Session 01 - Notation & Logic
Mathematics for Master Students
Introduction
About Me
- Field: Optimising & simulating complex systems
- Languages of choice: Julia, Python and Rust
- Interests: Modelling, simulations, machine learning
- Teaching: OR, algorithms, programming & mathematics
- More: tobiasvlcek.com
I really appreciate active participation and questions: the more you ask, the more useful these sessions become for you!
Which Price Would You Charge?
A small coffee roastery sells 1 kg bags of coffee.
- At €30 a bag, it sells \(600\) bags a week
- Every extra euro on the price loses \(20\) bags a week
- Each bag costs €5 to roast, plus €200 fixed costs per week
Which price earns the roastery the most profit per week: €20, €30 or €40? Vote by hands.
Keep your vote: in Session 3 you will compute the answer yourself.
Today’s Plan
- One question the six sessions answer, and how the tutorial works
- How to make good use of AI while studying
- Notation: the vocabulary of mathematics, enough to read one line of a transport model by the end of today
- Logic: the grammar of mathematical reasoning
- Your homework until the next session
Tutorial Structure: Sessions 1–4
| 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 |
Tutorial Structure: Sessions 5–7
| Session | Topic | Date | Time | Room |
|---|---|---|---|---|
| 5 | Single Variable Optimisation | Tue 6 Oct 2026 | 18:00–19:30 | Auditorium |
| 6 | Systems of Linear Equations | Tue 13 Oct 2026 | 18:00–19:30 | GF Lecture 1-3 |
| 7 | Mock Exam (voluntary!) | Tue 27 Oct 2026 | 18:00–19:30 | GF Lecture 1-3 |
Six sessions of 90 minutes plus a voluntary mock exam: a compact introduction to the mathematics used in your M.Sc. courses.
One Question, Six Sessions
Answering the roastery’s question takes every session:
- Notation & logic: write the roastery’s rules, like “at most 500 bags a week”, precisely
- Sets: collect every weekly amount it can roast into one set
- Functions: price → bags sold → revenue, cost and profit, as one chain
- Differentiation: what does a euro on the price do to revenue?
- Optimisation: find the point where one more euro stops helping
- Linear systems: blend two beans to hit a target cost per kg
Semester 1 Courses: Sessions 1–3
| Course | Notation | Logic | Sets | Functions |
|---|---|---|---|---|
| Statistics & Econometrics | ✓ | ✓ | ✓ | |
| Business Logistics & SCM | ✓ | ✓ | ||
| Fundamentals of Business Analytics | ✓ | ✓ | ||
| International Economics | ✓ | ✓ | ||
| Strategic Issues in SCM | ✓ | ✓ | ||
| Fundamentals of Data Science | ✓ | ✓ | ✓ | ✓ |
Semester 1 Courses: Sessions 4–6
| Course | Differentiation | Optimisation | Linear Systems |
|---|---|---|---|
| Statistics & Econometrics | ✓ | ||
| Business Logistics & SCM | ✓ | ✓ | |
| Fundamentals of Business Analytics | ✓ | ✓ | ✓ |
| International Economics | ✓ | ✓ | ✓ |
| Strategic Issues in SCM | ✓ | ||
| Fundamentals of Data Science | ✓ |
Semester 2 Courses: Sessions 1–3
| Course | Notation | Logic | Sets | Functions |
|---|---|---|---|---|
| Transportation & Distribution | ✓ | ✓ | ✓ | ✓ |
| Global Trends in Human Resources | ✓ | |||
| Economics of Digital Transformation | ✓ | |||
| Economics of Business Strategy | ✓ | ✓ | ||
| Business Strategy & Sustainability | ✓ |
Semester 2 Courses: Sessions 4–6
| Course | Differentiation | Optimisation | Linear Systems |
|---|---|---|---|
| Transportation & Distribution | ✓ | ✓ | ✓ |
| Global Trends in Human Resources | ✓ | ||
| Economics of Digital Transformation | ✓ | ||
| Economics of Business Strategy | ✓ | ✓ | ✓ |
| Business Strategy & Sustainability |
The Two Most Common Topics
Overall, Notation and Functions each appear in 8 of your 11 courses.
