Mathematics I · Week 1 — Sets and logic
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2026-07-19T00:00:00.000Z
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A concept-first note on sets, statements, quantifiers, and how to test a claim. # Week 1 — sets and logic This is a concept chapter, not an answer key.
Week 1 — sets and logic
This is a concept chapter, not an answer key. Its job is to make a question feel classifiable before you calculate or prove anything. The reliable order is: identify the objects, state the universe, translate the symbols into words, then decide whether the claim needs a proof, a witness, or a counterexample.
The one-page method
Collection question? Name the universal set and use membership. Truth question? identify proposition/predicate and the quantifier. “If” question? separate implication, converse, inverse, and contrapositive. Only then choose an example or proof method.
Learning objectives
By the end of this chapter, you should be able to:
- Distinguish an element from a subset, and state the universal set needed for a complement.
- Translate union, intersection, difference, complement, and Cartesian product into a membership rule.
- Tell a proposition from a predicate, and say why a domain changes a statement’s truth value.
- Choose the correct proof burden for
∀and∃: general argument, witness, counterexample, or impossibility argument. - Write an implication’s converse, inverse, and contrapositive—and identify the one that preserves truth.
Prerequisites
- Basic arithmetic: factors, divisibility, and a number line are enough for the running examples.
- Notation comfort: read brackets, braces, ordered pairs, and simple algebraic expressions without treating the symbols as a calculation by themselves.
- One honest habit: state the allowed domain before deciding whether an example proves anything.
Core content
Sets, elements, and universes
A set is a collection of distinct objects. Order and repetition do not matter:
{1, 2, 2} is the same set as {2, 1}. The objects inside are elements.| Symbol | Read it as | What it claims |
|---|---|---|
x ∈ A | x belongs to A | x is an element of A |
x ∉ A | x does not belong to A | x is not an element of A |
A ⊆ B | A is a subset of B | every element of A is in B |
A = B | A equals B | they contain exactly the same elements |
∅ | empty set | a set with no elements |
Do not confuse
∈ with ⊆. For example, 2 ∈ {1,2,3}, while {2} ⊆ {1,2,3}. The first compares an object with a set; the second compares two sets. That one distinction resolves a surprising number of early mistakes.The universal set
U is the environment under discussion. Complements only exist relative to it. If U = {1,2,3,4,5,6} and A = {2,4,6}, then Aᶜ = {1,3,5}. But if U were all integers, Aᶜ would be much larger. Never write a complement without silently knowing the universe.Set operations: classify before symbol-pushing
For two sets
A and B:- Union
A ∪ B: elements inAorB, including elements in both. - Intersection
A ∩ B: elements in bothAandB. - Difference
A \ B: elements inAbut not inB. - Complement
Aᶜ: elements inUbut not inA. - Cartesian product
A × B: ordered pairs(a,b)witha ∈ Aandb ∈ B.
Original worked example — membership is a checklist
Let
U = {1,2,3,4,5,6}, A = {1,2,4}, and B = {2,3,4,6}.| Question | Reasoning | Result |
|---|---|---|
A ∩ B | Keep only entries that pass both membership tests. | {2,4} |
A ∪ B | Keep every entry that passes either test once. | {1,2,3,4,6} |
A \ B | Keep entries in A that fail B’s test. | {1} |
Aᶜ | Start from U, remove A. | {3,5,6} |
The useful habit is not memorising the answers. It is asking, element by element, “Does this object meet the membership condition?” This same habit later becomes a truth table.
Make the membership test visible
The sandbox below is deliberately editable. Change a member of either set, choose an operation, then predict the result before looking at the highlighted region. The important move is to say the membership rule in a full sentence: “keep it if it is in A and in B”, not merely “intersection.”
Loading Visualizer...
A two-pass check
Before using a Venn diagram, list the universe and test one element that is in neither set. This prevents the common complement error: drawing only A and B hides the elements which still live in
U.Propositions, predicates, and truth values
A proposition is a declarative statement with a definite truth value. “13 is prime” is a proposition. A command, question, or vague phrase is not.
A predicate contains a variable and becomes a proposition once a value or domain is supplied.
P(x): x² > 9 is a predicate. If the domain is integers and x=4, then P(4) is a true proposition; if x=2, it is false.This domain matters. The claim “there exists an
x with x²=2” is false over integers and true over real numbers. Before judging a statement, locate the allowed inputs.Quantifiers decide the burden of proof
∀ means for every; ∃ means there exists. Read them in words before manipulating symbols.| Form | To establish it | To refute it |
|---|---|---|
∀x P(x) | give a general argument for an arbitrary permitted x | one permitted counterexample |
∃x P(x) | construct one permitted witness | prove no permitted input works |
Negation reverses the claim precisely:
- Not “every student submitted” is “at least one student did not submit.”
