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Which pairs are related
Think of your class. Thirty students, one register. Now think of the ways two students can be connected. Same house. Same birthday month. Taller than. Sits next to. Each of these is a RELATION. It is a rule that says, for any two students, yes they are connected, or no they are not.
Pairs, written as symbols
Mathematics writes the same idea with pairs. Take the set $A = \{1, 2, 3\}$. Pick some ordered pairs from it: $(1, 2)$, $(2, 2)$, $(3, 1)$. The pair $(1, 2)$ means “1 is connected to 2”. Put a bunch of such pairs together, and you have a relation.
Say the symbol aloud once. $A \times A$ reads “A cross A”. It is the list of every ordered pair you could build from $A$. A relation $R$ on $A$ is any subset of $A \times A$: you choose some of the pairs and leave the rest.
Two kinds of good behaviour
Nothing forces your choice to follow a pattern. Pick pairs at random and $R$ is still a relation. But a relation chosen at random is not good for much. The useful ones follow patterns, and this chapter is about the patterns that make a relation worth trusting. Two kinds of good behaviour matter.
The first is behaving like “equals”. “Same house as” sorts your class into four groups, with nobody in two groups and nobody left out. “Taller than” cannot sort anything. You will learn the three checks, REFLEXIVE, SYMMETRIC and TRANSITIVE, that tell these two apart. A relation that passes all three is an EQUIVALENCE RELATION, and it always sorts.
The second is behaving like “exactly one”. Every student has exactly one roll number. So “has roll number” is a FUNCTION: feed in a student, get out one number, never two. You met functions in Class 11. The new question is whether the function can be run backwards. Give me a roll number: can I get back the one student? Sometimes yes, sometimes no, and the two checks that decide, ONE-ONE and ONTO, are the second half of the chapter.
Try it yourself
Your turn: write any three ordered pairs on $A = \{1, 2, 3\}$. Whatever you wrote is already a relation. By the end of the chapter you will be able to say whether it is a useful one. It will take you under a minute.
Two everyday relations on four students, drawn as arrows: sharing a house runs both ways in pairs, taller than runs one way only.
A SCHOLAR INDIA REMEMBERS
Dignaga was born in south India, around the 5th or 6th century CE. Many thinkers of his time trusted several ways of knowing, including being told. He kept only two: what you see, and what you work out from what you see. Judge a relation his way. It has a pattern only when you can see it in the pairs, or work it out from its rule.
Dignaga · the kingfisher
Check yourself
A relation $R$ on a set $A$ is defined as \_\_\_\_\_\_.
any subset of $A \times A$
a subset of $A$ listing only the elements that take part in some pair
any list of ordered pairs, whether or not both coordinates come from $A$
a function that assigns each element of $A$ to exactly one element of $A$
Check your answer
✓ any subset of $A \times A$ — (A) A relation on $A$ is exactly any subset of $A \times A$, no more, no less is required of it.
a subset of $A$ listing only the elements that take part in some pair — The form used to write a relation down never decides which relation it is, listing “related elements only” throws away exactly the information (which element pairs with which) a relation exists to record.
any list of ordered pairs, whether or not both coordinates come from $A$ — A relation ON A specifically needs both coordinates of every pair to come from A itself, a pair with even one coordinate outside A is not a member of A times A.
a function that assigns each element of $A$ to exactly one element of $A$ — A function is a special relation where every element of A has exactly one image in A. A general relation on A carries no such requirement, an element can appear in many pairs, one pair, or none at all.
Why are reflexive, symmetric and transitive the three properties worth naming?
because reflexive, symmetric, and transitive are the same property viewed three ways
it earns the relation the name “equivalence relation”
because every relation on a set must have all three properties to be valid
because these are the only three properties a relation on a set can ever have
Check your answer
because reflexive, symmetric, and transitive are the same property viewed three ways — The three properties test three separate things, self-pairing, reversal, and chaining, and a relation can have any combination of them independently.
✓ it earns the relation the name “equivalence relation” — (B) All three together earn a relation the name equivalence relation. An equivalence relation sorts a set into classes, and that sorting is the payoff.
because every relation on a set must have all three properties to be valid — A relation needs none of the three properties to exist, it is simply any subset of $A \times A$. The properties are things a relation MIGHT have, never things it must have.
because these are the only three properties a relation on a set can ever have — These three are singled out because together they give something useful, not because no other property could ever be defined on a relation.
A function $f:A \to B$ is a special kind of relation. What makes it special?
every element of $B$ is related to exactly one element of $A$
every element of $A$ is related only to itself
every element of $A$ is related to exactly one element of $B$
every element of $A$ is related to at least one element of $B$, possibly more than one
Check your answer
every element of $B$ is related to exactly one element of $A$ — This describes elements of $B$ mapping back to $A$, the reverse direction, a function $f:A \to B$ fixes an image for each element of $A$, not the other way round.
every element of $A$ is related only to itself — Nothing about being a function requires an element to map to itself, that would describe only the identity function, a single special case.
✓ every element of $A$ is related to exactly one element of $B$ — (C) A function fixes exactly one image in $B$ for every element of $A$, the “exactly one” is what a general relation does not guarantee.
every element of $A$ is related to at least one element of $B$, possibly more than one — Allowing an element of $A$ to have more than one image in $B$ describes a general relation, not a function, “exactly one” is the requirement a function adds.
Most of this chapter builds on things you already met in Class 11. Try each check below. It takes a minute.
An ordered pair fixes a point (Class 9, Mathematics, Coordinate Geometry). Here: a relation’s pairs sit on the same axes you already plot points on. Check: are $(3, 0)$ and $(0, 3)$ the same point? No: order matters in a pair.
Set-builder form (Class 11, Mathematics (NCERT Class 11), page 3). Here: many relations in this chapter are written in this form. Check: how many elements does $V = \{a, e, i, o, u\}$ have? Five.
The Cartesian product of two sets (Class 11, Mathematics (NCERT Class 11), page 25). Here: a relation is simply a chosen subset of this product. Check: how many pairs are in $A \times B$, for $A = \{1, 2\}$ and $B = \{3, 4\}$? Four.
A relation as a subset (Class 11, Mathematics (NCERT Class 11), page 28). Here: this chapter decides which subsets earn the names reflexive, symmetric or transitive. Check: how many subsets does $A \times A$ have, for $A = \{1, 2\}$? $A \times A$ has $4$ pairs, so $16$ possible subsets.
Domain, range and codomain (Class 11, Mathematics (NCERT Class 11), page 28). Here: one-one and onto, coming soon, are both read off these three sets. Check: what is the domain of $R = \{(4, 2), (9, 3), (25, 5)\}$? $\{4, 9, 25\}$, of size $3$.
A function as a special relation (Class 11, Mathematics (NCERT Class 11), page 30). Here: one-one and onto both sharpen this same picture. Check: how many distinct images does $f(x) = x + 1$ give on $\{1, 2, 3\}$? Three, one each.
If any check felt new, go back and read that page first.
Three everyday relations, three arrow pictures. Run the same three questions on each and the verdicts come out different.
Start with the class again, and three everyday relations on it. For each one, let us ask three questions.
Three relations on your class
“Is the same age as.” Are you the same age as yourself? Yes. If you are the same age as a friend, is the friend the same age as you? Yes. If you are the same age as a friend, and the friend is the same age as a cousin, are you the same age as the cousin? Yes. Three questions, three yeses.
“Is taller than.” Are you taller than yourself? No. If you are taller than a friend, is the friend taller than you? No. If you are taller than a friend and the friend is taller than a cousin, are you taller than the cousin? Yes. Only the third survives.
“Lives within a kilometre of.” Do you live within a kilometre of yourself? Yes. If you live within a kilometre of a friend, does the friend live within a kilometre of you? Yes. If you live within a kilometre of a friend, and the friend within a kilometre of a cousin, do you live within a kilometre of the cousin? Not always. The friend could be 900 metres east of you, and the cousin 900 metres further east. That puts you 1800 metres from the cousin. The third fails.
Naming the three checks
Those three questions are the three checks this section is about. Each has a name.
Reflexive asks one thing of the picture: does every point have its own loop?
$R$ is REFLEXIVE when every element is related to itself: $(a, a) \in R$ for every $a$ in $A$. In the arrow picture, every point has its own loop. “Same age as” passes. “Taller than” fails.
Symmetric asks whether every arrow runs both ways. One arrow with a single head fails it.
$R$ is SYMMETRIC when every pair comes with its reverse: $(a, b) \in R$ forces $(b, a) \in R$. In the arrow picture, every arrow runs both ways. “Same age as” passes. “Taller than” fails. “Within a kilometre” passes.
Transitive asks whether every two-step chain has its one-step shortcut arrow.
You, the friend and the cousin: two gaps of 900 metres each pass the check, the whole 1,800 metres does not.
$R$ is TRANSITIVE when a chain of two pairs always closes into a third: $(a, b) \in R$ and $(b, c) \in R$ together force $(a, c) \in R$. Any two-step path has a one-step shortcut. “Same age as” passes. “Taller than” passes. “Within a kilometre” fails.
Relation
Reflexive
Symmetric
Transitive
Same age as
Yes
Yes
Yes
Taller than
No
No
Yes
Lives within a kilometre of
Yes
Yes
No
Look at how the three examples pass and fail in different combinations. None of the three checks implies another, so you run each one on its own, every time.
Three relations the board paper asks about, each drawn as arrows with its three verdicts. Check them before reading the list.
Five more relations to check
Here are five more, including the kind the board paper asks about. Check each yourself before reading the verdict.
“Follows on Instagram.” Reflexive? No, you cannot follow yourself. Symmetric? No, a follow goes one way. Transitive? No, you do not follow everyone your friends follow.
“Divides”, on the positive whole numbers: $a$ is related to $b$ when $a$ divides $b$ exactly. Reflexive? Yes: every $a$ divides itself. Symmetric? No, $2$ divides $6$ but $6$ does not divide $2$. Transitive? Yes: if $b = k a$ and $c = m b$, then $c = (m k) a$; $2$, $6$, $18$ is one instance.
*“$a \leq b$”,* on the real numbers. Reflexive? Yes. Symmetric? No. Transitive? Yes.
Two more from the board paper: within one fails the chain check, and a difference divisible by three passes all three.
The last two are the ones the board paper asks about most. One fails a single check; the other passes all three.
*“$|a - b| \leq 1$”,* on the real numbers: $a$ and $b$ are within one of each other. Reflexive? Yes, $|a - a| = 0$. Symmetric? Yes, $|a - b| = |b - a|$. Transitive? No. $1$ and $2$ are within one, $2$ and $3$ are within one, but $1$ and $3$ are not. This is the one-kilometre relation written in numbers.
*“$a - b$ is divisible by $3$”,* on the whole numbers. Reflexive? Yes, $a - a = 0$, and $3$ divides $0$. Symmetric? Yes, if $3$ divides $a - b$ it divides $b - a$. Transitive? Yes, if $3$ divides $a - b$ and $b - c$, it divides their sum, $a - c$. All three pass. This is “same age as” written in numbers.
The two extremes: no arrows at all passes symmetric and transitive with nothing to break them; every arrow drawn passes all three.
Two extreme cases
Two extreme relations are worth one look each, because the board paper asks about them often. The EMPTY relation, $R = \emptyset$, has no pairs at all. There is no pair to break the symmetric rule and no chain to break the transitive rule, so both hold. Reflexive fails the moment $A$ has even one element, because $(a, a)$ is missing. The UNIVERSAL relation, $R = A \times A$, has every pair, so no check can find anything missing, and all three pass.
Three checks, run separately. Does every element relate to itself? Does every pair have its reverse? Does every two-step chain have its shortcut?
row 1
AmanBinaChitraDev
row 2
EshaFaridGauri
row 3
HariIshaJaiKavya
Sits in the same row as: three checks pass, and the class is already sorted into rows.
Your turn: “sits in the same row as”, on your class. Run the three checks yourself before you read on. Check: all three pass, so the relation sorts the class into rows. That sorting is the next section.
A relation is just a choice of cells from the grid of all pairs. Move one cell, and all three patterns can be lost.
Check yourself
A relation $R$ on set $A$ is reflexive exactly when \_\_\_\_\_\_.
$(a, a) \in R$ for at least one $a \in A$
$(a, b) \in R$ always forces $(b, a) \in R$
$(a, a) \in R$ for every $a \in A$
$R$ contains no repeated pairs
Check your answer
$(a, a) \in R$ for at least one $a \in A$ — Reflexive means EVERY element of $A$ is paired with itself, not merely some, missing even one element breaks reflexivity.
$(a, b) \in R$ always forces $(b, a) \in R$ — “$(a,b) \in R$ forces $(b,a) \in R$” is the definition of symmetric. Reflexive is about an element paired with itself, not about reversing pairs.
