Tasks 01 - Notation & Logic

Mathematics for Master Students

These problems accompany Session 01 of the tutorial. Work through them at your own pace before the next session. We start Session 02 with your questions. On the website, every problem has a fully worked solution you can expand to check yourself. The PDF version contains only the problems, so you can print it and use it as a worksheet.

The problems are staged: the first two are warm-ups, the middle ones practice the core techniques, and the last two are at the level of the mock exam.

Problem 1: Classifying Numbers

For each of the following numbers, list all number sets among \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\) and \(\mathbb{R}\) that the number belongs to. Simplify first where possible. Remember the convention of this tutorial: \(\mathbb{N} = \{1, 2, 3, \dots\}\), and we write \(\mathbb{N}_0\) if zero is included.

  1. \(7\)
  2. \(-3\)
  3. \(\frac{12}{4}\)
  4. \(0.2\)
  5. \(\sqrt{2}\)
  6. \(0\)

Step 1 - Simplify where possible

Two of the numbers are disguised: \(\frac{12}{4} = 3\) and \(0.2 = \frac{2}{10} = \frac{1}{5}\). Always classify the value, not the way it is written.

Step 2 - Classify each number

  1. \(7\) is a counting number: \(7 \in \mathbb{N}\), and therefore also \(7 \in \mathbb{Z}\), \(7 \in \mathbb{Q}\), \(7 \in \mathbb{R}\): the sets are nested, \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\).
  2. \(-3\) is negative, so it is not a natural number, but \(-3 \in \mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\).
  3. \(\frac{12}{4} = 3\), so despite looking like a fraction it belongs to \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\).
  4. \(0.2 = \frac{1}{5}\) is a ratio of two integers, so \(0.2 \in \mathbb{Q}\) and \(\mathbb{R}\), but it is neither an integer nor a natural number.
  5. \(\sqrt{2}\) is irrational: it is a point on the number line, but no fraction of integers equals it exactly. So \(\sqrt{2} \in \mathbb{R}\) only.
  6. \(0 \in \mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\). In this tutorial \(0 \notin \mathbb{N}\). If we want to include it, we write \(0 \in \mathbb{N}_0\).

Problem 2: Intervals, Inequalities and Absolute Value

Translate between the different ways of describing a range of numbers.

  1. A supplier requires a minimum order quantity of \(20\) units and delivers at most \(500\) units per order. Write the condition on the order quantity \(q\) as a chained inequality and in interval notation.
  2. Pharmaceutical goods must be stored at a temperature \(T\) (in °C) of at least \(2\) and strictly below \(8\). Write this as a chained inequality and in interval notation. Is a measured temperature of exactly \(8.0\) °C acceptable?
  3. Write \(x \in (0, 12]\) as an inequality and state in words which endpoints are included.
  4. A filling machine is set to \(750\) ml per bottle. Quality control accepts a bottle if its content \(x\) satisfies \(|x - 750| \leq 5\). Describe the set of acceptable contents as an interval.

  1. “At least \(20\)” means \(q \geq 20\) and “at most \(500\)” means \(q \leq 500\). Chained: \(20 \leq q \leq 500\). Both endpoints are allowed, so the interval is closed: \(q \in [20, 500]\). (Strictly speaking, you can only order whole units, so additionally \(q \in \mathbb{N}\); combining conditions like this is exactly how real models are written.)

  2. “At least \(2\)” includes the endpoint, “strictly below \(8\)” excludes it: \(2 \leq T < 8\), i.e. the half-open interval \(T \in [2, 8)\). A temperature of exactly \(8.0\) °C is not acceptable: the round parenthesis excludes the endpoint \(8\).

  3. \(x \in (0, 12]\) means \(0 < x \leq 12\): the left endpoint \(0\) is excluded (round parenthesis), the right endpoint \(12\) is included (square bracket).

  4. \(|x - 750|\) is the distance between \(x\) and \(750\), so the condition says: the content deviates from \(750\) ml by at most \(5\) ml in either direction.

    Step 1 - Unpack the absolute value: \(|x - 750| \leq 5\) means \(-5 \leq x - 750 \leq 5\).

    Step 2 - Add \(750\) to all three parts: \(745 \leq x \leq 755\).

    The set of acceptable contents is the closed interval \([745, 755]\).

