Binary Operations

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  • Q1

    Consider the binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table. Using this table compute

    *

    1

    2

    3

    4

    5

    1

    1

    1

    1

    1

    1

    2

    1

    2

    1

    2

    1

    3

    1

    1

    3

    1

    1

    4

    1

    2

    1

    4

    1

    5

    1

    1

    1

    1

    5

    (i) (3*4)*5.
    (ii) 3*(4*5).
    (iii) (2*3)*(4*5).
    (iv) if * is commutative.

    Marks:3
    Answer:

    We have
    (i) (3*4)*5 = 1*5 [Since, 3*4=1]

    =1
    (ii)
    3*(4*5) = 3*1 [Since, 4*5=1]

    =1
    (iii) (2*3)*(4*5)
    = 1*1 [ 2*3=1 and 4*5=1]
    =1
    (iv) Since

    (2*5) = 1 and
    (5*2) =1
    (2*5) = (5*2)
    Operation * is commutative.

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  • Q2

    Let * be a binary operation defined by a*b=2a+b+ab+1. Find 3*5.

    Marks:1
    Answer:

    Given that

    a*b=2a+b+ab+1

    3*5=2×3+5+3×5+1

    =6+5+15+1

    =27

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  • Q3

    Marks:3
    Answer:


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  • Q4

    Let * be the binary operation on the set Q of rational numbers which are as follows:
    (i)a * b = a - b
    (ii)a * b = a2 + b2
    (iii)a * b = a + ab
    Find which of the binary operation are commutative and which are associative.

    Marks:5
    Answer:

    (i) Here a * b = a - b
    Now b *a = b - a, but a - bb - a
    a * bb * a
    * is not commutative

    a*(b*c) = a*(b - c) = a - (b - c) = a - b + c
    (a*b)*c = (a - b)*c = (a - b) - c
    Thus, a*(b*c)(a * b)*c
    * is not associative

    (ii) Here a*b = a2 + b2
    b*a = b2 + a2 = a2 + b2
    a*b = b*a
    * iscommutative

    a*(b*c) = a*(b2+ c2) = a2+ (b2+ c2)2
    (a*b)*c = (a2+ b2)*c = (a2 + b2)2 + c2
    Thus, a*(b*c)(a*b)*c
    * is not associative

    (iii) Here a * b = a+ ab
    Now b *a = b+ ba
    a * bb * a
    * is not commutative

    a*(b*c) = a*(b+bc) = a+ a(b+ bc) = a+ ab + abc
    (a*b)*c = (a+ ab)*c = a+ ab+ (a + ab)c = a + ab + ac + abc
    Thus, a*(b*c)(a * b)*c
    * is not associative

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  • Q5

    Let the * binary operation on N be defined by a*b = HCF of a and b. Is * commutative? Is * associative? Does there exist identity for this operation on N?

    Marks:3
    Answer:

    Here a*b = HCF of a and b.

    (a) We knowHCF of a, b = HCF of b, a

    a*b = b*a is commutative

    (b)a*(b*c) = a*(HCF of b, c)

    = HCF of a and HCF of b, c

    = HCF of a, b and c

    Now(a*b)*c = (HCF of a, b)*c

    = HCF of a, b and HCF of c

    = HCF of a, b and c = a*(b*c) = (a*b)*c

    * is associative

    (c)1*a = a*1 = HCF of a and 1

    i.e., 1*a = a*1 = 1
    1a

    There does not exist any identity for this operation

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