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Relations and Functions for CUET Maths

About 5 min to learn · 125 questions

Relations and functions open Section B1 of CUET Maths, the section you answer if you take Mathematics. CUET asks the same few things here every year: which properties a relation has, what an equivalence class is, and whether a function is one-one or onto. Learn each rule here, from reflexive, symmetric and transitive relations and equivalence classes to one-one, onto and bijective functions and counting them, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area
Section B1: Mathematics
On this page · 10 sections
  1. 01What a relation is
  2. 02Reflexive, symmetric and transitive relations
  3. 03Equivalence relations and classes
  4. 04One-one functions
  5. 05Onto and bijective functions
  6. 06How CUET asks this chapter
  7. 07Exam shortcuts
  8. 08Common mistakes
  9. 09Quick revision
  10. 10Check your understanding

What a relation is

Take the six students of a study group and pair up any two who were born in the same month. That pairing is a relation: a set of ordered pairs (a,b)(a, b) taken from A×AA \times A. It has three properties worth checking. Everyone is born in the same month as themselves. If Asha shares a month with Ravi, Ravi shares it with Asha. And if Asha shares with Ravi and Ravi with Meera, then Asha shares with Meera. A relation with all three is an equivalence relation, and it splits the group into classes, one per birth month.

What is a relation on a set?

A relation RR on a set AA is any subset of A×AA \times A. We write aRba \mathrel{R} b when (a,b)∈R(a, b) \in R. The empty relation has no pairs at all; the universal relation is the whole of A×AA \times A. A set with nn elements has n2n^{2} ordered pairs, so it carries 2n22^{n^{2}} relations.

Reflexive, symmetric and transitive relations

How do you check whether a relation is reflexive, symmetric or transitive?

  • Reflexive: (a,a)∈R(a, a) \in R for every aa in AA. One missing (a,a)(a, a) is enough to fail.
  • Symmetric: whenever (a,b)∈R(a, b) \in R, also (b,a)∈R(b, a) \in R.
  • Transitive: whenever (a,b)(a, b) and (b,c)(b, c) are in RR, so is (a,c)(a, c).

Symmetry and transitivity only make demands of pairs already in RR. That is why the empty relation is symmetric and transitive, though it is not reflexive.

Example 1 · A relation on a three-element set

On A={2,4,6}A = \{2, 4, 6\}, R={(2,2),(2,4),(4,2)}R = \{(2, 2), (2, 4), (4, 2)\}. Which properties does RR have?

  1. Reflexive? (4,4)(4, 4) and (6,6)(6, 6) are missing, so no.

  2. Symmetric? The only pair with a partner is (2,4)(2, 4), and (4,2)(4, 2) is there. Yes.

  3. Transitive? (4,2)(4, 2) and (2,4)(2, 4) are in RR, so (4,4)(4, 4) must be too. It is not, so no.

Answer

RR is symmetric only.

Watch out

Test transitivity on every chain, including the ones that turn back on themselves, such as (4,2)(4, 2) followed by (2,4)(2, 4). Those chains demand a pair (a,a)(a, a), and they are where most slips happen.

For a relation given by a rule on an infinite set, argue from the rule. "a≤ba \le b" on R\mathbb{R} is reflexive and transitive but not symmetric, since 1≤21 \le 2 while 2≤12 \le 1 is false. "∣a−b∣≤2\lvert a - b \rvert \le 2" is reflexive and symmetric but not transitive: 0 and 2 are related, 2 and 4 are related, but 0 and 4 are not.

Equivalence relations and classes

What is an equivalence class?

If RR is an equivalence relation on AA, the equivalence class of aa is the set of everything related to aa. Different classes never overlap, and together they cover AA: the classes split AA into separate pieces.

Example 2 · Classes of remainders

On Z\mathbb{Z}, aRba \mathrel{R} b when a−ba - b is divisible by 7. Find the class of 3, and the number of classes.

  1. RR is an equivalence relation: a−a=0a - a = 0 is divisible by 7; if 7 divides a−ba - b it divides b−ab - a; and if it divides a−ba - b and b−cb - c, it divides their sum a−ca - c.

  2. The class of 3 is every integer that leaves remainder 3 on division by 7: {…,−11,−4,3,10,17,…}\{\ldots, -11, -4, 3, 10, 17, \ldots\}.

Answer

The classes are the seven remainder classes [0],[1],…,[6][0], [1], \ldots, [6].

Key idea

An equivalence relation on a finite set is the same thing as a way of splitting the set into groups. That gives a quick count: {1,2,3}\{1, 2, 3\} can be split in 5 ways, so it has 5 equivalence relations.

One-one functions

When is a function one-one?

f:A→Bf : A \to B is one-one (injective) when different inputs always give different outputs: f(a1)=f(a2)f(a_1) = f(a_2) forces a1=a2a_1 = a_2. To show it, start from f(a1)=f(a2)f(a_1) = f(a_2) and solve; to disprove it, find two inputs with the same output.

Example 3 · One-one or not

Is f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=5−2xf(x) = 5 - 2x, one-one? Is g:R→Rg : \mathbb{R} \to \mathbb{R}, g(x)=x2−1g(x) = x^{2} - 1, one-one?

  1. 5−2a=5−2b5 - 2a = 5 - 2b gives a=ba = b, so ff is one-one.

  2. g(−3)=g(3)=8g(-3) = g(3) = 8, so gg is not one-one.

Answer

ff is one-one; gg is not.

Onto and bijective functions

When is a function onto, and what is a bijection?

f:A→Bf : A \to B is onto (surjective) when every element of BB is the image of something: the range is the whole codomain. A function that is both one-one and onto is a bijection. The codomain matters: x2x^{2} is onto [0,∞)[0, \infty) but not onto R\mathbb{R}.

Example 4 · The same rule on different sets

Classify f(n)=n+5f(n) = n + 5 as a map Z→Z\mathbb{Z} \to \mathbb{Z}, and as a map N→N\mathbb{N} \to \mathbb{N}.

  1. On Z\mathbb{Z}: one-one, and any integer mm is f(m−5)f(m - 5), so it is onto: a bijection.

  2. On N\mathbb{N}: still one-one, but 1, 2, 3, 4 and 5 have no natural number 5 below them, so it is not onto.

Answer

A bijection of Z\mathbb{Z}; one-one but not onto on N\mathbb{N}.

Example 5 · Counting functions

AA has 2 elements and BB has 3. How many functions from AA to BB are there, and how many are one-one?

  1. Each of the 2 elements has 3 choices: 32=93^{2} = 9 functions.

  2. For one-one, the second element must avoid the first one's image: 3×2=63 \times 2 = 6.

Answer

9 functions, 6 of them one-one, and none onto, since 2 images cannot cover 3 elements.

How CUET asks this chapter

  • Types of relations: a relation on a small set or by a rule, and which of the three properties it has; the empty and universal relations; how many relations of a kind a set carries.
  • Equivalence: whether a relation is one, its classes, how many classes or equivalence relations there are.
  • Functions: whether a given function is one-one, onto, both or neither, often the same rule on two different domains; how many functions, one-one functions or onto functions there are.
  • Formats: two statements, assertion and reason, match the list, and "which of these are true".

Exam shortcuts

  • Relations on an nn-element set: 2n22^{n^{2}}; reflexive ones: 2n2−n2^{n^{2} - n}; symmetric ones: 2n(n+1)22^{\frac{n(n + 1)}{2}}.
  • Functions from an mm-set to an nn-set: nmn^{m}; one-one: nPm{}^{n}P_{m} (none if m>nm > n); onto a 2-element set: 2m−22^{m} - 2.
  • A bijection between finite sets needs equal sizes; from a finite set to itself, one-one and onto come together.
  • "a−ba - b divisible by nn" on Z\mathbb{Z} is always an equivalence relation with nn classes.

Common mistakes

  • Calling the empty relation non-symmetric, or calling it reflexive.
  • Missing the chain that turns back on itself, such as (1,2),(2,1)(1, 2), (2, 1), which needs (1,1)(1, 1).
  • Judging one-one or onto from the rule alone, forgetting the domain and codomain.
  • Taking the range of exe^{x} or x2x^{2} to be all of R\mathbb{R}.

Quick revision

Relations

  • Reflexive: every (a,a)(a, a); symmetric: reverses; transitive: chains
  • Equivalence: all three; its classes split the set

Functions

  • One-one: equal outputs force equal inputs
  • Onto: range = codomain; bijection: both

Counting

  • nmn^{m} functions, nPm{}^{n}P_{m} one-one
  • 2n22^{n^{2}} relations on nn elements

Next in Section B1, functions that undo the trigonometric ones: inverse trigonometric functions.

