Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
A relation R from a set A to a set B is any subset of A × B. If (a, b) ∈ R we say a is related to b and write a R b. A relation in a set A is a subset of A × A.
The connection need not be a rule you can put into words: any collection of pairs is a relation.
| Property | Meaning | To prove it | To disprove it |
|---|---|---|---|
| Reflexive | (a, a) ∈ R for every a ∈ A | Take any a and show a R a | Find one a with (a, a) ∉ R |
| Symmetric | (a, b) ∈ R ⇒ (b, a) ∈ R | Assume a R b and show b R a | Find one pair with (a, b) ∈ R, (b, a) ∉ R |
| Transitive | (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R | Assume a R b, b R c and show a R c | Find a, b, c with (a, b), (b, c) ∈ R, (a, c) ∉ R |
A relation that is reflexive, symmetric and transitive is an equivalence relation.
List the pairs of the form (a, b) and (b, c). If there is no such pair, the relation is transitive vacuously. For example R = {(1, 2)} is transitive.
For an equivalence relation R on X, the equivalence class of a ∈ X is [a] = {b ∈ X : b R a}, the set of all elements related to a.
In Z with 'a − b is even': [0] = E, the set of all even integers, and [1] = O, the set of all odd integers.
An equivalence relation cuts X into mutually disjoint subsets A₁, A₂, … whose union is X, where all elements of one class are related and elements of different classes are not. And conversely, every such partition defines an equivalence relation.
f : X → Y is one-one if distinct elements of X have distinct images:
Otherwise f is many-one.
f : X → Y is onto if every element of Y is the image of some element of X: for every y ∈ Y there is an x ∈ X with f(x) = y.
f is onto if and only if the range of f = Y.
f is bijective (one-one and onto) if it is both. A bijective function pairs every element of X with exactly one element of Y.
| Function | One-one? | Onto? | Why |
|---|---|---|---|
| f : R → R, f(x) = 2x | yes | yes | x = y/2 |
| f : N → N, f(x) = 2x | yes | no | 1 is not 2x for any x ∈ N |
| f : R → R, f(x) = x² | no | no | f(−1) = f(1); −2 is not a square |
| f : R → R, f(x) = |x| | no | no | f(−1) = f(1); negatives are missed |
| f : R → R, f(x) = [x] | no | no | f(1.2) = f(1.5); 1/2 is missed |
| f : R → R, signum | no | no | only the values −1, 0, 1 |
| f : R• → R•, f(x) = 1/x | yes | yes | x = 1/y (R• = non-zero reals) |
A horizontal line y = c meets the graph of a one-one function at most once, and meets the graph of an onto function at least once for every c in the co-domain.
For a finite set X, a function f : X → X is one-one if and only if it is onto. So an onto f : {1, 2, 3} → {1, 2, 3} is always one-one, and a one-one one is always onto.
This fails for infinite sets: f(x) = 2x on N is one-one but not onto, and f(1) = f(2) = 1, f(x) = x − 1 for x > 2 on N is onto but not one-one. That is a characteristic difference between finite and infinite sets.
For f : A → B and g : B → C, the composition gof : A → C is
f : {2, 3, 4, 5} → {3, 4, 5, 9}, f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g : {3, 4, 5, 9} → {7, 11, 15}, g(3) = g(4) = 7, g(5) = g(9) = 11. Then
In general gof ≠ fog. For f(x) = cos x and g(x) = 3x²:
At x = 0 the first is 3 and the second is 1, so gof ≠ fog. Also, fog may not even be defined when gof is.
f : X → Y is invertible if there is g : Y → X with gof = I_X and fog = I_Y, where I is the identity function. Then g is the inverse of f, written f⁻¹.
f is invertible if and only if f is one-one and onto. So to prove a function is invertible, prove it is one-one and onto; you need not find the inverse.
