 :-  Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
(i) On Z+, define ∗ by a ∗ b = a – b
(ii) On Z+, define ∗ by a ∗ b = ab
(iii) On R, define ∗ by a ∗ b = ab²
(iv) On Z+, define ∗ by a ∗ b = |a – b|
(v) On Z+, define ∗ by a ∗ b = a

Solution :-
(i)	On  Z+, * is defined by a * b = a − b.
    e.g., (a, b) = (2, 3) under * is a * b = a − b ⇒ 2 * 3 = 2 − 3 = −1 ∉ Z+.
    It is not a binary operation. 
    
(ii) On Z+, * is defined by a * b = ab.
    It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+.
    This means that * carries each pair (a, b) to a unique element a * b = ab in Z+.
    e.g., (a, b) = (2, 4) under * is a * b = ab ⇒ 2 * 4 = 8 ∈ Z+.
    Therefore, * is a binary operation.
    
(iii) On R, * is defined by a * b = ab².
    It is seen that for each a, b ∈ R, there is a unique element ab² in R .
    This means that * carries each pair (a, b) to a unique element a * b = ab² in R .
    e.g., (a, b) = (1.2, 3) under * is a * b = ab² ⇒ 1.2 * 3 = 3.6 ∈ R.
    Therefore, * is a binary operation.
    
(iv) On Z+, * is defined by a * b = |a − b|.
    It is seen that for each a, b ∈ Z+, there is a unique element |a − b| in Z+.
    This means that * carries each pair (a, b) to a unique element a * b = |a − b| in Z+.
    e.g., (a, b) = (2, 3) under * is a * b = |a − b| ⇒ 2 * 3 = |2 − 3| = 1 ∈ Z+.
    Therefore, * is a binary operation.
    
(v)	 On Z+, * is defined by a * b = a.
     It is seen that for each a, b ∈ Z+, there is a unique element a in Z+.
     This means that * carries each pair (a, b) to a unique element a * b = a in Z+. 
     e.g., (a, b) = (3, 4) under * is a * b = a ⇒ 3 * 4 = 3 ∈ Z+.
     Therefore, * is a binary operation.
    

Question-2 :-  For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(i) On Z, define a ∗ b = a – b
(ii) On Q, define a ∗ b = ab + 1
(iii) On Q, define a ∗ b = ab/2
(iv) On Z+, define a ∗ b = 2ab
(v) On Z+, define a ∗ b = ab
(vi) On R – {– 1}, define a ∗ b = a/(b+1)

Solution :-
(i) On Z, define a ∗ b = a – b 
    For commutative:        ∴ [a * b = b * a]
    a ∗ b = a – b
    b ∗ a = b – a
    e.g., (a, b) = (2, 4) ∈ Z 
    2 * 4 = 2 - 4 = -2
    4 * 2 = 4 - 2 = 2
    Hence, the operation * is not commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a ∗ b) * c = (a–b) - c
    a * (b * c) = a - (b-c)
    e.g., (a, b, c) = (2, 4, 6) ∈ Z
    (2 ∗ 4) * 6 = (2–4) - 6 
                = -2 - 6 
                = -8
    2 * (4 * 6) = 2 - (4-6) 
                = 2 + 2 
                = 4
    Hence, the operation * is not associative.
    Therefore, it is neither commutative nor associative.
(ii) On Q, define a ∗ b = ab + 1
    For commutative:        ∴ [a * b = b * a]
    a ∗ b = ab + 1
    b ∗ a = ba + 1
    e.g., (a, b) = (1/2, 2/3) ∈ Q
    1/2 * 2/3 = 1/2 . 2/3 + 1 = 4/3
    2/3 * 1/2 = 2/3 . 1/2 + 1 = 4/3
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a ∗ b) * c = (ab + 1)c + 1
    a * (b * c) = a(bc + 1) + 1
    e.g., (a, b, c) = (1/2, 2/3, 3/4) ∈ Q 
    (1/2 * 2/3) * 3/4 = (1/2 . 2/3 + 1) x 3/4 + 1
                      = 4/3 . 3/4 + 1 
                      = 1 + 1 
                      = 2
    1/2 * (2/3 * 3/4) = 1/2 x (2/3 . 3/4 + 1) + 1 
                      = 1/2 . 3/2 + 1 
                      = 3/4 + 1 
                      = 7/3
    Hence, the operation * is not associative.
    Therefore, it is commutative but not associative.
 
