Sample term test

These mock exams are provided to help you prepare for Term/Final tests. The best way to use these practice tests is to try the problems as if you were taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand them.    

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 $a,b \in \mathbb{Z}.$ 

Then $a \otimes b= (a+b)(a+b)=a^2+2ab+b^2.$

Since · is commutative on $\mathbb{Z}, a^2, 2ab,b^2 \in \mathbb{Z}.$ Thus $a^2+2ab+b^2 \in \mathbb{Z}.$

Hence,$a \otimes b \in \mathbb{Z}$.

Thus, the binary operation is closed on $\mathbb{Z}.$ ⬜

2. is commutative,

Proof:

Let $a,b \in \mathbb{Z}.$ 

Then $a\otimes b =(a+b)(a+b) =a^2+2ab+b^2 (\mbox{ since · is commutative on} \mathbb{Z})=b^2+2ba+a^2 (\mbox{ since ·, + are commutative on } \mathbb{Z})=(b+a)(b+a).$

Thus $a\otimes b=b\otimes a.$

Thus, the binary operation is commutative on $\mathbb{Z}$. ⬜

 

3. is not 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 $a \otimes (b \otimes c) =2 \otimes [(3 + 4)(3+ 4)]=2 \otimes 49 =(2 +49)(2 + 49) =2601.$

Since $841 \ne 2601$, the binary operation is not associative on $\mathbb{Z}$. ⬜

4. Does not have an identity.

Proof by Contradiction:

Let e be the identity on ($\mathbb{Z}$, ).

Then  $a \otimes e=e \otimes a=a,a \in \mathbb{Z}$.

Now, $a = e \otimes a.$

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

$= e(e+a)+a(e+a ),$ since · is associative on $\mathbb{Z}$

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

$a = e^2+2ae+a^2, a \in \mathbb{Z}$., since · is commutative on$\mathbb{Z}$.

Then choose $a=0.$

Thus,$e^2 = 0 \implies e = 0.$

Hence $a^2 = a,$ for all $a\in \mathbb{Z}$

This is a contradiction.

Now choose, $e\ne 0$ then it won't work for $a=0$.

Thus, ($\mathbb{Z}$, \otimes) has no identity.⬜

Exercise $\PageIndex{2}$

For $a, b \in \mathbb{Z},$, the ominus of b from a is defined by $a \ominus b = ab + a -b$. The oplus of a by b is defined by $a⊕b = a + b + ab.$ The oslash of a by b is defined by $a \oslash b = (a + b)(a-b)$. Answer the following:

(a) Determine whether $\oslash$ is distributive over $\oplus$.

(b) Determine whether $\oslash$ is distributive over $\ominus.$

Answer

(a)

Counter Example:

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

Consider $2 \oslash (3 \oplus 4) = 2 \oslash (3+4+(3)(4)) = 2 \oslash 19 = (2+19)(2-19) =357.$

Now consider $(2 \oslash 3)\oplus (2 \oslash 4) = [(2+3)(2-3)]\oplus [(2+4)(2-4)] =(-5)\oplus (-12) =(-5)+(-12)+[(-5)(-12)] =41.$

Since $357 ≠ 41, \oslash$ is not distributive over $\oplus$.

(b)

Counter Example:

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

Consider $2 \oslash (3 \ominus 4)=2 \oslash [(3)(4)+3-4]=2 \oslash 11 = (2+11)(2-11) =117.$

Next consider, $(2 \oslash 3) \ominus (2 \oslash 4)=[(2+3)(2-3)] \ominus [(2+4)(2-4)] =(-5) \ominus (-12)=(-5)(-12)+(-5)-(-12) =67.$

Since $117 ≠ 67,  \oslash$ is not distributive over $\ominus.$

Exercise $\PageIndex{3}$

Let $m\in \mathbb{Z}_+.$ Let $a, b \in \mathbb{Z},$ define the relation $a \, R \, b$ iff $m \mid (a-b)$.

