Mathematics for Master Students

I really appreciate active participation and questions: the more you ask, the more useful these sessions become for you!
| Session | Topic | Date | Time | Room |
|---|---|---|---|---|
| 1 | Notation & Logic | Tue 8 Sep 2026 | 18:00–19:30 | Auditorium |
| 2 | Sets | Tue 15 Sep 2026 | 18:00–19:30 | Auditorium |
| 3 | Functions | Tue 22 Sep 2026 | 18:00–19:30 | Zoom |
| 4 | Differentiation | Tue 29 Sep 2026 | 18:00–19:30 | Zoom |
| 5 | Single Variable Optimization | Tue 6 Oct 2026 | 18:00–19:30 | Auditorium |
| 6 | Systems of Linear Equations | Tue 13 Oct 2026 | 18:00–19:30 | Auditorium |
| 7 | Mock Exam (voluntary!) | Tue 27 Oct 2026 | 14:30–16:00 | Auditorium |
Six sessions of 90 minutes plus a voluntary mock exam: a compact refresher of the mathematics used in your M.Sc. courses.
Semester 1
| Course | Notation | Logic | Sets | Functions | Differentiation | Linear Systems | Optimization |
|---|---|---|---|---|---|---|---|
| Statistics & Econometrics | ✓ | ✓ | ✓ | ✓ | |||
| Business Logistics & SCM | ✓ | ✓ | ✓ | ✓ | |||
| Fundamentals of Business Analytics | ✓ | ✓ | ✓ | ✓ | ✓ | ||
| International Economics | ✓ | ✓ | ✓ | ✓ | ✓ | ||
| Strategic Issues in SCM | ✓ | ✓ | ✓ | ||||
| Fundamentals of Data Science | ✓ | ✓ | ✓ | ✓ | ✓ |
Semester 2
| Course | Notation | Logic | Sets | Functions | Differentiation | Linear Systems | Optimization |
|---|---|---|---|---|---|---|---|
| Transportation & Distribution | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ |
| Global Trends in Human Resources | ✓ | ✓ | |||||
| Economics of Digital Transformation | ✓ | ✓ | |||||
| Economics of Business Strategy | ✓ | ✓ | ✓ | ✓ | ✓ | ||
| Business Strategy & Sustainability | ✓ |
Today’s topics, Notation and Logic, appear in almost every course.
Each session comes with four materials:
You work on the tasks between sessions. Every session starts with a recap and your questions.
Ask any question you have, no matter how simple it may seem. If you are lost, even after something was explained multiple times, ask again: that is exactly what this tutorial is for!
Use AI to learn, not to outsource your thinking. If the chatbot solves your homework, you have practised prompting, not mathematics.
Always question the maths output of LLMs!
For actual computation, tools like Wolfram Alpha are more reliable, as they use symbolic computation rather than pattern matching.
Mathematics is a language, with vocabulary and grammar:
\[\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\]
Some books include \(0\) in \(\mathbb{N}\). In this tutorial, \(\mathbb{N} = \{1, 2, 3, \dots\}\), and we write \(\mathbb{N}_0\) if zero is included.
Question
Which of these numbers are rational: \(\quad 0.75, \quad \sqrt{2}, \quad \frac{7}{3}, \quad \pi\)?
Common Mistake
\(\pi \approx \frac{22}{7}\), but \(\pi \neq \frac{22}{7}\): an approximation is not an equality!
Intervals describe connected ranges of real numbers:
Square bracket \([\) means the endpoint is included, round parenthesis \((\) means it is excluded. Think of the bracket as “grabbing” the endpoint. \(\infty\) always gets a parenthesis: it is not a number you could include.
Question
A warehouse accepts pallets weighing at least \(200\) kg and strictly less than \(1000\) kg. Write the admissible weight \(w\) as an interval.
\[w \in [200, 1000)\]
Common Mistake
“At least” includes the endpoint (square bracket), “strictly less than” excludes it (parenthesis). \((200, 1000]\) has the brackets on the wrong sides: it would reject a \(200\) kg pallet and accept a \(1000\) kg one.
| Symbol | Meaning | Business example |
|---|---|---|
| \(<\) | strictly less than | \(x < 100\) (below the threshold) |
| \(\leq\) | less than or equal | \(x \leq 500\) (capacity limit) |
| \(\geq\) | greater than or equal | \(q \geq 20\) (minimum order quantity) |
| \(\neq\) | not equal to | \(\text{supply} \neq \text{demand}\) |
Inequalities can be chained: \(\quad 0 \leq x \leq 500\)
means “\(x\) is at least \(0\) and at most \(500\)”, as an interval: \(x \in [0, 500]\).