Next · 09Quick revision

Check your understanding

125 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Types of relationsQ1 of 125Q1

Types of relations

45 questions

R={(1,1),(2,2),(3,3),(1,2),(2,1)}R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

Options for question 1

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AnswerB · Reflexive, symmetric and transitive

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is symmetric, since each pair's reverse is in RR, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check transitivity with care: it is transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if a−ba - b is divisible by 4. Which of these describes RR?

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AnswerD · Reflexive, symmetric and transitive

  1. a−a=0a - a = 0 is divisible by 4; if 4 divides a−ba - b it divides b−ab - a.

  2. If 4 divides a−ba - b and b−cb - c, it divides their sum a−ca - c.

Worth noting

Transitivity follows by adding the two differences.

Read the two statements and choose the correct option.

I. The empty relation on A={1,2}A = \{1, 2\} is symmetric and transitive.

II. The empty relation on A={1,2}A = \{1, 2\} is reflexive.

Options for question 3

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AnswerA · Only I

  1. Symmetry and transitivity only demand something of pairs already in RR; with none, nothing can fail. I is true.

  2. Reflexive needs (1,1)(1, 1) and (2,2)(2, 2), which are missing. II is false.

Worth noting

Reflexivity is the one property that asks for pairs to be present.

Which of these is the universal relation on A={1,2}A = \{1, 2\}?

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AnswerC · {(1,1),(1,2),(2,1),(2,2)}\{(1, 1), (1, 2), (2, 1), (2, 2)\}

  1. The universal relation relates every element to every element: it is all of A×AA \times A.

  2. A×AA \times A has 2×2=42 \times 2 = 4 pairs.

Worth noting

{(1,1),(2,2)}\{(1, 1), (2, 2)\} is the identity relation, which relates each element only to itself.

How many relations are there on a set with 3 elements?

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AnswerD · 512512

  1. A relation is any subset of A×AA \times A, which has 3×3=93 \times 3 = 9 pairs.

  2. 29=5122^{9} = 512 subsets.

Worth noting

23=82^{3} = 8 counts subsets of AA, not of A×AA \times A.

RR is the relation on R\mathbb{R} given by aRba \mathrel{R} b if a≤ba \le b. Which of these describes RR?

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AnswerA · Reflexive and transitive, but not symmetric

  1. a≤aa \le a always, and a≤ba \le b, b≤cb \le c give a≤ca \le c.

  2. But 1≤21 \le 2 while 2≤12 \le 1 is false, so it is not symmetric.

Worth noting

One counter-example, such as 1 and 2, is enough to rule out symmetry.

On A={1,2,3}A = \{1, 2, 3\}, the relation R={(1,1),(2,2),(3,3),(2,3)}R = \{(1, 1), (2, 2), (3, 3), (2, 3)\}.

Which of these describes this four-pair relation RR?

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AnswerD · Reflexive and transitive, but not symmetric

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is not symmetric, since (2,3)(2, 3) is in RR but (3,2)(3, 2) is not, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Symmetry fails on the one pair that is not of the form (a,a)(a, a).

How many relations can be defined on a set with 2 elements?

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AnswerC · 1616

  1. A×AA \times A has 2×2=42 \times 2 = 4 pairs, and a relation is any subset of them.

  2. 24=162^{4} = 16.

Worth noting

4 counts the pairs, not the sets of pairs.

On Z\mathbb{Z}, the relation RR is aRba \mathrel{R} b if ∣a∣=∣b∣\lvert a \rvert = \lvert b \rvert. Which of these describes RR?

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AnswerA · Reflexive, symmetric and transitive

  1. Equality of ∣a∣\lvert a \rvert and ∣b∣\lvert b \rvert is reflexive and symmetric.

  2. If ∣a∣=∣b∣\lvert a \rvert = \lvert b \rvert and ∣b∣=∣c∣\lvert b \rvert = \lvert c \rvert, then ∣a∣=∣c∣\lvert a \rvert = \lvert c \rvert: transitive.

Worth noting

A relation that says two things have equal values of something is always an equivalence.

R={(1,2),(2,3)}R = \{(1, 2), (2, 3)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. What is the least number of pairs that must be added to RR to make it reflexive and transitive?

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AnswerC · 44

  1. Add the 3 pairs (a,a)(a, a), then whatever the chains and reverses force.

  2. The pairs added: (1,1),(1,3),(2,2),(3,3)(1, 1), (1, 3), (2, 2), (3, 3), 4 in all.

Worth noting

3 adds only the pairs (a,a)(a, a); the chains still need their missing links.

Which of the following statements are true?

I. Every relation that is symmetric and transitive is reflexive.

II. The identity relation on a set is an equivalence relation.

III. The universal relation on a set is an equivalence relation.

IV. A relation can be both symmetric and not reflexive.

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AnswerA · II, III and IV only

  1. I is false: {(1,1)}\{(1, 1)\} on {1,2}\{1, 2\} is symmetric and transitive but lacks (2,2)(2, 2).

  2. II and III are true (both relate each element to itself, and are symmetric and transitive). IV is true: {(1,2),(2,1)}\{(1, 2), (2, 1)\}.

Worth noting

The 'proof' that symmetric and transitive give reflexive assumes every element is related to something.

R={(1,1),(2,2),(3,3),(1,3)}R = \{(1, 1), (2, 2), (3, 3), (1, 3)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

Options for question 12

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AnswerD · Reflexive and transitive, but not symmetric

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is not symmetric, since (1,3)(1, 3) is in RR but (3,1)(3, 1) is not, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check symmetry with care: it is not symmetric, since (1,3)(1, 3) is in RR but (3,1)(3, 1) is not.

RR is the relation on R\mathbb{R} given by aRba \mathrel{R} b if a<ba < b. Which of these describes RR?

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AnswerC · Transitive only

  1. a<aa < a is false, so it is not reflexive; 1<21 < 2 but not 2<12 < 1, so not symmetric.

  2. a<ba < b and b<cb < c give a<ca < c: it is transitive.

Worth noting

Unlike ≤\le, the strict << never relates a number to itself.

How many reflexive relations are there on a set with 3 elements?

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AnswerB · 6464

  1. A reflexive relation must contain the 3 pairs (a,a)(a, a); each of the other 9−3=69 - 3 = 6 pairs is in or out.

  2. 26=642^{6} = 64.

Worth noting

232^{3} lets only the diagonal pairs vary, but they are the ones that are fixed.

R={(1,2),(2,1)}R = \{(1, 2), (2, 1)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

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AnswerA · Symmetric only

  1. It is not reflexive, since (1,1)(1, 1) is missing.

  2. It is symmetric, since each pair's reverse is in RR, and not transitive, since (1,2)(1, 2) and (2,1)(2, 1) are in RR but (1,1)(1, 1) is not.

Worth noting

Check transitivity with care: it is not transitive, since (1,2)(1, 2) and (2,1)(2, 1) are in RR but (1,1)(1, 1) is not.

RR is the relation on N\mathbb{N} given by aRba \mathrel{R} b if aa divides bb. Which of these describes RR?

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AnswerB · Reflexive and transitive, but not symmetric

  1. Every aa divides itself, and if a∣ba \mid b and b∣cb \mid c then a∣ca \mid c.

  2. 2 divides 4 but 4 does not divide 2, so it is not symmetric.

Worth noting

Divisibility runs one way, from the smaller number to its multiple.

R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

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AnswerC · Reflexive and symmetric, but not transitive

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is symmetric, since each pair's reverse is in RR, and not transitive, since (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not.

Worth noting

Check transitivity with care: it is not transitive, since (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not.

RR is the relation on R\mathbb{R} given by aRba \mathrel{R} b if ∣a−b∣≤1\lvert a - b \rvert \le 1. Which of these describes RR?

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AnswerA · Reflexive and symmetric, but not transitive

  1. ∣a−a∣=0≤1\lvert a - a \rvert = 0 \le 1, and ∣a−b∣=∣b−a∣\lvert a - b \rvert = \lvert b - a \rvert: reflexive and symmetric.

  2. 0 and 1 are related, 1 and 2 are related, but ∣0−2∣=2>1\lvert 0 - 2 \rvert = 2 > 1: not transitive.

Worth noting

Small steps can add up to a big one, so 'close to' is not transitive.

R={(1,2),(2,3)}R = \{(1, 2), (2, 3)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

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AnswerD · Neither reflexive, symmetric nor transitive

  1. It is not reflexive, since (1,1)(1, 1) is missing.

  2. It is not symmetric, since (1,2)(1, 2) is in RR but (2,1)(2, 1) is not, and not transitive, since (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not.