✓ $(a, a) \in R$ for every $a \in A$ — (C) Reflexive means every single element of $A$ pairs with itself, no exceptions allowed.
$R$ contains no repeated pairs — Repeated pairs are not even meaningful in a set of pairs, and this has no connection to any of reflexive, symmetric, or transitive.
A relation $R$ on set $A$ is symmetric exactly when \_\_\_\_\_\_.
$(a, b) \in R$ always forces $(b, a) \in R$ too
$R$ contains exactly half of $A \times A$
$(a, a) \in R$ for every $a \in A$
$(a, b) \in R$ and $(b, a) \in R$ together force $a = b$
Check your answer
✓ $(a, b) \in R$ always forces $(b, a) \in R$ too — (A) Symmetric means the reverse of every pair in $R$ is also in $R$, swapping the two coordinates never leaves the set.
$R$ contains exactly half of $A \times A$ — Symmetry says nothing about how many pairs $R$ contains, a relation with 2 pairs or 200 can be symmetric or not, independent of size.
$(a, a) \in R$ for every $a \in A$ — “$(a,a) \in R$ for every $a$” is reflexive. Symmetric is about a pair’s reverse also being present, not about self-pairing.
$(a, b) \in R$ and $(b, a) \in R$ together force $a = b$ — Forcing $a = b$ whenever both $(a,b)$ and $(b,a)$ are present describes antisymmetric, a different property, symmetric requires the reverse pair to exist, not that it collapses the two elements together.
The empty relation $R = \emptyset$ on a non-empty set $A$ is \_\_\_\_\_\_.
symmetric only, since transitivity needs at least two pairs to chain
none of the three, because it contains no pairs to test
reflexive, symmetric, and transitive, all at once
symmetric and transitive, but not reflexive
Check your answer
symmetric only, since transitivity needs at least two pairs to chain — Transitive and symmetric are both “whenever a pair is present” statements, and both hold vacuously with zero pairs, there is no reason one holds and the other doesn’t.
none of the three, because it contains no pairs to test — A condition that starts “whenever a pair is present, ...” is true automatically when no pair is ever present, this is vacuous truth, not “the test doesn’t apply.”
reflexive, symmetric, and transitive, all at once — Reflexivity on a non-empty $A$ needs at least one pair, $(a,a)$ for every $a$. The empty relation has zero pairs, so it fails reflexivity on any set with an element in it.
✓ symmetric and transitive, but not reflexive — (D) With zero pairs, both “if a pair is present, its reverse is too” and “if two pairs chain, the closing pair is too” hold with nothing to check, vacuously true. Reflexivity is different , it demands a pair for every element, which zero pairs cannot supply on a non-empty set.
The universal relation $R = A \times A$ (every pair related) is \_\_\_\_\_\_.
not technically a relation, since it contains every possible pair
symmetric and transitive, but never reflexive
reflexive only
reflexive, symmetric, and transitive, all at once
Check your answer
not technically a relation, since it contains every possible pair — A relation on $A$ is any subset of $A \times A$, and $A \times A$ itself is a subset of itself, the universal relation is a relation like any other, just the largest possible one.
symmetric and transitive, but never reflexive — $(a, a)$ is one of the pairs in $A \times A$ for every $a \in A$, so the universal relation contains it, reflexive holds, it does not fail.
reflexive only — Every pair is in the universal relation, so every reversed pair and every chained pair is too, it cannot fail symmetric or transitive when nothing is ever missing.
✓ reflexive, symmetric, and transitive, all at once — (D) When every pair is present, no pair, no reversed pair, and no chained pair can ever be missing, all three properties hold at once, the extreme opposite of the empty relation.
The same relation is drawn three times, with only the question changing, which is why reflexive, symmetric and transitive each need a separate check.
Worked example
Check reflexive, symmetric, transitive for a relation on $\{1, 2, 3\}$
$R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ on $A = \{1, 2, 3\}$ This is the relation to test. Go through reflexive, symmetric and transitive, one at a time.
Reflexive: check $(1,1)$, $(2,2)$, $(3,3)$ Look for every element paired with itself. All three pairs are in $R$. Reflexive holds.
Symmetric: check $(1,2)$ and its reverse $(2,1)$ $(1,2)$ and $(2,1)$ are the only pairs not their own reverse, and they are each other’s reverse. Symmetric holds.
Transitive: chain $(1,2)$ and $(2,1)$ together A chain through a diagonal pair $(a,a)$ only forces a pair already in $R$: $(1,1)$ then $(1,2)$ forces $(1,2)$. The two pairs that can force something new are $(1,2)$ and $(2,1)$.
$(1,2) \in R$ and $(2,1) \in R$ force $(1,1) \in R$ Follow the transitive rule: if $(a,b)$ and $(b,c)$ are both in $R$, then $(a,c)$ must be too. Here $(1,1)$ is already in $R$.
$(2,1) \in R$ and $(1,2) \in R$ force $(2,2) \in R$ Run the same check the other way round. $(2,2)$ is already in $R$ too. Transitive holds.
*$R$ is reflexive, symmetric, and transitive* All three checks passed. $R$ is an equivalence relation on $A$.
Chain $(1,1)$ and $(1,2)$ together: do they force $(1,2)$? Check your transitive conclusion a second way. $(1,1)$ and $(1,2)$ force $(1,2)$, already in $R$: another chain that holds, agreeing with the conclusion above.
Check yourself
Take $R' = \{(1,1), (2,2), (3,3), (1,2)\}$ on $A = \{1,2,3\}$: the worked-check relation with $(2,1)$ removed. Which property does $R'$ now fail?
transitive: a chained pair is now missing
none: removing one pair from a passing relation cannot break any of the three properties
reflexive: one of $(1,1), (2,2), (3,3)$ is now missing
symmetric: $(1,2) \in R'$ but $(2,1) \notin R'$
Check your answer
transitive: a chained pair is now missing — Checking every pair-of-pairs in $R'$ still lands on a pair already present, removing $(2,1)$ happened not to break any transitive chain here.
none: removing one pair from a passing relation cannot break any of the three properties — One property specifically needed $(2,1)$ to pair with its reverse $(1,2)$, removing it is not harmless, it targets exactly the property that relied on it.
reflexive: one of $(1,1), (2,2), (3,3)$ is now missing — $(1,1),(2,2),(3,3)$ never depended on $(2,1)$, reflexivity is exactly as intact as it was in the original worked check.
✓ symmetric: $(1,2) \in R'$ but $(2,1) \notin R'$ — (D) Symmetric needs every pair’s reverse present too. Removing $(2,1)$ leaves $(1,2)$ without its reverse, so symmetric is exactly what breaks.
In verifying that $R = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}$ is transitive, why does checking the chain $(1,2)$, then $(2,1)$, matter, and not just $(1,2)$ alone?
because a relation needs at least two pairs to be called a relation at all
transitivity tests pairs of pairs, never a single pair alone
because $(1,2)$ by itself already proves $R$ is not reflexive
because transitivity only applies to pairs where the first and last element match
Check your answer
because a relation needs at least two pairs to be called a relation at all — A relation is any subset of $A \times A$, including a single pair or even zero pairs, nothing in the definition sets a minimum pair count.
✓ transitivity tests pairs of pairs, never a single pair alone — (B) Transitivity is always a claim about TWO pairs chaining into a third, checking one pair in isolation can never test it either way.
because $(1,2)$ by itself already proves $R$ is not reflexive — A single pair like $(1,2)$ says nothing about reflexivity, which needs $(a,a)$ for every element , the two checks are independent of each other.
because transitivity only applies to pairs where the first and last element match — Transitivity chains ANY two pairs $(a,b)$ and $(b,c)$, whatever $a$ and $c$ turn out to be, it is never restricted to cases where they happen to match.
Because $R = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}$ passes all three property tests, what is it called?
an equivalence relation
a function
the universal relation on $A$
a subset relation
Check your answer
✓ an equivalence relation — (A) Reflexive, symmetric, and transitive together earn a relation exactly one name , equivalence relation.
a function — Passing reflexive, symmetric, and transitive says nothing about whether every element of $A$ has exactly one image, that is a completely separate check, the one that defines a function.
the universal relation on $A$ — $R$ does not contain $(1,3)$ or $(3,1)$, so it is not $A \times A$, passing the three properties makes it an equivalence relation, not the universal relation specifically.
a subset relation — “Subset” here would mean a relation between two different sets, an unrelated idea to the reflexive/symmetric/transitive test just carried out on one set $A$.
The same relation twice. Listed in order, the arrows cross; listed by class, they fall into three separate pairs. Those pairs are the classes.
Assembly in house lines: every student stands in exactly one line, and no line shares a student.
Agni
AmanBinaChitra
Vayu
DevEshaFarid
Jal
GauriHari
Prithvi
IshaJaiKavya
Same house as, drawn as groups: four groups, every student in exactly one, none in two, none in none.
Go back to “same house as” on your class. It passed all three checks. Now look at what passing did for you. The relation sorted the class into four groups, one per house, and every student is in exactly one group. Nobody is in two houses, and nobody is in none.
Why the three checks were chosen
That sorting is the reward for passing all three checks, and it is the reason those three were chosen. Compare the checks with the equals sign. A thing equals itself: that is reflexive. If $a = b$ then $b = a$: symmetric. If $a = b$ and $b = c$ then $a = c$: transitive. The three checks are the three things “equals” does. A relation that does all three behaves like a kind of equals. So it can sort things into groups, the way “equals” does.
A relation that passes all three is called an EQUIVALENCE RELATION. Here is the sorting in symbols.
Equals is loops alone, nothing joined. A relation passing all three checks groups the set the way equals does, each with its own kind.
The class of one element
Take any element $a$ in $A$. Collect every element related to $a$. That collection is the EQUIVALENCE CLASS of $a$, written $[a]$. Formally, $[a] = \{x \in A : (x, a) \in R\}$, read as “every $x$ in $A$ that $R$ relates to $a$”. In the school example, the equivalence class of a student is their house.
*Is $a$ itself* always in $[a]$? Yes. $R$ is reflexive, so $(a, a) \in R$, which puts $a$ inside its own class.
Agni
AmanBinaChitra
Vayu
DevEshaFarid
Jal
GauriHari
Prithvi
IshaJaiKavya
The class of one student is everyone the relation ties to them: their whole house, and nothing outside it.
Can two classes, $[a]$ and $[b]$, overlap just a little? No. Say they share one element $c$.
$(c, a) \in R$ and $(c, b) \in R$, because $c$ is in both classes.
$(a, c) \in R$, by symmetric on $(c, a)$.
$(a, b) \in R$, by transitive on $(a, c)$ and $(c, b)$.
Take any $x \in [a]$: $(x, a) \in R$, and with $(a, b) \in R$, transitive gives $(x, b) \in R$. So $x \in [b]$.
Run the same argument back, swapping $a$ and $b$: every member of $[b]$ is in $[a]$.
Two classes are either exactly the same, or they share nothing at all.
Classes cut the set into a partition
Put those two facts together. Every element of $A$ sits in a class. No two classes partly overlap. So the classes cut $A$ into separate groups, with every element in exactly one. This cutting is called a PARTITION.
Two classes cannot share one member: symmetric and transitive would pull the two groups into one.
remainder 0
0369
remainder 1
14710
remainder 2
25811
Divisible-by-3 sorts 0 to 11 into three groups, one per remainder, every number in exactly one.
One hand position shows 1:00 and 13:00: the clock sorts hours by remainder on dividing by 12.
Numbers do the same thing. Take “$a - b$ is divisible by $3$” on the whole numbers from $0$ to $11$. The class of $0$ is $\{0, 3, 6, 9\}$. The class of $1$ is $\{1, 4, 7, 10\}$. The class of $2$ is $\{2, 5, 8, 11\}$. Three groups, every number in exactly one, and the groups are just the three possible remainders on dividing by $3$. A 12-hour clock does the same with $12$: 13:00 and 1:00 fall in the same class. That is why one hand position can show both.
You can go the other way too. Start from any partition, say the four houses, and declare two students related exactly when they share a house. The relation you get passes all three checks. A relation and its partition carry the same information, read two different ways.
Your turn: in the partition of a class by house, which group holds Hari? Check: Jal, the same house as Gauri.
Check yourself
For an equivalence relation $R$ on $A$, the equivalence class $[a]$ is \_\_\_\_\_\_.
$\{x \in A:(a, x) \notin R\}$, everything not related to $a$
the single ordered pair $(a, a)$
$\{x \in A:x = a\}$, just $a$ itself
$\{x \in A:(x, a) \in R\}$, everything related to $a$
Check your answer
$\{x \in A:(a, x) \notin R\}$, everything not related to $a$ — The class collects what IS related to $a$, the opposite of what is excluded from it, swapping “in” for “not in” reverses the definition.
the single ordered pair $(a, a)$ — $[a]$ is a SET of elements, potentially many of them, a single ordered pair is a different kind of object entirely.
