RIDDLES — NUMBER SETS
The Real Number System
RIDDLE 1
I belong to the very first family of numbers most people ever learn, built up one step at a time by repeatedly adding 1 to a starting value.
Zero and negative numbers are not members of me, which sets me apart from some of the larger number sets that contain me.
The numbers 1, 2, 3 and 4 are the smallest members of my set.
I am used whenever discrete objects are counted, such as the number of students in a class.
I am denoted by the symbol N.
Who am I?
ANSWER: Natural Numbers
SOLUTION
The natural numbers are the counting numbers {1, 2, 3, 4, ...}, formed by starting at 1 and repeatedly adding 1. They exclude zero and negative numbers entirely, which is what distinguishes them from the whole numbers and integers that later contain them. The set is denoted N, and its members are used to count discrete objects.
RIDDLE 2
I am extremely close in membership to the natural numbers, differing by only a single extra element.
I include everything in the set of natural numbers, plus zero.
I do not contain any negative numbers or fractions.
The numbers 0, 1, 2 and 3 are examples of my members.
I am denoted by the symbol W.
Who am I?
ANSWER: Whole Numbers
SOLUTION
The whole numbers are the natural numbers together with zero, {0, 1, 2, 3, ...}, denoted W. Adding that single extra element gives the whole numbers, but the set still excludes negative numbers and fractions.
RIDDLE 3
I extend the whole numbers by allowing negative values as full members, not just as results of a calculation.
The numbers -3, 0 and 7 are all examples of my members.
I am closed under addition, subtraction and multiplication, but not under division.
I do not include fractions or decimals.
I am denoted by the symbol Z, from the German word Zahlen, meaning numbers.
Who am I?
ANSWER: Integers
SOLUTION
The integers extend the whole numbers to include negative values, {..., -2, -1, 0, 1, 2, ...}, denoted Z after the German Zahlen. They are closed under addition, subtraction and multiplication, but not division, since dividing two integers can produce a fraction that is not itself an integer.
RIDDLE 4
Every integer you have ever used already belongs to me, though I contain far more numbers than just integers.
My decimal representation always either terminates or recurs, never continuing forever without settling into a pattern.
I am any number that can be expressed as a common fraction a/b, where a and b are integers and b is not zero.
The numbers 1/2, -3 and 0.75 are all examples of my members.
I am denoted by the symbol Q, standing for quotient.
Who am I?
ANSWER: Rational Numbers
SOLUTION
A rational number is any number expressible as a fraction a/b where a and b are integers and b is not zero, denoted Q for quotient. Every integer is rational, since it can be written as itself over 1, and a rational number's decimal representation always either terminates or falls into a repeating pattern.
RIDDLE 5
The ancient Greek thinker Hippasus is traditionally credited with discovering numbers like me, a discovery said to have unsettled his fellow Pythagoreans.
Pi and the square roots of non-perfect squares, such as √2, are examples of me.
My decimal representation is always non-terminating and non-recurring.
I cannot be expressed in the form a/b, where a and b are integers.
I am sometimes denoted by writing R minus Q.
Who am I?
ANSWER: Irrational Numbers
SOLUTION
Irrational numbers cannot be written as a ratio of two integers, so their decimal expansions never terminate and never settle into a repeating pattern; pi and square roots of non-perfect squares like √2 are classic examples. Hippasus's discovery of such numbers is said to have troubled the Pythagoreans, who believed all quantities were expressible as ratios of whole numbers.
RIDDLE 6
Every number you use in everyday life belongs to me, no matter how ordinary or unusual it looks.
Natural numbers, whole numbers, integers and rational numbers are all subsets of me, but I hold more numbers still.
I can be represented as points on a continuous number line, with no gaps between them.
I am the combination of the rational numbers and the irrational numbers.
I am denoted by the symbol R.
Who am I?
ANSWER: Real Numbers
SOLUTION
The real numbers combine the rational and irrational numbers into one continuous set, denoted R, and can be represented as every point on the number line with no gaps. Natural numbers, whole numbers, integers and rationals are therefore all subsets of the reals.
RIDDLE 7
I am a decimal number that ends after a finite number of digits.
The numbers 0.5 and 3.125 are examples of me.
I am always a rational number.
In its lowest terms, my equivalent fraction has a denominator whose only prime factors are 2 and 5.
I am the opposite of a decimal that continues without end.