Worth noting

Check transitivity with care: it is not transitive, since (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not.

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if a+ba + b is odd. Which of these describes RR?

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AnswerC · Symmetric only

  1. a+a=2aa + a = 2a is even, so no number is related to itself: not reflexive. a+b=b+aa + b = b + a, so it is symmetric.

  2. 1 and 2 are related, 2 and 3 are related, but 1+3=41 + 3 = 4 is even: not transitive.

Worth noting

An odd sum pairs an odd number with an even one, so a number is never related to itself.

R={(3,3)}R = \{(3, 3)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

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AnswerA · Symmetric and transitive, but not reflexive

  1. It is not reflexive, since (1,1)(1, 1) is missing.

  2. It is symmetric, since each pair's reverse is in RR, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check reflexivity with care: it is not reflexive, since (1,1)(1, 1) is missing.

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if a+ba + b is even. Which of these describes RR?

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AnswerB · Reflexive, symmetric and transitive

  1. a+a=2aa + a = 2a is even, and a+b=b+aa + b = b + a: reflexive and symmetric.

  2. An even sum means aa and bb have the same parity; same parity is passed on, so it is transitive.

Worth noting

Even sum means the same parity, and 'same parity' is an equivalence.

On A={1,2,3}A = \{1, 2, 3\}, the relation R={(1,1),(2,2),(3,3),(2,3)}R = \{(1, 1), (2, 2), (3, 3), (2, 3)\}.

What is the least number of pairs to add to this four-pair RR to make it an equivalence relation?

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AnswerB · 11

  1. It is already reflexive and transitive; it lacks only symmetry.

  2. Adding (3,2)(3, 2) is enough, and keeps it transitive: 1 pair.

Worth noting

2 also adds a pair linking 1, which nothing in RR requires.

On A={1,2,3}A = \{1, 2, 3\}, the relation R={(1,1),(2,2),(3,3),(2,3)}R = \{(1, 1), (2, 2), (3, 3), (2, 3)\}.

After that pair is added, into how many classes does the relation split AA?

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AnswerB · 22

  1. 2 and 3 are now related both ways; 1 is related only to itself.

  2. Classes {1}\{1\} and {2,3}\{2, 3\}: 2.

Worth noting

3 treats every element as its own class, forgetting that 2 and 3 now share one.

How many reflexive relations can be defined on a set with 4 elements?

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AnswerC · 4,0964{,}096

  1. The 4 pairs (a,a)(a, a) are fixed; each of the other 16−4=1216 - 4 = 12 pairs is in or out.

  2. 212=40962^{12} = 4096.

Worth noting

2162^{16} counts every relation, reflexive or not.

How many symmetric relations can be defined on a set with 2 elements?

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AnswerC · 88

  1. A symmetric relation on {a,b}\{a, b\} is fixed by (a,a)(a, a), (b,b)(b, b) and the block {(a,b),(b,a)}\{(a, b), (b, a)\}.

  2. 23=82^{3} = 8.

Worth noting

16 lets (a,b)(a, b) and (b,a)(b, a) be chosen separately.

How many relations on {1,2,3}\{1, 2, 3\} contain the pair (1,2)(1, 2)?

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AnswerB · 256256

  1. (1,2)(1, 2) is fixed in; each of the other 9−1=89 - 1 = 8 pairs is in or out.

  2. 28=2562^{8} = 256.

Worth noting

29−12^{9} - 1 removes only the empty relation; half of all relations lack (1,2)(1, 2).

On A={1,3,5}A = \{1, 3, 5\}, the relation R={(1,1),(1,3),(3,1),(3,3)}R = \{(1, 1), (1, 3), (3, 1), (3, 3)\}. Which of these describes RR?

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AnswerB · Symmetric and transitive, but not reflexive

  1. It is not reflexive, since (5,5)(5, 5) is missing.

  2. It is symmetric, since each pair's reverse is in RR, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check reflexivity with care: it is not reflexive, since (5,5)(5, 5) is missing.

On R−{0}\mathbb{R} - \{0\}, the relation RR is aRba \mathrel{R} b if ab>0ab > 0. Which of these describes RR?

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AnswerB · Reflexive, symmetric and transitive

  1. a⋅a=a2>0a \cdot a = a^{2} > 0 for a≠0a \ne 0, and ab=baab = ba: reflexive and symmetric.

  2. ab>0ab > 0 means aa and bb have the same sign, and 'same sign' passes along a chain: transitive.

Worth noting

With 0 removed, nothing can link a negative number to a positive one.

On A={1,3,5}A = \{1, 3, 5\}, the relation R={(1,1),(3,3),(5,5),(1,3),(3,5)}R = \{(1, 1), (3, 3), (5, 5), (1, 3), (3, 5)\}. Which of these describes RR?

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AnswerD · Reflexive only

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is not symmetric, since (1,3)(1, 3) is in RR but (3,1)(3, 1) is not, and not transitive, since (1,3)(1, 3) and (3,5)(3, 5) are in RR but (1,5)(1, 5) is not.

Worth noting

Check transitivity with care: it is not transitive, since (1,3)(1, 3) and (3,5)(3, 5) are in RR but (1,5)(1, 5) is not.

On N\mathbb{N}, the relation RR is aRba \mathrel{R} b if a+b=10a + b = 10. Which of these describes RR?

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AnswerD · Symmetric only

  1. 1+1≠101 + 1 \ne 10, so 1 is not related to itself: not reflexive. a+b=b+aa + b = b + a: symmetric.

  2. 1 and 9 are related, 9 and 1 are related, but 1+1≠101 + 1 \ne 10: not transitive.

Worth noting

Only 5 is related to itself, so the relation cannot be reflexive on all of N\mathbb{N}.

On the set of lines in a plane, L1RL2L_1 \mathrel{R} L_2 if L1L_1 is perpendicular to L2L_2. Which of these describes RR?

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AnswerB · Symmetric only

  1. No line is perpendicular to itself: not reflexive. If L1⊥L2L_1 \perp L_2, then L2⊥L1L_2 \perp L_1: symmetric.

  2. If L1⊥L2L_1 \perp L_2 and L2⊥L3L_2 \perp L_3, then L1L_1 is parallel to L3L_3, not perpendicular: not transitive.

Worth noting

Two right-angle turns bring a line back parallel to where it started.

R={(1,2),(2,3)}R = \{(1, 2), (2, 3)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. What is the least number of pairs that must be added to RR to make it an equivalence relation?

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AnswerD · 77

  1. Add the 3 pairs (a,a)(a, a), then whatever the chains and reverses force.

  2. The pairs added: (1,1),(1,3),(2,1),(2,2),(3,1),(3,2),(3,3)(1, 1), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), 7 in all.

Worth noting

3 adds only the pairs (a,a)(a, a); the chains still need their missing links.

Assertion (A): The relation R={(1,1),(2,2),(3,3)}R = \{(1, 1), (2, 2), (3, 3)\} on {1,2,3}\{1, 2, 3\} is transitive.

Reason (R): A relation in which every pair has the form (a,a)(a, a) can only chain a pair with itself.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. A is true: the only chains are (a,a),(a,a)(a, a), (a, a), which need (a,a)(a, a), already present.

  2. R is true and is exactly why A holds.

Match each relation on {1,2,3}\{1, 2, 3\} in List I with its description in List II.

List IList II
A. {(1,1),(2,2),(3,3)}\{(1, 1), (2, 2), (3, 3)\}I. Transitive only
B. {(1,2),(2,1)}\{(1, 2), (2, 1)\}II. An equivalence relation
C. {(1,2),(2,3),(1,3)}\{(1, 2), (2, 3), (1, 3)\}III. Reflexive and transitive, but not symmetric
D. {(1,1),(2,2),(3,3),(1,2)}\{(1, 1), (2, 2), (3, 3), (1, 2)\}IV. Symmetric only
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AnswerB · A-II, B-IV, C-I, D-III

  1. The identity relation is reflexive, symmetric and transitive; {(1,2),(2,1)}\{(1, 2), (2, 1)\} is symmetric but its chain needs (1,1)(1, 1).

  2. {(1,2),(2,3),(1,3)}\{(1, 2), (2, 3), (1, 3)\} is transitive only; adding (1,2)(1, 2) to the identity keeps it reflexive and transitive but not symmetric.

R={(2,3)}R = \{(2, 3)\} is a relation on A={1,2,3}A = \{1, 2, 3\}. Which of these describes RR?

