Quiz 2

Chatty deep dive · Maths 1 Week 1

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2026-08-16T00:00:00.000Z
Python Week 1: the first filter for runtime behavior
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Ultra-deep sets, number systems, relations, and functions spine from ChatGPT export. # Mathematics 1 — Week 1 ## Sets, Number Systems, Relations & Functions The goal isn't merely to remember definitions.

Mathematics 1 — Week 1

Sets, Number Systems, Relations & Functions

The goal isn't merely to remember definitions. By the end, you should be able to look at a question and think:
"Ah, this is that pattern."

0. The Week 1 Map

Think of the week as four layers:
Number Systems→Sets→Relations→Functions\boxed{\text{Number Systems}} \rightarrow \boxed{\text{Sets}} \rightarrow \boxed{\text{Relations}} \rightarrow \boxed{\text{Functions}}
And there is an important conceptual progression:
Numbers are objects → sets collect objects → relations connect objects → functions impose a special kind of connection.
That's the backbone.

1. Number Systems

Before sets, we need to know what kinds of numbers we're dealing with.

1.1 Natural numbers

Usually:
N={1,2,3,…}\mathbb N=\{1,2,3,\ldots\}
Some texts include 00:
N0={0,1,2,3,…}\mathbb N_0=\{0,1,2,3,\ldots\}
So always pay attention to the convention being used.

1.2 Integers

Z={…,−3,−2,−1,0,1,2,3,…}\mathbb Z=\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}
Contains:
  • negative integers
  • zero
  • positive integers

1.3 Rational numbers

A rational number can be written as:
pq\frac pq
where
p,q∈Z,q≠0p,q\in\mathbb Z,\qquad q\neq0
Examples:
12,−73,5,0.25\frac12,\quad -\frac73,\quad 5,\quad 0.25
Notice that integers are also rational:
5=515=\frac51
So:
Z⊆Q\mathbb Z\subseteq\mathbb Q

1.4 Irrational numbers

Numbers that cannot be expressed as pq\frac pq for integers p,qp,q with q≠0q\neq0.
Examples:
2,π,e\sqrt2,\quad \pi,\quad e
Their decimal representations are non-terminating and non-repeating.

1.5 Real numbers

Real numbers contain both rational and irrational numbers:
R=Q∪{irrational numbers}\mathbb R=\mathbb Q\cup\{\text{irrational numbers}\}
The containment hierarchy is therefore:
N⊆Z⊆Q⊆R\boxed{ \mathbb N\subseteq\mathbb Z\subseteq\mathbb Q\subseteq\mathbb R }
This little diagram is very high-value.

Recognition pattern

If a question asks:
Which of these numbers belongs to Q\mathbb Q?
ask:
Can I express it as pq\frac pq?
If yes → rational.

2. Set Theory

Now we move from individual objects to collections of objects.

2.1 What is a set?

A set is a collection of distinct objects.
Example:
A={1,2,3,4}A=\{1,2,3,4\}
The objects are called elements or members.
We write:
2∈A2\in A
meaning:
22 is an element of AA.
And:
7∉A7\notin A
meaning:
77 is not an element of AA.

3. Set Representation

There are two major forms you'll encounter.

3.1 Roster / listing form

Explicitly list elements:
A={2,4,6,8}A=\{2,4,6,8\}

3.2 Set-builder form

Describe a property:
A={x∈N:x<5}A=\{x\in\mathbb N:x<5\}
Read it as:
The set of all natural numbers xx such that x<5x<5.
So:
A={1,2,3,4}A=\{1,2,3,4\}

Pattern

Set-builder → roster
Translate the condition into actual elements.

4. Empty Set

The set containing no elements is:
∅\varnothing
or:
{}\{\}
Important distinction:
∅≠{∅}\varnothing\neq\{\varnothing\}
Why?

∅\varnothing

Contains zero elements.

{∅}\{\varnothing\}

Contains one element.
That one element happens to be the empty set.
This is a classic conceptual trap.

5. Cardinality

Cardinality means:
How many elements does the set contain?
Notation:
∣A∣|A|
Example:
A={2,4,6,8}A=\{2,4,6,8\}
Then:
∣A∣=4|A|=4
For the empty set:
∣∅∣=0|\varnothing|=0
But:
∣{∅}∣=1|\{\varnothing\}|=1

6. Subsets

This is one of the most important Week 1 concepts.
Suppose:
A={1,2,3}A=\{1,2,3\}
and:
B={1,2}B=\{1,2\}
Every element of BB is also in AA.
Therefore:
B⊆AB\subseteq A
Read:
BB is a subset of AA.