Problem 3: Sum and Product Notation

  1. Evaluate \(\sum_{i=1}^{5} (2i - 1)\).
  2. Evaluate \(\prod_{i=1}^{4} \frac{i+1}{i}\).
  3. Write \(x_3 + x_4 + \dots + x_{10}\) in sum notation. How many terms does the sum have?
  4. A company’s variable monthly costs \(c_t\) (in €) added up to \(\sum_{t=1}^{12} c_t = 36{,}000\) last year. A new contract gives a \(10\%\) discount on all variable costs but adds a fixed service fee of €500 per month. Evaluate the new annual total \(\sum_{t=1}^{12} (0.9 \, c_t + 500)\), without knowing the individual \(c_t\).

  1. Read \(\sum\) as a loop, one term for each \(i\) from \(1\) to \(5\): \[\sum_{i=1}^{5} (2i - 1) = 1 + 3 + 5 + 7 + 9 = 25\]

  2. Same loop idea, but multiply. Write out all four factors: \[\prod_{i=1}^{4} \frac{i+1}{i} = \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} = \frac{2 \cdot 3 \cdot 4 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4} = 5\] Almost everything cancels: each numerator reappears as the next denominator, leaving only \(\frac{5}{1} = 5\).

  3. The index starts at \(3\) and ends at \(10\): \[x_3 + x_4 + \dots + x_{10} = \sum_{i=3}^{10} x_i\] Number of terms: \(10 - 3 + 1 = 8\). (Careful: not \(10 - 3 = 7\), both endpoints count, so add one.)

  4. Step 1 - Split the sum over the addition: \[\sum_{t=1}^{12} (0.9 \, c_t + 500) = \sum_{t=1}^{12} 0.9 \, c_t + \sum_{t=1}^{12} 500\]

    Step 2 - Move the constant factor out and evaluate the constant sum. Adding €500 twelve times gives \(12 \cdot 500\): \[= 0.9 \sum_{t=1}^{12} c_t + 12 \cdot 500\]

    Step 3 - Insert the known total: \[= 0.9 \cdot 36{,}000 + 6{,}000 = 32{,}400 + 6{,}000 = 38{,}400\]

    The new annual total is €38,400. The two sum rules let us compute it without ever knowing a single monthly value.

Problem 4: Warehouse Flows with Double Indices

A company ships goods from two warehouses (\(i = 1, 2\)) to three customers (\(j = 1, 2, 3\)). Let \(x_{ij}\) be the quantity (in units) shipped from warehouse \(i\) to customer \(j\), and \(c_{ij}\) the transport cost per unit (in €) on that route:

\(x_{ij}\) Customer 1 Customer 2 Customer 3
Warehouse 1 30 10 20
Warehouse 2 0 25 15
\(c_{ij}\) Customer 1 Customer 2 Customer 3
Warehouse 1 2 4 3
Warehouse 2 5 1 2
  1. What does \(x_{12}\) mean in words, and what is its value?
  2. Write “the total quantity leaving warehouse 1” in sum notation and compute it.
  3. Write “the total quantity received by customer 2” in sum notation and compute it.
  4. Compute the total transport cost \(\sum_{i=1}^{2} \sum_{j=1}^{3} c_{ij} \, x_{ij}\).

  1. The first index is the origin, the second the destination: \(x_{12}\) is the quantity shipped from warehouse 1 to customer 2. From the table (first row, second column): \(x_{12} = 10\) units.

  2. Warehouse \(1\) is fixed (\(i = 1\)), and we add over all destinations \(j\): \[\sum_{j=1}^{3} x_{1j} = x_{11} + x_{12} + x_{13} = 30 + 10 + 20 = 60 \text{ units}\] In the table, this is the sum of the first row.

  3. Now the customer is fixed (\(j = 2\)), and we add over all origins \(i\): \[\sum_{i=1}^{2} x_{i2} = x_{12} + x_{22} = 10 + 25 = 35 \text{ units}\] In the table, this is the sum of the second column.

  4. The double sum visits every route once: multiply cost and quantity per route, then add everything up. Row by row:

    Warehouse 1: \(\quad 2 \cdot 30 + 4 \cdot 10 + 3 \cdot 20 = 60 + 40 + 60 = 160\)

    Warehouse 2: \(\quad 5 \cdot 0 + 1 \cdot 25 + 2 \cdot 15 = 0 + 25 + 30 = 55\)

    Total: \[\sum_{i=1}^{2} \sum_{j=1}^{3} c_{ij} \, x_{ij} = 160 + 55 = 215\]

    The total transport cost is €215. This is exactly the objective function of the transportation problem from the lecture, just with concrete numbers.