How This Tutorial Works
math-master.tobiasvlcek.com holds four materials for each session:
- Lecture: web page, slides and PDF
- Tasks: a printable worksheet, worked solutions on the website
- Self-Test: a short online quiz to check yourself
- Cheatsheet: a compact reference to keep at hand
You work on the tasks between sessions. From Session 2 on, every session starts with a recap and your questions.
No Grades, Just Preparation
- This tutorial is not graded: it is purely preparatory
- Nobody tracks your answers or your progress
- The only goal: you follow your M.Sc. courses with ease
- The mock exam in Session 7 is voluntary: Moodle marks it, and the score does not count
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!
Artificial Intelligence
Can You Trust Its Maths?
A chatbot says \(36 \cdot 24 = 872\). Can you spot the error in ten seconds, without a calculator?
The last digit gives it away: \(6 \cdot 4 = 24\), so it must end in \(4\) (it is \(864\)).
- Large language models (LLMs) predict plausible text: without a calculator tool, nothing guarantees an exact calculation
- They explain concepts well, and can confidently produce wrong numbers
For computation, Wolfram Alpha is more reliable: it uses symbolic computation.
The Chat Bubble & the Math Helper
- Chat bubble on every page: quick, direct answers about the tutorial material
- Math Helper: hints and follow-up questions until you solve it yourself
- Ask as specifically as possible: say what you tried and where you got stuck
- Both are set up by me for this tutorial
- Please don’t share personal information in the chat
How to Use AI in This Tutorial
- Feel free to use AI tools to support your understanding
- For homework, open the Math Helper: it is designed to guide your thinking, not to hand you solutions
- One rule:
Use AI to learn, not to outsource your thinking. If the chatbot solves your homework, you have practised prompting, not mathematics.
Notation
Can You Read This Line?
A constraint of the kind you meet in Transportation & Distribution:
\[\sum_{j=1}^{m} x_{ij} \leq s_i \qquad \forall\, i \in \{1, \dots, n\}\]
- Mathematics is a language: this line is precise, compact, and read the same worldwide
- In English it takes a full sentence, and you will be able to say that sentence by the end of today
- Reading it fluently is a skill, and skills can be trained
The Number Sets
- Counting trucks: \(\mathbb{N} = \{1, 2, 3, \dots\}\), the natural numbers
- Booking a loss of €2: we need negatives, \(\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}\), the integers
Rational and Real Numbers
- Splitting 3 pallets among 4 customers: we need fractions \(\frac{a}{b}\) with \(a, b \in \mathbb{Z}\) (\(\in\): “lies in”, so both are integers), \(b \neq 0\), the rational numbers \(\mathbb{Q}\)
- The diagonal of a \(1 \times 1\) m pallet is \(\sqrt{2}\) m, which no fraction equals: all points on the number line form the real numbers \(\mathbb{R}\)
Each set contains the one before: \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\)
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.
Your Turn: Rational or Not?
Which of these numbers are rational: \(\quad 0.75, \quad \sqrt{2}, \quad \frac{7}{3}, \quad \pi\)?
- \(0.75 = \frac{3}{4}\) and \(\frac{7}{3}\) are rational: both can be written as a fraction of integers
- \(\sqrt{2}\) and \(\pi\) are irrational: they are real, but no fraction equals them exactly
\(\pi \approx \frac{22}{7}\), but \(\pi \neq \frac{22}{7}\): an approximation is not an equality!
Inequality Notation
| Symbol | Meaning | Business example |
|---|---|---|
| \(<\) | strictly less than | \(w < 1000\) (pallet weight limit) |
| \(\leq\) | less than or equal | \(q \leq 500\) (capacity: bags per week) |
| \(\geq\) | greater than or equal | \(\text{order} \geq 20\) (a supplier’s minimum) |
| \(\neq\) | not equal to | \(\text{supply} \neq \text{demand}\) |
Inequalities can be chained: \(0 \leq q \leq 500\)
means “\(q\) is at least \(0\) and at most \(500\)”. Is there a shorter way to write this range?