(iii) On Q, define a ∗ b = ab/2
    For commutative:        ∴ [a * b = b * a]
    a ∗ b = ab/2
    b ∗ a = ba/2
    e.g., (a, b) = (2/3, 4/3) ∈ Q
    2/3 * 4/3 = (2/3 x 4/3)/2 = 8/18 
    4/3 * 2/3 = (2/3 x 4/3)/2 = 8/18 
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a ∗ b) * c = ((ab/2) x c)/2 = abc/4
    a * (b * c) = (a x (bc/2))/2 = abc/4
    e.g., (a, b, c) = (2/3, 4/3, 1/3) ∈ Q
    (2/3 ∗ 4/3) * 1/3 = (2/3 x 4/3 x 1/3)/4 
                      = 8/108 
    2/3 * (4/3 * 1/3) = (2/3 x 4/3 x 1/3)/4 
                      = 8/108 
    Hence, the operation * is associative.
    Therefore, it is commutative and associative.
(iv) On Z+, define a ∗ b = 2ab 
    For commutative:        ∴ [a * b = b * a]
    a ∗ b = 2ab
    b ∗ a = 2ba
    e.g., (a, b) = (2, 1) ∈ Z+
    2 * 1 = 22 = 4
    1 * 2 = 22 = 4
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a ∗ b) * c = 2ab * c
    a * (b * c) = a * 2bc
    e.g., (a, b, c) = (2, 1, 3) ∈ Z+
    (2 ∗ 1) * 3 = 22 * 3 
                = 4 * 3 
                = 212
    2 * (1 * 3) = 2 * 23 
                = 2 * 8 
                = 216
    Hence, the operation * is not associative.
    Therefore, it is commutative but not associative.
(v) On Z+, define a ∗ b = ab
    For commutative:        ∴ [a * b = b * a]
    a ∗ b = ab
    b ∗ a = ba
    e.g., (a, b) = (2, 1) ∈ Z+
    2 * 1 = 2¹ = 2
    1 * 2 = 1² = 1
    Hence, the operation * is not commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a ∗ b) * c = ab * c
    a * (b * c) = a * ba
    e.g., (a, b, c) = (2, 1, 3) ∈ Z+
    (2 ∗ 1) * 3 = 2¹ * 3 
                = 2 * 3 
                = 2³ 
                = 8
    2 * (1 * 3) = 2 * 1³ 
                = 2 * 1 
                = 2¹ 
                = 2
    Hence, the operation * is not associative.
    Therefore, it is neither commutative nor associative.
(vi) On R – {– 1}, define a ∗ b = a/(b+1)   
     For commutative:        ∴ [a * b = b * a]
    a ∗ b = a/(b+1)
    b ∗ a = b/(a+1)
    e.g., (a, b) = (3, -2) ∈ R – {– 1}
    3 * -2 = 3/(-2+1) = -3
    -2 * 3 = -2/(3+1) = -2/4 = -1/2
    Hence, the operation * is not commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a ∗ b) * c = a/(b+1) * c
    a * (b * c) = a * b/(c+1)
    e.g., (a, b, c) = (2, 1, 3) ∈ R – {– 1} 
    (2 ∗ 1) * 3 = 2/(1+1) * 3 
                = 1 * 3 
                = 1/(3+1) 
                = 1/4
    2 * (1 * 3) = 2 * 1/(3+1) 
                = 2 * 1/4 
                = 2/(1/4 + 1) 
                = 8/5 
    Hence, the operation * is not associative.
    Therefore, it is neither commutative nor associative. 
    

