
# Sample Test

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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.
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.

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.