A solution is usually a set, even when that set contains one number. A domain is a set. An equation is a statement whose truth depends on a variable. A theorem may claim that every object with one property has another property, while a counterexample shows that some object does not. Sets and logic are therefore not decoration placed around algebra; they are its grammar.
Compare these claims:
They are related but not interchangeable. The first is a predicate until \(x\) is specified. The second reports a complete set. The third asserts completeness. The fourth asserts existence but says nothing about how many solutions there are. Difficult algebra often depends on seeing exactly which claim is required.
After completing the chapter, you should be able to:
You should be able to distinguish \(=\), \(\Rightarrow\), and \(\Leftrightarrow\) and to audit a line of algebra. Review How to read and write an algebraic argument if a transformation’s logical direction is unclear.
A set is a collection whose membership is unambiguous. We write \(x\in A\) to mean “\(x\) is an element of \(A\)” and \(x\notin A\) for nonmembership. Order and repetition do not matter:
\[ \{1,2,3\}=\{3,2,1,1\}. \]
The empty set, written \(\varnothing\), has no elements. A singleton such as \(\{0\}\) has one element and is not the same object as \(0\).
Three common descriptions are:
In set-builder notation, the colon or vertical bar means “such that.” The universe or type restriction is essential. The notation \(\{x:x^2<2\}\) describes different sets when \(x\) ranges over integers, rationals, or reals.
We use \(\mathbb N\) for positive integers \(\{1,2,3,\ldots\}\) unless a page states another convention, \(\mathbb Z\) for integers, \(\mathbb Q\) for rationals, \(\mathbb R\) for reals, and \(\mathbb C\) for complex numbers. Because conventions for \(\mathbb N\) vary, write \(\mathbb N_0=\{0,1,2,\ldots\}\) when zero must be included explicitly.
Write the set of integer multiples of \(3\) whose absolute value is at most \(10\).
The phrase “integer multiples” suggests \(3k\) with \(k\in\mathbb Z\). The bound is \(|3k|\le10\), which allows \(k=-3,-2,-1,0,1,2,3\). Therefore
\[ \{3k:k\in\mathbb Z,|3k|\le10\} =\{-9,-6,-3,0,3,6,9\}. \]
The endpoints \(\pm10\) are not included merely because the inequality is nonstrict; they are not multiples of \(3\).
The statement \(A\subseteq B\) means every element of \(A\) is also an element of \(B\):
\[ A\subseteq B\quad\Longleftrightarrow\quad (\forall x)(x\in A\Rightarrow x\in B). \]
Some authors write \(A\subset B\) for a proper subset and others for an arbitrary subset. We avoid ambiguity by writing \(A\subsetneq B\) when \(A\subseteq B\) and \(A\ne B\).
Set equality is mutual containment:
\[ A=B\quad\Longleftrightarrow\quad A\subseteq B\text{ and }B\subseteq A. \]
This gives the standard proof method: choose an arbitrary element of one set, show it belongs to the other, then reverse the direction.
Let \(A=\{1,\{2\}\}\). Then \(1\in A\) and \(\{2\}\in A\). Also \(\{1\}\subseteq A\) and \(\{\{2\}\}\subseteq A\). But \(2\notin A\): the set \(\{2\}\) is an element, not its contained number. Braces change the object.
Prove
\[ \{x\in\mathbb R:x^2-5x+6=0\}=\{2,3\}. \]
Let \(S\) denote the left set. If \(x\in S\), then \(x^2-5x+6=(x-2)(x-3)=0\). The zero-product property gives \(x=2\) or \(x=3\), so \(x\in\{2,3\}\). Hence \(S\subseteq\{2,3\}\).
Conversely, substitution shows both \(2\) and \(3\) make the polynomial zero. Thus \(\{2,3\}\subseteq S\). Mutual containment proves equality. Notice that factoring found candidates, while the reverse containment verified completeness.