Options for question 36

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AnswerB · Transitive only

  1. It is not reflexive, since (1,1)(1, 1) is missing.

  2. It is not symmetric, since (2,3)(2, 3) is in RR but (3,2)(3, 2) is not, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check transitivity with care: it is transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if ab≥0ab \ge 0. Which of these describes RR?

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AnswerD · Reflexive and symmetric, but not transitive

  1. a⋅a=a2≥0a \cdot a = a^{2} \ge 0, and ab=baab = ba, so it is reflexive and symmetric.

  2. (−1)(0)≥0(-1)(0) \ge 0 and (0)(1)≥0(0)(1) \ge 0, but (−1)(1)<0(-1)(1) < 0: not transitive.

Worth noting

0 links a negative number and a positive one that are not linked to each other.

How many symmetric relations are there on a set with 3 elements?

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AnswerC · 6464

  1. A symmetric relation is fixed by its 3 diagonal pairs and the 3 pairs {(a,b),(b,a)}\{(a, b), (b, a)\}, each taken or left as a block.

  2. 23+3=26=642^{3 + 3} = 2^{6} = 64.

Worth noting

512 counts all relations; in a symmetric one (a,b)(a, b) and (b,a)(b, a) come together.

R={(1,2),(3,4)}R = \{(1, 2), (3, 4)\} is a relation on A={1,2,3,4}A = \{1, 2, 3, 4\}. What is the least number of pairs that must be added to RR to make it an equivalence relation?

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AnswerC · 66

  1. Add the 4 pairs (a,a)(a, a), then whatever the chains and reverses force.

  2. The pairs added: (1,1),(2,1),(2,2),(3,3),(4,3),(4,4)(1, 1), (2, 1), (2, 2), (3, 3), (4, 3), (4, 4), 6 in all.

Worth noting

4 adds only the pairs (a,a)(a, a); the chains still need their missing links.

R={(1,1),(2,2),(3,3),(4,4),(1,2),(2,3)}R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3)\} is a relation on A={1,2,3,4}A = \{1, 2, 3, 4\}. Which of these describes RR?

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AnswerC · Reflexive only

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is not symmetric, since (1,2)(1, 2) is in RR but (2,1)(2, 1) is not, and not transitive, since (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not.

Worth noting

Check transitivity with care: it is not transitive, since (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not.

RR is the relation on N\mathbb{N} given by aRba \mathrel{R} b if b=a+1b = a + 1. Which of these describes RR?

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AnswerD · Neither reflexive, symmetric nor transitive

  1. a=a+1a = a + 1 never holds, and (1,2)(1, 2) is in RR without (2,1)(2, 1).

  2. (1,2)(1, 2) and (2,3)(2, 3) are in RR but (1,3)(1, 3) is not, since 3≠1+13 \ne 1 + 1.

Worth noting

A step of 1 followed by another step of 1 is a step of 2, which the rule does not allow.

How many relations that are both reflexive and symmetric are there on a set with 4 elements?

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AnswerB · 6464

  1. The 4 diagonal pairs are fixed; the off-diagonal pairs come in 16−42=6\frac{16 - 4}{2} = 6 blocks {(a,b),(b,a)}\{(a, b), (b, a)\}.

  2. 26=642^{6} = 64.

Worth noting

2122^{12} counts the reflexive relations; symmetry then ties the off-diagonal pairs into blocks.

R={(1,2),(2,4),(1,4)}R = \{(1, 2), (2, 4), (1, 4)\} is a relation on A={1,2,3,4}A = \{1, 2, 3, 4\}. Which of these describes RR?

Options for question 43

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AnswerB · Transitive only

  1. It is not reflexive, since (1,1)(1, 1) is missing.

  2. It is not symmetric, since (1,2)(1, 2) is in RR but (2,1)(2, 1) is not, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check transitivity with care: it is transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

On A={1,2,3}A = \{1, 2, 3\}, the relation R={(1,1),(2,2),(3,3),(2,3)}R = \{(1, 1), (2, 2), (3, 3), (2, 3)\}.

How many reflexive relations on AA contain this four-pair RR?

Options for question 44

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AnswerB · 3232

  1. All 4 pairs of RR are fixed in; the other 9−4=59 - 4 = 5 pairs of A×AA \times A are free.

  2. 25=322^{5} = 32.

Worth noting

262^{6} counts the reflexive relations with no pair required besides the diagonal.

On A={1,2,3,4}A = \{1, 2, 3, 4\}, the relation R={(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,4),(4,3)}R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 4), (4, 3)\}. Which of these describes RR?

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AnswerA · Reflexive, symmetric and transitive

  1. It is reflexive, since every (a,a)(a, a) is in RR.

  2. It is symmetric, since each pair's reverse is in RR, and transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Worth noting

Check transitivity with care: it is transitive, since no pairs (a,b),(b,c)(a, b), (b, c) in RR lack (a,c)(a, c).

Equivalence relations and classes

26 questions

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if a−ba - b is divisible by 3. What is the equivalence class of 2?

Options for question 46

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AnswerD · {…,−4,−1,2,5,8,…}\{\ldots, -4, -1, 2, 5, 8, \ldots\}

  1. The class of 2 is every integer bb with 2−b2 - b divisible by 3: 2 plus or minus a multiple of 3.

  2. So it runs −4,−1,2,5,8-4, -1, 2, 5, 8 and on in both directions.

Worth noting

The multiples of 3 form the class of 0, not of 2.

Into how many equivalence classes does the relation 'a−ba - b is divisible by 6' split Z\mathbb{Z}?

Options for question 47

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AnswerC · 66

  1. Two integers are related exactly when they leave the same remainder.

  2. There are 6 possible remainders, so 6 classes.

Worth noting

The remainders on division by 6 are 0 to 5: six of them.

On A={1,2,…,10}A = \{1, 2, \ldots, 10\}, RR is given by aRba \mathrel{R} b if ∣a−b∣\lvert a - b \rvert is even. What is the equivalence class of 4?

Options for question 48

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AnswerD · {2,4,6,8,10}\{2, 4, 6, 8, 10\}

  1. The class of 4 is every element of AA related to 4.

  2. Here that is {2,4,6,8,10}\{2, 4, 6, 8, 10\}.

Worth noting

An even difference means the same parity, so the class of 4 is all the even numbers, 2 included.

On A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}, the relation RR is aRba \mathrel{R} b if a−ba - b is divisible by 3.

Which of these describes this remainder relation RR on AA?

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AnswerA · Reflexive, symmetric and transitive

  1. a−a=0a - a = 0 is divisible by 3, and if 3 divides a−ba - b it divides b−ab - a.

  2. If 3 divides a−ba - b and b−cb - c, it divides a−ca - c: an equivalence relation.

Worth noting

Adding the two differences proves transitivity.

On A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}, the relation RR is aRba \mathrel{R} b if a−ba - b is divisible by 3.

Into how many equivalence classes does this remainder relation split AA?

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AnswerB · 33

  1. The classes are {1,4}\{1, 4\}, {2,5}\{2, 5\} and {3,6}\{3, 6\}.

  2. 3 classes, one per remainder.

Worth noting

6 counts the elements, not the classes.

A club's 16 members belong to four houses with 3, 5, 2 and 6 members. Two members are related if they belong to the same house.

Which of these describes the same-house relation?

Options for question 51

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AnswerD · Reflexive, symmetric and transitive

  1. Everyone is in their own house; sharing a house works both ways.

  2. If A shares with B and B with C, all three are in one house: an equivalence.

Worth noting

'Same house' is 'equal value of something', which is always an equivalence.

On Z\mathbb{Z}, aRba \mathrel{R} b if ∣a∣=∣b∣\lvert a \rvert = \lvert b \rvert. What is the equivalence class of 3?

Options for question 52

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AnswerD · {−3,3}\{-3, 3\}

  1. The class of 3 is every integer with ∣b∣=3\lvert b \rvert = 3.

  2. That is {−3,3}\{-3, 3\}.

Worth noting

{3}\{3\} forgets −3-3, which has the same modulus.

A club's 16 members belong to four houses with 3, 5, 2 and 6 members. Two members are related if they belong to the same house.

How many equivalence classes does the same-house relation have?

Options for question 53

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AnswerB · 44

  1. Each class is one house.

  2. 4 classes.

Worth noting

16 counts the members.

On A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}, the relation RR is aRba \mathrel{R} b if a−ba - b is divisible by 3.

What is the equivalence class of 2 under this remainder relation?