Example: f(x) = 4x + 3 on N, with Y = {y : y = 4x + 3 for some x ∈ N}. Then x = (y − 3)/4, so g(y) = (y − 3)/4. Then gof(x) = (4x + 3 − 3)/4 = x and fog(y) = 4(y − 3)/4 + 3 = y.
| Property | Condition |
|---|---|
| Empty relation | R = ∅ |
| Universal relation | R = A × A |
| Reflexive | (a, a) ∈ R for all a |
| Symmetric | (a, b) ∈ R ⇒ (b, a) ∈ R |
| Transitive | (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R |
| One-one | f(x₁) = f(x₂) ⇒ x₁ = x₂ |
| Onto | range = co-domain |
| Invertible | one-one and onto |
| Composition | (gof)(x) = g(f(x)) |
Try each problem first, then open the solution to check your working.
In A = {1, 2, 3, 4}, let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. Is R reflexive, symmetric, transitive?
Given: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Use: Test each property on the listed pairs
Reflexive? (4, 4) ∉ R, so R is not reflexive.
Symmetric? (1, 2) and (2, 1) are both in R, and the pairs (a, a) are their own mirror images. So R is symmetric.
Transitive? The only pairs of the form (a, b), (b, c) with a ≠ c are (1, 2), (2, 1), which gives (1, 1) ∈ R, and (2, 1), (1, 2), which gives (2, 2) ∈ R. All other combinations give pairs already in R. So R is transitive.
Answer: symmetric and transitive, but not reflexive
Show that R = {(a, b) : 5 divides a − b} is an equivalence relation on Z. How many equivalence classes are there, and what is [2]?
Use: a R b if a − b = 5k for some integer k
Reflexive: a − a = 0 = 5 × 0, so a R a.
Symmetric: if a − b = 5k then b − a = 5(−k), so b R a.
Transitive: if a − b = 5k and b − c = 5m then a − c = 5(k + m), so a R c.
Every integer leaves remainder 0, 1, 2, 3 or 4 on division by 5, and two integers are related exactly when their remainders are equal. So there are 5 classes: [0], [1], [2], [3], [4].
[2] = {…, −8, −3, 2, 7, 12, …}, all integers of the form 5k + 2.
Answer: 5 classes; [2] = {5k + 2 : k ∈ Z}
On R, a R b means a − b is irrational. Is R reflexive, symmetric, transitive?
Use: Try the numbers √2 and 0
Reflexive? a − a = 0 is rational, so (a, a) ∉ R. Not reflexive.
Symmetric? If a − b is irrational, so is b − a = −(a − b). Symmetric.
Transitive? (√2, 0) ∈ R since √2 − 0 is irrational, and (0, √2) ∈ R since −√2 is irrational. But (√2, √2) ∉ R. Not transitive.
Answer: symmetric only
On N × N, define (a, b) R (c, d) if a + d = b + c. Show that R is an equivalence relation. List the pairs (c, d) with 1 ≤ c, d ≤ 5 related to (1, 3).
Use: a + d = b + c
Reflexive: a + b = b + a, so (a, b) R (a, b).
Symmetric: if a + d = b + c, then c + b = d + a, so (c, d) R (a, b).
Transitive: if a + d = b + c and c + f = d + e, adding gives a + d + c + f = b + c + d + e, so a + f = b + e, i.e. (a, b) R (e, f).
For (1, 3): 1 + d = 3 + c, so d = c + 2. With c, d from 1 to 5 the pairs are (1, 3), (2, 4), (3, 5).
Answer: (1, 3), (2, 4), (3, 5)
How many relations in {1, 2, 3} are reflexive? How many of these are also symmetric?
Given: A × A has 9 pairs
Use: A relation is a subset of A × A
Reflexive: the three pairs (1, 1), (2, 2), (3, 3) must be present. The other 6 pairs can each be in or out: 2⁶ = 64.
Reflexive and symmetric: the other 6 pairs come in 3 mirror pairs {(1, 2), (2, 1)}, {(1, 3), (3, 1)}, {(2, 3), (3, 2)}, and each mirror pair is in or out together: 2³ = 8.
Answer: 64; 8
On R, a R b means |a| = |b|. Show that R is an equivalence relation and find [5] and [0].