How many subsets does \(\{a,b,c\}\) have? Include the empty set and the whole set.
Each of the three elements is independently included or excluded, producing \(2^3=8\) subsets: \(\varnothing\), \(\{a\}\), \(\{b\}\), \(\{c\}\), \(\{a,b\}\), \(\{a,c\}\), \(\{b,c\}\), and \(\{a,b,c\}\).
For sets \(A\) and \(B\) in a universe \(U\):
\[ \begin{aligned} A\cup B&=\{x:x\in A\text{ or }x\in B\},\\ A\cap B&=\{x:x\in A\text{ and }x\in B\},\\ A\setminus B&=\{x:x\in A\text{ and }x\notin B\},\\ A^c&=U\setminus A. \end{aligned} \]
Mathematical “or” is inclusive unless stated otherwise: membership in both sets still gives membership in the union. Complements depend on the universe. The complement of \([0,1]\) in \(\mathbb R\) is \((-\infty,0)\cup(1,\infty)\); its complement in \([0,2]\) is \((1,2]\).
Let \(U=\{1,2,3,4,5,6\}\), \(A=\{1,2,4,6\}\), and \(B=\{2,3,6\}\). Then
\[ A\cup B=\{1,2,3,4,6\},\qquad A\cap B=\{2,6\}, \]
\[ A\setminus B=\{1,4\},\qquad A^c=\{3,5\}. \]
The element \(2\) appears only once in the union because multiplicity is not part of ordinary set membership.
Complement reverses union and intersection:
\[ (A\cup B)^c=A^c\cap B^c, \qquad (A\cap B)^c=A^c\cup B^c. \]
For an arbitrary \(x\in U\),
\[ \begin{aligned} x\in(A\cup B)^c &\Leftrightarrow x\notin A\cup B\\ &\Leftrightarrow \text{$x\notin A$ and $x\notin B$}\\ &\Leftrightarrow \text{$x\in A^c$ and $x\in B^c$}\\ &\Leftrightarrow x\in A^c\cap B^c. \end{aligned} \]
The equivalence holds for every \(x\in U\), so the sets are equal. The logical change from “not (in \(A\) or in \(B\))” to “not in \(A\) and not in \(B\)” drives the set identity.
For real \(a<b\):
| Interval | Inequality description | Endpoint membership |
|---|---|---|
| \((a,b)\) | \(a<x<b\) | neither endpoint |
| \([a,b]\) | \(a\le x\le b\) | both endpoints |
| \([a,b)\) | \(a\le x<b\) | \(a\) only |
| \((a,b]\) | \(a<x\le b\) | \(b\) only |
Infinity is not a real endpoint and is never included, so unbounded intervals use parentheses: \((-\infty,a)\), \((-\infty,a]\), \((a,\infty)\), and \([a,\infty)\).
An interval is bounded if it lies between two finite real bounds. It is open if every one of its points has a small real neighborhood inside the set, and closed if it contains all of its real limit points. For ordinary bounded intervals, round and square endpoint notation matches these ideas. However, \(\mathbb R\) and \(\varnothing\) are both open and closed. “Open” does not mean “unbounded,” and “closed” does not mean “finite.”
Solve and describe \(-2\le3x+1<10\).
Subtract \(1\) throughout and divide by positive \(3\):
\[ -3\le3x<9\Leftrightarrow-1\le x<3. \]
The solution set is
\[ \{x\in\mathbb R:-1\le x<3\}=[-1,3). \]
The left endpoint is included because equality is allowed; the right is excluded.
The condition \(|x|>2\) means the distance from \(0\) is greater than \(2\), so
\[ x<-2\text{ or }x>2, \]
which is the union \((-\infty,-2)\cup(2,\infty)\).
The simultaneous conditions \(x>-2\) and \(x\le5\) require intersection:
\[ (-2,\infty)\cap(-\infty,5]=(-2,5]. \]
Words such as “or” and “and” determine the set operation.
