Tasks 01-02 - Mathematical Foundations

Language, Sets, and Number Systems

Problem 1: Set Operations

Let \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) be the universal set.

Let \(A = \{2, 3, 5, 7\}\) and \(B = \{1, 2, 3, 4, 5\}\).

Find:

  1. \(A \cup B\)
  2. \(A \cap B\)
  3. \(A \setminus B\)
  4. \(B \setminus A\)
  1. \(A \cup B = \{1, 2, 3, 4, 5, 7\}\)
  2. \(A \cap B = \{2, 3, 5\}\)
  3. \(A \setminus B = \{7\}\)
  4. \(B \setminus A = \{1, 4\}\)

Problem 2: Properties of Operations

Determine if each statement is true or false. Provide a counterexample if false.

  1. Subtraction is associative: \((a - b) - c = a - (b - c)\)
  2. Division is commutative: \(\frac{a}{b} = \frac{b}{a}\)
  3. Multiplication distributes over subtraction: \(a(b - c) = ab - ac\)
  4. Addition distributes over multiplication: \(a + (b \times c) = (a + b) \times (a + c)\)
  1. False: \((5 - 3) - 2 = 0\) but \(5 - (3 - 2) = 4\)
  2. False: \(\frac{6}{2} = 3\) but \(\frac{2}{6} = \frac{1}{3}\)
  3. True: This is the distributive property extended to subtraction
  4. False: \(2 + (3 \times 4) = 14\) but \((2 + 3) \times (2 + 4) = 30\)

Problem 3: Logical Statements

Consider: “If a product is on sale, then its price is reduced by at least 20%”

  1. Write this using logical notation (define your propositions)
  2. Is the converse necessarily true?

Let \(p\): “product is on sale”, \(q\): “price reduced by at least 20%”

  1. \(p \Rightarrow q\)
  2. Converse (\(q \Rightarrow p\)) is not necessarily true - items can have 20% reduction without being “on sale”

Problem 4: Business Application - Customer Analysis

A company surveys 200 customers about their service preferences:

  • 120 use Service A
  • 85 use Service B
  • 60 use Service C
  • 40 use both A and B
  • 25 use both B and C
  • 35 use both A and C
  • 15 use all three services

Calculate:

  1. How many use exactly one service?
  2. How many use at least two services?
  3. How many use none of the services?

Step 1 - Set up the Venn diagram

  • Center (all three): 15
  • A and B only: 40 - 15 = 25
  • B and C only: 25 - 15 = 10
  • A and C only: 35 - 15 = 20
  • A only: 120 - 25 - 15 - 20 = 60
  • B only: 85 - 25 - 15 - 10 = 35
  • C only: 60 - 20 - 15 - 10 = 15

Step 2 - Answer the questions

  1. Exactly one service: 60 + 35 + 15 = 110
  2. At least two services: 25 + 10 + 20 + 15 = 70
  3. None: 200 - (60 + 35 + 15 + 25 + 10 + 20 + 15) = 200 - 180 = 20

Problem 5: Interest Rate Calculation

A bank account grows from €5,000 to €5,500 in one year.

  1. What is the percentage increase?
  2. If this rate continues, what will be the balance after 3 years?
  3. Express the growth using set notation (hint: think of the set of all possible balances)
  1. Percentage increase: \(\frac{5500 - 5000}{5000} \times 100\% = 10\%\)
  2. After 3 years: \(5000 \times (1.10)^3 = 5000 \times 1.331 = \text{€}6,655\)
  3. Set notation: Let \(B_n = \{b \in \mathbb{R} : b = 5000 \times (1.10)^n, n \in \mathbb{N}_0\}\) This represents all possible balances after n years.

Problem 6: Number Classification and Conversions

For each of the following numbers, classify them by listing ALL applicable number systems (\(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\)) and show your work for conversions where needed:

  1. \(0.\overline{27}\) (repeating decimal)
  2. \(\sqrt{16}\)
  3. \(\frac{-15}{3}\)
  4. \(0.101001000100001...\) (non-repeating, non-terminating)
  1. \(0.\overline{27}\): Let \(x = 0.272727...\)
    • \(100x = 27.272727...\)
    • \(100x - x = 27\), so \(99x = 27\)
    • Thus \(x = \frac{27}{99} = \frac{3}{11}\)
    • Classification: \(\mathbb{Q}, \mathbb{R}\)
  2. \(\sqrt{16} = 4\)
    • Classification: \(\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}\)
  3. \(\frac{-15}{3} = -5\)
    • Classification: \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}\)
  4. \(0.101001000100001...\) (pattern but non-repeating)
    • Classification: \(\mathbb{R}\) only (irrational)

Problem 7: Calculator – Fraction Calculations (x)

Use your calculator to evaluate the following expressions. Verify your answers are in simplified form.