Reflexive: |a| = |a|. Symmetric: |a| = |b| gives |b| = |a|. Transitive: |a| = |b| and |b| = |c| give |a| = |c|. So R is an equivalence relation.
[5] = {b : |b| = 5} = {5, −5}. [0] = {b : |b| = 0} = {0}.
Answer: [5] = {5, −5}, [0] = {0}
f : {1, 2, 3} → {p, q, r, s} is given by f(1) = p, f(2) = q, f(3) = s. Is f one-one? Is it onto?
The images p, q, s are all different, so f is one-one.
r is not the image of any element, so f is not onto (range = {p, q, s} ≠ Y).
Answer: one-one, not onto
Show that f : R → R given by f(x) = 5x − 7 is one-one and onto.
One-one: f(x₁) = f(x₂) ⇒ 5x₁ − 7 = 5x₂ − 7 ⇒ x₁ = x₂.
Onto: for any real y, x = (y + 7)/5 is real and f(x) = 5 × (y + 7)/5 − 7 = y.
Answer: bijective
Show that f : R → R given by f(x) = x² + 4x + 5 is neither one-one nor onto.
Use: Complete the square: f(x) = (x + 2)² + 1
f(−1) = 1 − 4 + 5 = 2 and f(−3) = 9 − 12 + 5 = 2, so f is not one-one.
f(x) = (x + 2)² + 1 ≥ 1 for every x, so 0 ∈ R is not an image and f is not onto.
Answer: neither one-one nor onto
Let f : Z → Z be given by f(n) = 2n + 1. Is f one-one? Is it onto?
f(n₁) = f(n₂) ⇒ 2n₁ + 1 = 2n₂ + 1 ⇒ n₁ = n₂, so f is one-one.
f(n) is always odd, so 0 is not an image (2n + 1 = 0 gives n = −1/2 ∉ Z). Not onto.
Answer: one-one, not onto
Show that f : R − {−1} → R − {1}, f(x) = x/(x + 1) is one-one and onto.
One-one: x₁/(x₁ + 1) = x₂/(x₂ + 1) ⇒ x₁x₂ + x₁ = x₁x₂ + x₂ ⇒ x₁ = x₂.
Onto: let y ≠ 1 and solve y = x/(x + 1): y(x + 1) = x, so x(1 − y) = y and x = y/(1 − y).
This x is defined since y ≠ 1, and it is never −1: y/(1 − y) = −1 would give y = y − 1, which is impossible. So x ∈ R − {−1} and f(x) = y.
Answer: bijective
f : {1, 2, 3, 4} → {1, 2, 3, 4} is given by f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 1. Show that f is both one-one and onto without checking onto directly. How many one-one functions are there from {1, 2, 3, 4} to itself?
Use: On a finite set, one-one ⇔ onto
The images 2, 3, 4, 1 are all different, so f is one-one. A one-one function from a finite set to itself is onto, since four different images use up all four elements.
One-one maps of a 4-element set to itself are the permutations of 4 symbols: 4! = 24.
Answer: bijective; 24 one-one functions
f : R → R, f(x) = x² is not bijective. Choose a domain and co-domain that make it bijective, and prove it.
Take f : [0, ∞) → [0, ∞), f(x) = x².
One-one: for x₁, x₂ ≥ 0, x₁² = x₂² gives (x₁ − x₂)(x₁ + x₂) = 0. If x₁ + x₂ = 0, both are 0 (they are non-negative), so x₁ = x₂; otherwise x₁ − x₂ = 0.
Onto: for any y ≥ 0, x = √y ≥ 0 satisfies x² = y.
Answer: f : [0, ∞) → [0, ∞), f(x) = x² is bijective
If f(x) = 2x + 1 and g(x) = x² on R, find gof and fog and show that gof ≠ fog.
Use: (gof)(x) = g(f(x))
gof(x) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1.
fog(x) = f(x²) = 2x² + 1.
At x = 1: gof(1) = 9 but fog(1) = 3, so gof ≠ fog.