For real \(a<b<c\), prove
\[ [a,c]\setminus(a,b) = \{a\}\cup[b,c]. \]
A point in \([a,c]\) remains after removing \((a,b)\) exactly when it is \(a\) or is at least \(b\). Thus it belongs to \(\{a\}\cup[b,c]\). Conversely, \(a\) and every point of \([b,c]\) lie in \([a,c]\) but not in \((a,b)\). Both containments hold. The singleton \(\{a\}\) prevents the common mistake of losing the left endpoint.
The Cartesian product is
\[ A\times B=\{(a,b):a\in A,\ b\in B\}. \]
Ordered pairs preserve order: generally \((a,b)\ne(b,a)\). Consequently, \(A\times B\) and \(B\times A\) need not agree.
If \(A=\{1,2\}\) and \(B=\{u,v,w\}\), then
\[ A\times B=\{(1,u),(1,v),(1,w),(2,u),(2,v),(2,w)\}. \]
There are \(2\cdot3=6\) pairs. In the coordinate plane, \([0,1]\times[2,4]\) is a closed rectangle, not a one-dimensional interval.
A graph of a real function is a special subset of a Cartesian product: it contains ordered pairs \((x,f(x))\) and never contains two pairs with the same first coordinate but different second coordinates.
A statement is a declarative sentence with a definite truth value. Questions, commands, and predicates with unspecified variables are not yet statements. Given statements \(P\) and \(Q\):
An implication is false only when its hypothesis \(P\) is true and its conclusion \(Q\) is false. A false hypothesis does not provide a counterexample.
“If an integer is divisible by \(4\), then it is even” is true. Divisibility by \(4\) is sufficient for evenness, and evenness is necessary for divisibility by \(4\). Evenness is not sufficient: \(6\) is even but not divisible by \(4\).
The converse “if an integer is even, then it is divisible by \(4\)” is false. Its counterexample must be even (true hypothesis) and not divisible by \(4\) (false conclusion), as \(6\) is.
For \(P\Rightarrow Q\):
The original implication is logically equivalent to its contrapositive, not in general to its converse or inverse.
Claim: if \(n^2\) is even for an integer \(n\), then \(n\) is even.
The contrapositive says: if \(n\) is odd, then \(n^2\) is odd. Write \(n=2k+1\). Then
\[ n^2=4k^2+4k+1=2(2k^2+2k)+1, \]
which is odd. Therefore the contrapositive, and hence the original implication, is true.
The universal quantifier \(\forall\) means “for every”; the existential quantifier \(\exists\) means “there exists at least one.” The order of quantifiers matters.
\[ (\forall x\in\mathbb R)(\exists y\in\mathbb R)(y>x) \]
is true: after \(x\) is given, choose \(y=x+1\). But
\[ (\exists y\in\mathbb R)(\forall x\in\mathbb R)(y>x) \]
is false: there is no single real number larger than every real number.
Negation switches the quantifier and negates the predicate:
\[ \neg(\forall x\,P(x))\Leftrightarrow\exists x\,\neg P(x), \]
\[ \neg(\exists x\,P(x))\Leftrightarrow\forall x\,\neg P(x). \]
In prose, “not every” means “at least one not,” not “none.”
Negate:
For every real \(x\), there exists an integer \(n\) such that \(n>x\).
Move inward one quantifier at a time:
\[ \neg(\forall x\in\mathbb R)(\exists n\in\mathbb Z)(n>x) \]
is equivalent to
\[ (\exists x\in\mathbb R)(\forall n\in\mathbb Z)(n\le x). \]
The inequality negates from \(n>x\) to \(n\le x\). The original statement is true (the integers are unbounded above), so its negation is false.
Claim: for every real \(x\), there exists a real \(y\) such that \(y^2=x\).