Who am I?
ANSWER: Terminating Decimal
SOLUTION
A terminating decimal ends after finitely many digits, such as 0.5 or 3.125. It is always rational, and in its lowest-terms fraction form, the denominator's only prime factors are 2 and 5, since those are exactly the prime factors of 10.
RIDDLE 8
I am a decimal number that continues endlessly, with or without repeating digits.
I never come to an end.
If my digits repeat in a pattern, such as 0.666..., I represent a rational number.
If my digits never settle into a repeating pattern, such as in pi, I represent an irrational number.
I include both recurring and non-recurring decimals as special cases.
Who am I?
ANSWER: Non-Terminating Decimal
SOLUTION
A non-terminating decimal never ends. If its digits eventually repeat in a pattern it represents a rational number, called a recurring decimal; if its digits never settle into a repeating pattern, such as in pi, it represents an irrational number instead.
RIDDLE 9
I am a decimal representation of a rational number in which one or more digits repeat infinitely. 0.333... and 5.252525... are examples of me.
My repeating digits are often marked with a bar placed above them for short.
Every one of my values can be converted back into an exact fraction, since I always represent a rational number.
For example, 0.333... is equal to the fraction 1/3.
Who am I?
ANSWER: Recurring Decimal
SOLUTION
A recurring decimal is a non-terminating decimal whose digits repeat infinitely, such as 0.333... = 1/3. Because algebraic manipulation can always convert the repeating pattern back into an exact fraction, every recurring decimal represents a rational number.
RIDDLE 10
A set is said to possess me under an operation if performing that operation on any two elements of the set always produces a result that is also a member of the set.
Whole numbers possess me under addition and multiplication, but not under subtraction or division.
For example, 3 minus 5 gives -2, which is not a whole number, so whole numbers fail me under subtraction.
Integers, however, do possess me under subtraction, since subtracting one integer from another always gives an integer.
I am one of the key properties, alongside the commutative, associative and distributive properties, used to describe how a set behaves under an operation.
Who am I?
ANSWER: Closure Property
SOLUTION
A set is closed under an operation if applying that operation to any two of its members always produces another member of the same set. Whole numbers are closed under addition and multiplication but not subtraction, since 3 - 5 = -2 falls outside the whole numbers, whereas integers are closed under subtraction as well.
Properties of Real Numbers
RIDDLE 11
I state that when two numbers are combined under an operation, the order in which they are combined does not change the result.
I am written as a + b = b + a for addition and ab = ba for multiplication.
For example, 3 + 5 gives the same result as 5 + 3.
I do not hold for subtraction or division, since 5 - 3 does not equal 3 - 5.
Vector addition and matrix addition both satisfy me, while function composition in general does not.
Who am I?
ANSWER: Commutative Property
SOLUTION
The commutative property says that swapping the order of two elements under an operation does not change the result: a + b = b + a and ab = ba. It holds for addition and multiplication but fails for subtraction and division, since order matters there.
RIDDLE 12
I state that no matter how real numbers are grouped when adding or multiplying them, the sum or product remains the same.
I am written as (a + b) + c = a + (b + c) for addition, with a similar form for multiplication.
For example, (2 + 3) + 4 gives the same result as 2 + (3 + 4).
I do not hold for subtraction or division, since (8 - 4) - 2 does not equal 8 - (4 - 2).
I only involve how terms are grouped, not the order in which they appear.
Who am I?
ANSWER: Associative Property
SOLUTION
The associative property says that how elements are grouped does not affect the result of repeated addition or multiplication: (a + b) + c = a + (b + c). It concerns grouping alone, not order, which is what distinguishes it from the commutative property.
RIDDLE 13
I define how multiplication is distributed over addition or subtraction.
I am written as a(b + c) = ab + ac.
For example, 2(3 + 4) equals 2 × 3 plus 2 × 4, which is 14 either way.
I also apply to subtraction, so a(b - c) = ab - ac.
Unlike the commutative and associative properties, I connect two different operations rather than describing just one.
Who am I?
ANSWER: Distributive Property
SOLUTION
The distributive property links multiplication with addition and subtraction: a(b + c) = ab + ac. It is the only one of these properties that connects two different operations rather than describing how a single operation alone behaves.
RIDDLE 14
I state that when a specific operation is performed on an element together with a special value, the result is the original element itself, left completely unchanged.