The absolute value \(|x|\) is the distance of \(x\) from zero:
\[|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}\]
The symbol \(\sum\) (capital sigma) compresses long sums:
\[\sum_{i=1}^{n} x_i = x_1 + x_2 + \dots + x_n\]
Let \(c_t\) be your company’s cost in month \(t\). The total annual cost is
\[\sum_{t=1}^{12} c_t = c_1 + c_2 + \dots + c_{12}\]
Two rules you will use constantly:
Read \(\sum\) as a small loop: “for each \(t\) from 1 to 12, add up …”.
Question
Compute \(\sum_{i=1}^{4} 2i\). What is the result?
Write out the loop, one term per index value:
\[\sum_{i=1}^{4} 2i = 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4 = 2 + 4 + 6 + 8 = 20\]
Or move the constant out first: \(2 \sum_{i=1}^{4} i = 2 \cdot 10 = 20\).
The symbol \(\prod\) (capital pi) works like \(\sum\), but with multiplication:
\[\prod_{i=1}^{n} x_i = x_1 \cdot x_2 \cdot \dots \cdot x_n\]
Business example: capital grows for three years at rates \(r_1, r_2, r_3\):
\[K_3 = K_0 \cdot \prod_{t=1}^{3} (1 + r_t) = K_0 (1+r_1)(1+r_2)(1+r_3)\]
Growth compounds: that is why it is a product, not a sum.
Question
Revenue grows by \(10\,\%\), then \(20\,\%\), then \(5\,\%\) over three years. By what total percentage has it grown?
\[\prod_{t=1}^{3} (1 + r_t) = 1.10 \cdot 1.20 \cdot 1.05 = 1.386\]
so the total growth is \(38.6\,\%\).
Common Mistake
\(10 + 20 + 5 = 35\,\%\) is wrong: each year’s growth applies to the already grown revenue. Rates multiply as factors, they do not add.
Many logistics quantities depend on two things at once:
Whenever you meet a variable, first ask: what do its indices mean? A well-defined model always states this explicitly.
With \(n\) warehouses and \(m\) customers:
\[\sum_{i=1}^{n} \sum_{j=1}^{m} c_{ij} \, x_{ij}\]
This is the objective function of the classic transportation problem. You will meet it again in Transportation & Distribution.
Question
Two warehouses, two customers. Compute \(\sum_{i=1}^{2} \sum_{j=1}^{2} c_{ij} \, x_{ij}\).
| \(c_{i1}\) | \(c_{i2}\) | \(x_{i1}\) | \(x_{i2}\) | ||
|---|---|---|---|---|---|
| \(i = 1\) | 2 | 5 | 10 | 0 | |
| \(i = 2\) | 4 | 3 | 5 | 20 |
\[2 \cdot 10 + 5 \cdot 0 + 4 \cdot 5 + 3 \cdot 20 = 20 + 0 + 20 + 60 = 100\]
Bonus: everything leaving warehouse \(2\) is \(\sum_{j=1}^{2} x_{2j} = 5 + 20 = 25\). Same data, different index fixed.
Greek letters are everywhere in your courses, a quick reference:
| Letter | Name | Typical use |
|---|---|---|
| \(\alpha\) | alpha | significance level |
| \(\beta\) | beta | regression coefficient |
| \(\delta, \Delta\) | delta | (small) change, difference |
| \(\varepsilon\) | epsilon | error term, small number |
| \(\lambda\) | lambda | arrival rate |
| Letter | Name | Typical use |
|---|---|---|
| \(\mu\) | mu | mean |
| \(\pi\) | pi | \(3.14159\dots\), profit |
| \(\rho\) | rho | correlation |
| \(\sigma, \Sigma\) | sigma | std. deviation, sum |
| \(\theta\) | theta | generic parameter |
A statement (proposition) is a declarative sentence that is either true or false, never both.
Logic does not care what a statement is about, only that it has exactly one truth value.
An open sentence becomes a statement once every variable is fixed or quantified: that is why defining your variables matters so much.
The negation \(\neg p\) (“not \(p\)”) flips the truth value of \(p\):
Common Mistake
The negation of “the warehouse is full” is not “the warehouse is empty”! Between full and empty there is a lot of room: the negation only claims “not full”.
\(p \wedge q\) (“\(p\) and \(q\)”) \(\qquad\qquad\) \(p \vee q\) (“\(p\) or \(q\)”)
| \(p\) | \(q\) | \(p \wedge q\) | \(p \vee q\) |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | F | T |
| F | F | F | F |
The mathematical “or” is inclusive: \(p \vee q\) is also true when both hold. “Dessert or coffee?” in maths means: taking both is fine, too.