To refute it, we need one real \(x\) for which every real \(y\) fails. Choose \(x=-1\). Every real square satisfies \(y^2\ge0\), so no real \(y\) has \(y^2=-1\). This identifies exactly why extending from \(\mathbb R\) to \(\mathbb C\) later is useful.
In \(P(x,y):x+y=0\), both variables are free. In \((\exists y)(x+y=0)\), \(y\) is bound by the quantifier while \(x\) remains free; the result is a predicate about \(x\). A variable’s symbol can be renamed inside its quantifier without changing meaning: \((\exists y)(x+y=0)\) and \((\exists t)(x+t=0)\) are equivalent.
Solving often means transforming a predicate into an explicit set description. Keep the universe visible and use logic to combine cases.
Describe
\[ S=\left\{x\in\mathbb R:\frac{x-1}{x+2}\ge0\right\}. \]
The critical values are the zero \(x=1\) and excluded denominator value \(x=-2\). On \((-\infty,-2)\), numerator and denominator are both negative, so the quotient is positive. On \((-2,1)\) their signs differ, so it is negative. On \((1,\infty)\) both are positive. Equality includes \(x=1\) but not \(x=-2\). Therefore
\[ S=(-\infty,-2)\cup[1,\infty). \]
The logical form is “\(x<-2\) or \(x\ge1\).”
For which real \(a\) does there exist a real \(x\) satisfying \(x^2=a\)?
The statement is
\[ (\exists x\in\mathbb R)(x^2=a). \]
Every real square is nonnegative, so existence implies \(a\ge0\). Conversely, if \(a\ge0\), the real number \(x=\sqrt a\) satisfies \(x^2=a\). Therefore existence is equivalent to \(a\in[0,\infty)\). The proof contains both necessity and sufficiency.
These statements happen to be equivalent, but \(x\subseteq A\) is usually meaningless unless \(x\) itself is a set. Track the type of each object.
The notation \([2,\infty]\) is not a real interval because \(\infty\) is not a real number. Write \([2,\infty)\).
The negation of \(x>0\) and \(x<1\) is \(x\le0\) or \(x\ge1\). De Morgan’s law switches “and” to “or.”
To refute \(P\Rightarrow Q\), an example must make \(P\) true and \(Q\) false. An object outside the hypothesis says nothing about the implication.
“Every student chose a book” allows different books. “There is a book every student chose” asserts one common book. The same difference drives parameter and existence problems.
Write \(\{x\in\mathbb Z:-3<x\le2\}\) in roster notation.
The allowed integers are \(-2,-1,0,1,2\), so the roster is \(\{-2,-1,0,1,2\}\). Strictness excludes \(-3\); nonstrictness includes \(2\).
Let \(A=(-3,4]\) and \(B=[1,7)\). Find \(A\cap B\), \(A\cup B\), and \(A\setminus B\).
The overlap is \([1,4]\). Together the intervals cover \((-3,7)\). Removing \(B\) from \(A\) leaves \((-3,1)\); the point \(1\) is removed because it belongs to \(B\).
Negate: “For every positive real \(x\), if \(x>1\), then \(x^2>x\).”
A negated universal implication asserts one counterexample:
There exists a positive real \(x\) such that \(x>1\) and \(x^2\le x\).
The negation is false, because for \(x>1\), multiplying by positive \(x\) gives \(x^2>x\).
Prove \(A\setminus(B\cup C)=(A\setminus B)\cap(A\setminus C)\).
For arbitrary \(x\),
\[ \begin{aligned} x\in A\setminus(B\cup C) &\Leftrightarrow x\in A\text{ and }x\notin B\cup C\\ &\Leftrightarrow x\in A\text{ and }x\notin B\text{ and }x\notin C\\ &\Leftrightarrow x\in A\setminus B\text{ and }x\in A\setminus C\\ &\Leftrightarrow x\in(A\setminus B)\cap(A\setminus C). \end{aligned} \]
Because this holds for every \(x\), the sets are equal.