$\{x \in A:x = a\}$, just $a$ itself — $a$ itself is only one member of $[a]$, reflexivity guarantees $a \in [a]$, but the class collects every $x$ related to $a$, which can be many elements.
✓ $\{x \in A:(x, a) \in R\}$, everything related to $a$ — (D) $[a]$ collects every element of $A$ that stands in relation $R$ to $a$, possibly many elements, not just $a$ alone.
If equivalence classes $[a]$ and $[b]$ of the same relation $R$ share even one element, what must be true?
exactly one of $[a]$, $[b]$ must be empty
$a$ and $b$ cannot both belong to $A$
$[a]$ and $[b]$ overlap partially, sharing only that one element
$[a]$ and $[b]$ are identical: they are the exact same class
Check your answer
exactly one of $[a]$, $[b]$ must be empty — Nothing about the fact says either class must be empty, both $[a]$ and $[b]$ can hold many elements and still be forced to be the SAME set once they share even one member.
$a$ and $b$ cannot both belong to $A$ — The fact is about what happens to $[a]$ and $[b]$ when they share an element, it places no separate condition on whether $a$ and $b$ themselves are related.
$[a]$ and $[b]$ overlap partially, sharing only that one element — Two classes are either identical or disjoint, never partly overlapping. So “sharing one element but staying distinct” is exactly the case that cannot happen.
✓ $[a]$ and $[b]$ are identical: they are the exact same class — (D) Sharing even one member forces the two classes to collapse into one and the same set, partial overlap is never possible for equivalence classes.
An equivalence relation $R$ on a 7-element set $A$ produces exactly two equivalence classes, $[a]$ with 3 elements and $[b]$ with 4 elements. What does this tell you about every element of $A$?
every element belongs to exactly one of $[a]$ or $[b]$
$[a]$ and $[b]$ must share at least one element, since $3 + 4 = 7$ leaves no room otherwise
the class sizes 3 and 4 must be miscounted, since equivalence classes are always equal in size
some elements of $A$ might belong to neither $[a]$ nor $[b]$
Check your answer
✓ every element belongs to exactly one of $[a]$ or $[b]$ — (A) Every element of $A$ sits in exactly one class, and the classes never overlap , 3 elements in $[a]$ plus 4 in $[b]$ accounts for all 7, with nothing shared and nothing left over.
$[a]$ and $[b]$ must share at least one element, since $3 + 4 = 7$ leaves no room otherwise — Disjointness does not depend on how the sizes add up, two classes of the same relation never share an element, whatever their individual counts are.
the class sizes 3 and 4 must be miscounted, since equivalence classes are always equal in size — Nothing in the setup gives a reason to doubt the stated class sizes, this just raises an unmotivated objection.
some elements of $A$ might belong to neither $[a]$ nor $[b]$ — Every element of $A$ belongs to exactly one class, there is no element left over that belongs to neither $[a]$ nor $[b]$.
A relation and its partition into equivalence classes “carry exactly the same information, read two different ways.” What does this mean?
the partition is only an approximation of the relation, useful when $R$ is too large to list
the relation determines the classes, but the classes cannot be used to recover the relation
every partition of $A$ corresponds to a unique relation that is not an equivalence relation
the relation and its classes determine each other exactly
Check your answer
the partition is only an approximation of the relation, useful when $R$ is too large to list — “Exactly the same information” rules out approximation, the classes are not a rough summary of $R$, they carry everything $R$ carries.
the relation determines the classes, but the classes cannot be used to recover the relation — “Read two different ways” means you can move in either direction, from $R$ to its classes, and from the classes back to exactly which pairs are in $R$.
every partition of $A$ corresponds to a unique relation that is not an equivalence relation — No rule says a partition corresponds to a relation that fails to be an equivalence relation. This option invents a fact.
✓ the relation and its classes determine each other exactly — (D) The relation and the partition determine each other exactly, from either one alone, you can rebuild the other with nothing lost and nothing added.
Congruent triangles sort into classes: inside a group every triangle matches, across groups none does.
Worked example
Show that congruence of triangles is an equivalence relation
$R$ relates two triangles when they are congruent This is the relation to test. Go through reflexive, symmetric and transitive, the same three checks as before.
Reflexive: every triangle is congruent to itself A triangle matches its own sides and its own angles exactly. So $(T, T) \in R$ for every triangle $T$.
Symmetric: if $T_1$ is congruent to $T_2$, then $T_2$ is congruent to $T_1$ Congruence is a side-for-side, angle-for-angle match. The match reads the same way in either direction.
Transitive: if $T_1$ is congruent to $T_2$, and $T_2$ is congruent to $T_3$ Check whether the match carries all the way through to $T_3$.
Every side and angle of $T_1$ matches $T_2$, and of $T_2$ matches $T_3$ Chain the two matches together. $T_1$ matches $T_3$, side for side and angle for angle.
*$R$ is reflexive, symmetric, and transitive* All three properties hold. Congruence is an equivalence relation on the set of all triangles in a plane.
Check: does “same area as” work the same way? Check it yourself. Equal-area triangles are also reflexive, symmetric and transitive: any “sameness” relation on shapes tends to pass the same three tests.
Check yourself
In the congruent-triangles worked example, why does “$T_1$ congruent to $T_2$ implies $T_2$ congruent to $T_1$” count as proving symmetry?
symmetry is exactly the swap-the-two-objects idea
because it shows three triangles chain together
because congruent triangles always have equal area
because it shows every triangle relates to itself
Check your answer
✓ symmetry is exactly the swap-the-two-objects idea — (A) Symmetry is precisely “swap the order, the relation still holds”, congruence favouring no direction is that exact idea, applied to two triangles.
because it shows three triangles chain together — Chaining three triangles ($T_1$ to $T_2$ to $T_3$) is the transitive check, symmetry only ever involves swapping the order of TWO triangles.
because congruent triangles always have equal area — Equal area is a true consequence of congruence, but it is not the reasoning the worked example uses to establish symmetry, that reasoning is specifically about direction.
because it shows every triangle relates to itself — “Every triangle is congruent to itself” is the reflexive check in the same worked example, a different line of reasoning from the swap-the-order idea that proves symmetry.
Congruence classes group triangles by shape and size, ignoring position. Why does position not matter?
a triangle keeps its class wherever it sits in the plane
position is the only thing that decides which class a triangle belongs to
the equivalence classes are defined only for triangles at the origin
triangles in different positions can never be congruent
Check your answer
✓ a triangle keeps its class wherever it sits in the plane — (A) A triangle’s class is fixed by its shape-and-size alone, sliding it to a new position in the plane never moves it to a different class.
position is the only thing that decides which class a triangle belongs to — Shape-and-size is the actual criterion; position is what gets ignored, swapping these two around reverses what the claim says.
the equivalence classes are defined only for triangles at the origin — Nothing restricts congruence classes to triangles sitting at the origin, the claim applies to every triangle in the plane, wherever it sits.
triangles in different positions can never be congruent — The whole point of “ignoring position” is that moving a triangle to a new spot does not change its shape-and-size, so it stays congruent to where it started, different position is exactly what does NOT block congruence.
Because congruence is an equivalence relation, every triangle in the plane belongs to \_\_\_\_\_\_.
as many classes as it has sides
exactly one congruence class, along with every other triangle of the same shape and size
exactly one congruence class per possible position it could be moved to
no class, since congruence classes are only defined for equal triangles, not similar ones
Check your answer
as many classes as it has sides — Nothing about congruence classes counts sides, every triangle has three sides regardless of which class it is in, so this cannot be what decides class membership.
✓ exactly one congruence class, along with every other triangle of the same shape and size — (B) Congruence being an equivalence relation guarantees every triangle sits in exactly one class, shared with every triangle of matching shape-and-size.
exactly one congruence class per possible position it could be moved to — A congruence class groups triangles by shape-and-size, not by how many positions one triangle could be moved to, those are different, unrelated ideas.
no class, since congruence classes are only defined for equal triangles, not similar ones — General equivalence-relation facts apply to every equivalence relation, congruence included, there is no stated exemption.
I checked $R = \{(1,1), (2,2), (1,2), (2,1)\}$ on $A = \{1, 2, 3\}$. It was symmetric. It was transitive. So I called it an equivalence relation.
Second look
$3$ never shows up in any pair of $R$, not once. $(3,3)$ is missing. $R$ is not reflexive after all.
So check reflexive on its own, element by element. $3$ was the one I never looked at.
Symmetric and transitive both pass. Three never appears in any pair, so nothing forces (3, 3), and reflexive fails.
Symmetric and transitive both talk about pairs already in $R$. So it looks as if they must deliver reflexive too, automatically.
What the two checks actually say
They do not. Symmetric and transitive only ever say something about elements that ALREADY appear in some pair of $R$. They say nothing about an element $R$ never mentions at all.
Check each property on $R = \{(1,1), (2,2), (1,2), (2,1)\}$ on $A = \{1, 2, 3\}$.
Symmetric: every pair’s reverse is present. Symmetric holds.
Transitive: $(1,2)$ and $(2,1)$ force $(1,1)$, already there. $(2,1)$ and $(1,2)$ force $(2,2)$, already there. Every other combination touches a diagonal pair and forces nothing new. Transitive holds.
Now check reflexive. Does $3$ pair with itself? $(3,3)$ should be in $R$. It is not. *$3$ never appears* in any pair of $R$ at all, not once. Reflexive fails.
Symmetric and transitive together only force reflexivity on elements $R$ actually relates to something. An element $R$ never mentions stays invisible to both checks, so reflexivity fails at that element.
Your turn: does $R$ above relate to $3$ in any way?
A SCHOLAR INDIA REMEMBERS
Take any $a$. Pick $b$ with $(a, b) \in R$. Symmetric gives $(b, a) \in R$. Transitive then gives $(a, a) \in R$, so $R$ looks reflexive. Which step is wrong? The wrong step is “pick $b$”: it assumes $a$ has a partner. In the relation above, $3$ has none, so the proof never starts for $3$.
Dignaga · the kingfisher
Check yourself
A student claims: “If a relation is symmetric and transitive, it must also be reflexive; the three properties basically come as a package.” Is the student right?
No: $R=\{(1,1),(2,2),(1,2),(2,1)\}$ on $\{1,2,3\}$ misses $(3,3)$
Yes: symmetric and transitive together always force reflexive on the whole set
Yes, but only when $A$ has exactly 3 elements
No: symmetric and transitive relations are never reflexive
Check your answer
✓ No: $R=\{(1,1),(2,2),(1,2),(2,1)\}$ on $\{1,2,3\}$ misses $(3,3)$ — (A) The counterexample is symmetric and transitive by inspection, yet $3$ is missing from every pair, so reflexivity fails, the “package deal” claim is false.
Yes: symmetric and transitive together always force reflexive on the whole set — This is exactly the trap being tested, stated as though it were correct, the counterexample $R = \{(1,1),(2,2),(1,2),(2,1)\}$ on $\{1,2,3\}$ shows it fails.
Yes, but only when $A$ has exactly 3 elements — The counterexample’s failure has nothing to do with pair count, it fails specifically because element $3$ never appears in any pair, not because of how many pairs exist.
No: symmetric and transitive relations are never reflexive — The correct answer is not “never” either, some symmetric-and-transitive relations ARE reflexive on the elements they mention; the counterexample only shows “always” is false.
Take $R = \{(1,1),(2,2),(1,2),(2,1)\}$ on $A = \{1,2,3,4\}$ this time: one more element than the original example. Is $R$ reflexive?
Yes: adding an element to $A$ makes the relation reflexive
No: both $3$ and $4$ are missing from every pair, so reflexivity fails twice over
No, but only because $A$ now has an even number of elements
Yes: $R$ was already reflexive on $\{1,2,3\}$, so it stays reflexive on any larger set
Check your answer
Yes: adding an element to $A$ makes the relation reflexive — Adding elements to $A$ that the relation still never mentions can only make reflexivity fail in MORE places, never fewer, a bigger set is not automatically friendlier to reflexivity.
✓ No: both $3$ and $4$ are missing from every pair, so reflexivity fails twice over — (B) The same mechanism as before, doubled, neither $3$ nor $4$ ever appears in a pair, so both dodge reflexivity, the way $3$ alone did originally.
No, but only because $A$ now has an even number of elements — Whether $A$ has an even or odd number of elements has nothing to do with which elements a relation mentions, reflexivity depends only on that.
Yes: $R$ was already reflexive on $\{1,2,3\}$, so it stays reflexive on any larger set — $R$ already failed reflexivity on the original set $\{1,2,3\}$, since element $3$ was already missing there too, this option’s premise is false from the start.