  1. \(\frac{3}{4} + \frac{5}{6} - \frac{1}{2}\)

  2. \(\frac{7}{12} \times \frac{8}{21}\)

  3. \(2\frac{1}{3} + 1\frac{3}{4}\)

  4. \(\frac{15}{8} \div \frac{5}{12}\)

  1. \(\frac{3}{4} + \frac{5}{6} - \frac{1}{2}\)
    • Find common denominator: \(\frac{9}{12} + \frac{10}{12} - \frac{6}{12} = \frac{13}{12}\)
    • Answer: \(\frac{13}{12}\) or \(1\frac{1}{12}\)
  2. \(\frac{7}{12} \times \frac{8}{21}\)
    • Multiply and simplify: \(\frac{56}{252} = \frac{2}{9}\)
    • Answer: \(\frac{2}{9}\)
  3. \(2\frac{1}{3} + 1\frac{3}{4}\)
    • Convert to improper: \(\frac{7}{3} + \frac{7}{4} = \frac{28}{12} + \frac{21}{12} = \frac{49}{12}\)
    • Answer: \(\frac{49}{12}\) or \(4\frac{1}{12}\)
  4. \(\frac{15}{8} \div \frac{5}{12}\)
    • Multiply by reciprocal: \(\frac{15}{8} \times \frac{12}{5} = \frac{180}{40} = \frac{9}{2}\)
    • Answer: \(\frac{9}{2}\) or \(4\frac{1}{2}\)

Problem 8: Calculator – Percentage Calculations (x)

Use your calculator’s percentage function to solve:

  1. What is 15% of 240?

  2. Increase 180 by 12%

  3. Decrease 500 by 8%

  4. A product costs 85€ after a 15% discount. What was the original price?

  1. 15% of 240
    • Enter: 240 × 15 SHIFT (%) =
    • Answer: 36
  2. Increase 180 by 12%
    • Enter: 180 + 180 × 12 SHIFT (%) =
    • Answer: 201.6
  3. Decrease 500 by 8%
    • Enter: 500 - 500 × 8 SHIFT (%) =
    • Answer: 460
  4. Original price if 85€ is 15% off
    • 85€ = 85% of original → Original = 85 ÷ 0.85
    • Answer: 100€

Problem 9: Calculator – Prime Factorization (x)

Use the FACT function (SHIFT + FORMAT → Primfakt.) to find:

  1. Prime factorization of 360

  2. Prime factorization of 1260

  3. Use prime factorization to simplify \(\frac{84}{126}\)

  1. 360
    • Enter: 360 = then SHIFT FORMAT → Primfakt.
    • Answer: \(2^3 \times 3^2 \times 5\)
  2. 1260
    • Enter: 1260 = then SHIFT FORMAT → Primfakt.
    • Answer: \(2^2 \times 3^2 \times 5 \times 7\)
  3. Simplify \(\frac{84}{126}\)
    • 84 = \(2^2 \times 3 \times 7\)
    • 126 = \(2 \times 3^2 \times 7\)
    • GCD = \(2 \times 3 \times 7 = 42\)
    • Answer: \(\frac{84}{126} = \frac{2}{3}\)

Problem 10: Calculator – GCD and LCM (x)

Use the GCD and LCM functions:

  1. Find GCD(48, 72)

  2. Find LCM(12, 18)

  3. Use GCD to simplify \(\frac{144}{180}\)

  4. Find the smallest number divisible by both 15 and 20

  1. GCD(48, 72)
    • Enter: GCD(48; 72) =
    • Answer: 24
  2. LCM(12, 18)
    • Enter: LCM(12; 18) =
    • Answer: 36
  3. Simplify \(\frac{144}{180}\)
    • GCD(144, 180) = 36
    • \(\frac{144}{180} = \frac{144 \div 36}{180 \div 36} = \frac{4}{5}\)
    • Answer: \(\frac{4}{5}\)
  4. Smallest number divisible by 15 and 20
    • LCM(15, 20) = 60
    • Answer: 60