Options for question 54

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AnswerD · {2,5}\{2, 5\}

  1. The elements of AA that differ from 2 by a multiple of 3.

  2. 2 and 5.

Worth noting

{2,4,6}\{2, 4, 6\} is the even numbers, the class for 'divisible by 2'.

Read the two statements and choose the correct option.

I. The equivalence classes of an equivalence relation on a set never overlap.

II. Every element of the set lies in exactly one equivalence class.

Options for question 55

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AnswerC · Both I and II

  1. If two classes shared an element, transitivity and symmetry would make them equal; so distinct classes are disjoint.

  2. Each aa is in its own class (reflexive), and in no other. Both are true.

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if a−ba - b is divisible by 5. What is the equivalence class of 3?

Options for question 56

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AnswerA · {…,−7,−2,3,8,13,…}\{\ldots, -7, -2, 3, 8, 13, \ldots\}

  1. The class of 3 is every integer bb with 3−b3 - b divisible by 5: 3 plus or minus a multiple of 5.

  2. So it runs −7,−2,3,8,13-7, -2, 3, 8, 13 and on in both directions.

Worth noting

The class runs both ways: −2-2 and −7-7 are in it too, since 3−(−2)=53 - (-2) = 5.

Into how many equivalence classes does the relation 'a−ba - b is divisible by 9' split Z\mathbb{Z}?

Options for question 57

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AnswerB · 99

  1. Two integers are related exactly when they leave the same remainder.

  2. There are 9 possible remainders, so 9 classes.

Worth noting

Each remainder 0 to 8 is one class, so there are 9, not 8.

RR is the relation on Z\mathbb{Z} given by aRba \mathrel{R} b if a−ba - b is divisible by 4. What is the equivalence class of 0?

Options for question 58

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AnswerB · {…,−8,−4,0,4,8,…}\{\ldots, -8, -4, 0, 4, 8, \ldots\}

  1. The class of 0 is every integer bb with 0−b0 - b divisible by 4: 0 plus or minus a multiple of 4.

  2. So it runs −8,−4,0,4,8-8, -4, 0, 4, 8 and on in both directions.

Worth noting

The class of 0 is the multiples of 4; the multiples of 2 also take in 2, 6 and 10, which are 2 away.

On A={1,2,…,12}A = \{1, 2, \ldots, 12\}, RR is given by aRba \mathrel{R} b if a−ba - b is divisible by 4. What is the equivalence class of 7?

Options for question 59

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AnswerA · {3,7,11}\{3, 7, 11\}

  1. The class of 7 is every element of AA related to 7.

  2. Here that is {3,7,11}\{3, 7, 11\}.

Worth noting

Stepping by 4 from 7 within 1 to 12 gives 3 as well as 11.

On A={1,2,…,9}A = \{1, 2, \ldots, 9\}, RR is given by aRba \mathrel{R} b if aa and bb leave the same remainder on division by 3. What is the equivalence class of 5?

Options for question 60

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AnswerC · {2,5,8}\{2, 5, 8\}

  1. The class of 5 is every element of AA related to 5.

  2. Here that is {2,5,8}\{2, 5, 8\}.

Worth noting

The class of 5 holds every number with remainder 2: 2 as well as 8.

Read the two statements and choose the correct option.

I. On {1,2,3}\{1, 2, 3\}, the smallest equivalence relation has 3 pairs.

II. On {1,2,3}\{1, 2, 3\}, the largest equivalence relation has 6 pairs.

Options for question 61

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AnswerA · Only I

  1. I: the identity relation {(1,1),(2,2),(3,3)}\{(1, 1), (2, 2), (3, 3)\} is the smallest. True.

  2. II: the largest is the universal relation, with 3×3=93 \times 3 = 9 pairs. False.

Worth noting

6 counts only the pairs (a,b)(a, b) with a≠ba \ne b, leaving out the 3 pairs (a,a)(a, a).

On A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}, the relation RR is aRba \mathrel{R} b if a−ba - b is divisible by 3.

How many ordered pairs does this remainder relation RR contain?

Options for question 62

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AnswerC · 1212

  1. Each class has 2 elements and gives 2×2=42 \times 2 = 4 pairs.

  2. 3 classes: 3×4=123 \times 4 = 12 pairs.

Worth noting

6 counts only the pairs (a,a)(a, a).

A club's 16 members belong to four houses with 3, 5, 2 and 6 members. Two members are related if they belong to the same house.

How many ordered pairs of members are related by the same-house relation?

Options for question 63

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AnswerC · 7474

  1. A house of nn members gives n2n^{2} ordered pairs.

  2. 9+25+4+36=749 + 25 + 4 + 36 = 74.

Worth noting

58 leaves out the 16 pairs of a member with themselves.

On R−{0}\mathbb{R} - \{0\}, aRba \mathrel{R} b if ab>0ab > 0. What is the equivalence class of −2-2?

Options for question 64

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AnswerA · (−∞,0)(-\infty, 0)

  1. −2⋅b>0-2 \cdot b > 0 exactly when b<0b < 0.

  2. So the class is all the negative real numbers.

Worth noting

{−2,2}\{-2, 2\} mixes this relation up with 'same modulus'; here 2 and −2-2 have opposite signs.

A club's 16 members belong to four houses with 3, 5, 2 and 6 members. Two members are related if they belong to the same house.

How many members are in the equivalence class of a member of the 5-member house?

Options for question 65

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AnswerC · 55

  1. The class of a member is everyone related to them, themselves included.

  2. That is the whole house: 5.

Worth noting

4 leaves the member out of their own class; the relation is reflexive.

Assertion (A): 'Is congruent to' is an equivalence relation on the set of all triangles in a plane.

Reason (R): Congruent triangles have equal areas.

Options for question 66

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AnswerB · Both A and R are true, but R is not the correct explanation of A.

  1. A is true: congruence is reflexive, symmetric and transitive.

  2. R is true, but it is a fact about areas; it does not show that congruence is reflexive, symmetric or transitive.

Worth noting

A true fact about congruent triangles is not, by itself, a reason that congruence is an equivalence relation.

How many equivalence relations are there on {1,2,3}\{1, 2, 3\}?

Options for question 67

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AnswerB · 55

  1. An equivalence relation is the same as a split of the set into classes.

  2. {1,2,3}\{1, 2, 3\}; {1,2},{3}\{1, 2\}, \{3\}; {1,3},{2}\{1, 3\}, \{2\}; {2,3},{1}\{2, 3\}, \{1\}; {1},{2},{3}\{1\}, \{2\}, \{3\}: 5 in all.

Worth noting

3 counts only the splits into a pair and a single; the whole set and three singles are splits too.

Which of these relations are equivalence relations?

I. On the students of a school, 'is in the same class as'.

II. On R\mathbb{R}, 'a−ba - b is an integer'.

III. On Z\mathbb{Z}, 'a≤b+1a \le b + 1'.

IV. On the triangles in a plane, 'is similar to'.

Options for question 68

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AnswerA · I, II and IV only

  1. I, II and IV are reflexive, symmetric and transitive; for II, differences of integers are integers.

  2. III is not symmetric: 0≤5+10 \le 5 + 1 but 5≤0+15 \le 0 + 1 is false.

Worth noting

II looks unusual, but adding two integer differences gives an integer difference.

How many equivalence relations on {1,2,3}\{1, 2, 3\} contain the pair (1,2)(1, 2)?

Options for question 69

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AnswerC · 22

  1. 1 and 2 must share a class; 3 either joins them or stands alone.

  2. {1,2,3}\{1, 2, 3\} or {1,2},{3}\{1, 2\}, \{3\}: 2 relations.

Worth noting

1 forgets that 3 can join the class of 1 and 2, giving the universal relation.

How many equivalence relations are there on {1,2,3,4}\{1, 2, 3, 4\}?

Options for question 70

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AnswerD · 1515

  1. Count the splits of 4 elements: one class (1), 3 + 1 (4), 2 + 2 (3), 2 + 1 + 1 (6), four singles (1).

  2. 1+4+3+6+1=151 + 4 + 3 + 6 + 1 = 15.

Worth noting

16 is 242^{4}, the number of subsets, not of splits.

How many equivalence relations on {1,2,3}\{1, 2, 3\} contain both (1,2)(1, 2) and (2,3)(2, 3)?

Options for question 71

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AnswerB · 11

  1. 1 must share a class with 2, and 2 with 3, so all three are in one class.

  2. Only the universal relation does that: 1.

Worth noting

2 counts the relations with (1,2)(1, 2) alone; adding (2,3)(2, 3) forces the single class.

One-one functions

27 questions

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=3x−5f(x) = 3x - 5?

