Mathematical Language & Sets
Translate ideas precisely, distinguish expressions from sentences, and work with sets.
Learning goals
- Translate ordinary language into mathematical notation.
- Distinguish open and closed sentences.
- Describe sets and perform basic set operations.
Precise, concise, and powerful
Mathematical notation is useful because it lets us state exactly which objects and relationships we mean. A small change in wording or symbols can change the claim. We will practice moving between words and notation before using that precision to describe sets.
Mathematical language makes distinctions explicit. Write \(\pi\approx3.14\), not an exact equality. An absolute value is nonnegative because zero is possible. A rational number is a quotient of integers with a nonzero denominator.
Translate and solve
Show worked solution
Define the variable before writing the equation.
Expressions and sentences
Mathematical writing becomes easier to read when we notice the job each piece is doing. An expression names something; a sentence makes a claim. Before asking whether a statement is true, make sure it is actually a statement.
An expression names a mathematical object; a sentence states a complete relationship. An open sentence contains an unassigned variable and becomes true or false after the domain and values are supplied. A closed sentence has a definite truth value.
| Notation | Type | Interpretation |
|---|---|---|
| \(x+x^2\) | Expression | Sum of a number and its square |
| \(3+7=10\) | Closed sentence | True |
| \(x^2=x\) | Open sentence | \(x=0\text{ or }x=1\text{ over }\mathbb R\) |
| \(x+0=x\) | Open sentence | True for every real assignment |
An open sentence can hold for every assignment, some assignments, or none. An explicit quantifier closes the statement.
Translation and equivalent expressions
We have distinguished expressions from claims. Now let us translate ordinary language without changing its meaning. Pay attention to which quantity is being compared with which, especially in phrases involving subtraction or division. Read your mathematical expression back in words; if that reading differs from the original sentence, revise the translation.
Rectangle area
Show worked solution
Let \(w\) be its positive width. Length is \(w+4\), so \(A=w(w+4)=w^2+4w\).
Consecutive even integers
Show worked solution
Coins
Show worked solution
Let \(x\) count the five-peso coins. There are \(20-x\) ten-peso coins.
\[V=5x+10(20-x)=200-5x\]Here \(x\in\{0,1,\ldots,20\}\).
Sets and membership
We now have a way to express relationships precisely. Sets give us a language for grouping the objects involved. Keep the distinction between an object and a collection of objects in mind; it explains many of the notation choices that follow.
A set is a well-defined collection. Its elements satisfy an objective membership rule. Write \(a\in A\) or \(a\notin A\). Order and repetitions do not change a set.
| Number set | Description |
|---|---|
| \(\mathbb N\) | Natural numbers; specify whether zero is included |
| \(\mathbb Z\) | Integers |
| \(\mathbb Q\) | Rational numbers |
| \(\mathbb R\) | Real numbers |
Roster notation
Do these two rosters describe the same set? Explain what happens to order and repeated entries.
\(\{1,1,2,3\}=\{3,2,1\}\)Show worked solution
Both contain exactly the same three distinct elements.
Set-builder notation and cardinality
A membership test rather than a list
Set-builder notation names a variable, specifies its allowed universe, and gives a condition for membership. The cardinality of a finite set is its number of distinct elements.
To decide whether an object belongs, check both the universe and the condition; cardinality then counts the distinct qualifying objects.
A long list is not always the clearest way to describe a set. A membership condition can describe it more efficiently, but the universe and restrictions must be clear. Once membership is settled, cardinality counts distinct elements; repeated entries in a written list do not create new members.
\[A=\{x\in\mathbb Z\mid x^2=25\}=\{-5,5\}\]The cardinality is \(|A|=2\). The empty set \(\varnothing\) has no elements; \(\{\varnothing\}\) is a singleton containing the empty set.
A domain changes a set
List the elements of this set of real numbers and determine its cardinality.
\(\{x\in\mathbb R\mid x^2<-1\}\)Show worked solution
Squares of real numbers are nonnegative, so the set is \(\varnothing\) and its cardinality is zero.
Sets can themselves be the objects we count
The power set \(\mathcal P(A)\) is the set of all subsets of A. If A has n distinct elements, then \(|\mathcal P(A)|=2^n\): for each element independently choose “include” or “exclude.” Every string of choices determines exactly one subset, and every subset determines exactly one string. This proves the count, including the empty set and A itself.
Build a set of choices
An activity offers three optional tools: ruler, compass and protractor. Treat a toolkit as a subset of these tools. How many nonempty toolkits omit at least one tool?
Show worked solution
There are \(2^3=8\) subsets. Exclude the empty toolkit and the full toolkit, leaving \(6\). These two exclusions are distinct because the original set is nonempty. The resulting objects are sets of tools, not individual tools.
Do not confuse \(\varnothing\) with \(\{\varnothing\}\): their sizes are zero and one. Also \(\mathcal P(\varnothing)=\{\varnothing\}\) has one element; blindly subtracting two from \(2^0\) would count the same excluded subset twice.