For addition, that special preserving value is zero, since any number plus zero stays the same.
For multiplication, that special preserving value is one, since any number multiplied by one stays the same.
Every real number has exactly one such preserving value for each of these two operations.
This preserving value is different from a number's partner under the inverse property, which produces a fixed result rather than leaving the original number unchanged.
Who am I?
ANSWER: Identity Property
SOLUTION
The identity property says that combining an element with a special value under an operation leaves the element unchanged, with 0 filling that role for addition and 1 for multiplication. It differs from the inverse property, which instead pairs an element with a partner that produces the identity value as the result of the combination.
RIDDLE 15
I am the value that, when added to any number, leaves that number unchanged.
Zero is my value.
I am written in the form a + 0 = a for any real number a.
Adding me to a number is different from finding that number's additive inverse, which produces zero instead of leaving the number unchanged.
I am unique among real numbers, since only zero has this property under addition.
Who am I?
ANSWER: Additive Identity
SOLUTION
The additive identity is zero, since a + 0 = a for every real number a. It is the unique value that leaves any number unchanged when added, unlike the additive inverse, which instead produces zero when added to a number.
RIDDLE 16
I am the value that, when multiplied by any number, leaves that number unchanged.
One is my value.
I am written in the form a × 1 = a for any real number a.
Unlike the additive identity, I do not equal zero, since multiplying any number by zero always gives zero rather than leaving it unchanged.
I am unique among real numbers, since only one has this property under multiplication.
Who am I?
ANSWER: Multiplicative Identity
SOLUTION
The multiplicative identity is one, since a x 1 = a for every real number a. It cannot be zero, because multiplying by zero always gives zero rather than preserving the original number, which is what separates it from the additive identity.
RIDDLE 17
Under an operation of addition or multiplication, I connect an element with a special partner element, so that combining the two produces the identity value of that operation.
For addition, combining a number with its partner produces zero; for multiplication, combining a number with its partner produces one.
For example, 6 and -6 form one such pair under addition, while 6 and 1/6 form one such pair under multiplication.
Unlike the identity property, I describe a relationship between two related numbers rather than a single fixed value shared by everything.
Every real number has a partner of mine under addition, but zero alone has none under multiplication.
Who am I?
ANSWER: Inverse Property
SOLUTION
The inverse property pairs each element with a partner that combines with it to produce the identity value: additive inverses sum to zero, multiplicative inverses multiply to one. Every real number has an additive inverse, but zero has no multiplicative inverse, since division by zero is undefined.
RIDDLE 18
When I am added to a number, the sum of the two is zero.
The additive inverse of 5 is -5.
I am found by simply changing the sign of a number, from positive to negative or negative to positive.
Zero is the only number that is its own additive inverse.
Finding me is the basis for how subtraction is defined in terms of addition.
Who am I?
ANSWER: Additive Inverse
SOLUTION
The additive inverse of a number is found by reversing its sign so that the two sum to zero; the additive inverse of 5 is -5. Subtraction is defined in terms of this inverse, since a - b is the same as adding the additive inverse of b to a.
RIDDLE 19
Finding me is the basis for how division is defined in terms of multiplication.
Zero is the only real number that has no partner like me, since division by zero is undefined.
When I am combined with a number under multiplication, the result is always one.
For the number 4, my value is 1/4.
I am found by swapping the numerator and denominator of a fraction.
Who am I?
ANSWER: Multiplicative Inverse
SOLUTION
The multiplicative inverse (reciprocal) of a number is found by swapping the numerator and denominator so that the two multiply to give one; the multiplicative inverse of 4 is 1/4. Division is defined as multiplying by the multiplicative inverse, and zero alone has none, since division by zero is undefined.
Sets, Subsets & Venn Diagrams
RIDDLE 20
I describe a relationship between two sets where every element of the smaller set is guaranteed to also belong to the larger one.
Every set holds this relationship with itself, and the empty set holds this relationship with every set without exception.
For example, if A = {1, 2} and B = {1, 2, 3}, then A holds this relationship with B, since both 1 and 2 also belong to B.
I differ from a stricter version of this relationship, which additionally requires the two sets not to be equal to each other.
This relationship between A and B is written using the symbol ⊆.
Who am I?
ANSWER: Subset
SOLUTION
A set A is a subset of B, written A ⊆ B, if every element of A also belongs to B. Every set is a subset of itself, and the empty set is a subset of every set, since it has no elements that could ever fail to belong.