Answer: gof(x) = 4x² + 4x + 1, fog(x) = 2x² + 1
f : {1, 2, 3} → {4, 5, 6} has f(1) = 5, f(2) = 4, f(3) = 6, and g : {4, 5, 6} → {7, 8, 9} has g(4) = 8, g(5) = 9, g(6) = 7. Find gof. Is it one-one and onto?
gof(1) = g(5) = 9, gof(2) = g(4) = 8, gof(3) = g(6) = 7.
The images 9, 8, 7 are different and cover {7, 8, 9}, so gof is one-one and onto.
Answer: gof = {(1, 9), (2, 8), (3, 7)}, bijective
Find the inverse of f : R → R, f(x) = (2x + 5)/3, and verify it.
Use: y = f(x), solve for x
f is one-one and onto, so it is invertible. From y = (2x + 5)/3 we get 3y = 2x + 5, so x = (3y − 5)/2.
f⁻¹(y) = (3y − 5)/2.
Check: f⁻¹(f(x)) = (2x + 5 − 5)/2 = x and f(f⁻¹(y)) = (3y − 5 + 5)/3 = y.
Answer: f⁻¹(y) = (3y − 5)/2
Let f : N → Y be f(x) = 5x − 2, where Y is the range of f. Show that f is invertible and find f⁻¹.
Y = {3, 8, 13, 18, …}. f is onto Y by definition of Y, and one-one since 5x₁ − 2 = 5x₂ − 2 gives x₁ = x₂. So f is invertible.
From y = 5x − 2, x = (y + 2)/5, so g(y) = (y + 2)/5 on Y.
Check: g(f(x)) = (5x − 2 + 2)/5 = x and f(g(y)) = (y + 2) − 2 = y.
Answer: f⁻¹(y) = (y + 2)/5
Show that f : R → R, f(x) = x³ + 1 is invertible and find f⁻¹. Compute f⁻¹(9).
One-one: x₁³ + 1 = x₂³ + 1 ⇒ x₁³ = x₂³ ⇒ x₁ = x₂. Onto: for any y, x = ∛(y − 1) is real and f(x) = y. So f is invertible.
f⁻¹(y) = ∛(y − 1). Then f⁻¹(9) = ∛8 = 2, and indeed f(2) = 9.
Answer: f⁻¹(y) = ∛(y − 1); f⁻¹(9) = 2
If f(x) = x + 3 and (gof)(x) = 2x + 7 for all real x, find g.
Use: Write g(f(x)) = 2x + 7 and put y = f(x)
Let y = x + 3, so x = y − 3. Then g(y) = 2(y − 3) + 7 = 2y + 1.
Check: g(f(x)) = 2(x + 3) + 1 = 2x + 7.
Answer: g(y) = 2y + 1
Let f : R − {1} → R − {1} be f(x) = (x + 2)/(x − 1). Show that fof is the identity and hence find f⁻¹.
f(f(x)) = [ (x + 2)/(x − 1) + 2 ] / [ (x + 2)/(x − 1) − 1 ]. Multiplying numerator and denominator by (x − 1):
= [ (x + 2) + 2(x − 1) ] / [ (x + 2) − (x − 1) ] = 3x / 3 = x.
So fof = I. This means f is its own inverse: f⁻¹ = f.
Answer: fof(x) = x, so f⁻¹(y) = (y + 2)/(y − 1)
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(i) y = 3x, so R = {(1, 3), (2, 6), (3, 9), (4, 12)}.
Not reflexive: (1, 1) ∉ R (3 × 1 − 1 ≠ 0). Not symmetric: (1, 3) ∈ R but (3, 1) ∉ R. Not transitive: (1, 3), (3, 9) ∈ R but (1, 9) ∉ R.
(ii) R = {(1, 6), (2, 7), (3, 8)}.
Not reflexive: (1, 1) ∉ R. Not symmetric: (1, 6) ∈ R but (6, 1) ∉ R. Transitive: there is no pair (x, y), (y, z) in R (no y in {6, 7, 8} is a first coordinate), so the condition holds vacuously.