Question-3 :-  Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}. Write the operation table of the operation ∧ .

Solution :-
     Let A = {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}.
     binary
    

Question-4 :-  Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
(ii) Is ∗ commutative?
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
(Hint: use the following table) binary

Solution :-
(i) (2 * 3) * 4 = 1 * 4 = 1
     2 * (3 * 4) = 2 * 1 = 1
(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have a * b = b * a. 
     Therefore, the operation * is commutative.
(iii) (2 * 3) = 1 and (4 * 5) = 1
     ∴(2 * 3) * (4 * 5) = 1 * 1 = 1
    

Question-5 :- Let ∗′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a ∗′ b = H.C.F. of a and b. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer.

Solution :-
     The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b. 
     The operation table for the operation *′ can be given as:
     binary
    

Question-6 :-  Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find
(i) 5 * 7, 20 * 16
(ii) Is * commutative?
(iii) Is * associative?
(iv) Find the identity of * in
(v) Which elements of N are invertible for the operation *?

Solution :-
    The binary operation * on N is defined as a * b = L.C.M. of a and b. 
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
    20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:
    L.C.M of a and b = L.C.M of b and a where a, b ∈ N.
    ∴a * b = b * a
    Thus, the operation * is commutative.

(iii) For a, b, c ∈ N, we have:
    (a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c
    a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
    ∴(a * b) * c = a * (b * c)
    Thus, the operation * is associative.

(iv) It is known that:
    L.C.M. of a and 1 = a = L.C.M. 1 and a where a ∈ N  
 ⇒ a * 1 = a = 1 * a, where a ∈ N 
    Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation * if 
    there exists an element b in N, such that a * b = e = b * a.
    Here, e = 1
    This means that:
    L.C.M of a and b = 1 = L.C.M of b and a
    This case is possible only when a and b are equal to 1.
    Thus, 1 is the only invertible element of N with respect to the operation *.
    

Question-7 :- Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.

Solution :-
     The operation * on the set A = {1, 2, 3, 4, 5} is defined as
     a * b = L.C.M. of a and b.
     Then, the operation table for the given operation * can be given as:
     binary
     It can be observed from the obtained table that:
     3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A
     3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A
     Hence, the given operation * is not a binary operation.
    

Question-8 :-  Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N ?

Solution :-
    The binary operation * on N is defined as:
    a * b = H.C.F. of a and b
    It is known that:
    H.C.F. of a and b = H.C.F. of b and a where a, b ∈ N.
   ∴a * b = b * a
    Thus, the operation * is commutative.
    
    For a, b, c ∈ N, we have:
    (a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c
    a *(b * c)= a *(H.C.F. of b and c) = H.C.F. of a, b, and c
    ∴(a * b) * c = a * (b * c)
    Thus, the operation * is associative.
   
    Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a where a ∈ N.
    But this relation is not true for any a ∈ N.
    Thus, the operation * does not have any identity in N.
    

Question-9 :- Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a − b,   (ii) a * b = a² + b²,  (iii) a * b = a + ab,  (iv) a * b = (a − b)²,   (v) a * b = ab/4,  (vi) a * b = ab²
Find which of the binary operations are commutative and which are associative.