Problem 5: Truth Tables and De Morgan’s Laws

Let \(p\) and \(q\) be two statements.

  1. Complete the truth table:

    \(p\) \(q\) \(\neg p\) \(p \wedge q\) \(p \vee q\)
    T T
    T F
    F T
    F F

  2. Extend the table by two columns: \(\neg(p \wedge q)\) and \(\neg p \vee \neg q\). Compare them. Which law do you recognize?

  3. Using that law, negate in plain English: “The shipment arrived on time and the invoice was correct.”

  4. Negate in plain English: “The container is damaged or it was misrouted.”

  1. Work column by column: \(\neg p\) flips \(p\); \(p \wedge q\) is true only in the first row; \(p \vee q\) is false only in the last row (the mathematical “or” is inclusive):

    \(p\) \(q\) \(\neg p\) \(p \wedge q\) \(p \vee q\)
    T T F T T
    T F F F T
    F T T F T
    F F T F F

  2. Negate the \(p \wedge q\) column for the first new column; combine the \(\neg p\) column with \(\neg q\) via “or” for the second:

    \(p\) \(q\) \(\neg(p \wedge q)\) \(\neg p \vee \neg q\)
    T T F F
    T F T T
    F T T T
    F F T T

    The two columns are identical in every row. This is De Morgan’s law: \(\neg(p \wedge q) \Leftrightarrow \neg p \vee \neg q\). Negation flips “and” into “or”.

  3. Let \(p\): “the shipment arrived on time”, \(q\): “the invoice was correct”. The statement is \(p \wedge q\), so its negation is \(\neg p \vee \neg q\): “The shipment was late or the invoice was incorrect.” One failure is enough to break the original statement, and the “or” is inclusive, so both may have gone wrong.

  4. This time negation flips “or” into “and” (\(\neg(p \vee q) \Leftrightarrow \neg p \wedge \neg q\)): “The container is not damaged and it was not misrouted.” In other words: nothing went wrong on either count.

Problem 6: Implications in Quality Management

A logistics company works with the rule: “If a supplier is certified, then it has passed the quality audit.”

  1. Define suitable propositions \(p\) and \(q\) and write the rule in logical notation.
  2. State the converse, the inverse and the contrapositive of the rule, each in notation and in plain English. Which of the three is equivalent to the original rule?
  3. For each situation, state what you can conclude from the rule, if anything:
    1. Supplier A has passed the quality audit.
    2. Supplier B has not passed the quality audit.
    3. Supplier C is not certified.
  4. Fill in necessary or sufficient: “Passing the audit is a ______ condition for being certified. Being certified is a ______ condition for having passed the audit.”

  1. Let \(p\): “the supplier is certified” and \(q\): “the supplier has passed the quality audit”. The rule is the implication \(p \Rightarrow q\).

  2. The three derived statements:

    Derived statement Form In plain English Equivalent?
    Converse \(q \Rightarrow p\) “If a supplier passed the audit, it is certified.” ✗
    Inverse \(\neg p \Rightarrow \neg q\) “If a supplier is not certified, it did not pass the audit.” ✗
    Contrapositive \(\neg q \Rightarrow \neg p\) “If a supplier did not pass the audit, it is not certified.” ✓

    Only the contrapositive is equivalent to \(p \Rightarrow q\): flip and negate. Flipping alone (converse) or negating alone (inverse) breaks the logic.

    1. Nothing follows. Concluding “A is certified” from \(p \Rightarrow q\) and \(q\) would be the converse error: a supplier can pass the audit without (or before) being certified.
    2. B is not certified. This is the contrapositive \(\neg q \Rightarrow \neg p\), which is valid: if B were certified, it would have passed the audit.
    3. Nothing follows. Concluding “C did not pass” would use the inverse \(\neg p \Rightarrow \neg q\), which is not valid: C might have passed the audit but simply not (yet) hold a certificate.

  4. Passing the audit is a necessary condition for being certified: without it, certification is impossible. Being certified is a sufficient condition for having passed the audit: knowing a supplier is certified already guarantees it.

Problem 7: Negating Quantified Statements

For each statement below, first write it in quantifier notation (introduce a suitable set and predicate), then write the negation both formally and in plain English.