RIDDLE 21
I describe a relationship between two sets that is stricter than an ordinary containment relationship, since it additionally forbids the two sets from being equal to each other.
Unlike that ordinary relationship, a set can never hold this stricter one with itself.
For example, if A = {1, 2} and B = {1, 2, 3}, then A is related to B in this stricter way, since B contains an extra element A lacks.
A set with n elements has 2n subsets in total, but only 2n - 1 of these satisfy this stricter condition.
This relationship between A and B is written as A ⊂ B.
Who am I?
ANSWER: Proper Subset
SOLUTION
A is a proper subset of B, written A ⊂ B, if A ⊆ B but A does not equal B, meaning B contains at least one extra element that A lacks. A set can never be a proper subset of itself, which is exactly what makes this relationship stricter than an ordinary subset.
RIDDLE 22
I describe the reverse relationship of an ordinary containment relationship, applying to the larger of two sets rather than the smaller one.
For example, if A = {2, 4} and B = {1, 2, 3, 4}, then B holds this reverse relationship with A, since every element of A is contained within B.
The universal set always holds this reverse relationship with every other set under consideration in a problem, and every set holds it with itself as well.
If B holds this reverse relationship with A, then A is simultaneously related to B the ordinary way, since the two relationships describe the same pair of sets viewed from opposite directions.
This relationship between B and A is written as B ⊇ A.
Who am I?
ANSWER: Superset
SOLUTION
B is a superset of A, written B ⊇ A, exactly when A is a subset of B, since the two relationships describe the same pair of sets viewed from opposite directions. The universal set is therefore a superset of every set under consideration in a problem.
RIDDLE 23
I am a pictorial representation of the relationship between two or more sets.
I was suggested by the English mathematician and philosopher John Venn.
I typically use overlapping circles enclosed within a rectangle that represents the universal set.
The regions of overlap show the intersection of the sets involved, while the non-overlapping regions show elements unique to each set.
I was introduced in an 1880 paper, though similar diagrams had appeared earlier in the work of Leonhard Euler.
Who am I?
ANSWER: Venn Diagram
SOLUTION
A Venn diagram pictures the relationships between sets using overlapping circles inside a rectangle representing the universal set; overlapping regions show intersections and non-overlapping regions show elements unique to each set. It was popularised by John Venn in an 1880 paper, building on similar diagrams that had appeared earlier in the work of Leonhard Euler.
RIDDLE 24
In relation to a universal set U, I refer to all the elements in U that are not members of set A.
I am written as A′, and I am sometimes written as Ac instead.
For example, if U = {1, 2, 3, 4, 5} and A = {1, 2}, then I would be {3, 4, 5}.
Together, set A and I make up the entire universal set, with no elements shared between us.
I am used together with De Morgan's laws to simplify expressions involving unions and intersections.
Who am I?
ANSWER: Complement of a Set
SOLUTION
The complement of A, written A′ or Ac, is everything in the universal set U that is not in A. A and its complement together make up all of U with no overlap between them, and the idea is central to applying De Morgan's laws.
RIDDLE 25
I am the set containing all the elements under consideration in a particular problem, usually denoted by the symbol U.
Every other set discussed is a subset of me.
In a Venn diagram, I am represented by the rectangle that encloses all the circles.
My choice depends entirely on the context of the problem; for a problem about students, I might be the set of every student in a school.
Every element that is not inside a particular subset A is, by definition, inside the complement of A within me.
Who am I?
ANSWER: Universal Set
SOLUTION
The universal set U contains every element relevant to a given problem, and every other set discussed is treated as a subset of it. In a Venn diagram it is the enclosing rectangle, and its choice depends entirely on context, such as every student in a school for a problem about students.
Set Operations & Three-Set Problems
RIDDLE 26
I combine all the elements of two or more sets into a single set, keeping every element that appears in at least one of the original sets.
Shared elements are listed only once in my result, never duplicated.
For example, combining A = {1, 2, 3} and B = {3, 4, 5} this way gives {1, 2, 3, 4, 5}.
I am the opposite counting operation to the one that keeps only elements shared by every set involved.
I appear in the counting formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B), and I am written using the symbol ∪.
Who am I?
