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Mathematics LibreTexts

Sample Test

  • Page ID
    18224
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    Exercise \(\PageIndex{1}\)

    For \(a, b \in \mathbb{Z},\) define an operation \( \otimes\), by \( a \otimes b= (a+b)(a+b).\) Determine whether \( \otimes \) on \(\mathbb{Z}\).

    1. is closed,
    2. is commutative,
    3. is associative, and
    4. has an identity.
    Answer
    1. is closed,

    Proof:

    Let a,b ∈ Z.

    Then consider,\( a \otimes b= (a+b)(a+b).\)

    (a+b)(a+b)=a2+2ab+b2 , since · is commutative on Z.

    a2, 2ab, b2, a2+2ab, and (a2+2ab)+b2Z , since ·, + are closed on Z.

    Hence, a2+2ab+b2Z.

    Hence,\( a \otimes b \in \mathbb{z}\).

    Thus, the binary operation is closed on Z. ⬜

    1. is commutative,

    Proof:

    Let \(a,b ∈Z.\)

    Then consider, \(a\otimes b=(a+b)(a+b)\)

    \( =a^2+2ab+b^2 \) (since · is commutative on Z.)

    \( =b^2+2ba+a^2\) (since ·, + are commutative on Z.)

    \( =(b+a)(b+a)\).

    Thus \( ab=ba.\)

    Thus, the binary operation is commutative on Z. ⬜

    1. is associative,

    Counterexample:

    Choose a=2, b=3, c=4.

    Then consider, \(( a \otimes b) \otimes c=\)

    \( =[(2 + 3)(2 + 3)] \otimes 4\)

    \( =25 \otimes 4\)

    \( =(25 + 4)(25 + 4)\)

    =841.

    Now consider a(bc)=2(3 + 4)(3+ 4)

    =249

    =(2 +49)(2 + 49)

    =2601.

    Since 841 ≠ 2601, the binary operation is not associative on Z. ⬜

    has an identity.

    Proof by Contradiction:

    Let e be the identity on (Z, ).

    Then consider, ae=ea=a,a ∈ Z.

    Now, a = ea.

    = (e+a)(e+a).

    = e(e+a)+a(e+a ),since · is associative on Z.

    = e^2+ea+ae+a^2.

    a = e^2+2ae+a^2, a∈ Z, since · is commutative on Z.

    Then choose a=0.

    Thus, e^2 = 0 and e = 0.

    Hence a^2 = a, ∀a∈ Z.

    This is a contradiction.

    Thus, (Z, ) has no identity. ⬜