Sample Final

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

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

\(a², 2ab, b², a²+2ab,\) and a²+2ab+b² ∈\mathbb{Z} , since ·, + are closed on \mathbb{Z}.

Hence, a²+2ab+b² ∈\mathbb{Z}.

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

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

2. is commutative,

Proof:

Let \(a,b ∈\mathbb{Z}.\)

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

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

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

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

Thus \( ab=ba.\)

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

 

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 \(\mathbb{Z}\). ⬜

has an identity.

Proof by Contradiction:

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

 

Then consider, \(ae=ea=a,a \in \mathbb{Z}\).

 

Now, \( a = ea.\)

 

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

 

\(= e(e+a)+a(e+a ),\) 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,\( e^2 = 0\) and e = 0.

 

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

 

This is a contradiction.

 

Thus, (\(\mathbb{Z}\), ) 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 \(2 \oslash(3⊕4)=2\oslash(3+4+(3)(4))\)

\(=2 \oslash19\)

\( =(2+19)(2-19)\)

\( =357.\)

Now consider (23)⊕(24) = [(2+3)(2-3)]⊕[(2+4)(2-4)]

=(-5)⊕(-12)

=(-5)+(-12)+[(-5)(-12)]

=41.

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

(b)

Counter Example:

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

Consider 2\(\oslash\) (3 \(\ominus\)4)=2 \(\oslash\) [(34)+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 \( a, b \in \mathbb{ Z}, \)define the relation \(a R b \) iff \(4 \mid (3a+b)\).

Determine whether this relation

(a) is reflexive,

(b) is symmetric,

(c) is antisymmetric,

(d) is transitive.

(e) If \(R\) is an equivalence relation, describe the equivalence classes of \(\mathbb{Z}\).

 

Answer

TBA

Exercise \(\PageIndex{4}\)

(a) Prove that \(n^2 + 1\) not divisible by \(3\) for any integer \(n.\)

(b) Let \(a, b ∈ ℤ+.\) If \(a | b,\) is it necessarily true that \(a^3 | b^5\)?

Answer

TBA

Exercise \(\PageIndex{5}\)

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

TBA

Exercise \(\PageIndex{6}\)

(a) In a \(101\)-digit multiple of \(7\), the first \(50\) digits are all \(8\)s and the last \(50\) digits are all \(1\)s. What is the middle digit?

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

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{7}\)

(a) An elementary art teacher has \(4\) art classes with \(21, 32, 35\) and \(29\) students, respectively. The teacher wants to order some equipment that can be used by equal-sized groups in each class. What is the largest number of students in a group in each class so that each group has the same number of students?

(b) Given that April 11, 2018, is a Wednesday, what day of the week is Jan 1, 2025?

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{8}\)

(a) Using Euclidean algorithm, find gcd(2017, 7021) and lcm(2017, 7021).

(b) Using prime factorization,find gcd(2017, 7021) and lcm(2017, 7021).

Answer

TBA

Exercise \(\PageIndex{9}\)

(a) Solve the Diophantine equation \(2058x - 1935y = 54\).

(b) When Mrs.Brown cashed her cheque, the absent-minded teller gave her as many cents as she should have dollars, and as many dollars as she should have cents. Equally absent-minded Mrs, Brown left with the cash without noticing the discrepancy. It was only after she spent 25 cents that she noticed now she had twice as much money as she should. What was the amount of her cheque?

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{10}\)

The Quadripark town uses only numbers which are 1 more than some of the multiples of 4.

1. Construct a 5X5 Quadripark multiplication table with the numbers 1, 5, 9, 13 and 17.
2. Find the smallest ten prime numbers in Quadripark.
3. Find a number with two different Quadripark prime factorizations.
4. Does the Prime divisibility Theorem hold for the Quadripark number system? Explain

Answer

2.

5, 9, 13, 17, 21, 29, 33, 37, 41, 49