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Jamb Mathematics - Lesson Notes on Binary operation for UTME Candidate

Feb 11 2025 10:13 PM

Osason

Jamb Updates

Binary operation | Jamb Mathematics

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Cadet, listen up! Your next mission is to master the art of progression—arithmetic, geometric, and beyond—because precision and strategy are key to victory. Drill through formulas, decode patterns, and sharpen your problem-solving skills, because only the well-prepared survive the battlefield of mathematics. Fall in and get to work! 🚀🔥
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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Binary operation? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Binary operation together and move one step closer to achieving your exam success! Blissful learning.
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Closure Property Questions
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1. Check if the set of natural numbers (N)(\mathbb{N}) is closed under subtraction.
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Solution:
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A set is closed under an operation if performing the operation on any two elements of the set results in an element still in the set.
  • Let a,bNa, b \in \mathbb{N}.
  • Subtraction: aba - b
Example:
  • If a=5a = 5, b=3b = 3, then 53=2N5 - 3 = 2 \in \mathbb{N} (Valid)
  • If a=3a = 3, b=5b = 5, then 35=2N3 - 5 = -2 \notin \mathbb{N} (Invalid)
Since subtraction sometimes results in a number not in N\mathbb{N}, N\mathbb{N} is not closed under subtraction.
Answer: Not closed under subtraction.

2. Is the set of whole numbers (W)(\mathbb{W}) closed under division?
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Solution:
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Whole numbers (W)(\mathbb{W}) are 0,1,2,3,4,0, 1, 2, 3, 4, \dots.
  • Division: a÷ba \div b
  • If a=4a = 4, b=2b = 2, then 4÷2=2W4 \div 2 = 2 \in \mathbb{W} (Valid)
  • If a=5a = 5, b=2b = 2, then 5÷2=2.5W5 \div 2 = 2.5 \notin \mathbb{W} (Invalid)
Since division does not always produce a whole number, W\mathbb{W} is not closed under division.
Answer: Not closed under division.

3. Show that the set of integers (Z)(\mathbb{Z}) is closed under addition.
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Solution:
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For any two integers a,bZa, b \in \mathbb{Z}:
  • Addition: a+ba + b
Examples:
  • If a=3a = -3, b=5b = 5, then 3+5=2Z-3 + 5 = 2 \in \mathbb{Z}
  • If a=7a = -7, b=2b = -2, then 7+(2)=9Z-7 + (-2) = -9 \in \mathbb{Z}
Since adding any two integers always results in another integer, Z\mathbb{Z} is closed under addition.
Answer: Closed under addition.

Commutativity Questions
4. Show that multiplication of real numbers (R)(\mathbb{R}) is commutative.
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Solution:
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A binary operation \star is commutative if: ab=baa \star b = b \star a
For multiplication: a×b=b×aa \times b = b \times a
Examples:
  • If a=4a = 4, b=7b = 7, then 4×7=284 \times 7 = 28 and 7×4=287 \times 4 = 28.
  • If a=3a = -3, b=2b = 2, then 3×2=6-3 \times 2 = -6 and 2×(3)=62 \times (-3) = -6.
Since multiplication always satisfies a×b=b×aa \times b = b \times a, multiplication is commutative in R\mathbb{R}.
Answer: Commutative.

5. Is subtraction commutative for integers (Z)(\mathbb{Z})?
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Solution:
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Subtraction is commutative if: ab=baa - b = b - a
Examples:
  • If a=6a = 6, b=3b = 3, then 63=36 - 3 = 3 but 36=33 - 6 = -3 (Not Equal)
  • If a=5a = -5, b=2b = -2, then 5(2)=3-5 - (-2) = -3 but 2(5)=3-2 - (-5) = 3 (Not Equal)
Since abbaa - b \neq b - a in general, subtraction is not commutative.
Answer: Not commutative.

Associativity Questions
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6. Prove that addition is associative in real numbers (R)(\mathbb{R}).
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Solution:
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A binary operation is associative if: (ab)c=a(bc)(a \star b) \star c = a \star (b \star c)
For addition: (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)
Examples:
  • Let a=2,b=3,c=4a = 2, b = 3, c = 4: (2+3)+4=5+4=9(2 + 3) + 4 = 5 + 4 = 9 2+(3+4)=2+7=92 + (3 + 4) = 2 + 7 = 9
Since both expressions are equal, addition is associative.
Answer: Associative.