6.1 Proper subset

If:
B⊆AB\subseteq A
and
B≠AB\neq A
then BB is a proper subset of AA.
Often written:
B⊂AB\subset A
depending on the notation convention.

6.2 Important pattern

To determine whether:
B⊆AB\subseteq A
don't ask:
"Does BB look smaller?"
Ask:
Is every element of BB contained in AA?
Example:
A={1,2,3,4}A=\{1,2,3,4\} B={2,4}B=\{2,4\}
Yes:
B⊆AB\subseteq A
But:
C={2,5}C=\{2,5\}
No, because:
5∉A5\notin A
Therefore:
C⊈AC\not\subseteq A

7. Set Equality

Two sets are equal if they contain exactly the same elements.
Example:
A={1,2,3}A=\{1,2,3\} B={3,2,1}B=\{3,2,1\}
Then:
A=BA=B
because order doesn't matter in a set.
Likewise:
{1,1,2,2,3}={1,2,3}\{1,1,2,2,3\}=\{1,2,3\}
because duplicate elements don't create new set elements.

Contrast

A list:
text
[1, 2, 3]
and
text
[3, 2, 1]
may be different.
A set:
{1,2,3}={3,2,1}\{1,2,3\}=\{3,2,1\}

8. Power Set

The power set of AA is the set containing all subsets of AA.
Notation:
P(A)\mathcal P(A)
Suppose:
A={a,b}A=\{a,b\}
Its subsets are:
∅,{a},{b},{a,b}\varnothing,\{a\},\{b\},\{a,b\}
Therefore:
P(A)={∅,{a},{b},{a,b}}\mathcal P(A) = \{\varnothing,\{a\},\{b\},\{a,b\}\}
If:
∣A∣=n|A|=n
then:
∣P(A)∣=2n\boxed{|\mathcal P(A)|=2^n}

Why?

Each element has two possibilities:
included / not included.
For nn elements:
2×2×⋯×2=2n2\times2\times\cdots\times2=2^n
This is one of those patterns that later connects directly to counting in Statistics.

9. Set Operations

Now the fun part.
Suppose:
A={1,2,3,4}A=\{1,2,3,4\} B={3,4,5,6}B=\{3,4,5,6\}

9.1 Union

Union means:
Everything in AA or BB.
A∪BA\cup B
Therefore:
A∪B={1,2,3,4,5,6}A\cup B=\{1,2,3,4,5,6\}

Mental model

OR\boxed{\text{OR}}

10. Intersection

Intersection means:
Things common to both.
A∩BA\cap B
Therefore:
A∩B={3,4}A\cap B=\{3,4\}

Mental model

AND\boxed{\text{AND}}
This OR/AND interpretation is extremely useful.

11. Difference

A−BA-B
means:
Elements in AA that are not in BB.
Therefore:
A−B={1,2}A-B=\{1,2\}
Notice:
A−B≠B−AA-B\neq B-A
Here:
B−A={5,6}B-A=\{5,6\}
So set difference is not commutative.

12. Complement

If the universal set is UU, then the complement of AA is everything in UU that isn't in AA.
Ac=U−AA^c=U-A
Example:
U={1,2,3,4,5,6}U=\{1,2,3,4,5,6\} A={1,2,3}A=\{1,2,3\}
Then:
Ac={4,5,6}A^c=\{4,5,6\}

Important

You cannot determine a complement without knowing the universe.

13. The Core Set Pattern

Given:
A={1,2,3,4}A=\{1,2,3,4\} B={3,4,5,6}B=\{3,4,5,6\}
Then:
ExpressionResultInterpretation
A∪BA\cup B{1,2,3,4,5,6}\{1,2,3,4,5,6\}OR
A∩BA\cap B{3,4}\{3,4\}AND
A−BA-B{1,2}\{1,2\}A but not B
B−AB-A{5,6}\{5,6\}B but not A
This is a pattern, not four facts to memorize.

14. Set Identities

Some useful identities:

Commutativity

A∪B=B∪AA\cup B=B\cup A A∩B=B∩AA\cap B=B\cap A
But:
A−B≠B−AA-B\neq B-A
in general.