What is the precise reach of the fact “symmetric and transitive together force reflexivity”?
only on elements that appear in at least two pairs
only on the elements the relation actually relates
on every element of $A$, without exception
never: symmetric and transitive relations are always non-reflexive
Check your answer
only on elements that appear in at least two pairs — Nothing about how many pairs an element appears in decides this, an element that appears in even one pair, self-paired or not, is covered differently from one that never appears at all.
✓ only on the elements the relation actually relates — (B) The force only reaches elements $R$ actually pairs with something, an element left out of every pair is left out of the guarantee too.
on every element of $A$, without exception — This restates the original trap as if it were the precise scope, it is exactly the claim the counterexample refutes.
never: symmetric and transitive relations are always non-reflexive — The correct scope is not “never”, symmetric and transitive DO force reflexivity on elements the relation actually relates, just not on elements it never mentions.
A function is a rule that gives every input exactly one output. Your class register is one. Feed in a student, get out a roll number. In symbols, $f: A \to B$ sends each element of $A$ to one element of $B$. Class 11 stopped there. Class 12 asks two more questions about a function.
The register: no two students share a roll number, so a roll number names one student. One-one.
First question: do two inputs ever share an output? On the register, no. No two students have the same roll number. So if I tell you a roll number, you can name the one student. Now compare “has blood group”. Feed in a student, get out one of eight groups. Many students share a group, so if I tell you “B positive”, you cannot name the student.
Blood group: three students share B positive, so the group cannot name the student. Not one-one.
$f$ is ONE-ONE, also called injective, when different inputs always give different outputs. Said the other way round: $f(x_1) = f(x_2)$ forces $x_1 = x_2$. The roll-number function is one-one. The blood-group function is not.
A full bus uses every seat. With fewer passengers, some seats are nobody’s output: not onto.
Does every output get used?
Second question: does every possible output actually get used? Think of a bus with $40$ seats and $40$ passengers, each given a seat. Every seat is somebody’s. Now let us run the same bus with $25$ passengers. Fifteen seats are nobody’s output.
$f$ is ONTO, also called surjective, when every element of $B$ is somebody’s output. For every $b \in B$, some $a \in A$ gives $f(a) = b$. Nothing in $B$ is left out. The full bus is onto. The bus with fifteen empty seats is not.
A function that is both one-one and onto is called BIJECTIVE. The full bus, one passenger per seat, is bijective: no seat is shared, and no seat is empty.
Function
One-one?
Onto?
Roll-number register
Yes
-
Blood-group function
No
-
Full bus, 40 seats/40 passengers
-
Yes
Bus with 25 passengers, 40 seats
-
No
A “-” means that property was not asked of this function.
Doubling never repeats an output, so it is one-one, but no whole number doubles to 3. One-one and not onto.
Two independent questions
The two questions are independent. One-one asks whether two inputs ever give the same output. Onto asks whether every output is reached. On an infinite set, the answer to one tells you nothing about the other. $f(x) = 2x$ on the whole numbers is one-one, since doubling two different numbers gives two different results. But it is not onto, since no whole number doubles to $3$. You check each on its own.
Same size both sides: no shared output means every target is used, and one shared output leaves one target unused.
On a finite set mapped to a set of the same size, the two come as a pair. $40$ passengers, $40$ seats, nobody sharing: then every seat is used, automatically. One-one gives onto for free, and onto gives one-one for free. You will use that shortcut later in the chapter.
Birthdays: two students share a day, and 30 cannot cover 365. Neither one-one nor onto.
Why do these two questions matter? Because together they decide whether the function can be run backwards. The roll-number function can: from the number you get the one student. The blood-group function cannot: from “B positive” you cannot pick the student. That is where the chapter is going.
One-one: no two inputs share an output. Onto: no output is left unused.
Take “has a birthday on”, from your class to the $365$ days of the year. One-one? Probably not: in a class of $30$, two students sharing a birthday is more likely than not. Onto? Certainly not: $30$ students cannot cover $365$ days.
The full bus: no seat is shared and no seat is empty, so this seating is one-one and onto, which is bijective.
Four small diagrams, one per combination: both, one-one only, onto only, neither.
A SCHOLAR INDIA REMEMBERS
Say what one-one means, without using the word. “Each input gets one output” is not it. Every function does that. One-one asks for more: two different inputs never share an output. Now say onto the same way. Every element of $B$ is somebody’s output. Now each word is a test you can run.
Dignaga · the kingfisher
Check yourself
A function $f:A \to B$ is one-one (injective) exactly when \_\_\_\_\_\_.
every element of $B$ is somebody’s output
$A$ and $B$ have the same number of elements
$f(x_1) = f(x_2)$ always forces $x_1 = x_2$
$f(x_1) = f(x_2)$ for at least one pair $x_1 \neq x_2$
Check your answer
every element of $B$ is somebody’s output — “Every element of $B$ is somebody’s output” is onto. One-one is about distinct inputs never sharing an output, a completely different requirement.
$A$ and $B$ have the same number of elements — One-one says nothing about how many elements $A$ and $B$ have, it is a statement about distinct inputs and their outputs, not about set sizes.
✓ $f(x_1) = f(x_2)$ always forces $x_1 = x_2$ — (C) One-one means no two distinct inputs ever share the same output, equal outputs force equal inputs.
$f(x_1) = f(x_2)$ for at least one pair $x_1 \neq x_2$ — One-one says $f(x_1) = f(x_2)$ FORCES $x_1 = x_2$, the reverse direction, “$x_1 = x_2$ forces $f(x_1) = f(x_2)$,” is just true of every function and proves nothing about injectivity.
A function $f:A \to B$ is onto (surjective) exactly when \_\_\_\_\_\_.
$f(a) = a$ for every $a \in A$
distinct inputs always give distinct outputs
for every $b \in B$, some $a \in A$ has $f(a) = b$
for every $a \in A$, some $b \in B$ has $f(a) = b$
Check your answer
$f(a) = a$ for every $a \in A$ — “$f(a) = a$ for every $a$” describes only the identity function, one very specific case, onto covers far more functions than that single one.
distinct inputs always give distinct outputs — “$f(x_1) = f(x_2)$ forces $x_1 = x_2$” is one-one. Onto is about every element of $B$ being somebody’s output, a completely different requirement.
✓ for every $b \in B$, some $a \in A$ has $f(a) = b$ — (C) Onto means nothing in $B$ is left out, every element of $B$ has at least one element of $A$ mapping to it.
for every $a \in A$, some $b \in B$ has $f(a) = b$ — Every function already assigns some output to every input by definition, that test is true of every $f$ and can never tell onto functions apart from any other kind.
One-one and onto are called independent properties. What does that mean?
a function can have one, both, or neither
a function is automatically onto whenever it is one-one
a function must have at least one of the two properties
the two properties always occur together in every function
Check your answer
✓ a function can have one, both, or neither — (A) Neither property implies or excludes the other, all four combinations (both, either alone, neither) are possible for some function.
a function is automatically onto whenever it is one-one — If being one-one guaranteed onto, the two properties would move together, not independently, the worked example’s restricted $g(x) = x^2$ needed a SEPARATE onto check even after passing one-one.
a function must have at least one of the two properties — Independence does not require a function to have at least one of the two properties, the worked check’s original $f(x) = x^2$ on $\mathbb{R} \to \mathbb{R}$ has NEITHER, which independence allows.
the two properties always occur together in every function — “Always occur together” is the opposite of independent, a function having one but not the other, like $f(x) = x^2$ on $\mathbb{R} \to \mathbb{R}$ (neither), shows they do not.
A function that is both one-one and onto is called \_\_\_\_\_\_.
symmetric
reflexive
bijective
transitive
Check your answer
symmetric — Symmetric describes whether a relation’s pairs reverse. It is a property of relations, not a type of function.
reflexive — Reflexive describes whether a relation pairs every element with itself, a property of relations on one set, not a name for any type of function.
✓ bijective — (C) One-one AND onto together earn a function exactly one name, bijective.
transitive — Transitive describes chaining pairs in a relation, also a relation property, not something that names a kind of function.
Squaring is neither one-one nor onto over all real numbers, but restricting both sides to non-negative numbers makes it both.
Worked example
Check one-one and onto for $f(x) = x^2$ under two domain choices
$f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ Start with this function on the whole real line. Test one-one first.
$f(-2) = 4$ and $f(2) = 4$ Two different inputs, $-2$ and $2$. Same output, $4$.
$f$ is not one-one Different inputs gave the same output. That breaks the one-one rule straight away.
No real $x$ gives $f(x) = -1$ $x^2$ is never negative. So $-1$ has no input that reaches it.
$f$ is not onto Every negative number in $\mathbb{R}$ is left with no input reaching it.
$g: [0, \infty) \to [0, \infty)$, $g(x) = x^2$ Same formula. Now cut the domain and the codomain down to the non-negative numbers only.
$x_1^2 = x_2^2$ with $x_1, x_2 \geq 0$ forces $x_1 = x_2$ Both inputs are non-negative now. Equal squares force equal inputs. $g$ is one-one.
Every $y \geq 0$ has an input $\sqrt{y} \geq 0$ $\sqrt{y}$ sits inside the restricted domain, and $g(\sqrt{y}) = y$. $g$ is onto.
*$g$ is bijective, though $f$ was neither* Cutting the domain and codomain changed the verdict completely, with no change to the formula at all.
Check: try $x = 3$ in $g$ Check it yourself. $g(3) = 9$, and $\sqrt{9} = 3$ sends you straight back. One input, one output, both ways: exactly what bijective means.
Worked example
The same test, on a genotype-to-phenotype function
Genotypes TT and Tt (Biology 12, Chapter 4) both produce a tall plant A dominant allele hides a recessive one in a heterozygote. So the homozygous-dominant genotype and the heterozygous genotype look the same from outside.
$f(\text{TT}) = \text{tall}$ and $f(\text{Tt}) = \text{tall}$ Two different genotypes give the same phenotype.
$f$ is not one-one Different inputs gave the same output. That is exactly what breaks one-one: the same pattern as $f(-2) = f(2) = 4$ for $f(x) = x^2$, above.
Compare with $f(-2) = f(2) = 4$ for $f(x) = x^2$ Check it yourself. Numbers or genotypes, the test is identical: two different inputs sharing one output breaks one-one every time.
Check yourself
Let $f:\mathbb{R} \to \mathbb{R}$, $f(x) = x^3$. Using the same reasoning as the worked check on $x^2$, is $f$ one-one?
cannot be determined without restricting the domain
No: $f(-2) = f(2) = 8$, just like the $x^2$ example
No: cubes are always negative for negative inputs, so no real $y$ is ever reached
Yes: distinct reals always give distinct cubes
Check your answer
cannot be determined without restricting the domain — Unlike $x^2$, the cube function never needs its domain restricted to become one-one, distinct reals already always give distinct cubes.
No: $f(-2) = f(2) = 8$, just like the $x^2$ example — Copying the $x^2$ example’s collision without recomputing fails here, $(-2)^3 = -8$ and $2^3 = 8$, two DIFFERENT outputs, so no collision actually occurs.
No: cubes are always negative for negative inputs, so no real $y$ is ever reached — Cubes of negative reals are negative, so negative outputs ARE reached, and this option also answers the wrong question, onto rather than one-one.
✓ Yes: distinct reals always give distinct cubes — (D) Cubing preserves order on the reals, $x_1 \neq x_2$ always gives $x_1^3 \neq x_2^3$, so no collision like $x^2$’s $(-2, 2)$ pair ever happens.
The worked check restricts $f(x) = x^2$ from $\mathbb{R} \to \mathbb{R}$ to $g(x) = x^2$ on $[0, \infty) \to [0, \infty)$, and both answers flip: one-one becomes yes, onto becomes yes. What actually changed?
only the codomain changed; the domain stayed all of $\mathbb{R}$
the formula itself was rewritten to $g(x) = \sqrt{x}$
restricting a domain always makes a function onto, regardless of the codomain
only the domain and codomain changed, not the formula
Check your answer
only the codomain changed; the domain stayed all of $\mathbb{R}$ — Both the domain AND the codomain moved to $[0, \infty)$ together, restricting only one of the two would not have flipped both answers.
the formula itself was rewritten to $g(x) = \sqrt{x}$ — The rule $g(x) = x^2$ is exactly the same formula as $f(x) = x^2$, nothing about the formula itself changed, only which inputs and outputs were allowed.
restricting a domain always makes a function onto, regardless of the codomain — This one worked example does not license “restricting a domain always makes a function onto” as a general rule, it shows what happened for this specific function.
✓ only the domain and codomain changed, not the formula — (D) Same rule, $x^2$, throughout, restricting where it is allowed to look, both as input and as output, was the entire cause of both flips.
Take $h(x) = x^2$ but now restricted to domain $(-\infty, 0]$ with codomain $[0, \infty)$ (non-positive inputs only). Is $h$ one-one?