  1. “Every delivery last week arrived on time.”
  2. “Some warehouse operates above \(90\%\) of its capacity.”
  3. “There exists a natural number \(n\) with \(n^2 = 20\).” Additionally: which is true, the statement or its negation? Justify your answer.

Remember the rule: negation swaps the quantifier and negates the inside: \(\neg(\forall x: P(x)) \Leftrightarrow \exists x: \neg P(x)\) and \(\neg(\exists x: P(x)) \Leftrightarrow \forall x: \neg P(x)\).

  1. Step 1 - Formalize. Let \(D\) be the set of last week’s deliveries and \(T(d)\): “delivery \(d\) arrived on time”. The statement is \[\forall d \in D: T(d)\]

    Step 2 - Negate. Swap \(\forall\) to \(\exists\) and negate the predicate: \[\exists d \in D: \neg T(d)\]

    In plain English: “At least one delivery last week did not arrive on time.” (Not: “no delivery arrived on time”; that is far too strong. A single late delivery already refutes the original claim.)

  2. Step 1 - Formalize. Let \(W\) be the set of warehouses and \(C(w)\): “warehouse \(w\) operates above \(90\%\) of its capacity”. The statement is \[\exists w \in W: C(w)\]

    Step 2 - Negate. Swap \(\exists\) to \(\forall\) and negate the predicate: \[\forall w \in W: \neg C(w)\]

    In plain English: “Every warehouse operates at at most \(90\%\) of its capacity.” Note that “above \(90\%\)” negates to “at most \(90\%\)”: exactly \(90\%\) belongs to the negation, since it is not above \(90\%\).

  3. Step 1 - Formalize. The statement is \[\exists n \in \mathbb{N}: n^2 = 20\]

    Step 2 - Negate: \[\forall n \in \mathbb{N}: n^2 \neq 20\]

    In plain English: “No natural number squared equals \(20\).”

    Step 3 - Which is true? The negation is true. The squares of natural numbers are \(1, 4, 9, 16, 25, \dots\): for \(n \leq 4\) we have \(n^2 \leq 16 < 20\), and for \(n \geq 5\) we have \(n^2 \geq 25 > 20\). The squares jump right over \(20\), so no natural number works. (\(\sqrt{20} \approx 4.47\) is a real number, but not a natural one.)

Problem 8: True or False, One Number Decides

A fellow student preparing for the mock exam writes down the four claims below. For each claim, decide whether it is true or false and justify your decision with a single well-chosen number (or pair of numbers). Remember from the lecture: to disprove a \(\forall\) statement, one counterexample suffices; to prove an \(\exists\) statement, one witness suffices.

  1. \(\forall x \in \mathbb{R}: x^2 \geq x\)
  2. \(\forall a, b \in \mathbb{R}: |a + b| = |a| + |b|\)
  3. \(\exists n \in \mathbb{N}: n^2 + n = 12\)
  4. “Every rational number is an integer.”

  1. False. Counterexample: \(x = \frac{1}{2}\), because \(x^2 = \frac{1}{4} < \frac{1}{2} = x\). The claim feels true if you only test integers (it holds for all \(x \leq 0\) and all \(x \geq 1\)), but every number strictly between \(0\) and \(1\) gets smaller when squared. Always test values from different regions before believing a \(\forall\) claim.

  2. False. Counterexample: \(a = 5\), \(b = -3\), because \[|a + b| = |5 + (-3)| = |2| = 2 \quad \text{but} \quad |a| + |b| = 5 + 3 = 8\] Absolute values do not simply add up when the signs differ: a gain of \(5\) and a loss of \(3\) leave a net change of \(2\), while the total movement is \(8\). (If \(a\) and \(b\) have the same sign, equality does hold, but a \(\forall\) claim must hold for all choices, so one failing pair settles it.)

  3. True. Witness: \(n = 3\), because \(3^2 + 3 = 9 + 3 = 12\). For an \(\exists\) statement, exhibiting one working number is a complete justification. No further checking is needed.

  4. False. In quantifier form the claim is \(\forall q \in \mathbb{Q}: q \in \mathbb{Z}\). Counterexample: \(q = \frac{1}{2}\) is rational (a ratio of two integers) but clearly not an integer. The correct relationship runs the other way: every integer is rational, i.e. \(\mathbb{Z} \subset \mathbb{Q}\); compare the hierarchy \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\) from Problem 1.