7. Is division associative in R\mathbb{R}?
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Solution:
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Division is associative if: (a÷b)÷c=a÷(b÷c)(a \div b) \div c = a \div (b \div c)
Let a=8,b=4,c=2a = 8, b = 4, c = 2: (8÷4)÷2=2÷2=1(8 \div 4) \div 2 = 2 \div 2 = 1
8÷(4÷2)=8÷2=48 \div (4 \div 2) = 8 \div 2 = 4
Since 141 \neq 4, division is not associative.
Answer: Not associative.

Distributivity Questions
8. Prove that multiplication distributes over addition in R\mathbb{R}.
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Solution:
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Multiplication is distributive over addition if: a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c)
Examples:
  • Let a=2,b=3,c=4a = 2, b = 3, c = 4: 2×(3+4)=2×7=142 \times (3 + 4) = 2 \times 7 = 14 (2×3)+(2×4)=6+8=14(2 \times 3) + (2 \times 4) = 6 + 8 = 14
Since both sides are equal, multiplication distributes over addition.
Answer: Distributive.

9. Is division distributive over addition?
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Solution:
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For distributivity: a÷(b+c)=(a÷b)+(a÷c)a \div (b + c) = (a \div b) + (a \div c)
Let a=12,b=4,c=2a = 12, b = 4, c = 2: 12÷(4+2)=12÷6=212 \div (4 + 2) = 12 \div 6 = 2
(12÷4)+(12÷2)=3+6=9(12 \div 4) + (12 \div 2) = 3 + 6 = 9
Since 292 \neq 9, division is not distributive over addition.
Answer: Not distributive.

10. Verify if exponentiation is distributive over multiplication.
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Solution:
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Exponentiation is distributive if: (a×b)c=(ac)×(bc)(a \times b)^c = (a^c) \times (b^c)
Examples:
  • If a=2,b=3,c=2a = 2, b = 3, c = 2: (2×3)2=62=36(2 \times 3)^2 = 6^2 = 36 (22)×(32)=4×9=36(2^2) \times (3^2) = 4 \times 9 = 36
Since both are equal, exponentiation distributes over multiplication.
Answer: Distributive.
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Closure Property Questions
11. Is the set of rational numbers (Q)(\mathbb{Q}) closed under division?
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Solution:
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Rational numbers Q\mathbb{Q} are fractions of integers where the denominator is nonzero.
  • If a=34a = \frac{3}{4} and b=25b = \frac{2}{5}, a÷b=34÷25=34×52=158Qa \div b = \frac{3}{4} \div \frac{2}{5} = \frac{3}{4} \times \frac{5}{2} = \frac{15}{8} \in \mathbb{Q}
  • If a=56a = \frac{5}{6} and b=0b = 0, then a÷0a \div 0 is undefined.
Since division by zero is not allowed, Q\mathbb{Q} is not closed under division.
Answer: Not closed under division.

12. Show that the set of even integers is closed under addition.
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Solution:
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Let a=2ma = 2m and b=2nb = 2n, where m,nm, n are integers.
a+b=2m+2n=2(m+n)a + b = 2m + 2n = 2(m+n)
Since 2(m+n)2(m+n) is always even, the sum of any two even numbers is also even.
Answer: Closed under addition.

13. Is the set of odd integers closed under multiplication?
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Solution:
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Let a=2m+1a = 2m + 1 and b=2n+1b = 2n + 1, where m,nm, n are integers.
a×b=(2m+1)×(2n+1)a \times b = (2m + 1) \times (2n + 1)
Expanding: =4mn+2m+2n+1= 4mn + 2m + 2n + 1
=2(2mn+m+n)+1= 2(2mn + m + n) + 1
Since this expression is odd, the product of two odd numbers is always odd.
Answer: Closed under multiplication.

  1. Check if the set of prime numbers is closed under subtraction.**
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    Solution: Consider prime numbers p1=5p_1 = 5 and p2=3p_2 = 3.
53=2P5 - 3 = 2 \in \mathbb{P}
However, for p1=7p_1 = 7 and p2=5p_2 = 5:
75=2P7 - 5 = 2 \in \mathbb{P}
But for p1=11p_1 = 11 and p2=7p_2 = 7:
117=4P11 - 7 = 4 \notin \mathbb{P}
Since subtraction can result in a non-prime, the set of prime numbers is not closed under subtraction.
Answer: Not closed under subtraction.

Commutativity Questions
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15. Is exponentiation commutative for real numbers?
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Solution:
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Exponentiation is commutative if: ab=baa^b = b^a
Counterexample:
  • If a=2,b=3a = 2, b = 3, then 23=82^3 = 8 and 32=93^2 = 9.
Since 898 \neq 9, exponentiation is not commutative.
Answer: Not commutative.