Yes: non-positive inputs never collide under squaring
cannot be one-one, because the codomain $[0, \infty)$ contains $0$
No: $h(-2) = h(2) = 4$ still collide, since $-2$ is in the new domain
No: squaring never preserves distinctness
Check your answer
✓ Yes: non-positive inputs never collide under squaring — (A) With inputs restricted to $x \leq 0$, each output has exactly one non-positive square root, the collision that broke one-one-ness on all of $\mathbb{R}$ cannot happen here.
cannot be one-one, because the codomain $[0, \infty)$ contains $0$ — Whether $0$ sits in the codomain has nothing to do with one-one-ness, which is about inputs never colliding, not about which outputs are reachable.
No: $h(-2) = h(2) = 4$ still collide, since $-2$ is in the new domain — $2$ is outside the domain $(-\infty, 0]$, only $x \leq 0$ is allowed, so the collision $h(-2) = h(2)$ never actually happens inside this restricted function.
No: squaring never preserves distinctness — Squaring fails to be one-one on all of $\mathbb{R}$ because BOTH signs are allowed, restricted to non-positive inputs only, there is exactly one non-positive square root of any value, so distinctness is preserved.
Two functions in a row: dollars to rupees, then GST. The pair is one new function, dollars in and final price out.
Two functions in a row. A shop takes a price in dollars and runs it through two steps. First it converts to rupees at $83$ to the dollar, then it adds $18$ percent GST. Feed in $100$ dollars. Step one gives $8300$ rupees. Step two gives $9794$ rupees. The two steps together are one new function: dollars in, final price out.
g after f: the two arrows join into one.
That is COMPOSITION. Take $f: A \to B$ and $g: B \to C$. Build a new function that sends $x$ to $g(f(x))$: apply $f$ first, then feed the result into $g$. It is written $g \circ f$ and read “g after f”. *$f$, on the right,* happens first. It works because $f$’s output falls in $B$, exactly where $g$ expects its input.
100 dollars
→
×83
dollars to rupees
→
+18%
GST
→
₹9,794
100 dollars
→
+18%
GST
→
×83
dollars to rupees
→
₹9,794
both steps multiply: order does not matter
Both steps multiply, so either order gives 9,794. That is the exception, not the rule.
₹8,300
→
− ₹100
discount
→
+18%
GST
→
₹9,676
₹8,300
→
+18%
GST
→
− ₹100
discount
→
₹9,694
a flat discount and GST, in both orders
Add a flat discount and the order shows: 9,676 one way, 9,694 the other. Composition is not commutative.
Does the order matter?
Does the order matter? Not when both steps multiply, as the figure above shows: either order gives $9794$ rupees. Add a flat discount of $100$ rupees as a third step, and order starts to matter, as the next figure shows: discount then GST gives $9676$; GST then discount gives $9694$.
So composition is NOT commutative in general. $g(f(x))$ and $f(g(x))$ can give different answers. Sometimes one of them is not even defined, because $g$’s output does not fall inside $f$’s domain the other way round.
5
→
g after f
+1, then ×2
→
h
−3
→
9
5
→
f
+1
→
h after g
×2, then −3
→
9
group the three functions either way
Group f with g, or g with h: both mean f, then g, then h, and both give 9. The brackets can go.
Composition is ASSOCIATIVE. Compose $f$, then $g$, then $h$, in any grouping you like: $(h \circ g) \circ f$ and $h \circ (g \circ f)$ give the same function. Both mean “do $f$, then $g$, then $h$”. So the brackets can be dropped in a chain.
2
→
+1
→
square
→
9
2
→
square
→
+1
→
5
square after add-one, add-one after square
Square after add-one gives 9; add-one after square gives 5. Order matters, and the next example proves it.
Your turn: for $f(x) = x + 1$ and $g(x) = x^2$, is $g(f(3))$ the same as $f(g(3))$? Work both out and compare. You will meet the same check again soon.
Check yourself
Given $f:A \to B$ and $g:B \to C$, the composition $(g ∘ f)(x)$ means \_\_\_\_\_\_.
apply $f$ and $g$ separately and take whichever gives a defined answer
add the two functions: $f(x) + g(x)$
apply $f$ first, then apply $g$ to the result: $g(f(x))$
apply $g$ first, then apply $f$ to the result: $f(g(x))$
Check your answer
apply $f$ and $g$ separately and take whichever gives a defined answer — Composition fixes a specific order, $f$ then $g$ (for $g \circ f$), it is not a choice between whichever function happens to be defined.
add the two functions: $f(x) + g(x)$ — Composition is applying one function to the OUTPUT of another, not adding their two outputs together , a completely different operation.
✓ apply $f$ first, then apply $g$ to the result: $g(f(x))$ — (C) $(g ∘ f)(x)$ is read right-to-left in application order, $f$ acts on $x$ first, then $g$ acts on that result.
apply $g$ first, then apply $f$ to the result: $f(g(x))$ — Applying $g$ first and $f$ to the result is $(f \circ g)(x) = f(g(x))$, the OTHER order, $(g \circ f)(x)$ applies $f$ first.
Composition of functions is associative but not, in general, commutative. What is the difference between these two claims?
associative applies only to addition and multiplication, never to functions
associative and commutative mean the same thing for functions, so this is a contradiction
grouping (associative) vs order (commutative): composition has only the first
associative means order does not matter; commutative means grouping does not matter: composition has neither
Check your answer
associative applies only to addition and multiplication, never to functions — The claim is specifically about function composition having these properties, nothing restricts associativity to only addition and multiplication.
associative and commutative mean the same thing for functions, so this is a contradiction — If the two words meant the same thing, the claim “associative but not commutative” would be a contradiction , they describe genuinely different properties, which is exactly why composition can have one without the other.
✓ grouping (associative) vs order (commutative): composition has only the first — (C) Grouping three functions (associative) and swapping the order of two (commutative) are different questions, composition answers yes to the first, no to the second in general.
associative means order does not matter; commutative means grouping does not matter: composition has neither — This swaps which word means what, and also denies that composition IS associative, both parts contradict the stated fact.
$f(g(x))$ might not be defined, even when $g(f(x))$ is. When does this happen?
whenever $f$ and $g$ are both one-one
composition is always defined both ways, so this never happens
when $g$’s codomain and $f$’s domain do not line up
only when $A$, $B$, and $C$ are all the same set
Check your answer
whenever $f$ and $g$ are both one-one — Whether $f$ or $g$ is one-one has no bearing on whether one function’s outputs are valid inputs for the other, that is purely about matching codomain to domain.
composition is always defined both ways, so this never happens — $f(g(x))$ can fail to be defined even when $g(f(x))$ is. Composition is not guaranteed to work both ways.
✓ when $g$’s codomain and $f$’s domain do not line up — (C) $f(g(x))$ needs $g$’s outputs to be valid inputs for $f$. If $g$’s codomain does not sit inside $f$’s domain, the composition breaks down.
only when $A$, $B$, and $C$ are all the same set — If $A$, $B$, and $C$ are all the same set, there is no domain/codomain mismatch to worry about at all, this is the case where composition both ways is safest, not the case where it fails.
The two composite curves meet only once, at x = 0, which solving (x+1) squared equals x squared plus 1 confirms.
Worked example
Compute $g(f(x))$ and $f(g(x))$ for $f(x) = x+1$, $g(x) = x^2$
$f(x) = x + 1$, $g(x) = x^2$, both $\mathbb{R} \to \mathbb{R}$ Two given functions. Compute both orders of composition and compare.
$g(f(x)) = g(x+1) = (x+1)^2$ Apply $f$ first. Feed its output, $x+1$, into $g$.
$g(f(x)) = x^2 + 2x + 1$ Expand the square.
$f(g(x)) = f(x^2) = x^2 + 1$ Apply $g$ first this time. Feed its output, $x^2$, into $f$.
*$g(f(x)) = x^2+2x+1 \neq x^2+1 = f(g(x))$* Compare the two results. They are genuinely different functions.
Check: try $x = 1$ in both Check it yourself. $g(f(1)) = 4$, and $f(g(1)) = 2$. Two different numbers, confirming order matters here too.
Check yourself
Let $f(x) = x - 2$ and $g(x) = x^2$, both $\mathbb{R} \to \mathbb{R}$. What is $g(f(x))$?
$x^2 - 4x - 4$, a sign slip in the expansion
$x - 4$, dropping the squaring entirely
$x^2 - 2$
$(x-2)^2 = x^2 - 4x + 4$
Check your answer
$x^2 - 4x - 4$, a sign slip in the expansion — Expanding $(x-2)^2$ gives $x^2 - 4x + 4$, the constant term’s sign is flipped here, a common expansion slip.
$x - 4$, dropping the squaring entirely — This skips squaring $g$’s input entirely, as though $g(y) = y$ rather than $g(y) = y^2$, $g$ must actually square whatever $f(x)$ produces.
$x^2 - 2$ — $x^2 - 2$ is $f(g(x))$, computed the other way round, $g(x) = x^2$ applied first, then $f$ subtracts 2. This question asks for $g(f(x))$ specifically.
✓ $(x-2)^2 = x^2 - 4x + 4$ — (D) $g(f(x))$ squares whatever $f(x) = x - 2$ produces, $(x-2)^2$, which expands to $x^2 - 4x + 4$.
Using the same $f(x) = x - 2$, $g(x) = x^2$ from the previous question, what is $f(g(x))$?
$x^2 - 2$
$f(g(x))$ is undefined, since $g$’s output cannot be an input to $f$
$x^2 - 4$
$(x-2)^2 = x^2 - 4x + 4$
Check your answer
✓ $x^2 - 2$ — (A) $f(g(x))$ subtracts 2 from whatever $g(x) = x^2$ produces, $x^2 - 2$, genuinely different from $g(f(x))$.
$f(g(x))$ is undefined, since $g$’s output cannot be an input to $f$ — Both $f$ and $g$ map all of $\mathbb{R}$ to $\mathbb{R}$ here, so $g$’s output is always a valid input for $f$, nothing blocks this composition from being defined.
$x^2 - 4$ — $f$ subtracts 2 from whatever it is given, dropping that subtraction treats $f$ as though it does nothing, which is not what $f(x) = x - 2$ says.
$(x-2)^2 = x^2 - 4x + 4$ — $(x-2)^2 = x^2 - 4x + 4$ is $g(f(x))$, the order computed in the previous question, this question asks for $f(g(x))$ instead.
In the worked example, $g(f(x)) = x^2 + 2x + 1$ while $f(g(x)) = x^2 + 1$. The two are genuinely different. What does this confirm?
$f$ and $g$ must not actually be functions, since composing them two ways gives different results
composition order matters: $g ∘ f$ and $f ∘ g$ need not be the same function, even when both are defined
$x^2 + 2x + 1$ and $x^2 + 1$ are actually the same expression, just written differently
one of the two calculations must contain an arithmetic mistake, since composing the same two functions should always give the same answer
Check your answer
$f$ and $g$ must not actually be functions, since composing them two ways gives different results — $f$ and $g$ are ordinary, well-defined functions, getting two different results from composing them in two orders says nothing about whether they qualify as functions.
✓ composition order matters: $g ∘ f$ and $f ∘ g$ need not be the same function, even when both are defined — (B) Two correctly-computed, genuinely different results from the same pair of functions is exactly the point, order is not a formality in composition.
$x^2 + 2x + 1$ and $x^2 + 1$ are actually the same expression, just written differently — $x^2 + 2x + 1$ and $x^2 + 1$ differ by exactly $2x$, which is nonzero whenever $x \neq 0$, they are genuinely different expressions, not two ways of writing the same one.
one of the two calculations must contain an arithmetic mistake, since composing the same two functions should always give the same answer — The whole worked example exists to show $g(f(x))$ and $f(g(x))$ need NOT agree, assuming they must contradicts what was just computed.
The register runs both ways: student to number, number to student. The second function undoes the first.
The class register runs both ways. Give me a student, I give you the roll number. Give me a roll number, I give you the student. The second function undoes the first. The blood-group function has no such partner: give me “B positive” and I cannot hand you back one student.
a
→
f
→
g
→
a
b
→
g
→
f
→
b
both ways come back where they started
Undoing means both ways come back where they started: a through f and g is a again, b through g and f is b again.
A function $f: A \to B$ is INVERTIBLE when some other function, $g: B \to A$, undoes it completely. Undoing means two things, both at once: $g(f(a)) = a$ for every $a \in A$, and $f(g(b)) = b$ for every $b \in B$. Student to number to the same student, and number to student to the same number. Undoing in one direction only is not enough.
A reverse rule fails two ways: two inputs share an output, or an output is never reached. Remove both and it works.
When a reverse rule fails
*When does such a $g$ exist? Let us look at the two ways a reverse rule can fail. If two inputs share an output, the reverse rule has two inputs to choose between, and no way to pick one: that is one-one failing. If some output is never reached, the reverse rule has no input to pair with it: that is onto failing. Remove both failures* and the reverse works.