Options for question 72

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AnswerD · One-one and onto

  1. 3a−5=3b−53a - 5 = 3b - 5 gives a=ba = b: one-one.

  2. For any yy, x=y+53x = \frac{y + 5}{3} gives f(x)=yf(x) = y: onto.

Worth noting

A non-constant linear function on R\mathbb{R} is a bijection.

f:{1,2,3}→{a,b,c}f : \{1, 2, 3\} \to \{a, b, c\} is f={(1,b),(2,c),(3,a)}f = \{(1, b), (2, c), (3, a)\}. Which of these describes ff?

Options for question 73

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AnswerB · One-one and onto

  1. No two elements share an image: one-one.

  2. Every element of the codomain is an image: onto.

AA has 3 elements and BB has 4. How many functions are there from AA to BB?

Options for question 74

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AnswerC · 6464

  1. Each of the 3 elements of AA has 4 choices of image.

  2. 43=644^{3} = 64.

Worth noting

34=813^{4} = 81 counts maps from BB to AA, the other way round.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x2f(x) = x^{2}?

Options for question 75

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AnswerA · Neither one-one nor onto

  1. f(−2)=f(2)=4f(-2) = f(2) = 4: not one-one.

  2. No real xx has x2=−1x^{2} = -1: not onto.

Worth noting

Squaring loses the sign and never gives a negative number.

A={1,2,3,4}A = \{1, 2, 3, 4\}, B={a,b,c,d,e}B = \{a, b, c, d, e\}, and f:A→Bf : A \to B is f={(1,a),(2,c),(3,e),(4,b)}f = \{(1, a), (2, c), (3, e), (4, b)\}.

How many elements of BB are not in the range of this ff?

Options for question 76

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AnswerB · 11

  1. The range is {a,b,c,e}\{a, b, c, e\}.

  2. Only dd is left out: 1 element.

Worth noting

4 is the size of the range, not of what it misses.

A={1,2,3,4}A = \{1, 2, 3, 4\}, B={a,b,c,d,e}B = \{a, b, c, d, e\}, and f:A→Bf : A \to B is f={(1,a),(2,c),(3,e),(4,b)}f = \{(1, a), (2, c), (3, e), (4, b)\}.

Which of these describes this ff from AA to BB?

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AnswerA · One-one but not onto

  1. The four images a,c,e,ba, c, e, b are all different: one-one.

  2. dd is no element's image: not onto.

Worth noting

With 4 elements mapping into 5, at least one element of BB must be missed.

How many one-one functions are there from a set with 2 elements to a set with 4 elements?

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AnswerC · 1212

  1. 4 choices for the first element's image, then 3 left for the second.

  2. 4×3=124 \times 3 = 12.

Worth noting

16 allows both elements the same image.

Which of these functions from R\mathbb{R} to R\mathbb{R} is one-one?

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AnswerA · f(x)=2x3+5f(x) = 2x^{3} + 5

  1. 2a3+5=2b3+52a^{3} + 5 = 2b^{3} + 5 gives a3=b3a^{3} = b^{3}, so a=ba = b: one-one.

  2. The others repeat a value: f(−1)=f(1)=2f(-1) = f(1) = 2, f(−1)=f(0)=1f(-1) = f(0) = 1 and f(0)=f(6)=0f(0) = f(6) = 0.

Worth noting

x4+x2x^{4} + x^{2} has only even powers, so xx and −x-x give the same value.

f:{1,2,3}→{4,5,6,7}f : \{1, 2, 3\} \to \{4, 5, 6, 7\} has f(1)=4f(1) = 4 and f(2)=6f(2) = 6. How many choices of f(3)f(3) make ff one-one?

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AnswerB · 22

  1. For one-one, f(3)f(3) must differ from f(1)=4f(1) = 4 and f(2)=6f(2) = 6.

  2. That leaves 5 and 7: 2 choices.

Worth noting

4 lets f(3)f(3) take an image already used by 1 or 2.

AA has 5 elements and BB has 3. How many one-one functions are there from AA to BB?

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AnswerA · 00

  1. A one-one function needs 5 different images, but BB has only 3 elements.

  2. So there are none.

Worth noting

60 is 5P3{}^{5}P_{3}, which counts one-one maps from a 3-element set into a 5-element one, the other way round.

Read the two statements and choose the correct option.

I. If f:A→Bf : A \to B is one-one and AA, BB are finite, then AA has no more elements than BB.

II. A function f:R→Rf : \mathbb{R} \to \mathbb{R} that is strictly increasing is one-one.

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AnswerC · Both I and II

  1. I: distinct elements need distinct images, so ∣A∣≤∣B∣|A| \le |B|. True.

  2. II: if a<ba < b then f(a)<f(b)f(a) < f(b), so f(a)≠f(b)f(a) \ne f(b). True.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x3f(x) = x^{3}?

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AnswerD · One-one and onto

  1. a3=b3a^{3} = b^{3} gives a=ba = b: one-one.

  2. Every real yy has the real cube root y3\sqrt[3]{y}: onto.

Worth noting

Unlike x2x^{2}, the cube keeps the sign, so it is one-one.

f:{1,2,3}→{a,b,c,d}f : \{1, 2, 3\} \to \{a, b, c, d\} is f={(1,a),(2,c),(3,d)}f = \{(1, a), (2, c), (3, d)\}. Which of these describes ff?

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AnswerD · One-one but not onto

  1. No two elements share an image: one-one.

  2. b is no element's image: not onto.

AA has 3 elements and BB has 5. How many one-one functions are there from AA to BB?

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AnswerB · 6060

  1. The images must all differ: 5 choices for the first element, 4 for the second, 3 for the third.

  2. 5P3=5×4×3=60{}^{5}P_{3} = 5 \times 4 \times 3 = 60.

Worth noting

125 counts every function, repeats allowed.

Which of these describes f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=2nf(n) = 2n?

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AnswerA · One-one but not onto

  1. 2a=2b2a = 2b gives a=ba = b: one-one.

  2. No natural number maps to 3, an odd number: not onto.

Worth noting

On N\mathbb{N}, doubling reaches only the even numbers.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=∣x∣f(x) = \lvert x \rvert?

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AnswerB · Neither one-one nor onto

  1. f(−3)=f(3)=3f(-3) = f(3) = 3: not one-one.

  2. ∣x∣\lvert x \rvert is never negative, so −1-1 is not reached: not onto.

Worth noting

The modulus, like the square, loses the sign.

A={1,2,3,4}A = \{1, 2, 3, 4\}, B={a,b,c,d,e}B = \{a, b, c, d, e\}, and f:A→Bf : A \to B is f={(1,a),(2,c),(3,e),(4,b)}f = \{(1, a), (2, c), (3, e), (4, b)\}.

How many onto functions are there from this AA to this BB?

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AnswerA · 00

  1. An onto function needs all 5 elements of BB as images.

  2. 4 elements give at most 4 images, so there are none.

Worth noting

120 counts the one-one functions.

A={1,2,3,4}A = \{1, 2, 3, 4\}, B={a,b,c,d,e}B = \{a, b, c, d, e\}, and f:A→Bf : A \to B is f={(1,a),(2,c),(3,e),(4,b)}f = \{(1, a), (2, c), (3, e), (4, b)\}.

How many one-one functions are there in all from this AA to this BB?

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AnswerC · 120120

  1. 5, 4, 3 and 2 choices for the images of 1, 2, 3 and 4.

  2. 5×4×3×2=1205 \times 4 \times 3 \times 2 = 120.

Worth noting

625 is 545^{4}, every function, repeats allowed.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x3+xf(x) = x^{3} + x?

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AnswerD · One-one and onto

  1. f′(x)=3x2+1>0f'(x) = 3x^{2} + 1 > 0, so ff is strictly increasing: one-one.

  2. It runs from −∞-\infty to ∞\infty without a break: onto.

Worth noting

An odd-degree polynomial that only rises takes every real value exactly once.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x2+2xf(x) = x^{2} + 2x?

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AnswerD · Neither one-one nor onto

  1. f(−2)=f(0)=0f(-2) = f(0) = 0: not one-one.

  2. x2+2x=(x+1)2−1≥−1x^{2} + 2x = (x + 1)^{2} - 1 \ge -1, so −2-2 is never reached: not onto.

Worth noting

A quadratic on R\mathbb{R} is never one-one, and never onto R\mathbb{R}.

Which of these describes f:Z→Zf : \mathbb{Z} \to \mathbb{Z}, f(n)=2n+1f(n) = 2n + 1?