Interval Notation
A chain like \(0 \leq q \leq 500\) describes a connected range. Intervals write it with just the two endpoints: \(q \in [0, 500]\).
- Closed: \([a, b]\) means \(a \leq x \leq b\), endpoints included
- Open: \((a, b)\) means \(a < x < b\), endpoints excluded
- Half-open: \([a, b)\) means \(a \leq x < b\)
- Unbounded: \([0, \infty)\) means \(x \geq 0\), all non-negative reals
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.
Your Turn: Write the Interval
A warehouse accepts pallets weighing at least \(200\) kg and strictly less than \(1000\) kg. Write the admissible weight \(w\) as an interval.
\[w \in [200, 1000)\]
“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.
Absolute Value
\(|5| = 5\) and \(|-5| = 5\): the absolute value \(|x|\) is the distance of \(x\) from zero, and distance is never negative.
- \(|a - b|\) is the distance between \(a\) and \(b\)
- Business use: the forecast error \(|\text{demand} - \text{forecast}|\): we care about the size of the miss, not its direction
\[|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}\]
\(|x| \leq 5\) means \(-5 \leq x \leq 5\): at most 5 from zero, on either side.
Your Turn: Within Tolerance
The forecast is \(400\) units, and the forecast error \(|d - 400|\) is at most \(50\). Which values can demand \(d\) take?
| (a) \(d \in (350, 450)\) | (b) \(d \in [400, 450]\) |
| (c) \(d \in [350, 450]\) | (d) \(d \in [-50, 50]\) |
(c): \(-50 \leq d - 400 \leq 50\), add \(400\) to all three parts: \(350 \leq d \leq 450\). “At most” keeps both endpoints.
\(|x - a| \leq \varepsilon\) always means “within \(\varepsilon\) of \(a\)”: the interval \([a - \varepsilon, a + \varepsilon]\).
Sum Notation
Let \(c_t\) be the roastery’s cost in month \(t\). Twelve monthly costs, written out without shortcuts: \(c_1 + c_2 + \dots + c_{12}\)
The symbol \(\sum\) (capital sigma) compresses this long sum; read it as a loop, “for each \(t\) from 1 to 12, add \(c_t\)”: \(\displaystyle\sum_{t=1}^{12} c_t = c_1 + c_2 + \dots + c_{12}\)
- \(t\) is the index: it runs from the lower limit \(1\) to the upper limit \(12\)
- \(c_t\) is the summand: the expression added up for each \(t\)
- In general: \(\sum_{i=1}^{n} x_i = x_1 + \dots + x_n\)
- The index name is arbitrary: \(\sum_{i=1}^{n} x_i = \sum_{k=1}^{n} x_k\)
Two Rules for Sums
Each week costs €200 fixed plus €5 per bag: \(200 + 5q_t\) for \(q_t\) bags in week \(t\). Four weeks:
\[\sum_{t=1}^{4} (200 + 5q_t) = \sum_{t=1}^{4} 200 + 5 \sum_{t=1}^{4} q_t = 4 \cdot 200 + 5 \sum_{t=1}^{4} q_t\]
- Sums split over addition: €200 fixed once a week gives \(4 \cdot 200\)
- Constant factors move out: €5 per bag, whatever the week
Sales of \(400, 450, 500, 450\) bags: \(800 + 5 \cdot 1800 =\) €9,800, without computing a single weekly cost.
Your Turn: Evaluate a Sum
Compute \(\sum_{i=3}^{6} 2i\). What is the result?
Write out the loop, one term per index value:
\[\sum_{i=3}^{6} 2i = 2 \cdot 3 + 2 \cdot 4 + 2 \cdot 5 + 2 \cdot 6 = 6 + 8 + 10 + 12 = 36\]
Or move the constant out first: \(2 \sum_{i=3}^{6} i = 2 \cdot 18 = 36\). Four terms: \(6 - 3 + 1\), both endpoints count.
