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

1. Let \(m\in \mathbb{Z}_+.\)

(a) \(R\) is reflexive:

Let \(a \in \mathbb{Z}\). Observe that \(a - a = 0 = m(0)\). Since \(0 \in \mathbb{Z}\),  \(m \mid (a - a)\). Thus, \(a \, R \, a\), so \(R\) is reflexive on \(\mathbb{Z}\).

 

(b) \(R\) is symmetric:

Let \(a, b \in \mathbb{Z}\) such that \(a \, R \, b\). By definition, \(m \mid (a - b)\), which means \(a - b = mk\) for some \(k \in \mathbb{Z}\). Multiplying both sides by \(-1\) yields: \(b - a = m(-k). \) Since \(-k \in \mathbb{Z}\), \(m \mid (b - a)\), which implies \(b \, R \, a\). Thus, \(R\) is symmetric on \(\mathbb{Z}\).

(c) \(R\) is not antisymmetric: 

 Counterexample: Choose \(m = 2\), \(a = 5\), and \(b = 7\). Then \(5 \, R \, 7\) because \(5 - 7 = -2 = 2(-1)\). So \(2 \mid (5 - 7)\).  Also,  \(7 \, R \, 5\) because \(7 - 5 = 2 = 2(1)\) . So \(2 \mid (7 - 5)\).  However, \(5 \neq 7\). Thus, \(a \, R \, b\) and \(b \, R \, a\) do not imply \(a = b\), so \(R\) is not antisymmetric on \(\mathbb{Z}\).

 

(d) \(R\) is transitive.

Let \(a, b, c \in \mathbb{Z}\) such that \(a \, R \, b\) and \(b \, R \, c\). By definition, \(m \mid (a - b)\) and \(m \mid (b - c)\), which means there exist \(k_1, k_2 \in \mathbb{Z}\) such that \( a - b = m k_1 \) and \( \quad b - c = m k_2.\) Adding these two equations gives: \( (a - b) + (b - c) = m k_1 + m k_2 \implies a - c = m(k_1 + k_2) \). Since \((k_1 + k_2) \in \mathbb{Z}\), \(m \mid (a - c)\). Which implies \(a \, R \, c\). Thus, \(R\) is transitive on \(\mathbb{Z}\).

2. Since \(R\) is reflexive, symmetric, and transitive, \(R\) is an equivalence relation on \(\mathbb{Z}\) (commonly known as congruence modulo \(m\)). By the Division Algorithm, for any integer \(a \in \mathbb{Z}\), there exist unique integers \(q\) and \(r\) such that \(a = mq + r\) with \(0 \le r < m\). Thus, \(a - r = mq\), meaning \(a \, R \, r\). Therefore, every integer belongs to exactly one equivalence class determined by its remainder modulo \(m\). The distinct equivalence classes are: \[ [r] = \{x \in \mathbb{Z} : x \, R \, r\} = \{x \in \mathbb{Z} : x \equiv r \pmod m\} = \{mq + r : q \in \mathbb{Z}\} \] for \(r \in \{0, 1, 2, \dots, m - 1\}\). Explicitly, the set of equivalence classes partitions \(\mathbb{Z}\) into \(m\) distinct sets: \begin{align*} [0] &= \{\dots, -2m, -m, 0, m, 2m, \dots\} \\ [1] &= \{\dots, -2m+1, -m+1, 1, m+1, 2m+1, \dots\} \\ &\;\;\vdots \\ [m - 1] &= \{\dots, -m-1, -1, m-1, 2m-1, 3m-1, \dots\} \end{align*}

 

 

 

 

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) First, reduce each factor modulo \(13\): \begin{align*} 201 &= 13(15) + 6 &\implies 201 &\equiv 6 \pmod{13} \\ 203 &= 13(15) + 8 &\implies 203 &\equiv 8 \pmod{13} \\ 207 &= 13(15) + 12 &\implies 207 &\equiv 12 \pmod{13} \\ 209 &= 13(16) + 1 &\implies 209 &\equiv 1 \pmod{13} \end{align*} Using the multiplicative property of congruences, we evaluate the product step-by-step: \begin{align*} 201 \cdot 203 \cdot 207 \cdot 209 &\equiv 6 \cdot 8 \cdot 12 \cdot 1 \pmod{13} \\ &\equiv 48 \cdot 12 \pmod{13} \end{align*} Since \(48 = 13(3) + 9\), we have \(48 \equiv 9 \pmod{13}\). Substituting this back gives: \begin{align*} 9 \cdot 12 &= 108 \end{align*} Finally, reducing \(108\) modulo \(13\) yields \(108 = 13(8) + 4\). Thus: \[ 201 \cdot 203 \cdot 207 \cdot 209 \equiv 4 \pmod{13} \] The remainder is \(4\).

(b) 

Observe the base modulo \(8\): \( 7 \equiv -1 \pmod 8 \) Using the power rule for congruences, we raise both sides to the exponent \(1453\): \( 7^{1453} \equiv (-1)^{1453} \pmod 8 .\) Since \(1453\) is an odd integer, \((-1)^{1453} = -1\). Therefore, \( 7^{1453} \equiv -1 \equiv 7 \pmod 8. \) Alternatively, observing the pattern of powers of \(7 \pmod 8\): \begin{align*} 7^1 &\equiv 7 \pmod 8 \\ 7^2 &\equiv 1 \pmod 8 \\ 7^3 &\equiv 7 \pmod 8 \\ 7^4 &\equiv 1 \pmod 8. \end{align*}  In general, \(7^k \equiv 7 \pmod 8\) for all odd integers \(k \ge 1\). Since \(1453\) is odd, \(7^{1453} \equiv 7 \pmod 8\). The remainder is \(7\).

Exercise \(\PageIndex{5}\)

a) Let \(a, b \in \mathbb{Z}_+\) such that \(7 \mid (a + 2b + 5)\) and \(7 \mid (b - 9)\). Prove that \(7 \mid (a + b)\).

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

Answer

a)  Proof: Let \(a, b \in \mathbb{Z}_+\). By definition of divisibility, there exist integers \(k_1, k_2 \in \mathbb{Z}\) such that: \( a + 2b + 5 = 7k_1 \) and \( b - 9 = 7k_2 .\)

Subtracting the second equation from the first yields: \( (a + 2b + 5) - (b - 9) = 7k_1 - 7k_2. \)

Simplifying the left-hand side gives: \( a + b + 14 = 7(k_1 - k_2). \)

Rearranging terms yields: \( a + b = 7(k_1 - k_2) - 14 = 7(k_1 - k_2 - 2) .\) Since \((k_1 - k_2 - 2) \in \mathbb{Z}\), it follows by definition that \(7 \mid (a + b)\).

b) Proof:

Let \(a, b \in \mathbb{Z}_+\) such that \(a \mid b\). By definition of divisibility, there exists an integer \(k \in \mathbb{Z}_+\) such that \(b = ak\). Raising both sides to the fourth power gives: \[ b^4 = (ak)^4 = a^4 k^4 \] Factoring out \(a^3\) yields: \[ b^4 = a^3 (a k^4) \] Since \(a, k \in \mathbb{Z}_+\), the product \(m = a k^4\) is an integer in \(\mathbb{Z}_+\). Thus: \[ b^4 = a^3 m \quad \text{where } m \in \mathbb{Z}_+ \] Therefore, \(a^3 \mid b^4\).