If the reverse rule fails because
What goes wrong
Two inputs share an output
one-one fails: two inputs to choose between, no way to pick
Some output is never reached
onto fails: no input to pair with it
*Such a $g$ exists* exactly when $f$ is bijective. When it exists it is unique: each output has exactly one input to go back to, so the reverse rule is forced. It is written $f^{-1}$.
3
→
×3
→
÷3
→
3
0
→
°C→°F
→
°F→°C
→
0
4
→
square
→
root
→
4
three undo pairs
Three undo pairs: times three and divide by three, Celsius and Fahrenheit, square and root on the non-negatives.
This is the reason it matters now. Every “undo” in mathematics is an inverse function.
Multiplying by $3$ is undone by dividing by $3$.
Converting Celsius to Fahrenheit is undone by converting back.
Squaring is undone by the square root, but only on $[0, \infty)$. On all of $\mathbb{R}$, squaring is not one-one: $(-2)^2$ and $2^2$ give the same output.
The next chapter undoes $\sin$, $\cos$ and $\tan$. None of them is one-one until you cut the domain down.
One height, many angles: sine is not one-one until the domain is cut to one window. Chapter 2 starts here.
The fixed method: solve, then swap
Here is the fixed method for finding $f^{-1}$ when you have a formula. Write $y = f(x)$. Solve for $x$ in terms of $y$. Then swap the two names, $x$ and $y$. What you get is $f^{-1}(x)$.
Try it yourself, before the worked example below. Take $f(x) = x + 5$. Write $y = x + 5$. Solve for $x$. Swap the names. You should get $f^{-1}(x) = x - 5$.
f(x) = 3x − 2
→
write y
→
solve for x
→
swap
→
f⁻¹(x) = (x + 2)/3
the fixed method: write, solve, swap
Write y equals f of x, solve for x, swap the names. The worked example runs the same three steps.
Check yourself
A function $f:A \to B$ is invertible when \_\_\_\_\_\_.
some $g:B \to A$ undoes $f$ both ways, for every $a \in A$ and every $b \in B$
some function $g:B \to A$ exists with $f(g(b)) = b$ for every $b \in B$, with no further condition needed
some function $g:B \to A$ exists with $g(f(a)) = a$ for every $a \in A$, with no further condition needed
some function $g:B \to A$ exists with $g(f(a)) = a$ for some $a \in A$
Check your answer
✓ some $g:B \to A$ undoes $f$ both ways, for every $a \in A$ and every $b \in B$ — (A) Invertibility needs $g$ to undo $f$ in BOTH directions, for every element of both $A$ and $B$, either direction alone is not enough.
some function $g:B \to A$ exists with $f(g(b)) = b$ for every $b \in B$, with no further condition needed — This keeps only the “$f$ undoes $g$” direction. The definition needs both directions to hold, not just this one.
some function $g:B \to A$ exists with $g(f(a)) = a$ for every $a \in A$, with no further condition needed — This keeps only the “$g$ undoes $f$” direction. Invertibility also needs the reverse, $f(g(b)) = b$ for every $b \in B$.
some function $g:B \to A$ exists with $g(f(a)) = a$ for some $a \in A$ — “For some $a$” is far weaker than the actual requirement. $g$ must undo $f$ for every $a \in A$, with no exceptions.
Invertibility is possible if and only if $f$ is \_\_\_\_\_\_.
onto only
bijective
reflexive
one-one only
Check your answer
onto only — Onto alone is not enough either, again, both properties are required together, not just this one.
✓ bijective — (B) The section states the “if and only if” condition precisely, bijective, meaning both properties hold together, is exactly what invertibility requires.
reflexive — Reflexive is a property of relations. It has no bearing on whether a function is invertible.
one-one only — One-one alone is not enough, the section states invertibility needs BOTH one-one and onto together, i.e. bijective.
When $f^{-1}$ exists, the method is: write $y = f(x)$, solve for $x$ in terms of $y$, then swap the names. Why is “swap the names” the last step, not the first?
the swap step is optional cosmetic: the unswapped formula and $f^{-1}(x)$ are two different functions
because $y = f(x)$ can only be solved for $x$ after the swap
because $f^{-1}$ only exists after the names are swapped, not before
solving finds the rule; swapping just relabels it as $f^{-1}(x)$
Check your answer
the swap step is optional cosmetic: the unswapped formula and $f^{-1}(x)$ are two different functions — The unswapped formula (x in terms of y) and $f^{-1}(x)$ (after swapping) are the same rule, swapping just relabels which letter is the input, it does not create a different function.
because $y = f(x)$ can only be solved for $x$ after the swap — The method solves $y = f(x)$ for $x$ FIRST, then swaps the names, doing the swap first would leave nothing solved for yet.
because $f^{-1}$ only exists after the names are swapped, not before — Whether $f^{-1}$ exists depends only on whether $f$ is bijective, it exists (or does not) regardless of when or whether you relabel the letters.
✓ solving finds the rule; swapping just relabels it as $f^{-1}(x)$ — (D) The solving step finds the rule. The swap only relabels that rule so it reads as a function of $x$, the way every other function here is written.
Because invertibility requires bijectivity, which of the following facts guarantees a function cannot be invertible?
the function’s domain and codomain are both $\mathbb{R}$
the function was defined using a formula rather than a table
the function is one-one and onto
the function is one-one but not onto
Check your answer
the function’s domain and codomain are both $\mathbb{R}$ — Both domain and codomain being $\mathbb{R}$ says nothing about whether a specific function on them happens to be one-one or onto, it does not by itself decide invertibility either way.
the function was defined using a formula rather than a table — How a function is presented, formula or table, has no bearing on whether it is bijective, and therefore none on whether it is invertible.
the function is one-one and onto — One-one AND onto together is the bijective case, exactly the case where invertibility DOES hold, the opposite of what the question asks for.
✓ the function is one-one but not onto — (D) Missing onto, even while being one-one, already fails the “both together” requirement, so invertibility is ruled out.
Reflecting f’s graph across y = x swaps each point’s coordinates, turning (0, -2) on f into (-2, 0) on f inverse.
Worked example
Find the inverse of $f(x) = 3x - 2$
$f: \mathbb{R} \to \mathbb{R}$, $f(x) = 3x - 2$ Before finding an inverse, check that $f$ is bijective.
$f$ is strictly increasing A strictly increasing function never repeats an output. So $f$ is one-one.
Every real $y$ has some real $x$ with $3x-2=y$ Solve for $x$ and you always get a real answer. So $f$ is onto.
$f$ is bijective One-one and onto together. An inverse exists.
$y = 3x - 2$ Write the function as $y$ in terms of $x$.
$x = (y+2)/3$ Solve for $x$.
$f^{-1}(x) = (x+2)/3$ Swap the names $x$ and $y$. Now it reads as a function of $x$ again.
$f(f^{-1}(x)) = 3 \cdot (x+2)/3 - 2 = x + 2 - 2 = x$ Check it yourself. Put your inverse back into $f$. You should get back plain $x$, and you did.
Check yourself
Let $f:\mathbb{R} \to \mathbb{R}$, $f(x) = 2x + 5$, which is bijective by the same reasoning as $f(x) = 3x - 2$. Using solve-and-swap, what is $f^{-1}(x)$?
$2x - 5$
$(x+5)/2$
$(x-2)/5$
$(x-5)/2$
Check your answer
$2x - 5$ — “Flip the signs and swap positions” is not the method, the worked example specifically solves $y = f(x)$ for $x$ step by step, it does not use a sign-flipping shortcut.
$(x+5)/2$ — Solving $y = 2x + 5$ for $x$ gives $x = (y - 5)/2$, subtracting 5, not adding it, before dividing by 2.
$(x-2)/5$ — This puts the constant where the coefficient should be and vice versa, the actual solve gives $(x-5)/2$, dividing the WHOLE numerator by 2.
✓ $(x-5)/2$ — (D) $y = 2x + 5$ solved for $x$ gives $x = (y-5)/2$, swap the names for $f^{-1}(x) = (x-5)/2$.
To check that $f^{-1}(x) = (x+2)/3$ is really the inverse of $f(x) = 3x - 2$, the worked example computes $f(f^{-1}(x))$. What must it equal?
$x$: applying $f$ to $f^{-1}(x)$ must return exactly the original input
$3x - 2$, the original formula for $f$, unchanged
$0$, since a function composed with its own inverse always cancels to zero
$f^{-1}(x)$ again: checking should return the inverse unchanged
Check your answer
✓ $x$: applying $f$ to $f^{-1}(x)$ must return exactly the original input — (A) A genuine inverse undoes $f$ completely, applying $f$ to $f^{-1}(x)$ must come back to exactly $x$, confirming the two functions really cancel.
$3x - 2$, the original formula for $f$, unchanged — $3x - 2$ is $f$’s own formula, a different expression from the identity result $x$ that the check is designed to confirm.
$0$, since a function composed with its own inverse always cancels to zero — A function composed with its genuine inverse cancels down to $x$, the identity, not to $0$. Zero would only happen for one specific input, not for every $x$.
$f^{-1}(x)$ again: checking should return the inverse unchanged — The check applies $f$ to $f^{-1}(x)$, if it returned $f^{-1}(x)$ unchanged, that would mean $f$ did nothing at all, which is not what the check is verifying.
For $f(x) = 2x + 5$ and $f^{-1}(x) = (x-5)/2$, what does $f(f^{-1}(x))$ simplify to?
$2x$
$x$
$x - 5$
$x + 5$
Check your answer
$2x$ — Dropping the $+5$ term from $f(x) = 2x+5$ entirely leaves only $2 \cdot (x-5)/2 = x - 5$, not the correct check.
✓ $x$ — (B) $f(f^{-1}(x)) = 2 \cdot (x-5)/2 + 5 = (x - 5) + 5 = x$, the check confirms the inverse is correct.
$x - 5$ — The computation needs the $+5$ to cancel the $-5$ inside the formula, subtracting 5 without that cancellation is a different, incorrect path through the arithmetic.
$x + 5$ — $2 \cdot (x-5)/2 + 5 = (x-5) + 5 = x$, the $+5$ and $-5$ cancel completely; leaving $x + 5$ forgets to let them cancel.
✕MISCONCEPTION
Each positive y here has two preimages, so restricting the domain to non-negative numbers is what makes the rule invertible, not a new formula.
Every output is reached, so onto holds. Send 4 back and the rule must choose −2 or 2: not a function.
Onto guarantees that nothing in $B$ is left out. Every element of $B$ has something that reaches it. So it looks as if onto alone is enough to build a reverse function.
What onto does not guarantee
It is not enough. Onto only guarantees a preimage exists. It says nothing about whether that preimage is the ONLY one.
Here is the gap, made concrete. Take $f: \mathbb{R} \to [0, \infty)$, $f(x) = x^2$. Is $f$ onto? Yes: every non-negative number is reached by some real $x$. Is $f$ one-one? No. $f(-2) = f(2) = 4$. Two different inputs reach the same output, $4$.
Why the reverse rule fails
Let us try to build a reverse rule anyway. Send $4$ back to what? $-2$, or $2$? Both are valid preimages. A function allows only ONE output per input. So this “reverse” rule is not a function at all.
Invertibility needs both properties together, never onto alone. One-one guarantees the reverse rule never has to choose between two answers. Onto guarantees the reverse rule has an answer for every single input. Bijective means carrying both at once, and nothing weaker will do.
Cubing never repeats an output and reaches every real number, so the reverse works. Invertible.
Your turn: is $f(x) = x^3$ on $\mathbb{R} \to \mathbb{R}$ invertible?
A SCHOLAR INDIA REMEMBERS
Run solve and swap on $f(x) = x^2$, from $\mathbb{R}$ to $[0, \infty)$. You write $y = x^2$, then $x = \sqrt{y}$. Which step is wrong? Solving $y = x^2$ gives two answers for every $y > 0$, $\sqrt{y}$ and $-\sqrt{y}$. You kept one and dropped the other, and nothing in $f$ told you to. The method ran, yet there is no inverse. Test one-one before you solve.
Dignaga · the kingfisher
Check yourself
A student claims: “If every element of $B$ is somebody’s output (onto), $f$ must be invertible; nothing in $B$ is left out.” Is the student right?
Yes, but only for functions defined by a formula
Yes: reaching every element of $B$ is exactly what invertibility requires
No: $x^2$ from $\mathbb{R}$ to $[0, \infty)$ is onto but not one-one
No: onto functions are never invertible
Check your answer
Yes, but only for functions defined by a formula — Nothing about the counterexample depends on how the function is defined, the failure is about one-one-ness, not about formula versus any other presentation.
Yes: reaching every element of $B$ is exactly what invertibility requires — This is exactly the trap being tested, stated as though correct, $f(x) = x^2$ from $\mathbb{R}$ to $[0, \infty)$ is the counterexample that refutes it.