Subsets and equality
Membership asks whether one object lies in a set. Being a subset asks whether every member of one set also lies in another. Try explaining the difference with a collection containing just one object. To show two sets are equal, we must account for membership in both directions, not simply notice a few shared elements.
\(A\subseteq B\) means every element of \(A\) belongs to \(B\). A proper subset also requires \(A\ne B\). Equality means both inclusion directions hold.
Element or subset?
Decide whether each statement is true and explain the different roles of the two symbols.
\(2\in\{1,2,3\},\qquad\{2\}\subseteq\{1,2,3\}\)Show worked solution
Both are true. The first compares an object with a set; the second compares two sets.
The empty set is a subset of every set. It is an element only of sets that explicitly contain it.
Specify the universe before taking a complement
For a specified universe U and a subset A, the complement is \(A^c=U\setminus A=\{x\in U:x\notin A\}\). It means everything allowed by the universe that is not in A. Unlike \(A\setminus B\), a complement notation needs the surrounding universe to be understood.
The same set, two complements
Let \(A=\{2,4\}\). Find its complement first in \(U=\{1,2,3,4\}\), then in \(V=\{1,2,3,4,5,6\}\). Are the answers contradictory?
Show worked solution
In U the complement is \(\{1,3\}\). In V it is \(\{1,3,5,6\}\). Both are correct: the universe changes what “not in A” is allowed to include. A membership test must check inclusion in the universe as well as exclusion from A.
Union, intersection, and difference
Combining membership conditions
Union includes objects in either set or both; intersection keeps only objects shared by both. The difference of the first set and the second keeps objects belonging to the first but not the second, so order matters.
A region diagram is a picture of those membership tests, not a substitute for specifying the universe.
Once the sets are clear, imagine checking one object at a time. Does it belong to either set, to both, or to one but not the other? These membership questions give meaning to the operations and help you verify a diagram or a list.
| Operation | Membership rule |
|---|---|
| \(A\cup B\) | In either set, possibly both |
| \(A\cap B\) | In both sets |
| \(A\setminus B\) | In A but not B |
Apply the three rules
Find the union, intersection, and both set differences for these sets.
\(A=\{1,2,4,5\},\quad B=\{2,3\}\)Show worked solution
Practice
Closed or open?
Classify this sentence as open or closed, then determine whether any real assignment makes it true.
\(x^2+2x+1<0\)Show worked solution
It is open, with no real solution because \((x+1)^2\ge0\).
Simplify a translation
Show worked solution
Nested sets
How many elements does E have? Count each inner set as one object.
\(E=\{\{1,2,3\},\{4,5,6\}\}\)Show worked solution
There are two elements, each itself a set: \(|E|=2\).
An interval of integers
Write F in roster notation and find its cardinality.
\(F=\{x\in\mathbb Z\mid -2<x\le3\}\)Show worked solution
Additional practice: language and sets
Classify mathematical sentences
Work over the real numbers. Decide whether each sentence is open or closed. For closed sentences give its truth value; for open sentences state when it is true.
- \(3+7=10\)
- \(x+3=3+x\)
- \(x^2=x\)
- \(x+0=x\)
- \(x^2+2x+1<0\)
Show worked solution
- Closed and true.
- Open; true for every real \(x\) by commutativity.
- Open; \(x(x-1)=0\), so \(x=0\) or \(x=1\).
- Open; true for every real \(x\).
- Open; never true over the reals because \((x+1)^2\ge0\).
Different names, same number
Write an expression equal to five using the specified operation. Many answers are possible.
- Addition
- Subtraction
- Multiplication
- Division
Show worked solution
- \(2+3=5\)
- \(8-3=5\)
- \(1\times5=5\)
- \(10\div2=5\)
Well-defined collections
Explain whether each description determines a set.
- All students enrolled in Math 10.
- All beautiful students.
Show worked solution
- Yes. Enrollment records determine membership.
- Not without a stated criterion: beauty is subjective.
Describe each set
List the elements when possible. If a finite roster is impossible, describe the set. Then give its cardinality.
- \(A=\{x\in\mathbb Z\mid3^2+4^2=x^2\}\)
- \(B=\{x\in\mathbb R\mid0<x<1\}\)
- \(C=\{x\in\mathbb R\mid x^2<-1\}\)
- \(D=\{x\in\mathbb Z_{>0}\mid1<x<3\}\)
- \(E=\{2,4,6,8,10\}\)
- \(F=\{1,2,3,\ldots,100\}\)
- \(G=\{\{1,2,3\},\{4,5,6\}\}\)
- \(H=\varnothing\)
- \(I=\{\varnothing\}\)
Show worked solution
- \(A=\{-5,5\}\); cardinality \(2\). Both roots satisfy the equation.
- \(B=(0,1)\); infinitely many real numbers.
- \(C=\varnothing\); cardinality \(0\), since a real square is nonnegative.
- \(D=\{2\}\); cardinality \(1\).
- \(|E|=5\).
- \(|F|=100\).