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AnswerA · One-one but not onto

  1. 2a+1=2b+12a + 1 = 2b + 1 gives a=ba = b: one-one.

  2. Every value is odd, so 4 is never reached: not onto.

Worth noting

On Z\mathbb{Z}, 2n+12n + 1 reaches only the odd integers.

On which of these domains is f(x)=x2−4xf(x) = x^{2} - 4x one-one?

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AnswerD · [2,∞)[2, \infty)

  1. f(a)=f(b)f(a) = f(b) gives (a−b)(a+b−4)=0(a - b)(a + b - 4) = 0, so a repeat needs a+b=4a + b = 4 with a≠ba \ne b: two points on either side of 2.

  2. On [2,∞)[2, \infty), a+b=4a + b = 4 forces a=b=2a = b = 2: one-one. Each other domain holds such a pair: f(1)=f(3)=−3f(1) = f(3) = -3, f(0)=f(4)=0f(0) = f(4) = 0.

Worth noting

The graph of x2−4xx^{2} - 4x turns at x=2x = 2, not at 0, so [0,∞)[0, \infty) still holds 1 and 3, which share the value −3-3.

Which of the following statements are true?

I. f:Z→Zf : \mathbb{Z} \to \mathbb{Z}, f(n)=n2f(n) = n^{2}, is one-one.

II. f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=n2+nf(n) = n^{2} + n, is one-one.

III. f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x3−xf(x) = x^{3} - x, is one-one.

IV. f:Z→Zf : \mathbb{Z} \to \mathbb{Z}, f(n)=3n−7f(n) = 3n - 7, is one-one.

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AnswerA · II and IV only

  1. I is false: f(−2)=f(2)=4f(-2) = f(2) = 4. III is false: f(−1)=f(0)=f(1)=0f(-1) = f(0) = f(1) = 0.

  2. II is true: n2+nn^{2} + n grows as nn grows on N\mathbb{N}. IV is true: 3a−7=3b−73a - 7 = 3b - 7 gives a=ba = b.

Worth noting

n2n^{2} is one-one on N\mathbb{N}, but Z\mathbb{Z} brings in negatives, and −2-2 and 2 have the same square.

Assertion (A): f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=sin⁡xf(x) = \sin x, is not one-one.

Reason (R): sin⁡0=sin⁡π\sin 0 = \sin \pi.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. A is true: sin⁡x\sin x repeats its values.

  2. R is true and is a pair of different inputs with the same output, which is exactly what makes A true.

Which of these describes f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=n2f(n) = n^{2}?

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AnswerD · One-one but not onto

  1. On N\mathbb{N} there are no negatives, so a2=b2a^{2} = b^{2} gives a=ba = b: one-one.

  2. 2 is not a square: not onto.

Worth noting

On N\mathbb{N}, squaring is one-one, though on R\mathbb{R} it is not.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=∣x∣+xf(x) = \lvert x \rvert + x?

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AnswerB · Neither one-one nor onto

  1. f(−1)=f(−2)=0f(-1) = f(-2) = 0: not one-one.

  2. f(x)=2xf(x) = 2x for x≥0x \ge 0 and 0 otherwise, so it is never negative: not onto.

Worth noting

Every negative xx is sent to 0.

How many one-one functions f:{1,2,3}→{1,2,3,4,5,6}f : \{1, 2, 3\} \to \{1, 2, 3, 4, 5, 6\} satisfy f(1)<f(2)<f(3)f(1) < f(2) < f(3)?

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AnswerA · 2020

  1. A one-one ff takes 3 different values, and any 3 different values can be put in increasing order in exactly one way.

  2. So each choice of 3 images out of 6 gives one such ff: 6C3=20{}^{6}C_{3} = 20.

Worth noting

120 is 6P3{}^{6}P_{3}, every one-one function; only 1 of the 3!=63! = 6 orders of each set of images is increasing.

Onto and bijective functions

27 questions

Which of these describes f:Z→Zf : \mathbb{Z} \to \mathbb{Z}, f(n)=n+3f(n) = n + 3?

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AnswerB · One-one and onto

  1. a+3=b+3a + 3 = b + 3 gives a=ba = b: one-one.

  2. Any integer mm is f(m−3)f(m - 3), and m−3m - 3 is an integer: onto.

Worth noting

On Z\mathbb{Z} there is always a number 3 below; on N\mathbb{N} there would not be.

f:{1,2,3}→{x,y,z}f : \{1, 2, 3\} \to \{x, y, z\} is f={(1,x),(2,x),(3,y)}f = \{(1, x), (2, x), (3, y)\}. Which of these describes ff?

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AnswerC · Neither one-one nor onto

  1. Two elements both go to x: not one-one.

  2. z is no element's image: not onto.

f:R→Rf : \mathbb{R} \to \mathbb{R} is f(x)=2x+35f(x) = \frac{2x + 3}{5}. To show that ff is onto, which xx gives f(x)=yf(x) = y for any real yy?

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AnswerD · x=5y−32x = \frac{5y - 3}{2}

  1. Solve 2x+35=y\frac{2x + 3}{5} = y: 2x+3=5y2x + 3 = 5y, so x=5y−32x = \frac{5y - 3}{2}.

  2. This xx is real for every real yy, and f(x)=yf(x) = y, so ff is onto.

Worth noting

Moving 3 across the equals sign changes its sign: 2x=5y−32x = 5y - 3, not 5y+35y + 3.

AA has 4 elements and BB has 4. How many bijections are there from AA to BB?

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AnswerC · 2424

  1. A bijection from a 4-element set to itself is an arrangement of the 4 images.

  2. 4!=244! = 24.

Worth noting

256 counts every function from the set to itself.

For which codomain is f(x)=sin⁡xf(x) = \sin x, defined on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], a bijection?

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AnswerD · [−1,1][-1, 1]

  1. On [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], sin⁡x\sin x is strictly increasing, so it is one-one.

  2. It runs from −1-1 to 1, so it is onto exactly when the codomain is [−1,1][-1, 1].

Worth noting

[−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] is the domain; the codomain has to be the set of values, [−1,1][-1, 1].

Which of these describes f:R→[0,∞)f : \mathbb{R} \to [0, \infty), f(x)=x2f(x) = x^{2}?

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AnswerA · Onto but not one-one

  1. Every y≥0y \ge 0 is f(y)f(\sqrt{y}): onto.

  2. But f(−1)=f(1)f(-1) = f(1): not one-one.

Worth noting

Shrinking the codomain to [0,∞)[0, \infty) makes x2x^{2} onto, but not one-one.

f:{1,2,3,4}→{p,q,r}f : \{1, 2, 3, 4\} \to \{p, q, r\} is f={(1,p),(2,q),(3,r),(4,q)}f = \{(1, p), (2, q), (3, r), (4, q)\}. Which of these describes ff?

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AnswerA · Onto but not one-one

  1. Two elements both go to q: not one-one.

  2. Every element of the codomain is an image: onto.

Which of these describes f:[0,∞)→Rf : [0, \infty) \to \mathbb{R}, f(x)=x2f(x) = x^{2}?

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AnswerC · One-one but not onto

  1. On [0,∞)[0, \infty), a2=b2a^{2} = b^{2} gives a=ba = b: one-one.

  2. Negative numbers are not reached: not onto.

Worth noting

Cutting down the domain removes the repeats; the codomain still has numbers never reached.

Which of these describes f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=n+3f(n) = n + 3?

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AnswerD · One-one but not onto

  1. a+3=b+3a + 3 = b + 3 gives a=ba = b: one-one.

  2. 1, 2 and 3 have no natural number 3 below them: not onto.

Worth noting

The same rule as on Z\mathbb{Z} loses the numbers 1, 2 and 3 on N\mathbb{N}.

f:A→Bf : A \to B is onto and AA has 5 elements. What is the largest number of elements BB can have?

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AnswerB · 55

  1. Every element of BB must be the image of at least one element of AA.

  2. 5 elements give at most 5 different images.

Worth noting

32 is 252^{5}, the number of subsets of AA; it has nothing to do with the images.

How many onto functions are there from a set with 3 elements to a set with 3 elements?

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AnswerB · 66

  1. From a 3-element set onto a 3-element set, onto means one-one too.

  2. So it is the number of arrangements: 3!=63! = 6.

Worth noting

27 counts every function, including those that miss an element.

Which of these describes f:R−{0}→R−{0}f : \mathbb{R} - \{0\} \to \mathbb{R} - \{0\}, f(x)=1xf(x) = \frac{1}{x}?

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AnswerD · One-one and onto

  1. 1a=1b\frac{1}{a} = \frac{1}{b} gives a=ba = b: one-one.