✓ No: $x^2$ from $\mathbb{R}$ to $[0, \infty)$ is onto but not one-one — (C) $f(x) = x^2$ reaches every non-negative output, yet two inputs share each one, a reverse rule has no way to choose between them, so onto alone fails to deliver invertibility.
No: onto functions are never invertible — The correct answer is not “never”, an onto function that is ALSO one-one (bijective) is perfectly invertible; the counterexample only shows onto ALONE is not enough.
Let $f:\mathbb{R} \to [0, \infty)$, $f(x) = |x|$. This function is onto (every non-negative number is reached). Using the same reasoning as $x^2$, is $f$ invertible?
Yes: onto is enough on its own, so $f$ must be invertible
No: $|x|$ is never onto, so this does not even apply
Yes: $|x|$ is one-one because absolute values are always positive
No: $f(-3) = f(3) = 3$, so $f$ is not one-one either
Check your answer
Yes: onto is enough on its own, so $f$ must be invertible — This reapplies the same “onto is enough” trap to a new function, $f(-3) = f(3) = 3$ shows $|x|$ fails one-one just as $x^2$ did.
No: $|x|$ is never onto, so this does not even apply — The question already establishes $f$ is onto, claiming otherwise contradicts the setup rather than answering it.
Yes: $|x|$ is one-one because absolute values are always positive — Every output of $|x|$ being non-negative says nothing about whether DIFFERENT inputs collide, non-negativity and one-one-ness are separate properties.
✓ No: $f(-3) = f(3) = 3$, so $f$ is not one-one either — (D) The same collision pattern as $x^2$, two different inputs, one shared output , blocks one-one-ness, so onto alone still cannot deliver invertibility here either.
Onto alone is not enough for an inverse. What does invertibility actually require?
neither property: invertibility depends only on the formula’s shape
one-one alone, with onto being irrelevant
onto alone, as the student originally claimed
both properties together: bijective, not just onto
Check your answer
neither property: invertibility depends only on the formula’s shape — Invertibility depends on whether the function is actually bijective, not on what its formula looks like, this invents an unrelated condition.
one-one alone, with onto being irrelevant — One-one alone is just as insufficient as onto alone, a one-one function that misses part of its codomain is not onto, and so not invertible either.
onto alone, as the student originally claimed — This is the very claim the $x^2$ counterexample refutes: onto alone is not enough.
✓ both properties together: bijective, not just onto — (D) Onto by itself is not enough. Invertibility needs bijectivity: one-one and onto holding together.
The whole chapter on one page: five ideas, each with its rule, a picture and a worked number, and the two traps students fall into.
EQUIVALENCE RELATION: reflexive, symmetric and transitive together. $R=\{(1,1),(2,2),(1,2),(2,1)\}$ on $A=\{1,2,3\}$ is symmetric and transitive, but not reflexive: $3$ never appears.
EQUIVALENCE CLASS: $[a] = \{x \in A : (x,a) \in R\}$. For “same remainder on dividing by $5$” on $ZZ$, $[0]$ is every multiple of $5$.
ONE-ONE AND ONTO: $f(x)=x^2$ on $\mathbb{R} \to \mathbb{R}$ is neither: $f(-2)=f(2)=4$ breaks one-one, and no $x$ gives $-1$.
BIJECTIVE AND INVERTIBLE: a function is invertible exactly when it is bijective. $f(x)=3x-2$ has inverse $f^{-1}(x) = (x+2)/3$.
Two ideas, running through all four
You have used every one of these rules already, in the pages above. Two ideas run through all of them: SAMENESS and REVERSIBILITY. An equivalence relation turns “related to” into a rigorous kind of “the same as”. It splits a set into classes that never overlap. A bijective function turns “sends to” into a rigorous kind of “reversible”. It guarantees one unique way back. The one skill every later section drills is this: check which properties a relation or function actually has. Never assume one from the shape of a definition alone.
Check yourself
Two ideas run through relations and functions: sameness and reversibility. Which pairing is correct?
sameness pairs with equivalence relations; reversibility with bijective functions
both ideas describe reflexive relations only, not equivalence relations
sameness pairs with bijective functions; reversibility pairs with equivalence relations
sameness and reversibility are two names for the same one idea, applied twice
Check your answer
✓ sameness pairs with equivalence relations; reversibility with bijective functions — (A) Equivalence relations turn “related to” into exact sameness. Bijective functions turn “maps to” into exact reversibility. Each idea belongs to one half of the topic.
both ideas describe reflexive relations only, not equivalence relations — The two ideas belong to equivalence relations and bijective functions, not to reflexive relations alone. Reflexivity is only one of the three properties equivalence needs, and it says nothing about reversibility.
sameness pairs with bijective functions; reversibility pairs with equivalence relations — Sameness goes with equivalence relations, which group elements as “the same as” each other. Reversibility goes with bijective functions, which guarantee one way back. This option swaps them.
sameness and reversibility are two names for the same one idea, applied twice — Sameness and reversibility are two different ideas. Treating them as one name for the same idea loses what each one does.
Every later section on relations and functions drills one skill. What is it?
memorizing the formula for finding an inverse
checking which properties actually hold, never assuming them
proving every relation given in the exercises is an equivalence relation
drawing a diagram for every relation before testing it
Check your answer
memorizing the formula for finding an inverse — Memorizing the inverse formula is one narrow technique. The skill is the broader habit of checking properties directly, for relations, functions and operations alike.
✓ checking which properties actually hold, never assuming them — (B) The skill is a habit, not a formula: check each property directly against its definition, rather than assuming it from how a relation or function looks.
proving every relation given in the exercises is an equivalence relation — Exercise 1.1 explicitly includes relations that FAIL to be equivalence relations, naming which property fails, the skill is checking honestly, not forcing every relation into being one.
drawing a diagram for every relation before testing it — Drawing a diagram can help, but it is not the skill. The skill is checking properties against their definitions.
Which single fact ties together equivalence relations and invertible functions?
both require the set $A$ to be finite
both turn an informal notion into a rigorous one, once confirmed
both concepts are new in Class 12, with no connection to Class 11
both are defined only for numbers, never for shapes like triangles
Check your answer
both require the set $A$ to be finite — Neither equivalence relations nor invertible functions require $A$ to be finite, the congruence example works on the infinite set of all triangles in a plane.
✓ both turn an informal notion into a rigorous one, once confirmed — (B) Neither idea is automatic, “related to” only becomes rigorous sameness, and “maps to” only becomes guaranteed reversibility, once the right properties are actually checked and confirmed.
both concepts are new in Class 12, with no connection to Class 11 — Functions are already familiar from Class 11. This option’s claim of no prior connection contradicts that.
both are defined only for numbers, never for shapes like triangles — The congruence-of-triangles worked example is exactly a case where the set is NOT numbers, equivalence relations are defined on any set, triangles included.
Check yourself: the whole chapter
A relation $R$ on the set $A = \{1, 2, 3\}$ is written as $R = \{(1, 2), (3, 1)\}$. Is this a valid relation on $A$?
No: a relation must relate every element of $A$ to something.
Yes: any set of ordered pairs from $A \times A$ is a relation on $A$.
No: a relation must be given by a formula or rule, not just a list of pairs.
Yes, but only if it is also reflexive, symmetric or transitive.
Check your answer
No: a relation must relate every element of $A$ to something. — Nothing in the definition asks every element to appear. $R$ leaves 2 unrelated to anything, and it is still a relation.
✓ Yes: any set of ordered pairs from $A \times A$ is a relation on $A$. — (B) A relation on $A$ is any subset of $A \times A$, any collection of ordered pairs. $R$ qualifies, with no further conditions.
No: a relation must be given by a formula or rule, not just a list of pairs. — A rule is one convenient way to describe a relation, not a requirement. A bare list of pairs is equally a relation.
Yes, but only if it is also reflexive, symmetric or transitive. — Reflexive, symmetric and transitive are patterns a relation MIGHT have. A relation does not need any of them to exist.
On the non-empty set $A = \{1, 2\}$, take the empty relation $R = \emptyset$: no pair is related to any other. Which properties does $R$ have?
None of the three: an empty relation cannot have any property.
All three: there is no pair that breaks any condition.
Symmetric and transitive, but not reflexive.
Reflexive only: every element trivially relates to itself.
Check your answer
None of the three: an empty relation cannot have any property. — Symmetric and transitive are conditions of the form “if a pair is present, then…”. With no pairs present, neither can be broken, so both hold.
All three: there is no pair that breaks any condition. — Reflexivity is different in kind: it DEMANDS that $(1, 1)$ and $(2, 2)$ be present. An empty relation supplies neither, so it fails.
✓ Symmetric and transitive, but not reflexive. — (C) Symmetric and transitive hold vacuously, no pair exists to violate them. Reflexivity fails, because it demands $(1, 1)$ and $(2, 2)$ actually be in $R$.
Reflexive only: every element trivially relates to itself. — Nothing relates to itself automatically. Reflexivity is the condition that requires those pairs, and here they are absent.
An equivalence relation on a set $A$ splits it into equivalence classes. Two classes $[a]$ and $[b]$ share at least one element. What follows?
They are the same class: $[a] = [b]$.
They overlap partially, sharing some elements but not others.
One is a smaller part of the other, with members left over.
Nothing follows without knowing the relation.
Check your answer
✓ They are the same class: $[a] = [b]$. — (A) One shared element is enough to force $[a] = [b]$. Equivalence classes are identical or disjoint, which is why they partition $A$ cleanly.
They overlap partially, sharing some elements but not others. — Partial overlap is exactly what an equivalence relation rules out. Classes are identical or completely disjoint, never in between.
One is a smaller part of the other, with members left over. — Containment is not an option either. Sharing one element forces the two classes to coincide entirely.
Nothing follows without knowing the relation. — It follows for every equivalence relation, no matter which one, that is what makes the classes a partition.
A function $f: A \to B$ is known to be one-one (injective). What does that tell you about whether it is onto (surjective)?
It must also be onto: one-one and onto always come together.
It cannot be onto: the two properties exclude each other.
It is onto only if $A$ and $B$ are infinite.
Nothing: the two properties are independent.
Check your answer
It must also be onto: one-one and onto always come together. — They are separate conditions. $f: \mathbb{N} \to \mathbb{N}$ with $f(x) = 2x$ is one-one but misses every odd number, so it is not onto.
It cannot be onto: the two properties exclude each other. — They are not opposites either, a bijection is precisely a function that is both at once.
It is onto only if $A$ and $B$ are infinite. — Size is not the deciding factor. Being one-one simply says nothing either way about being onto.
✓ Nothing: the two properties are independent. — (D) The two are independent: a function can be one-one without being onto, onto without being one-one, both, or neither.
For $f(x) = x + 1$ and $g(x) = x^2$, both from $\mathbb{R}$ to $\mathbb{R}$, compare $g(f(x))$ and $f(g(x))$ at $x = 2$.
They agree: composition gives the same answer either way.
They differ: $g(f(2)) = 9$ while $f(g(2)) = 5$.
They differ: $g(f(2)) = 5$ while $f(g(2)) = 9$.
Only $g(f(2))$ is defined; $f(g(2))$ is not.
Check your answer
They agree: composition gives the same answer either way. — Order changes the result. Squaring after adding is not the same as adding after squaring.
✓ They differ: $g(f(2)) = 9$ while $f(g(2)) = 5$. — (B) $g(f(2))$ applies $f$ first: $f(2) = 3$, then $g(3) = 9$. $f(g(2))$ applies $g$ first: $g(2) = 4$, then $f(4) = 5$. Order matters.
They differ: $g(f(2)) = 5$ while $f(g(2)) = 9$. — The order is the other way round: $g(f(x))$ applies $f$ FIRST, so $f(2) = 3$ and then $g(3) = 9$.
Only $g(f(2))$ is defined; $f(g(2))$ is not. — Both are defined here, $f$ and $g$ both map $\mathbb{R}$ to $\mathbb{R}$. The point is that they give different answers.
You have probably used most of these ideas already, just without their names.
Grouping matching records. A spreadsheet groups two customer rows together whenever they share one phone number. The maths: each group is one equivalence class. Classes never overlap. $120$ customers sharing $40$ distinct numbers split into exactly $40$ classes.
A seating chart is a function. A wedding planner gives every guest exactly one seat. A seat can stay empty. The maths: no guest ever gets two seats. That single rule is exactly what makes it a function. $150$ guests mapped to $150$ distinct seats is a one-one function.
A boarding pass can be reversed. Read an airline’s seat map backwards, and a seat number always names one passenger. The maths: a function you can read backwards with no ambiguity, with every seat taken, is invertible. Seat $1$ gives passenger A. Passenger A gives back seat $1$.
Composing two price rules. A shop first converts dollars to rupees, then adds GST. Doing both steps, in order, is one combined rule. The maths: feeding one function’s output into a second function is composition. $100$ dollars at $83$ rupees each, then $18\%$ GST, gives $9794$ rupees.