Product Notation
The roastery’s €1000 reserve grows \(2\,\%\) in year 1, \(3\,\%\) in year 2, \(4\,\%\) in year 3. How much after three years?
Written out, the factors multiply: \(1000 \cdot 1.02 \cdot 1.03 \cdot 1.04 = 1092.62\). Growth compounds: that is why it is a product, not a sum.
The symbol \(\prod\) (capital pi) works like \(\sum\), but with multiplication. With \(r_t\) the rate in year \(t\) as a decimal (\(2\,\% = 0.02\)), \(K_0 =\) €1000 today and \(K_3\) the amount after three years:
\[K_3 = K_0 \cdot \prod_{t=1}^{3} (1 + r_t)\]
When the Rate Is Constant
Same rate every year? Then all factors are equal and the product is a power: \(\prod_{t=1}^{3} 1.05 = 1.05^3 \approx 1.1576\), so €1000 becomes €1157.63.
Your Turn: Compounded Growth
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 \(38.6\,\%\).
\(10 + 20 + 5 = 35\,\%\) is wrong: each year’s growth applies to the already grown revenue. Rates multiply as factors, they do not add.
Double Indices
Many logistics quantities depend on two things at once:
- \(x_{ij}\) = quantity shipped from warehouse \(i\) to customer \(j\)
- \(c_{ij}\) = transport cost per unit from \(i\) to \(j\)
| \(C1\) | \(C2\) | \(C3\) | |
|---|---|---|---|
| \(W1\) | |||
| \(W2\) | \(x_{23}\) |
The first index is the origin (row, where it leaves), the second the destination (column, where it arrives).
Double Sums
With \(n\) warehouses and \(m\) customers:
- Everything leaving warehouse \(i\): \(\sum_{j=1}^{m} x_{ij}\)
- Everything reaching customer \(j\): \(\sum_{i=1}^{n} x_{ij}\)
- Total transport cost over all routes: cost per unit \(\times\) units, on every route
\[\sum_{i=1}^{n} \sum_{j=1}^{m} c_{ij} \, x_{ij}\]
The Transportation Problem
The double sum is the total cost the classic transportation problem minimises. You will meet it again in Transportation & Distribution.
And \(\sum_{j=1}^{m} x_{ij}\) from the line at the start is everything leaving warehouse \(i\): only \(\forall\) is left to read.
Whenever you meet a variable, first ask: what do its indices mean? A well-defined model always states this explicitly.
Your Turn: Total Transport Cost
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 in Your Courses
| 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 |
Logic
Who Gets the Discount?
“Customers get 5 % off if they are members or order over €500 and pay in advance.”
A member orders for €200 and pays by card. Discount: yes or no?
- Reading 1: member, or (over €500 and prepaid) \(\to\) yes
- Reading 2: (member or over €500) and prepaid \(\to\) no
- English hides the brackets. Logic makes every piece true or false and every bracket visible
Statements and Truth Values
A statement is a sentence that is either true or false, never both.
- “The customer is a member.”: a statement, true or false for each customer
- “\(2 + 2 = 5\)”: a false statement
- “Pay in advance!”: a command, no truth value
- “\(x + 2 = 2x\)”: an open sentence, true for \(x = 2\), false otherwise
An open sentence becomes a statement once every variable is fixed.
Negation
The negation \(\neg p\) (“not \(p\)”) flips the truth value of \(p\):
- \(p\): “The warehouse is full.” \(\quad \neg p\): “The warehouse is not full.”
- If \(p\) is true, \(\neg p\) is false, and vice versa
- Double negation cancels: \(\neg(\neg p)\) is just \(p\)
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”.
Conjunction and Disjunction
| \(p\) | \(q\) | \(p \wedge q\) (“\(p\) and \(q\)”) | \(p \vee q\) (“\(p\) or \(q\)”) |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | F | T |
| F | F | F | F |
Back to the discount, member (\(m\)), over €500 (\(o\)), prepaid (\(a\)): \(m \vee (o \wedge a)\) is T, \((m \vee o) \wedge a\) is F. Same words, two statements.
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.