For real parameters \(a\) and \(b\), determine exactly when
\[ (-\infty,a]\cup[b,\infty)=\mathbb R. \]
Interpret first. The left set covers everything at or below \(a\) and everything at or above \(b\). A gap exists only if \(a<b\), namely \((a,b)\).
Necessity. Suppose the union equals \(\mathbb R\). If \(a<b\), the midpoint \(m=(a+b)/2\) satisfies \(a<m<b\). It belongs to neither interval, contradicting full coverage. Therefore \(a\ge b\).
Sufficiency. Suppose \(a\ge b\) and let \(x\in\mathbb R\). If \(x\le a\), then \(x\) belongs to the first interval. If \(x>a\), then \(x>a\ge b\), so \(x\ge b\) and belongs to the second. Every real \(x\) belongs to the union.
Thus
\[ (-\infty,a]\cup[b,\infty)=\mathbb R \quad\Longleftrightarrow\quad a\ge b. \]
Strategy lesson. Draw a qualitative number line, test the possible orderings of parameters, use a midpoint to prove a gap, and prove the no-gap case for an arbitrary element. The diagram discovers the condition; the two-direction argument proves it.
Write the odd integers between \(-6\) and \(8\) in roster and set-builder form.
Decide whether each is true for \(A=\{0,\{1\},2\}\): \(1\in A\), \(\{1\}\in A\), \(\{0,2\}\subseteq A\).
Translate \(x<-4\) or \(2\le x<7\) into interval notation.
Write the complement of \([-2,5)\) in \(\mathbb R\).
If \(A=\{a,b\}\) and \(B=\{1,2,3\}\), list \(A\times B\).
State the converse and contrapositive of “if \(x>5\), then \(x>0\).”
For \(A=(-4,2]\) and \(B=[-1,6)\), find union, intersection, and both differences.
Prove \(A\cap B\subseteq A\cup B\) directly from membership definitions.
Give a counterexample to “if \(xy=0\), then \(x=0\)” and repair the statement.
Negate “there exists a real \(x\) such that \(x^2+1=0\).”
Negate “for every integer \(n\), \(n\) is even or \(n\) is odd.”
Determine whether the converse of “if an integer is divisible by \(6\), then it is divisible by \(3\)” is true.
Express the real solution of \(|x-2|\le5\) in set-builder and interval notation.
Determine whether \([0,1)\cup[1,2]=[0,2]\) and justify each endpoint.
Prove \((A\cap B)\cup(A\setminus B)=A\).
Solve \((x-2)(x+1)\le0\) and express the logical sign conditions before giving interval notation.
For which real \(a\) is \((-\infty,a)\cap[3,\infty)\) nonempty?
Compare \((A\setminus B)\times C\) with \((A\times C)\setminus(B\times C)\). Prove equality or give a counterexample.
Translate into quantifiers: “Every real polynomial of odd degree has at least one real zero.” Then write its formal negation.
Prove that \(A\subseteq B\) if and only if \(A\cap B=A\).
Determine all real \(a,b\) for which \([a,b]=\varnothing\) under the convention that \([a,b]=\{x\in\mathbb R:a\le x\le b\}\).
Prove that \(A\times B=B\times A\) need not imply \(A=B\) unless additional nonemptiness conditions are imposed. State and prove the strongest simple corrected claim.
Negate and decide the truth of \((\forall x\in\mathbb R)(\exists! y\in\mathbb R)(x+y=0)\), where \(\exists!\) means “there exists exactly one.”
Let \(S_a=\{x\in\mathbb R:|x-a|<1\}\). Determine and prove exactly when \(S_a\cap S_b=\varnothing\).
Let \(x\in(A\cap B)\cup(A\setminus B)\). In the first case \(x\in A\); in the second case \(x\in A\) as well. Thus the union is a subset of \(A\).