Solution :-
(i) a * b = a − b 
    For Commutative :       ∴ [a * b = b * a]
    a * b = a − b
    b * a = b - a
    e.g., (a, b) = (1/2, 1/3) ∈ Q 
    1/2 * 1/3 = 1/2 - 1/3 = 1/6
    1/3 * 1/2 = 1/3 - 1/2 = -1/6
    Hence, the operation * is not commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = (a − b) - c
    a * (b * c) = a - (b - c)
    e.g., (a, b, c) = (1/2, 1/3, 1/4) ∈ Q 
    (1/2 * 1/3) * 1/4 = (1/2 - 1/3) - 1/4 
                      = 1/6 - 1/4
                      = -1/12
    1/2 * (1/3 * 1/4) = 1/2 - (1/3 - 1/4) 
                      = 1/2 - 1/12 
                      = 5/12
    Hence, the operation * is not associative.
(ii) a * b = a² + b²
    For Commutative :       ∴ [a * b = b * a]
    a * b = a² + b²
    b * a = b² + a² 
    e.g., (a, b) = (1/2, 1/3) ∈ Q 
    1/2 * 1/3 = (1/2)² + (1/3)² 
              = 1/4 + 1/9 
              = 13/36
    1/3 * 1/2 = (1/3)² + (1/2)²
              = 1/9 + 1/4 
              = 13/36
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = (a² + b²) + c²  
    a * (b * c) = a² + (b² + c²) 
    e.g., (a, b, c) = (1, 2, 3) ∈ Q 
    (1 * 2) * 3 = (1² + 2²) * 3
                = (1 + 4) * 3 
                = 5 * 3 
                = 5² + 3² 
                = 25 + 9
                = 34 
    1 * (2 * 3) = 1 * (2² + 3²) 
                = 1 * (4 + 9) 
                = 1 * 13 
                = 1² + 13²
                = 1 + 169
                = 170
    Hence, the operation * is associative.
(iii) a * b = a + ab = a(1 + b) 
    For Commutative :       ∴ [a * b = b * a]
    a * b = a(1 + b)
    b * a = b(1 + a)
    e.g., (a, b) = (1/2, 1/3) ∈ Q 
    1/2 * 1/3 = 1/2 x (1 + 1/3) 
              = 1/2 x 4/3 
              = 2/3
    1/3 * 1/2 = 1/3 x (1 + 1/2)
              = 1/3 x 3/2
              = 1/2 
    Hence, the operation * is not commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = (a − b) - c
    a * (b * c) = a - (b - c)
    e.g., (a, b, c) = (1/2, 1/3, 1/4) ∈ Q 
    (1/2 * 1/3) * 1/4 = (1/2 - 1/3) - 1/4 
                      = 1/6 - 1/4
                      = -1/12
    1/2 * (1/3 * 1/4) = 1/2 - (1/3 - 1/4)
                      = 1/2 - 1/12 
                      = 5/12
    Hence, the operation * is not associative.
(iv) a * b = (a − b)²
    For Commutative :       ∴ [a * b = b * a]
    a * b = (a − b)²
    b * a = (b − a)²
    e.g., (a, b) = (3, 2) ∈ Q 
    3 * 2 = (3 - 2)² = 1
    2 * 3 = (2 - 3)² = 1
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = (a − b)² * c
    a * (b * c) = a * (b - c)²
    e.g., (a, b, c) = (3, 2, 1) ∈ Q
     (3 * 2) * 1 = (3 - 2)² * 1 
                 = 1 * 1
                 = (1 - 1)²
                 = 0
    3 * (2 * 1)  = 3 * (2 - 1)²
                 = 3 * 1 
                 = (3 - 1)²
                 = 4
    Hence, the operation * is not associative.
(v)  a * b  = ab/4
    For Commutative :       ∴ [a * b = b * a]
    a * b = ab/4
    b * a = ba/4
    e.g., (a, b) = (4, 2) ∈ Q 
    4 * 2 = (4 x 2)/4 = 2
    2 * 4 = (2 x 4)/4 = 2
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = (ab/4) * c
    a * (b * c) = a * (bc/4)
    e.g., (a, b, c) = (4, 2, 1) ∈ Q
     (4 * 2) * 1 = (8/4) * 1 
                 = 2 * 1
                 = 2/4 
                 = 1/2
    4 * (2 * 1)  = 4 * (2/4)
                 = 4 * 1/2 
                 = 4/8
                 = 1/2
    Hence, the operation * is associative.
(vi) a * b = ab²
    For Commutative :       ∴ [a * b = b * a]
    a * b = ab²
    b * a = ba²
    e.g., (a, b) = (3, 2) ∈ Q 
    3 * 2 = 3 x 2² = 12
    2 * 3 = 2 x 3² = 18
    Hence, the operation * is not commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = ab² * c
    a * (b * c) = a * bc²
    e.g., (a, b, c) = (3, 2, 1) ∈ Q
     (3 * 2) * 4 = (3 x 2²) * 4 
                 = 12 * 4
                 = 12 x 4²
                 = 192
    3 * (2 * 4)  = 3 * (2 x 4²)
                 = 3 * 32 
                 = 3 x (32)²
                 = 3072
    Hence, the operation * is not associative.

