Maths1_Week1_Sets_Relations

Maths 1 Week 1: Sets, Relations & Functions

0. Prerequisites

NOTE

What you need to know before starting:

• Basic Arithmetic: Addition, subtraction, multiplication, division.
• Number Line: Understanding positive and negative integers.
• Basic Algebra: Solving simple equations like $2x+3=7$.

Quick Refresher

• Natural Numbers ($\mathbb{N}$): Counting numbers $\{1,2,3,\dots \}$.
• Integers ($\mathbb{Z}$): Whole numbers plus negatives $\{\dots ,-2,-1,0,1,2,\dots \}$.
• Rational Numbers ($\mathbb{Q}$): Fractions $p/q$ where $q\ne 0$.
• Real Numbers ($\mathbb{R}$): All numbers on the number line, including $\sqrt{2},\pi$.

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1. Core Concepts

1.1 Sets

A Set is a well-defined collection of distinct objects.

• Notation: $A=\{1,2,3\}$.
• Cardinality ($|A|$): The number of elements in set $A$. If $A=\{a,b,c\}$, then $|A|=3$.
• Subset ($A\subseteq B$): Every element of $A$ is also in $B$.
• Power Set ($P(A)$): The set of all subsets of $A$. If $|A|=n$, then $|P(A)|=2^n$.

Operations

• Union ($A\cup B$): Everything in $A$ OR $B$.
• Intersection ($A\cap B$): Everything in BOTH $A$ AND $B$.
• Difference ($A\setminus B$): Everything in $A$ that is NOT in $B$.
• Cartesian Product ($A\times B$): All pairs $(a,b)$ where $a\in A,b\in B$.
  • Size: $|A\times B|=|A|\cdot |B|$.

1.2 Relations

A Relation ($R$) from $A$ to $B$ is a subset of $A\times B$. It connects elements. If $R$ is on a set $A$ (subset of $A\times A$), we check three properties:

1. Reflexive: Every element relates to itself. $\forall a\in A,(a,a)\in R$.
   • Visual: Every node has a self-loop.
2. Symmetric: If $a$ relates to $b$, then $b$ relates to $a$. $(a,b)\in R⟹(b,a)\in R$.
   • Visual: If there’s an arrow $a\to b$, there must be $b\to a$.
3. Transitive: If $a\to b$ and $b\to c$, then $a\to c$.
   • Visual: Shortcuts must exist.

Equivalence Relation: A relation that is Reflexive, Symmetric, AND Transitive.

1.3 Functions

A Function ($f:A\to B$) is a relation where every input in $A$ has exactly one output in $B$.

• Domain: The set of all inputs ($A$).
• Codomain: The set of potential outputs ($B$).
• Range: The set of actual outputs produced.

Types of Functions

1. One-to-One (Injective): No two inputs give the same output.
   • Test: Horizontal Line Test (graph intersects at most once).
   • Algebra: $f(x)=f(y)⟹x=y$.
2. Onto (Surjective): Every element in Codomain is covered. Range = Codomain.
3. Bijective: Both One-to-One and Onto. (Invertible).

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2. Pattern Analysis & Goated Solutions

Pattern 1: Finding Cardinality (Inclusion-Exclusion)

Context: You are given sizes of sets and their intersections/unions and asked to find a missing value.

TIP

Mental Algorithm:

1. Identify the sets involved (e.g., Students taking Math, Physics).
2. Write down the known values: $|M|$, $|P|$, $|M\cap P|$, etc.
3. Use the formula: $|A\cup B|=|A|+|B|-|A\cap B|$.
4. For 3 sets: $|A\cup B\cup C|=\sum |A|-\sum |A\cap B|+|A\cap B\cap C|$.

Example (Detailed Solution)

Problem: In a class of 50 students, 30 like Tea, 25 like Coffee, and 10 like both. How many like neither? Solution:

1. Define Sets:
   • $U$ (Total students) = 50
   • $T$ (Tea) = 30
   • $C$ (Coffee) = 25
   • $T\cap C$ (Both) = 10
2. Goal: Find students who like Neither. This is $|U|-|T\cup C|$.
3. Step 1: Find Union ($T\cup C$).
   • Formula: $|T\cup C|=|T|+|C|-|T\cap C|$
   • Calculation: $30+25-10=45$.
   • Why? When we add Tea (30) and Coffee (25), we count the people who like both twice. Subtracting 10 fixes this double counting.
4. Step 2: Subtract from Total.
   • Neither = Total - Union
   • Calculation: $50-45=5$. Answer: 5 students like neither.

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Pattern 2: Checking Equivalence Relations

Context: You are given a rule (e.g., $aRb$ if $a-b$ is even) and asked if it’s an equivalence relation.

TIP

Mental Algorithm (The “RST” Check):

1. R (Reflexive): Check if $xRx$ is always true. Plug in the same value twice.
2. S (Symmetric): Assume $xRy$ is true. Does this force $yRx$ to be true?
3. T (Transitive): Assume $xRy$ AND $yRz$. Does this force $xRz$?