Conversely, let \(x\in A\). Either \(x\in B\) or \(x\notin B\). In the first case \(x\in A\cap B\); in the second \(x\in A\setminus B\). Therefore \(x\) belongs to the union. Mutual containment proves equality.
Suppose \(A\subseteq B\). An element of \(A\cap B\) lies in \(A\), so \(A\cap B\subseteq A\). Conversely, if \(x\in A\), the subset assumption gives \(x\in B\), hence \(x\in A\cap B\). Thus \(A\cap B=A\).
For the reverse direction, suppose \(A\cap B=A\). If \(x\in A\), then \(x\in A\cap B\) by the equality, so \(x\in B\). Hence \(A\subseteq B\).
If \(A=\varnothing\) and \(B\) is any nonempty set, then \(A\times B=\varnothing=B\times A\) although \(A\ne B\). This disproves the claim without nonemptiness.
Now suppose \(A\) and \(B\) are nonempty and \(A\times B=B\times A\). Choose \(b_0\in B\). For any \(a\in A\), \((a,b_0)\in A\times B=B\times A\), so its first coordinate \(a\) lies in \(B\). Thus \(A\subseteq B\). By symmetry, \(B\subseteq A\), and therefore \(A=B\).
We have \(S_a=(a-1,a+1)\) and \(S_b=(b-1,b+1)\). Assume without loss of generality that \(a\le b\). They are disjoint exactly when the right endpoint of the first is at or left of the left endpoint of the second: \(a+1\le b-1\), or \(b-a\ge2\). Equality still gives disjoint open intervals because the touching endpoint belongs to neither. Restoring symmetry yields \(|a-b|\ge2\).
A learner claims
\[ \left\{x\in\mathbb R:\frac{x^2-1}{x-1}=x+1\right\}=\mathbb R. \]
Identify the error and write the correct set.
The rational expression is undefined at \(x=1\). For \(x\ne1\), factoring and cancelling proves equality. Therefore the solution set is \(\mathbb R\setminus\{1\}=(-\infty,1)\cup(1,\infty)\). Formula simplification did not restore an excluded domain point.
Decide whether the following is true, and prove or refute it:
For every real \(x\), if there exists a real \(y\) with \(xy=1\), then \(x\ne0\); and if \(x\ne0\), there exists exactly one such \(y\).
The statement is true. If \(xy=1\) and \(x=0\), then \(0=1\), impossible, so \(x\ne0\). Conversely, for \(x\ne0\), choose \(y=1/x\); then \(xy=1\). If both \(y_1\) and \(y_2\) satisfy the equation, \(xy_1=xy_2\) and division by nonzero \(x\) gives \(y_1=y_2\). Thus existence and uniqueness both hold.
Answer from definitions rather than pattern recognition. The generated notebook contains the same questions and readable solutions.
Continue with The real number system and its extensions, which uses this chapter’s set and quantifier language to study closure, rationality, irrationality, density, exact comparison, and approximation.
SymPy represents mathematical sets explicitly. Use these objects to check a manual result and inspect endpoint behavior.
(Interval(1, 4), Interval.open(-3, 7), Interval.open(-3, 1))The output should represent \([1,4]\), \((-3,7)\), and \((-3,1)\), respectively. Names such as Lopen describe the open left endpoint; always compare the rendered set with your inequality interpretation.
Use solveset rather than assuming that a generic list of roots carries domain information:
({2, 3}, Union(Interval.open(-oo, -2), Interval(1, oo)))The expected exact sets are \(\{2,3\}\) and \((-\infty,-2)\cup[1,\infty)\). The excluded denominator value is absent from the second result.
Logic can also be represented symbolically, but a computer simplification does not replace a proof that explains the universe and predicates:
\(\displaystyle \text{True}\)
The result True verifies the truth-table equivalence between an implication and its contrapositive. Your written work must still translate \(P\) and \(Q\) back to the mathematical claims under discussion.