Associativity

(A∪B)∪C=A∪(B∪C)(A\cup B)\cup C=A\cup(B\cup C) (A∩B)∩C=A∩(B∩C)(A\cap B)\cap C=A\cap(B\cap C)

Distributivity

A∩(B∪C)=(A∩B)∪(A∩C)A\cap(B\cup C) = (A\cap B)\cup(A\cap C)
and:
A∪(B∩C)=(A∪B)∩(A∪C)A\cup(B\cap C) = (A\cup B)\cap(A\cup C)

15. De Morgan's Laws

Very high-value.
(A∪B)c=Ac∩Bc\boxed{(A\cup B)^c=A^c\cap B^c}
and:
(A∩B)c=Ac∪Bc\boxed{(A\cap B)^c=A^c\cup B^c}
Mental translation:
NOT (A OR B) = NOT A AND NOT B
NOT (A AND B) = NOT A OR NOT B
This same logic appears later in:
  • Statistics
  • CT
  • Python conditionals
So this isn't just a Maths 1 trick.

16. Relations

Now we move from collections to connections.
Suppose:
A={1,2,3}A=\{1,2,3\} B={a,b}B=\{a,b\}
The Cartesian product is:
A×BA\times B
which contains ordered pairs:
A×B={(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)}A\times B= \{ (1,a),(1,b), (2,a),(2,b), (3,a),(3,b) \}
In general:
∣A×B∣=∣A∣∣B∣|A\times B|=|A||B|
if both sets are finite.

17. Why Ordered Pairs Matter

Notice:
(1,a)≠(a,1)(1,a)\neq(a,1)
because order matters.
This is different from sets:
{1,a}={a,1}\{1,a\}=\{a,1\}
So:
Set: order irrelevant
Ordered pair: order matters
That's a common exam trap.

18. Relation

A relation from AA to BB is essentially a subset of A×BA\times B.
Suppose:
R={(1,a),(2,b),(3,a)}R=\{(1,a),(2,b),(3,a)\}
Then RR is a relation from AA to BB.
Think:
text
A                  B

1 ───────────────→ a
2 ───────────────→ b
3 ───────────────→ a
A relation simply specifies which connections exist.

19. Function

Now comes the crucial restriction.
A function from AA to BB is a relation where:
Every input has exactly one output.
Notation:
f:A→Bf:A\to B
Think:
text
input → exactly one output

20. Function Recognition Pattern

Suppose:
text
1 → a
2 → b
3 → a
Valid function.
Why?
Each input has exactly one arrow leaving it.
It is perfectly okay for multiple inputs to have the same output.
But:
text
1 → a
1 → b
2 → a
is not a function.
Why?
Input 11 has two outputs.
And:
text
1 → a
2 → b
3
is not a function from the entire domain {1,2,3}\{1,2,3\}, because 33 has no output.

Golden rule

Every input → exactly one output\boxed{\text{Every input → exactly one output}}

21. Domain, Codomain & Range

Suppose:
f:A→Bf:A\to B
Then:

Domain

The set of allowed inputs:
AA

Codomain

The set in which outputs are supposed to lie:
BB

Range

The outputs that are actually produced.
These are not necessarily the same.
Example:
f:{1,2,3}→{a,b,c,d}f:\{1,2,3\}\to\{a,b,c,d\}
with:
1→a1\to a 2→b2\to b 3→a3\to a
Then:
Domain={1,2,3}\text{Domain}=\{1,2,3\} Codomain={a,b,c,d}\text{Codomain}=\{a,b,c,d\}
but:
Range={a,b}\text{Range}=\{a,b\}
This distinction is very testable.

22. One-to-One / Injective

A function is one-to-one if:
Different inputs always produce different outputs.
Formally:
f(a)=f(b)⇒a=bf(a)=f(b)\Rightarrow a=b
Example:
1→a1\to a 2→b2\to b 3→c3\to c
is injective.
But:
1→a1\to a 2→a2\to a
is not injective.

Mental model

No two inputs collide.

23. Onto / Surjective

A function is onto if:
Every element of the codomain is actually hit.
Suppose:
A={1,2,3}A=\{1,2,3\} B={a,b,c}B=\{a,b,c\}
and:
1→a,2→b,3→c1\to a,\quad2\to b,\quad3\to c
Then every codomain element gets used.
So:
f:A→Bf:A\to B
is onto.