- Not “there exists a number with property P” is “every permitted number fails P.”
Checking three examples does not prove “for every.”
Examples are excellent for finding a pattern or a counterexample. A universal claim needs reasoning that covers an arbitrary allowed input.
Original worked example — counterexample versus witness
Consider the statement: “Every even integer is divisible by 4.” The integer
6 is allowed, even, and not divisible by 4. One valid counterexample settles a universal claim.Now consider: “There exists an even integer that is not divisible by 4.” The same
6 is a witness. One witness settles an existential claim. The value is the same; the role comes from the quantifier.Conditionals: four statements that look alike but are not
Let
P be “the integer is divisible by 4” and Q be “the integer is even.”| Name | Form | Status for this P and Q |
|---|---|---|
| Implication | P → Q | true |
| Converse | Q → P | false: 6 is a counterexample |
| Inverse | ¬P → ¬Q | false: 6 is a counterexample |
| Contrapositive | ¬Q → ¬P | true; always equivalent to P → Q |
“P is sufficient for Q” means
P → Q. “Q is necessary for P” means the same direction. The contrapositive is often easier to prove because it preserves truth exactly; the converse does not inherit truth automatically.Animate the one direction that survives
An implication and its contrapositive are logically equivalent. The sequence below is not a proof by animation; use it as a translation rehearsal, then state each line in ordinary language.
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The animation shows a valid equivalence, not a shortcut.
You still need to identify
P and Q correctly. Reversing the arrow makes a converse, and a true implication does not automatically make its converse true.Key formulas and translation sheet
| Symbolic form | Read it in words | What you must check |
|---|---|---|
x ∈ A | x is an element of A | x is an object listed or described by A. |
A ⊆ B | every member of A belongs to B | no member of A is outside B. |
Aᶜ | everything in U that is not in A | the universal set U has been stated. |
A × B | all ordered pairs with first item from A and second from B | order matters; (a,b) need not equal (b,a). |
∀x P(x) | every permitted x has property P | one valid counterexample refutes it. |
∃x P(x) | at least one permitted x has property P | one valid witness establishes it. |
P → Q | if P, then Q | the contrapositive ¬Q → ¬P has the same truth value. |
A decision routine for any Week 1 claim
- Name the universe. What objects are permitted?
- Translate. Replace symbols with “belongs,” “every,” “there exists,” “if,” or “not.”
- Classify. Is this a set operation, an equality of sets, a predicate, or a conditional?
- Choose evidence. Universal: general proof or one counterexample to refute. Existential: witness or impossibility proof.
- Audit the example. Is it inside the stated domain? Does it meet the required conditions exactly?
Common pitfalls
Pitfall 1 — treating an element as a one-element set
2 ∈ A and {2} ⊆ A make different kinds of claim. The first talks about an object; the second talks about a collection. Check the shape of both sides before deciding which symbol belongs.Pitfall 2 — forgetting the universe in a complement
Aᶜ is not “whatever is not visibly in A.” It means the members of the stated U that fail membership in A. Change U and the complement changes.Pitfall 3 — using examples as a universal proof
Three examples can suggest a pattern. One counterexample can refute
∀. But to establish ∀, the argument must work for an arbitrary permitted input—not merely the friendly examples you happened to test.Pitfall 4 — promoting the converse by accident
From
P → Q, you may always use ¬Q → ¬P. You may not silently reverse the direction to Q → P. Test the reversed statement with a value such as “even but not divisible by 4.”Pitfall 5 — letting the domain drift
A witness or counterexample outside the allowed universe has no force. Before you write one down, name the domain and verify your chosen value belongs to it.
Retrieval ladder — close the note before answering
- Give an example showing the difference between
x ∈ Aand{x} ⊆ A. - With
U={1,2,3,4,5}, inventAandB, then describeA ∪ B,A ∩ B, andAᶜin plain English. - Negate: “Every real number has an integer square root.” State the negation in words first.
- For a conditional of your choice, write its converse and contrapositive. Which one must match the original truth value, and why?
- In the sandbox, choose
A \ B. Change one element so the result changes, then explain whyB \ Awould not be the same task.
Five-minute exit check
You are ready to move on when you can say, without looking: the universe defines the complement; a counterexample refutes “every”; a witness establishes “there exists”; and the contrapositive—not the converse—shares an implication’s truth value.