- \(|G|=2\); count the two sets, not their individual members.
- \(|H|=0\).
- \(|I|=1\); the empty set is itself the single element.
Equal sets
Determine whether the two sets are equal. Explain your answer.
- \(\{1,2,3,4,5\},\quad\{3,4,2,1,5\}\)
- \(\{1,2,3,4,5\},\quad\{1,1,2,2,3,3,4,4,5,5\}\)
Show worked solution
- Yes. Changing the order does not change membership.
- Yes. Repeating an element does not create a new member.
Additional practice: mixed exercises
Translate precisely
Define a variable when needed, then write and simplify the expression.
- The square of three added to the product of five and two.
- Five is added to twice a number. Dividing the result by three gives three. Find the number.
- Write three added to itself fifteen times as a short expression.
- The sum of a number and its square.
- The age of a woman fifteen years ago.
- Describe \(\{x\in\mathbb R\mid x\text{ is prime}\}\).
- Describe the set of people taking a Math 10 course.
Show worked solution
- \(5\cdot2+3^2=10+9=19\).
- Let \(x\) be the number. \((2x+5)/3=3\) gives \(2x+5=9\), hence \(x=2\).
- \(15\cdot3=45\). Multiplication abbreviates repeated addition.
- Let \(x\) be the number: \(x+x^2\).
- Let \(a\) be her present age in years, with \(a\ge15\). Her age then was \(a-15\).
- Under the usual definition, primes are positive integers greater than \(1\) with exactly two positive divisors. The set is \(\{2,3,5,7,11,\ldots\}\), an infinite subset of \(\mathbb R\).
- Membership is determined by enrollment in the specified term. A roster would require those enrollment records; the description alone does not give the students’ names or the cardinality.
Nested sets and subsets
Distinguish a number from a set containing that number. Here the proper-subset symbol means a subset that is not equal.
- Are \(\{1,2,3,4,5\}\) and \(\{\{1\},\{2\},\{3\},\{4\},\{5\}\}\) equal?
- Determine the cardinality of \(\{\{1\},\{2,3\},\{4\},\{5\},\{6\}\}\).
- Is \(\{2\}\subsetneq\{1,2,3\}\) true?
- Is \(2\subseteq\{1,2,3\}\) the correct way to say that two belongs to this set?
- Is \(\{2\}\subseteq\{\{1\},\{2\},\{3\}\}\) true?
- Is \(\{2,2\}\subseteq\{1,\{2\},\{3\}\}\) true?
- Is \(\varnothing\subseteq\{1,2,3\}\) true?
Show worked solution
- No. For instance, \(1\) belongs to the first set, while the second contains the singleton \(\{1\}\) instead of the number \(1\).
- There are five elements, each itself a set. Thus the cardinality is \(5\), not \(6\).
- Yes. Its only element \(2\) belongs to the larger set, and the two sets are unequal.
- No. In this elementary notation, \(2\) is a number, so use membership: \(2\in\{1,2,3\}\). A subset comparison instead uses \(\{2\}\).
- No. Its element is the number \(2\), which is not an element of the right-hand set. However, \(\{2\}\in\{\{1\},\{2\},\{3\}\}\) is true.
- No. Repetition does not change the left set: it is \(\{2\}\). The number \(2\) is absent from the right set, even though the singleton \(\{2\}\) is present.
- Yes. The empty set has no element that could fail the subset requirement; it is a subset of every set.
Takeaways and connections
- Define variables and domains.
- Count distinct elements, including nested sets as single objects.
- Distinguish membership from inclusion.
Reasoning, connections and deeper practice
Set notation makes a claim about individual membership. To prove two sets equal, explain why an arbitrary object belongs to one exactly when it belongs to the other. A diagram can guide this reasoning but does not replace it.
Application · Reconstruct the survey groups
In a class of \(40\) students, \(24\) join a music club, \(18\) join a science club and \(7\) join neither. Find the number in both, music only and science only, explaining each subtraction.
Hint
First find the union from the class total and the number in neither.
Show worked solution
The union has \(40-7=33\) students. Adding club totals counts their overlap twice, so the overlap is \(24+18-33=9\). Music only has \(24-9=15\), and science only has \(18-9=9\). Check the disjoint regions: \(15+9+9+7=40\). The overlap must be subtracted once from the sum of totals, not from the class size.
Advanced / Honors · A difference law by membership
Prove \(A\setminus(B\cup C)=(A\setminus B)\cap(A\setminus C)\). Then give a counterexample to replacing the intersection on the right by a union.
Hint
Translate belonging to a difference as belonging to the first set and not to the second.
Show worked solution
For any object \(x\), membership in the left side means \(x\in A\), \(x\notin B\) and \(x\notin C\). This is exactly membership in both \(A\setminus B\) and \(A\setminus C\), proving equality in both directions. For the incorrect union version take \(A=\{1\}, B=\{1\}, C=\varnothing\). The left side is empty, whereas the proposed right side is \(\varnothing\cup\{1\}=\{1\}\).