24. Bijective

A function is bijective if it is both:
injective + surjective\boxed{\text{injective + surjective}}
So:
no collisions + nothing left unused.
This is the strongest one-to-one correspondence between two sets.

25. The Function Taxonomy

Keep this mental picture:
text
                    FUNCTIONS
                       │
             ┌─────────┴─────────┐
             │                   │
        Injective?           Surjective?
             │                   │
             └─────────┬─────────┘
                       │
                  Both = Bijective
And remember:
  • Injective: no output collision.
  • Surjective: no unused codomain element.
  • Bijective: both.

26. Week 1 Pattern Bank — First Version

We're going to maintain this and expand it as we study.
IDPatternDifficulty
W1-N1Identify number-system membershipEasy
W1-S1Set membershipEasy
W1-S2Set-builder ↔ rosterEasy
W1-S3CardinalityEasy
W1-S4Subset checkingEasy
W1-S5Power-set cardinalityEasy
W1-S6Union/intersection/differenceEasy
W1-S7Multi-operation set expressionMedium
W1-S8ComplementEasy
W1-S9Set identitiesMedium
W1-S10De Morgan transformationsMedium
W1-R1Cartesian productEasy
W1-R2Relation identificationEasy
W1-F1Function identificationEasy
W1-F2Domain/codomain/rangeEasy
W1-F3Injective checkingMedium
W1-F4Surjective checkingMedium
W1-F5Bijective checkingMedium
W1-F6Mixed representationHard

27. Now I want you to do something

Don't just read all that and go:
"yeah yeah makes sense"
That's the exact failure mode we're trying to avoid.
Let's do a diagnostic round before I give you the next layer.
No calculator. Don't search. Don't overthink.

Diagnostic — Week 1

Q1 — Number systems

Classify each as natural, integer, rational, irrational, real where applicable:
−7,0,34,2,π-7,\quad 0,\quad \frac34,\quad \sqrt2,\quad \pi

Q2 — Sets

Let:
A={1,2,3,4,5}A=\{1,2,3,4,5\} B={2,4,6}B=\{2,4,6\}
Find:
A∩BA\cap B A∪BA\cup B A−BA-B

Q3 — Subsets

Is:
{1,3}⊆{1,2,3,4}?\{1,3\}\subseteq\{1,2,3,4\}?
Is:
{1,5}⊆{1,2,3,4}?\{1,5\}\subseteq\{1,2,3,4\}?
Give the reason, not merely yes/no.

Q4 — Power set

If:
∣A∣=5|A|=5
what is:
∣P(A)∣?|\mathcal P(A)|?
And why?

Q5 — Cartesian product

If:
A={1,2}A=\{1,2\}
and:
B={x,y,z}B=\{x,y,z\}
what is:
∣A×B∣?|A\times B|?

Q6 — Function recognition

Is this a function?
text
1 → a
2 → b
3 → a
Why?

Q7 — Function recognition

Is this a function?
text
1 → a
1 → b
2 → c
Why?

Q8 — Domain / codomain / range

Suppose:
f:{1,2,3}→{a,b,c,d}f:\{1,2,3\}\to\{a,b,c,d\}
with:
f(1)=a,f(2)=b,f(3)=af(1)=a,\qquad f(2)=b,\qquad f(3)=a
Give:
  • Domain
  • Codomain
  • Range
  • Is it injective?
  • Is it surjective?
  • Is it bijective?

Q9 — Slightly nastier

Let:
A={1,2,3,4}A=\{1,2,3,4\} B={3,4,5}B=\{3,4,5\} C={4,5,6}C=\{4,5,6\}
Find:
A∩(B∪C)A\cap(B\cup C)
Do not calculate it mentally in one jump. Show the intermediate set.

Q10 — Pattern recognition

Suppose somebody gives you:
"A function has domain {1,2,3,4}\{1,2,3,4\} and codomain {a,b,c}\{a,b,c\}. Each element of the domain maps to exactly one element of the codomain, and every codomain element is used."
What property does the function definitely have?
And what cannot you conclude from this information alone?
Reply with your answers however you naturally would — messy is fine. I care more about your reasoning than formatting.
Then I'll diagnose each answer as:
Concept gap / Pattern gap / Execution error / Careless error / Correct\boxed{\text{Concept gap / Pattern gap / Execution error / Careless error / Correct}}
and we'll use that to decide whether to push deeper into Week 1 or move to Week 2.
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