Example (Detailed Solution)

Problem: Let $R$ be a relation on Integers defined by $xRy$ if $x-y$ is a multiple of 5. Is $R$ an equivalence relation? Solution:

1. Reflexive Check ($xRx$):
   • Is $x-x$ a multiple of 5?
   • $x-x=0$.
   • Is 0 a multiple of 5? Yes ($5\times 0=0$).
   • $⟹$ Reflexive: YES.
2. Symmetric Check ($xRy⟹yRx$):
   • Assume $x-y=5k$ (multiple of 5).
   • We need to check $y-x$.
   • $y-x=-(x-y)=-(5k)=5(-k)$.
   • Since $-k$ is an integer, $5(-k)$ is a multiple of 5.
   • $⟹$ Symmetric: YES.
3. Transitive Check ($xRy,yRz⟹xRz$):
   • Assume $x-y=5k$ and $y-z=5m$.
   • We need to check $x-z$.
   • Trick: Add the two equations.
   • $(x-y)+(y-z)=5k+5m$.
   • $x-z=5(k+m)$.
   • Since $k+m$ is integer, this is a multiple of 5.
   • $⟹$ Transitive: YES. Conclusion: Since R, S, and T are all YES, it is an Equivalence Relation.

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Pattern 3: Finding Domain of Functions

Context: Given a function like $f(x)=\sqrt{x-2}+\frac{1}{x-5}$, find valid $x$ values.

TIP

Mental Algorithm: Look for “Illegal Operations”:

1. Division by Zero: Denominator $\ne 0$.
2. Square Root of Negative: Inside of $\sqrt{\dots }\ge 0$.
3. Log of Non-Positive: Inside of $\log (\dots )>0$. Solve each condition and find the Intersection (overlap) of all valid regions.

Example (Detailed Solution)

Problem: Find the domain of $f(x)=\frac{\sqrt{9-x^2}}{x-1}$. Solution:

1. Condition 1: Square Root ($\sqrt{9-x^2}$)
   • Inside must be non-negative: $9-x^2\ge 0$.
   • $x^2\le 9$.
   • This means $x$ is between -3 and 3: $x\in [-3,3]$.
2. Condition 2: Denominator ($x-1$)
   • Denominator cannot be zero: $x-1\ne 0$.
   • $x\ne 1$.
3. Combine (Intersection):
   • We need $x$ to be in $[-3,3]$ AND $x\ne 1$.
   • Draw number line:
     • Valid: $[-3,3]$
     • Hole at: $1$
   • Final Domain: $[-3,1)\cup (1,3]$.

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3. Practice Exercises

1. Cardinality: In a group of 100, 60 like Football, 50 like Cricket, 20 like neither. How many like both?
   • Hint: Use $|A\cup B|=|U|-|\text{Neither}|$. Then use Inclusion-Exclusion.
2. Relations: $S=\{1,2,3\}$. $R=\{(1,1),(2,2),(3,3),(1,2)\}$. Is it Symmetric?
   • Hint: We have $(1,2)$. Do we have $(2,1)$?
3. Domain: Find domain of $f(x)=\sqrt{x^2-4}$.
   • Hint: $x^2-4\ge 0⟹x^2\ge 4$. Be careful with inequalities!

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🧠 Level Up: Advanced Practice

Question 1: The Domain Trap

Problem: Find the domain of $f(x)=\sqrt{\frac{x-1}{x-2}}$. Common Mistake: Thinking $\frac{x-1}{x-2}\ge 0$ means $x-1\ge 0$ and $x-2>0$. Correct Logic:

• The fraction is positive if BOTH numerator and denominator are positive OR if BOTH are negative.
• Case 1 (+/+): $x-1\ge 0$ AND $x-2>0⟹x\ge 1$ AND $x>2⟹x>2$.
• Case 2 (-/-): $x-1\le 0$ AND $x-2<0⟹x\le 1$ AND $x<2⟹x\le 1$.
• Union: $(-\infty ,1]\cup (2,\infty )$.

Question 2: Relation Cardinality

Problem: Set $A$ has $n$ elements. How many relations on $A$ are Reflexive? Logic:

1. Total pairs in $A\times A$ is $n^2$.
2. Reflexive means all $(a,a)$ must be present ($n$ specific pairs).
3. Remaining pairs: $n^2-n$.
4. Each remaining pair can be either IN or OUT (2 choices).
5. Answer: $2^{n^2-n}$.

Question 3: Subset Logic

Problem: If $A\subset B$, what is $A\cup (B\cap C)$? Logic:

• Draw Venn Diagram.
• Since $A$ is inside $B$, $A\cup (B\cap C)$ depends on $C$.
• Actually, simplify: $B\cap C$ is the part of $C$ inside $B$.
• Adding $A$ (which is already inside $B$) doesn’t simplify neatly unless we know more about $C$.
• Trap: Don’t assume $A\cup (B\cap C)=B$.
• Test: $A=\{1\},B=\{1,2\},C=\{3\}$. $B\cap C=\varnothing$. $A\cup \varnothing =\{1\}=A$.