Negation flips “and” into “or”, and vice versa:
\[\neg(p \wedge q) \Leftrightarrow \neg p \vee \neg q \qquad\qquad \neg(p \vee q) \Leftrightarrow \neg p \wedge \neg q\]
Business example: “The shipment is on time and complete.”
Its negation: “The shipment is late or incomplete”: one failure is enough.
You will meet the same pattern again in a moment: negating a statement swaps \(\forall\) and \(\exists\), just like it swaps \(\wedge\) and \(\vee\).
\(p \Rightarrow q\): “if \(p\), then \(q\)”, \(p\) is the premise, \(q\) the conclusion.
| \(p\) | \(q\) | \(p \Rightarrow q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Think of a promise
“If your order exceeds €500 (\(p\)), shipping is free (\(q\)).” The promise is only broken when the order exceeds €500 but shipping is charged: the row \(p\) = T, \(q\) = F. If \(p\) is false, nothing was promised.
Question
“If the order exceeds €500, shipping is free.” Which of these cases breaks the promise?
| (a) Order €600, shipping charged | (b) Order €300, shipping charged |
| (c) Order €300, shipping free | (d) Order €600, shipping free |
Only (a): premise true, conclusion false, the one row where \(p \Rightarrow q\) is F.
Common Mistake
Cases (b) and (c) do not break the promise: for an order of €300, nothing was promised at all. An implication with a false premise is true, whatever the conclusion.
\(p \Leftrightarrow q\): “\(p\) if and only if \(q\)”: both have the same truth value, i.e. \(p \Rightarrow q\) and \(q \Rightarrow p\).
| \(p\) | \(q\) | \(p \Leftrightarrow q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
Common Mistake
\(\Rightarrow\) and \(\Leftrightarrow\) are not interchangeable: \(x = 2 \Rightarrow x^2 = 4\) holds, but \(x^2 = 4 \Rightarrow x = 2\) does not (it could be \(-2\)).
Given \(p \Rightarrow q\):
“If someone is a millionaire (\(p\)), then they are rich (\(q\)).”
Question
“A container clears the port only if its customs declaration is valid.” Is a valid declaration a necessary or a sufficient condition for clearance?
Necessary. The sentence says: clearance \(\Rightarrow\) valid declaration. Without a valid declaration, clearance is impossible.
Common Mistake
It is not sufficient: a valid declaration alone does not guarantee clearance, an inspection or unpaid duties can still stop the container. “Only if” points to a necessary condition, “if” to a sufficient one.
Two symbols turn open sentences into statements:
In business language:
Negation swaps the quantifier and negates the inside:
\[\neg\,(\forall x: P(x)) \;\Leftrightarrow\; \exists x: \neg P(x)\]
\[\neg\,(\exists x: P(x)) \;\Leftrightarrow\; \forall x: \neg P(x)\]
Question
Negate: “All shipments arrive on time.”
“At least one shipment does not arrive on time.”
Common Mistake
The negation is not “no shipment arrives on time”: that is far too strong. One single late shipment already makes the original statement false.
We know: “If a shipment is delayed (\(p\)), the customer complains (\(q\)).”
Today, a customer complained. Was the shipment delayed?
Common Mistake
Concluding \(p\) from \(p \Rightarrow q\) and \(q\) is the converse error (“affirming the consequent”), one of the most frequent reasoning errors in business and in exams.
From \(p \Rightarrow q\) (“delayed \(\Rightarrow\) complaint”):
| Derived statement | Form | Valid? |
|---|---|---|
| No complaint \(\Rightarrow\) not delayed | contrapositive \(\neg q \Rightarrow \neg p\) | ✓ |
| Complaint \(\Rightarrow\) delayed | converse \(q \Rightarrow p\) | ✗ |
| Not delayed \(\Rightarrow\) no complaint | inverse \(\neg p \Rightarrow \neg q\) | ✗ |
The contrapositive is always equivalent to the original implication: flip and negate. Flipping alone (converse) or negating alone (inverse) breaks the logic.
That’s it for today.
Any Questions?
The tasks are where the actual learning happens: the lecture only opens the door.
See you there, and bring your questions!
Mathematics also has formal proof techniques: direct proof, proof by contradiction, and induction. They are beyond the scope of this tutorial, but the textbooks above cover them if you are curious.
Session 01 - Notation & Logic | Dr. Tobias Vlcek | Home