Your Turn: A Broken Promise?
“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): the order is over €500, yet shipping is charged.
Cases (b) and (c) do not break the promise: for an order of €300, nothing was promised at all.
Implication
\(p \Rightarrow q\): “if \(p\), then \(q\)”, \(p\) is the premise, \(q\) the conclusion. Your vote:
| \(p\): over €500 | \(q\): shipping free | \(p \Rightarrow q\) | Case |
|---|---|---|---|
| T | T | T | (d) |
| T | F | F | (a) |
| F | T | T | (c) |
| F | F | T | (b) |
\(p \Rightarrow q\) is a rule, and a rule is false only when a counterexample exists. A €300 order cannot be one, whatever happens to its shipping: the rule is not broken, so it holds.
A Customer Complained
We know: “If a shipment is delayed (\(p\)), the customer complains (\(q\)).”
Today, a customer complained. Was the shipment delayed?
- Tempting, but we cannot conclude that
- Complaints can have other causes: damaged goods, wrong invoice, …
- From \(p \Rightarrow q\) and \(q\), nothing follows about \(p\)
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.
Your Turn: Which Slips to Turn?
You audit “If the order exceeds €500, shipping is free.” Each slip shows one side only. Which slips must you turn over?
| (a) Order €600 | (b) Order €300 |
| (c) Shipping free | (d) Shipping charged |
(a) and (d). Either could hide the one broken row: €600 with shipping charged. (b) and (c) cannot break the rule, whatever is on the back.
Turning over (d) checks the rule flipped and negated: shipping charged \(\Rightarrow\) order at most €500.
What You May Conclude
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 | denying the premise \(\neg p \Rightarrow \neg q\) | ✗ |
The contrapositive is always equivalent to the original implication: flip and negate. Flipping alone (converse) or negating alone (denying the premise) breaks the logic: that is why the auditor turns over (d), never (c).
Necessary and Sufficient Conditions
Given \(p \Rightarrow q\):
- \(p\) is sufficient for \(q\): \(p\) alone guarantees \(q\)
- \(q\) is necessary for \(p\): without \(q\), no \(p\)
- Millionaire \(\Rightarrow\) rich: being a millionaire is sufficient; being rich is necessary but not sufficient (you can be rich without the million)
- “\(p\) only if \(q\)” also means \(p \Rightarrow q\)
Both Directions at Once
\(p \Leftrightarrow q\), “\(p\) if and only if \(q\)”: both \(p \Rightarrow q\) and \(q \Rightarrow p\), same truth value in every row.
\(x = 2 \Rightarrow x^2 = 4\) holds, but \(x^2 = 4 \Rightarrow x = 2\) does not (\(x = -2\)). So \(\Rightarrow\) is not \(\Leftrightarrow\).
Your Turn: Necessary or Sufficient?
“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.
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.
De Morgan’s Laws
“The contract promises delivery on time and complete.” When may the customer complain?
Late or incomplete: one failure is enough.
Negation flips “and” into “or”, and vice versa:
\[\neg(p \wedge q) \Leftrightarrow \neg p \vee \neg q \qquad\qquad \neg(p \vee q) \Leftrightarrow \neg p \wedge \neg q\]
Negation swaps \(\wedge\) and \(\vee\) here. In a moment it swaps “all” and “some”, and next week the same law returns for sets.
Quantifiers
“Suppliers deliver late.” All of them? Some? The open sentence \(P(x)\): “supplier \(x\) delivers late” needs a count word.
- “Every supplier delivers late”: \(\forall x: P(x)\), for all
- “Some supplier delivers late”: \(\exists x: P(x)\), there exists (at least one)
- To prove “some supplier delivers late”, one late supplier suffices: a witness
- In symbols: \(\forall x \in \mathbb{R}: x^2 \geq 0 \quad\) and \(\quad \exists n \in \mathbb{N}: n > 10^6\)
Negating Quantified Statements
What does it take to prove “all routes are profitable” wrong?