16. Show that set intersection ()(\cap) is commutative.
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Solution: Intersection is commutative if: AB=BAA \cap B = B \cap A
Example:
  • If A={1,2,3}A = \{1, 2, 3\} and B={2,3,4}B = \{2, 3, 4\},
AB={2,3}=BAA \cap B = \{2, 3\} = B \cap A
Since the order does not matter, set intersection is commutative.
Answer: Commutative.

17. Is matrix multiplication commutative?
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Solution:
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Matrix multiplication is commutative if: A×B=B×AA \times B = B \times A
Counterexample: Let A=[1201]A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} and B=[3045]B = \begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix}.
A×BB×AA \times B \neq B \times A
Since matrix multiplication does not always commute, it is not commutative.
Answer: Not commutative.

Associativity Questions
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#18. Show that multiplication is associative in real numbers (R)(\mathbb{R}).
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Solution:
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Multiplication is associative if: (a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times c)
Example:
  • If a=2,b=3,c=4a = 2, b = 3, c = 4:
(2×3)×4=6×4=24(2 \times 3) \times 4 = 6 \times 4 = 24
2×(3×4)=2×12=242 \times (3 \times 4) = 2 \times 12 = 24
Since both are equal, multiplication is associative.
Answer: Associative.

19. Is exponentiation associative?
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Solution:
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Exponentiation is associative if: (ab)c=a(bc)(a^b)^c = a^{(b^c)}
Counterexample:
  • Let a=2,b=3,c=2a = 2, b = 3, c = 2:
(23)2=82=64(2^3)^2 = 8^2 = 64
2(32)=29=5122^{(3^2)} = 2^9 = 512
Since 6451264 \neq 512, exponentiation is not associative.
Answer: Not associative.

Distributivity Questions
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20. Verify if modulus operation (mod)(\mod) distributes over addition.
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Solution: Distributivity means: (a+b)modm=(amodm+bmodm)modm(a + b) \mod m = (a \mod m + b \mod m) \mod m
Example:
  • Let a=7,b=5,m=4a = 7, b = 5, m = 4:
(7+5)mod4=12mod4=0(7 + 5) \mod 4 = 12 \mod 4 = 0
(7mod4+5mod4)mod4=(3+1)mod4=4mod4=0(7 \mod 4 + 5 \mod 4) \mod 4 = (3 + 1) \mod 4 = 4 \mod 4 = 0
Since both are equal, modulus distributes over addition.
Answer: Distributive.

21. Is set union ()(\cup) distributive over intersection ()(\cap)?
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Solution:
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Distributivity means: A(BC)=(AB)(AC)A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
Example:
  • If A={1,2}A = \{1, 2\}, B={2,3}B = \{2, 3\}, C={2,4}C = \{2, 4\}:
BC={2}B \cap C = \{2\}
A(BC)={1,2}A \cup (B \cap C) = \{1, 2\}
(AB)={1,2,3},(AC)={1,2,4}(A \cup B) = \{1, 2, 3\}, (A \cup C) = \{1, 2, 4\}
(AB)(AC)={1,2}(A \cup B) \cap (A \cup C) = \{1, 2\}
Since both sides are equal, set union distributes over intersection.
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Calculation problems involving identity and inverse element

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Identity Element Questions
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1. Find the identity element for addition in R\mathbb{R} (set of real numbers).
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Solution:
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The identity element ee for an operation \star satisfies: ae=ea=aa \star e = e \star a = a
For addition: a+e=aa + e = a
Solving for ee: e=0e = 0
Answer: The identity element for addition in R\mathbb{R} is 0.

2. Find the identity element for multiplication in R\mathbb{R}.
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Solution:
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For multiplication: a×e=e×a=aa \times e = e \times a = a
Solving for ee: e=1e = 1
Answer: The identity element for multiplication in R\mathbb{R} is 1.

3. Is there an identity element for subtraction in R\mathbb{R}?
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Solution:
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For subtraction: ae=aa - e = a
Solving for ee: e=0e = 0
Checking: a0=a,0aaa - 0 = a, \quad 0 - a \neq a
Since eaae - a \neq a, subtraction does not have an identity element.
Answer: No identity element for subtraction.

4. Is 0 an identity element for division in R\mathbb{R}?
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Solution:
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For division: a÷e=aa \div e = a
Setting e=0e = 0: a÷0a \div 0
Since division by zero is undefined, 0 is not an identity element for division.
Answer: No identity element for division.