  2. Any non-zero yy is f(1y)f\left(\frac{1}{y}\right): onto.

Worth noting

With 0 removed from both sides, 1x\frac{1}{x} undoes itself, so it is a bijection.

Which of these describes f:R→(0,∞)f : \mathbb{R} \to (0, \infty), f(x)=exf(x) = e^{x}?

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AnswerA · One-one and onto

  1. exe^{x} is strictly increasing: one-one.

  2. Every y>0y > 0 is elog⁡ye^{\log y}: onto (0,∞)(0, \infty).

Worth noting

The same exe^{x} is not onto R\mathbb{R}; here the codomain is its exact range.

Which of these functions from R\mathbb{R} to R\mathbb{R} is onto?

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AnswerC · f(x)=x3−3xf(x) = x^{3} - 3x

  1. For any real yy, x3−3x−y=0x^{3} - 3x - y = 0 is a cubic equation, and a cubic always has a real root: onto. (It is not one-one, as f(0)=f(3)=0f(0) = f(\sqrt{3}) = 0, but onto does not need that.)

  2. The others miss values: x2−3x≥−94x^{2} - 3x \ge -\frac{9}{4}, 0<11+x2≤10 < \frac{1}{1 + x^{2}} \le 1 and 3x>03^{x} > 0.

Worth noting

A quadratic has a lowest value: x2−3x=(x−32)2−94x^{2} - 3x = \left(x - \frac{3}{2}\right)^{2} - \frac{9}{4} never goes below −94-\frac{9}{4}.

How many bijections f:{1,2,3,4,5}→{1,2,3,4,5}f : \{1, 2, 3, 4, 5\} \to \{1, 2, 3, 4, 5\} have f(1)≠1f(1) \ne 1?

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AnswerB · 9696

  1. There are 5!=1205! = 120 bijections; those with f(1)=1f(1) = 1 arrange the other 4 elements: 4!=244! = 24.

  2. 120−24=96120 - 24 = 96; or 4 choices for f(1)f(1), then 4!=244! = 24 for the rest: 4×24=964 \times 24 = 96.

Worth noting

119 removes only the identity map; every bijection with f(1)=1f(1) = 1, 24 of them, has to go.

How many onto functions f:{1,2,3,4}→{p,q}f : \{1, 2, 3, 4\} \to \{p, q\} have f(1)=pf(1) = p?

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AnswerB · 77

  1. With f(1)=pf(1) = p, each of 2, 3 and 4 goes to pp or qq: 23=82^{3} = 8 ways.

  2. Onto needs qq to be reached, so drop the one way that sends all three to pp: 8−1=78 - 1 = 7.

Worth noting

8 keeps the function that sends every element to pp, which misses qq.

Which of the following statements are true?

I. A function from a finite set to itself is one-one exactly when it is onto.

II. f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=n+1f(n) = n + 1, is one-one and onto.

III. A bijection from AA to BB exists only if AA and BB have the same number of elements, when both are finite.

IV. Every onto function from a 3-element set to a 2-element set is one-one.

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AnswerD · I and III only

  1. I and III are true. II is false: nothing maps to 1, so it is not onto.

  2. IV is false: 3 elements sharing 2 images must repeat one.

Worth noting

II fails at 1: n+1≥2n + 1 \ge 2 for every natural nn.

Match each function in List I with its description in List II.

List IList II
A. f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=2x+1f(x) = 2x + 1I. One-one but not onto
B. f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=3nf(n) = 3nII. Neither one-one nor onto
C. f:R→[0,∞)f : \mathbb{R} \to [0, \infty), f(x)=∣x∣f(x) = \lvert x \rvertIII. One-one and onto
D. f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x2+1f(x) = x^{2} + 1IV. Onto but not one-one
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AnswerB · A-III, B-I, C-IV, D-II

  1. 2x+12x + 1 is a bijection of R\mathbb{R}; 3n3n is one-one but misses 1 and 2.

  2. ∣x∣\lvert x \rvert onto [0,∞)[0, \infty) repeats ±x\pm x; x2+1x^{2} + 1 repeats and never reaches 0.

Which of these describes f:N→Nf : \mathbb{N} \to \mathbb{N}, f(n)=n+12f(n) = \frac{n + 1}{2} for odd nn and f(n)=n2f(n) = \frac{n}{2} for even nn?

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AnswerA · Onto but not one-one

  1. f(1)=1f(1) = 1 and f(2)=1f(2) = 1: not one-one.

  2. Any mm is f(2m)f(2m): onto.

Worth noting

Each mm is hit twice, by 2m−12m - 1 and 2m2m.

AA has 4 elements and BB has 2. How many onto functions are there from AA to BB?

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AnswerC · 1414

  1. Of the 24=162^{4} = 16 functions, only the 2 constant ones miss an element of BB.

  2. 16−2=1416 - 2 = 14.

Worth noting

16 includes the two functions that send everything to one element.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=exf(x) = e^{x}?

Options for question 119

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AnswerC · One-one but not onto

  1. exe^{x} is strictly increasing, so it is one-one.

  2. ex>0e^{x} > 0 always, so 0 and negatives are not reached: not onto.

Worth noting

The range of exe^{x} is (0,∞)(0, \infty), smaller than the codomain R\mathbb{R}.

AA has 4 elements and BB has 3. How many onto functions are there from AA to BB?

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AnswerC · 3636

  1. Two elements of AA share an image: choose them, 4C2=6{}^{4}C_{2} = 6 ways.

  2. Then match the 3 groups to the 3 images: 6×3!=366 \times 3! = 36.

Worth noting

81 counts all functions, including those that miss an image.

Which of these describes f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=[x]f(x) = [x], the greatest integer function?

Options for question 121

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AnswerB · Neither one-one nor onto

  1. f(1.2)=f(1.7)=1f(1.2) = f(1.7) = 1: not one-one.

  2. Only integers are reached, so 0.5 is not: not onto.

Worth noting

Its values are integers only, so it misses most of R\mathbb{R}.

How many onto functions are there from a set with 5 elements to a set with 2 elements?

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AnswerB · 3030

  1. Of the 25=322^{5} = 32 functions, 2 are constant and miss an element.

  2. 32−2=3032 - 2 = 30.

Worth noting

31 removes only one constant function; there are two.

Which of these describes f:[−1,1]→[−1,1]f : [-1, 1] \to [-1, 1], f(x)=x∣x∣f(x) = x\lvert x \rvert?

Options for question 123

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AnswerD · One-one and onto

  1. f(x)=−x2f(x) = -x^{2} for x<0x < 0 and x2x^{2} for x≥0x \ge 0: strictly increasing, so one-one.

  2. It runs from f(−1)=−1f(-1) = -1 to f(1)=1f(1) = 1: onto [−1,1][-1, 1].

Worth noting

Unlike x2x^{2}, x∣x∣x\lvert x \rvert keeps the sign of xx, so it never repeats.

Assertion (A): f:R→Rf : \mathbb{R} \to \mathbb{R}, f(x)=x1+∣x∣f(x) = \frac{x}{1 + \lvert x \rvert}, is not onto.

Reason (R): ∣f(x)∣<1\lvert f(x) \rvert < 1 for every real xx.

Options for question 124

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. ∣f(x)∣=∣x∣1+∣x∣<1\lvert f(x) \rvert = \frac{\lvert x \rvert}{1 + \lvert x \rvert} < 1, since the denominator is larger than the numerator. R is true.

  2. So 1, for one, is never reached, and ff is not onto R\mathbb{R}: A is true, and R is exactly why.

Worth noting

ff is one-one, but that does not make it onto: its values stay inside (−1,1)(-1, 1).

f:N→Zf : \mathbb{N} \to \mathbb{Z} is given by f(n)=n2f(n) = \frac{n}{2} when nn is even and f(n)=−n−12f(n) = -\frac{n - 1}{2} when nn is odd. Which of these describes ff?

Options for question 125

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AnswerD · One-one and onto

  1. Even nn give the positive integers 1,2,3,…1, 2, 3, \ldots; odd nn give 0,−1,−2,…0, -1, -2, \ldots. The two lists never meet, and within each, different nn give different values: one-one.

  2. Any m>0m > 0 is f(2m)f(2m), and any m≤0m \le 0 is f(1−2m)f(1 - 2m): onto. So ff is a bijection.

Worth noting

Z\mathbb{Z} seems to hold twice as many numbers as N\mathbb{N}, but ff lists them as 0,1,−1,2,−2,…0, 1, -1, 2, -2, \ldots and reaches every integer.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.