    

Question-10 :- Find which of the operations given above has identity.

Solution :-
     Let the identity be I. 
(i) a * I = a - I 
(ii) a * I = a² - I²
(iii) a * I = a + aI
(iv) a * I = (a - I)²
(v)  a * I = aI/4
(vi) a * I = aI²
    Therefore, none of the operations given above has identity. 
    

Question-11 :-  Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d) Show that * is commutative and associative. Find the identity element for * on A, if any.

Solution :-
    A = N x N and * is a binary operation defined on A.
    (a, b) * (c, d) = (a+c, b+d)
                    = (c+a, d+b)
                    = (c, d) * (a, b)
     Hence, The operation is commutative.
    Again, [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) 
                                      = (a+c+e, b+d+f)
    And    (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) 
                                      = (a+c+e, b+d+f) 
    Here, [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)]
    Hence, The operation is associative.  
    Let identity function be (e, f), then (a, b) * (e, f) = (a+e, b+f)
    For identity function a = a + e
                          e = 0
    and for b + f = b
                f = 0
    As 0 ≠ N, therefore, identity-element does not exist.
    

Question-12 :-  State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N, a * a = a ∀ a ∈ N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a

Solution :-
(i)  * being a binary operation on N, is defined as a * a =  a∀a ∈ N
     Hence operation * is not defined, therefore, the given statement is false. 

(ii) * being a binary operation on N.
     c * b = b * c and (c * b) * a = (b * c) * a = a * (b * c) 
     Thus, a * (b * c) = (c * b) * a,  therefore the given statement is true.
    

Question-13 :-  Consider a binary operation ∗ on N defined as a ∗ b = a³ + b³. Choose the correct answer.
(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?

Solution :-
    Binary operation ∗ on N defined as a ∗ b = a³ + b³
    For Commutative :       ∴ [a * b = b * a]
    a * b = a³ + b³
    b * a = b³ + a³ 
    e.g., (a, b) = (2, 3) ∈ Q 
    2 * 3 = (2)³ + (3)³ 
          = 8 + 27 
          = 35
    3 * 2 = (3)³ + (2)³
          = 27 + 8 
          = 35
    Hence, the operation * is commutative.
    For associative:        ∴ [(a * b) * c = a * (b * c)]
    (a * b) * c = (a³ + b²) + c³  
    a * (b * c) = a³ + (b³ + c³) 
    e.g., (a, b, c) = (1, 2, 3) ∈ Q 
    (1 * 2) * 3 = (1³ + 2³) * 3
                = (1 + 8) * 3 
                = 9 * 3 
                = 9³ + 3³ 
                = 729 + 27
                = 756 
    1 * (2 * 3) = 1 * (2³ + 3³) 
                = 1 * (8 + 27) 
                = 1 * 35 
                = 1³ + 35³
                = 1 + 42875
                = 42876
    Hence, the operation * is not associative.
    Therefore, option (B) is correct.
    
CLASSES

Connect with us:

Copyright © 2015-16 by a1classes.

www.000webhost.com