- To disprove “all routes are profitable”, one loss-making route suffices
- To disprove “some supplier delivers late”, you must check every supplier
Negation Rules for Quantifiers
With three suppliers A, B, C: \(\forall x: P(x)\) is \(P(A) \wedge P(B) \wedge P(C)\), and \(\exists x: P(x)\) is \(P(A) \vee P(B) \vee P(C)\).
Negating each, De Morgan gives:
- “All”: \(\neg(P(A) \wedge P(B) \wedge P(C)) \Leftrightarrow \neg P(A) \vee \neg P(B) \vee \neg P(C)\)
- “Some”: \(\neg(P(A) \vee P(B) \vee P(C)) \Leftrightarrow \neg P(A) \wedge \neg P(B) \wedge \neg P(C)\)
In general, for any number of cases:
\[\neg\,(\forall x: P(x)) \;\Leftrightarrow\; \exists x: \neg P(x)\]
\[\neg\,(\exists x: P(x)) \;\Leftrightarrow\; \forall x: \neg P(x)\]
Your Turn: Negate a Statement
Negate: “All shipments arrive on time.”
“At least one shipment does not arrive on time.”
The negation is not “no shipment arrives on time”: that is far too strong. One single late shipment already makes the original statement false.
Your Turn: Is the Plan Feasible?
The line from the start, with stocks \(s_1 = 15\) and \(s_2 = 20\): \(\sum_{j=1}^{2} x_{ij} \leq s_i \;\; \forall\, i \in \{1, 2\}\). Plan from “Total Transport Cost”: \(x_{11} = 10\), \(x_{12} = 0\), \(x_{21} = 5\), \(x_{22} = 20\). Is it feasible?
| (a) Yes, both warehouses comply | (b) No, warehouse 1 ships too much |
| (c) No, warehouse 2 ships too much | (d) No, both ship too much |
(c): warehouse 1 ships \(10 + 0 = 10 \leq 15\), warehouse 2 ships \(5 + 20 = 25 > 20\).
One warehouse is enough to break a “for all”: \(\exists\, i: \sum_{j} x_{ij} > s_i\). Read aloud, the line says: every warehouse ships at most what it has in stock.
Closing
Key Takeaways: Notation
- \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\): \(0.75 = \frac{3}{4}\) is rational, \(\pi\) is not, however close \(\frac{22}{7}\) comes
- Brackets decide membership: \([200, 1000)\) takes the \(200\) kg pallet and rejects the \(1000\) kg one
- \(\sum\) adds, \(\prod\) multiplies: \(+10\,\%\), \(+20\,\%\), \(+5\,\%\) compound to \(38.6\,\%\), not \(35\,\%\)
- The line from the start, read aloud: every warehouse \(i\) ships at most its stock \(s_i\), for all \(i\); one warehouse over stock breaks it
Key Takeaways: Logic
- “Over €500 \(\Rightarrow\) free shipping” breaks only when an order over €500 pays shipping; free shipping alone says nothing about the order value
- Negation swaps: “on time and complete” fails by late or incomplete, “all shipments on time” fails by one late shipment
- Member, €200, card: \(m \vee (o \wedge a)\) says yes, \((m \vee o) \wedge a\) says no, so brackets decide
- “Clears only if declaration valid”: valid is necessary, not sufficient
Until the Next Session
- Work through the Tasks: problems with worked solutions
- Check yourself with the Self-Test quiz
- Keep the Cheatsheet next to you while practising
- Note down anything unclear: we start next session with your questions
The tasks are where the actual learning happens: the lecture only opens the door.
Preview: Session 02: Sets
- What sets are and how to describe them
- And, or, not return as intersection, union, complement: today’s De Morgan becomes a picture
- Venn diagrams as a thinking tool
- Why sets are the foundation for functions in Session 3
Of \(200\) customers, \(120\) ordered online, \(90\) in store, \(40\) both. How many ordered nothing? Bring a guess.
See you there, and bring your questions!
Literature & Further Reading
- These sessions cover the essentials: textbooks offer more depth and practice
- Start with Sydsæter et al. (2012) or Jacques (2015); full recommendations on the tutorial’s literature page
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.