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 \(\mathbb{Z}\).
 is closed,
 is commutative,
 is associative, and
 has an identity.
 Answer


is closed,
Proof:
Let a,b ∈ Z.
Then consider,\( a \otimes b= (a+b)(a+b).\)
(a+b)(a+b)=a^{2}+2ab+b^{2} , since · is commutative on Z.
a^{2}, 2ab, b^{2}, a^{2}+2ab, and (a^{2}+2ab)+b^{2} ∈Z , since ·, + are closed on Z.
Hence, a^{2}+2ab+b^{2} ∈Z.
Hence,\( a \otimes b \in \mathbb{z}\).
Thus, the binary operation is closed on Z. ⬜

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. ⬜

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. ⬜