5. Find the identity element in matrix addition.
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Solution:
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For an n×nn \times n matrix AA, the identity element II satisfies: A+I=AA + I = A
The identity matrix for addition is the zero matrix OO: O=[0000]O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
Answer: The zero matrix is the identity element for matrix addition.

Inverse Element Questions
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6. Find the additive inverse of 7 in R\mathbb{R}.
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Solution:
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The additive inverse a1a^{-1} satisfies: a+a1=0a + a^{-1} = 0
For a=7a = 7: 7+(7)=07 + (-7) = 0
Answer: The additive inverse of 7 is -7.

7. Find the multiplicative inverse of 5 in R\mathbb{R}.
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Solution:
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The multiplicative inverse a1a^{-1} satisfies: a×a1=1a \times a^{-1} = 1
For a=5a = 5: 5×15=15 \times \frac{1}{5} = 1
Answer: The multiplicative inverse of 5 is 15\frac{1}{5}.

8. Find the additive inverse of -12 in R\mathbb{R}.
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Solution:** 12+12=0-12 + 12 = 0
Answer: The additive inverse of -12 is 12.

9. Find the multiplicative inverse of 37\frac{3}{7} in Q\mathbb{Q} (set of rational numbers).
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Solution:** The multiplicative inverse is: (37)1=73\left(\frac{3}{7}\right)^{-1} = \frac{7}{3}
37×73=1\frac{3}{7} \times \frac{7}{3} = 1
Answer: The multiplicative inverse of 37\frac{3}{7} is 73\frac{7}{3}.

10. Does 0 have a multiplicative inverse in R\mathbb{R}?
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Solution:
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A number a1a^{-1} satisfies: 0×a1=10 \times a^{-1} = 1
Since 0×x=00 \times x = 0 for any xx, no number exists that satisfies 0×x=10 \times x = 1.
Answer: 0 has no multiplicative inverse.

11. Find the inverse of matrix A=[2002]A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}.
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Solution:
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The inverse of a matrix AA is: A1=1det(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A)
Determinant: det(A)=2×20×0=4\det(A) = 2 \times 2 - 0 \times 0 = 4
A1=14[2002]A^{-1} = \frac{1}{4} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}
=[120012]= \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}
Answer: [120012]\begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}

12. Find the inverse of 3 in modulo 7 arithmetic.
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Solution:
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The inverse of aa in modm\mod m satisfies: a×x1(modm)a \times x \equiv 1 \pmod{m}
Finding xx such that: 3x1(mod7)3x \equiv 1 \pmod{7}
Checking values: 3×5=151(mod7)3 \times 5 = 15 \equiv 1 \pmod{7}
Answer: The inverse of 3 in mod7\mod 7 is 5.

13. Find the inverse of -4 in mod 9 arithmetic.
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Solution:
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Finding xx such that: 4×x1(mod9)-4 \times x \equiv 1 \pmod{9}
Equivalent to solving: 5x1(mod9)5x \equiv 1 \pmod{9}
Checking values: 5×2=101(mod9)5 \times 2 = 10 \equiv 1 \pmod{9}
Answer: The inverse of -4 in mod 9 is 2.

14. Find the inverse of 2 in modulo 5 arithmetic.
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Solution:
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Solving: 2x1(mod5)2x \equiv 1 \pmod{5}
Checking values: 2×3=61(mod5)2 \times 3 = 6 \equiv 1 \pmod{5}
Answer: The inverse of 2 in mod 5 is 3.

15. Find the additive inverse of 9 in mod 11 arithmetic.
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Solution:
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9+x0(mod11)9 + x \equiv 0 \pmod{11}
Solving: x=92(mod11)x = -9 \equiv 2 \pmod{11}
Answer: The additive inverse of 9 in mod 11 is 2.

16. Find the inverse of the identity matrix.
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Solution:
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The inverse of the identity matrix II satisfies: I×I1=II \times I^{-1} = I
Since I×I=II \times I = I, we get: I1=II^{-1} = I
Answer: The inverse of the identity matrix is itself.

17. Find the inverse of 4 in mod 7 arithmetic.
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Solution:
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4x1(mod7)4x \equiv 1 \pmod{7}
Checking values: 4×2=81(mod7)4 \times 2 = 8 \equiv 1 \pmod{7}
Answer: The inverse of 4 in mod 7 is 2.

18. Find the additive inverse of 17 in mod 19 arithmetic.
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Solution:
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17+x0(mod19)17 + x \equiv 0 \pmod{19}
Solving: x=172(mod19)x = -17 \equiv 2 \pmod{19}
Answer: The additive inverse of 17 in mod 19 is 2.

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