Grouping teams by points. A league table groups every team with the same points into one block. The maths: sharing the same points is an equivalence relation. Each block is one equivalence class. $4$ teams tied on $9$ points sit together in one block.
One balance per account. A bank’s app always gives one account number exactly one balance, never two. The maths: that one-balance rule is exactly what makes it a function. Account $401$ maps to the single balance $12000$, never a second figure beside it.
Your turn. A courier company gives every parcel ID exactly one delivery address. Could two different parcels share the same address?
seat 1
→
passenger of
→
seat of
→
seat 1
a boarding pass runs both ways
Seat 1 gives passenger A, and A gives back seat 1. That loop back to seat 1 is what makes the boarding pass invertible.
LATER
Where equivalence relations go next
You will meet equivalence relations again, in university Discrete Mathematics courses. Splitting a set into equivalence classes is exactly how a “quotient structure” gets defined there.
You have already seen one example of this idea. Modular arithmetic sorts integers into classes by their remainder: the same partition idea, given a new name.
practice Let $R = \{(a,b): a,b \in ZZ, a-b \text{is even}\}$ on the set of integers. Check whether $R$ is reflexive, symmetric, and transitive.
practice On the set $A = \{1,2,3,4\}$, let $R = \{(1,2),(2,1),(1,1),(2,2)\}$. Is $R$ transitive? Justify.
practice Which of these relations on $\{1,2,3\}$ is reflexive but NOT symmetric?
$\{(1,1),(2,2),(3,3),(1,2)\}$
$\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$\{(1,2),(2,1)\}$
$\{(1,1),(2,2)\}$
practice Let $R$ be the relation on the set of all triangles in a plane defined by $R = \{(T_1,T_2): T_1 \text{is similar to} T_2\}$. Is $R$ an equivalence relation?
practice On $ZZ$, define $R = \{(a,b): a \leq b\}$. Which property fails, and why?
Answers
Equivalence relation: all three hold.
Yes, transitive.
(A) $\{(1,1),(2,2),(3,3),(1,2)\}$.
Yes: equivalence relation.
Symmetry fails.
Exercise 1.1 gives you a relation on a set. Test it for reflexive, symmetric and transitive, one at a time. Then say whether it is an equivalence relation. If it is not, name the properties that fail.
Exercise 1.1: further practice, relations
practice Let $R$ be the relation on $ZZ$ defined by $R = \{(a,b): a b > 0\}$. Check whether $R$ is reflexive, symmetric, and transitive.
practice On $A = \{1,2,3,4\}$, let $R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,3)\}$. Is $R$ transitive? Justify.
practice Which of these relations on $\{1,2,3\}$ is symmetric but NOT reflexive?
$\{(1,1),(2,2),(3,3)\}$
$\{(1,2),(2,1)\}$
$\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$\{(1,2),(2,3)\}$
practice Let $R$ be the relation on the set of all people defined by $R = \{(x,y): x \text{has the same age as} y\}$. Is $R$ an equivalence relation?
practice On $ZZ$, define $R = \{(a,b): a + b \text{is odd}\}$. Which properties fail, and why?
practice How many relations are there on $A = \{1, 2, 3\}$?
practice How many reflexive relations on $A = \{1, 2, 3\}$ contain the pair $(1, 2)$?
practice Let $A = \{1, 2, 3\}$. How many equivalence relations on $A$ contain the pair $(1, 2)$?
$1$
$2$
$3$
$4$
practice Let $L$ be the set of all lines in a plane, and let $R$ relate $L_1$ to $L_2$ when $L_1$ is perpendicular to $L_2$. Show that $R$ is symmetric but neither reflexive nor transitive.
practice Let $L$ be the set of all lines in a plane, and let $R$ relate $L_1$ to $L_2$ when $L_1$ is parallel to $L_2$ (a line counts as parallel to itself). Show that $R$ is an equivalence relation, and find all the lines related to $y = 2x + 4$.
Answers
Not reflexive ($0 \cdot 0 = 0$ is not $> 0$); symmetric and transitive.
No: $(1,2)$ and $(2,3)$ force $(1,3)$, which is missing.
(B) $\{(1,2),(2,1)\}$.
Yes: equivalence relation.
Reflexive and transitive both fail; only symmetric holds. $a + a = 2 a$ is always even, and $a$, $c$ each share $b$’s opposite parity, so $a + c$ is even too.
$2^9 = 512$.
$2^5 = 32$.
(B) $2$.
Symmetric only. No line is perpendicular to itself, and the $x$-axis, the $y$-axis and the line $y = 1$ break transitivity.
An equivalence relation. The lines related to $y = 2x + 4$ are the lines $y = 2x + c$, $c \in \mathbb{R}$.
practice $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$. Is $f$ one-one? Is it onto?
practice $f: [0, \infty) \to [0, \infty)$, $f(x) = x^2$. Is $f$ one-one and onto?
practice $f: \mathbb{R} \to \mathbb{R}$, $f(x) = 2x+1$. Is $f$ one-one? Is it onto?
practice $f: ZZ \to ZZ$, $f(x) = x^2$. Which of these is true?
one-one and onto
one-one but not onto
onto but not one-one
neither one-one nor onto
Answers
Neither.
Both: bijective.
Both: bijective.
(D) neither one-one nor onto.
Exercise 1.2 gives you a function. Test whether it is one-one, onto, both, or neither. Several problems reuse the same formula under a different domain and codomain. Watch the same rule answer differently each time.
Exercise 1.2: further practice, functions
practice $f: \mathbb{R} \to \mathbb{R}$, $f(x) = 5 - 2 x$. Is $f$ one-one? Is it onto?
practice $f: \mathbb{N} \to \mathbb{N}$, $f(x) = x + 2$. Is $f$ one-one? Is it onto?
practice Let $f: \{1,2,3,4\} \to \{1,2,3,4\}$ be a bijection. How many such functions are there?
practice $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^5$. Which of these is true?
one-one and onto
one-one but not onto
onto but not one-one
neither one-one nor onto
practice In a tall (dominant) and dwarf (recessive) pea-plant cross (Biology 12, Chapter 4), the second generation has three genotypes, TT, Tt and tt. TT and Tt both produce a tall plant; tt produces a dwarf plant. Let $f$ send each genotype to the phenotype it produces. Is $f$ one-one? Is it onto the set of phenotypes?
practice Now let $g$ send each genotype to itself, $g(\text{TT}) = \text{TT}$, $g(\text{Tt}) = \text{Tt}$, $g(\text{tt}) = \text{tt}$, instead of to its phenotype. Is $g$ one-one? Is it onto?
practice The signum function $f: \mathbb{R} \to \mathbb{R}$ is $f(x) = 1$ for $x > 0$, $f(0) = 0$ and $f(x) = -1$ for $x < 0$. Is $f$ one-one? Is it onto?
practice The greatest integer function $f: \mathbb{R} \to \mathbb{R}$ is $f(x) = [x]$, the greatest integer less than or equal to $x$. Is $f$ one-one? Is it onto?
practice Let $f: \mathbb{N} \to \mathbb{N}$ be $f(n) = (n+1)/2$ when $n$ is odd and $f(n) = n/2$ when $n$ is even. Is $f$ one-one? Is it onto?
practice Let $A = \{1, 2, 3\}$ and $B = \{a, b\}$. How many functions from $A$ to $B$ are onto? Is any of them one-one?
practice How many one-one functions are there from $A = \{1, 2\}$ to $B = \{a, b, c\}$? Is any of them onto?
Answers
Both: bijective.
One-one, not onto.
$24$
(A) one-one and onto.
$f$ is not one-one (TT and Tt both give tall) but is onto (tall and dwarf are both reached).
$g$ is one-one and onto: bijective.
Neither. $f(1) = f(2) = 1$, and $2$ is never an output.
Neither. $f(1.2) = f(1.9) = 1$, and $0.7$ is never an output.
Onto, but not one-one. $f(1) = f(2) = 1$, and $f(2m) = m$ for every $m$.
practice Show that $R = \{(a,b): a,b \in ZZ, \left|a-b\right| \leq 1\}$ on $ZZ$ is reflexive and symmetric but not transitive.
practice Let $f: \mathbb{R} \to \mathbb{R}$, $f(x) = \left|x\right|$. Is $f$ one-one? Is it onto?
practice Let $f: \mathbb{R} \to \mathbb{R}$, $f(x) = 2x+3$. Find $f^{-1}(x)$.
practice Which statement about $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$, is true?
one-one and onto
one-one but not onto
onto but not one-one
neither one-one nor onto
practice On $A = \{1, 2, 3, 4, 5, 6\}$, the relation $R = \{(a, b) : \left|a - b\right| \text{is even}\}$ is an equivalence relation. Write down the equivalence classes $[1]$ and $[2]$.
practice Let $g: \mathbb{R} \to \mathbb{R}$, $g(x) = 3x$. Find $g^{-1}(x)$ and verify $g(g^{-1}(x)) = x$.
practice “Is a friend of” on the students in a class is symmetric (if $A$ is a friend of $B$, $B$ is a friend of $A$) but nobody counts as their own friend. Is this relation an equivalence relation? Why or why not?
Answers
Transitivity fails.
Neither.
$f^{-1}(x) = (x-3)/2$
(A) one-one and onto.
$[1] = \{1, 3, 5\}$; $[2] = \{2, 4, 6\}$.
$g^{-1}(x) = x/3$
No: reflexivity fails.
The miscellaneous exercise mixes every section of this chapter: relations, functions, composition and inverses, all in one set. Most problems do not tell you which technique to use. Working that out is part of the exercise.
Miscellaneous: further practice
practice Let $f: \mathbb{R} \to \mathbb{R}$, $f(x) = 4 x + 9$. Find $f^{-1}(x)$ and verify $f(f^{-1}(x)) = x$.
practice On $\mathbb{N}$, define $R = \{(a,b): a \text{divides} b\}$. Is $R$ reflexive? Is it symmetric? Is it transitive?
practice If $f(x) = 2x+3$ and $g(x) = x^2-1$, find the composition $(f \circ g)(x)$ and $(g \circ f)(x)$.
Answers
$f^{-1}(x) = (x-9)/4$
Reflexive and transitive; not symmetric: $2$ divides $4$ but $4$ does not divide $2$.
practice The function $f$ from $ZZ$ to $ZZ$ defined by $f(x) = x+2$ is ______. (1 mark)
one-one but not onto
onto but not one-one
one-one and onto
neither one-one nor onto
practice Assertion (A). The relation $R = \{(1,1), (2,2), (1,2), (2,1)\}$ on $A = \{1, 2, 3\}$ is an equivalence relation. Reason (R). $R$ is symmetric and transitive. (1 mark)
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
practice Give an example of a relation on $A = \{1, 2, 3\}$ that is reflexive and symmetric but not transitive. Justify your answer. (2 marks)
practice Show that the function $f$ from $\mathbb{R}$ to $\mathbb{R}$ defined by $f(x) = 3x-4$ is a bijective function. (3 marks)
practice Let $f$ be the function from $\mathbb{N}$ to $\mathbb{N}$ given by $f(x) = x^2+x+1$. Show that $f$ is one-one but not onto. (5 marks)
practice A school allots every Class 12 student a unique four-digit roll number for the year. Let $A$ be the set of $40$ students in Section A, and let $B$ be the set of the $40$ roll numbers allotted to Section A. Define $f$, from $A$ to $B$, by sending each student to the roll number allotted to them. (i) Is $f$ a function from $A$ to $B$? Give a reason. (1 mark) (ii) Is $f$ one-one? Give a reason. (1 mark) (iii) Using $|A| = |B| = 40$, decide whether $f$ is onto, and justify. (2 marks)
practice Let $R$ be the relation on $ZZ$ defined by $R = \{(a, b): 5 \text{divides} (a-b)\}$. Show that $R$ is an equivalence relation, and write the equivalence class of $0$. (3 marks)
Answers
(C) one-one and onto.
(D) A is false but R is true.
$R = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}$ is reflexive and symmetric; $(1,2)$ and $(2,3)$ are in $R$ but $(1,3)$ is not, so $R$ is not transitive.
$f$ is bijective: one-one (equal outputs force equal inputs) and onto (every real $y$ has preimage $(y+4)/3$).
$f$ is one-one (the factoring forces $x_1=x_2$, since $x_1+x_2+1$ is never $0$ on $\mathbb{N}$) but not onto ($1$ and $2$ in the codomain have no preimage).
(i) Yes, a function: each student gets exactly one roll number. (ii) Yes, one-one: no roll number repeats. (iii) Yes, onto: a one-one function between finite sets of the same size is automatically onto.
$R$ is an equivalence relation (reflexive, symmetric, transitive, all shown from $a-b$ being a multiple of $5$). The equivalence class of $0$ is the set of all multiples of $5$.