 

1. Determine whether the relation is

a) reflexive,

b) symmetric

 c) antisymmetric

d) transitive.

2.  If $R$ is an equivalence relation, describe the equivalence classes of $\mathbb{Z}$.

Answer

(a) is reflexive,

Proof:

Let $m \in \mathbb{Z}_+.$ 

Let $a \in \mathbb{Z}.$ 

Since $a - a = 0 = m(0), m \mid (a - a).$

Hence, $a \, R \, a.$ Thus $R$ is reflexive.◻

 

(b) is symmetric,

Proof:

Let $a, b \in \mathbb{Z},$ such that $a \, R \, b.$ Thus $m \mid (a-b)$.

We shall show that $b \, R \, a.$

Since $m \mid (a-b),  a - b = m(k), k \in \mathbb{Z}.$

Now,$(b - a) =m(-k), -k \in \mathbb{Z}.$

Thus,$m \mid (b-a) \implies  b \, R \, a.$, therefore, $R$ is symmetric on $\mathbb{Z}.$◻

 

(c) is antisymmetric,

Counter Example:

We shall show that m | (a - b) and m | (b - a) but a ≠ b.

Choose m = 2, a = 5 and b = 7.

Then 2  | (5 - 7) since -2 = 2(-1) and 2 | (7 - 5) since 2 = 2(1).

But, 5 ≠ 7, thus m | (a - b) is not antisymmetric on ℤ.◻

 

(d) is transitive.

Proof:

Let a, b, c, m ∈ ℤ+  s.t. m | (a - b) and m | (b - c).

We shall show that m | (a - c).

Since m | (a - b), (a - b) = m(k1), k1 ∈ ℤ.

Since m | (b - c), (b - c) = m(k2), k2 ∈ ℤ.

Now consider (a - b) + ( b - c) = mk1 + mk2.

Thus (a - c) = m(k1 + k2), k1 + k2 ∈ ℤ.

Hence m | (a - c), thus a R c, therefore R is transitive on ℤ.◻

 

 

Exercise $\PageIndex{4}$

Find the remainder

(a) when $201 \times 203 \times 207 \times 209$ is divided by $13.$

(b) when $7^{3453}$ is divided by $8.$

Answer

(a) 201 ≡ 6 (mod 13), 203 ≡ 8 (mod 13), 207 ≡ 12 (mod 13), 209 ≡ 1 (mod 13)

6*8*12*1 = 48*12*1 ≡ 9*12*1 (mod 13)

9*12 = 108 ≡ 4 (mod 13)

(b) 

71=7 (mod 8) Since 1453 is odd, 71453 ≡ 7 (mod 8)

72=1 (mod 8)

73=7 (mod 8)

74=1 (mod 8)

Exercise $\PageIndex{5}$

(a) Let a and b be positive integers such that 7|(a+2b+5) and 7|(b-9).  Prove that 7|(a+b).

(b) Let $a, b \in \mathbb{Z}_+.$ If $a | b,$ is it necessarily true that $a^3 | b^4$?

Answer

a) Let a, b ∈ ℤ_+ such that  7|(a+2b+5),  and 7|(b-9),

 Then 7|((a+2b+5)+(b-9), 7|(a+3b-4)

If 7|(a+b) then 7|(a+3b-4)-(a+b)), 7|(2b-4)

7|(a+b), such that a=7x+5, b=7x+9

 

  7|(a+2b+5)                                     7|(b-9)                             7|(a+b)

=>7|(7x+5 + 2(7x+9) +5)              =>7|(7x+9-9)                    =>7|(7x+5+7x+9)

=>7|(21x + 28)                                 =>7|(7x)                             =>7|(14x+14)

(Do not use equal sign for relations!!!!!!)

b) Proof:

Let a,b ∈ ℤ+ s.t. a | b.

Since a | b, ∃ k ∈ ℤ+ such that b=a(k).

Now consider b4 = (a(k))4,

   = a3(ak4).

Since ak4 ∈ ℤ+, a3 | b4.◻