ANSWER: Union
SOLUTION
The union of two sets, A ∪ B, collects every element that appears in at least one of them, with shared elements listed only once. It is the counting operation opposite to intersection, and it appears in n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
RIDDLE 27
I find the elements that are common to two or more sets, keeping only what every set involved shares.
Only elements that appear in all the sets under consideration survive into my result.
For example, combining A = {1, 2, 3} and B = {3, 4, 5} this way gives {3}.
If two sets share no elements at all, my result is the empty set, and the sets involved are described as disjoint.
I am the opposite counting operation to the one that keeps every element from both sets, and I am written using the symbol ∩.
Who am I?
ANSWER: Intersection
SOLUTION
The intersection of two sets, A ∩ B, keeps only the elements common to both. If two sets share no elements, their intersection is the empty set and the sets are called disjoint.
RIDDLE 28
I find the elements that belong to one set but not to another, isolating what is unique to the first set alone.
My result contains every element of set A that is not also an element of set B.
For example, applying me to A = {1, 2, 3} and B = {2, 3, 4} in that order gives {1}.
Unlike the union and intersection, I am not commutative, since applying me to A and B does not generally give the same result as applying me to B and A.
I can be written either as A minus B, or as A ∩ B′, combining the intersection and complement operations.
Who am I?
ANSWER: Set Difference
SOLUTION
The set difference A - B, equivalently A ∩ B′, contains elements of A that are not in B. Unlike union and intersection, it is not commutative, since A - B generally differs from B - A.
RIDDLE 29
I state that the complement of the union of two sets equals the intersection of their complements, written as (A ∪ B)′ = A′ ∩ B′.
I am named after a British mathematician and logician.
For example, if U = {1, 2, 3, 4, 5}, A = {1, 2} and B = {2, 3}, then both sides of my equation give the set {4, 5}.
I have a partner law that instead deals with the complement of an intersection.
I am useful for simplifying set expressions and for rewriting logical statements that combine "and", "or" and "not".
Who am I?
ANSWER: De Morgan's First Law (De Morgan's Law for Union)
SOLUTION
De Morgan's first law states (A ∪ B)′ = A′ ∩ B′: the complement of a union equals the intersection of the complements. Named after Augustus De Morgan, it is useful for simplifying set expressions and rewriting logical statements built from "and", "or" and "not".
RIDDLE 30
I state that the complement of the intersection of two sets equals the union of their complements, written as (A ∩ B)′ = A′ ∪ B′.
I am named after a British mathematician and logician.
For example, if U = {1, 2, 3, 4, 5}, A = {1, 2} and B = {2, 3}, then both sides of my equation give the set {1, 3, 4, 5}.
I have a partner law that instead deals with the complement of a union.
Together, my partner and I show how the operations of union, intersection and complement relate to each other.
Who am I?
ANSWER: De Morgan's Second Law (De Morgan's Law for Intersection)
SOLUTION
De Morgan's second law states (A ∩ B)′ = A′ ∪ B′: the complement of an intersection equals the union of the complements. Together with the first law, it shows exactly how union, intersection and complement interact with one another.
RIDDLE 31
I relate the number of elements in the union of two sets to the number of elements in each set and their intersection.
I am written as n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
I subtract the intersection once because simply adding n(A) and n(B) would count shared elements twice.
For example, if n(A) = 10, n(B) = 7 and n(A ∩ B) = 3, then n(A ∪ B) = 14.
If sets A and B are disjoint, with no shared elements, I simplify to n(A ∪ B) = n(A) + n(B).
Who am I?
ANSWER: Two-Set Problem Equation
SOLUTION
The two-set counting formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B) corrects for double-counting: simply adding n(A) and n(B) counts every shared element twice, so the intersection is subtracted once to compensate. When A and B are disjoint, n(A ∩ B) = 0 and the formula reduces to plain addition.
RIDDLE 32
I extend the two-set counting formula to three sets, accounting for every possible overlap between A, B and C, including the region common to all three.
I am written as n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C).
I add back the triple intersection because it is subtracted three times when the pairwise intersections are removed.
This kind of counting technique is known more generally as the inclusion-exclusion principle.
I am commonly applied in Venn diagram problems involving three overlapping regions, such as surveys of students who study different subjects.
Who am I?
ANSWER: Three-Set Problem Equation
SOLUTION
The three-set counting formula extends inclusion-exclusion to three sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C). The triple intersection is added back because subtracting each of the three pairwise intersections removes it three times over, so it must be restored once to correct the count.