(iii) Reflexive: x divides x. Not symmetric: (1, 2) ∈ R (2 is divisible by 1) but (2, 1) ∉ R. Transitive: if y = kx and z = my, then z = (km)x, so z is divisible by x.
(iv) x − y is always an integer, so R = Z × Z. It is reflexive, symmetric and transitive.
(v)(a) Working at the same place is reflexive, symmetric and transitive (an equivalence relation).
(b) Living in the same locality is reflexive, symmetric and transitive.
(c) Not reflexive (x is not 7 cm taller than x). Not symmetric (if x is 7 cm taller than y, y is not taller than x). Not transitive: if x is 7 cm taller than y and y is 7 cm taller than z, then x is 14 cm taller than z.
(d) Not reflexive. Not symmetric: if x is the wife of y, then y is the husband of x, not the wife. For transitivity no pair (x, y), (y, z) can exist, and the NCERT key records R as not transitive since x being the wife of z does not follow.
(e) Not reflexive. Not symmetric. Not transitive: if x is the father of y and y is the father of z, then x is the grandfather of z, not the father.
Answer: (i) none of the three (ii) transitive only (iii) reflexive and transitive (iv) all three (v) (a) and (b) all three; (c), (d), (e) none of the three
Not reflexive: take a = 1/2. Then b² = 1/4 and 1/2 ≤ 1/4 is false, so (1/2, 1/2) ∉ R.
Not symmetric: (1, 4) ∈ R since 1 ≤ 16, but (4, 1) ∉ R since 4 ≤ 1 is false.
Not transitive: (10, 4) ∈ R since 10 ≤ 16, and (4, 3) ∈ R since 4 ≤ 9, but (10, 3) ∉ R since 10 ≤ 9 is false.
Answer: neither reflexive, nor symmetric, nor transitive
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.
Not reflexive: (1, 1) ∉ R. Not symmetric: (1, 2) ∈ R but (2, 1) ∉ R.
Not transitive: (1, 2), (2, 3) ∈ R but (1, 3) ∉ R.
Answer: none of the three
Reflexive: a ≤ a for every real a.
Transitive: a ≤ b and b ≤ c give a ≤ c.
Not symmetric: (1, 2) ∈ R since 1 ≤ 2, but (2, 1) ∉ R since 2 ≤ 1 is false.
Answer: reflexive and transitive, not symmetric
Not reflexive: a = 1/2 gives b³ = 1/8, and 1/2 ≤ 1/8 is false.
Not symmetric: (1, 2) ∈ R since 1 ≤ 8, but (2, 1) ∉ R since 2 ≤ 1 is false.
Not transitive: (3, 3/2) ∈ R since 3 ≤ 27/8 = 3.375, and (3/2, 6/5) ∈ R since 3/2 ≤ 216/125 = 1.728, but (3, 6/5) ∉ R since 3 ≤ 1.728 is false.
Answer: none of the three
Symmetric: (1, 2) ∈ R and (2, 1) ∈ R.
Not reflexive: (1, 1) ∉ R (and (2, 2), (3, 3) ∉ R).
Not transitive: (1, 2), (2, 1) ∈ R but (1, 1) ∉ R.
Answer: symmetric but neither reflexive nor transitive
Reflexive: a book has the same number of pages as itself.
Symmetric: if x and y have the same number of pages, so do y and x.
Transitive: if x and y have the same number of pages and y and z have the same number of pages, then x and z have the same number of pages.
So R is reflexive, symmetric and transitive: an equivalence relation.
Answer: equivalence relation
Reflexive: |a − a| = 0 is even.
Symmetric: |a − b| = |b − a|.
Transitive: if a − b and b − c are even, then a − c = (a − b) + (b − c) is even, so |a − c| is even. Hence R is an equivalence relation.
Any two of 1, 3, 5 differ by 2 or 4, which is even, so they are related. |2 − 4| = 2 is even, so 2 and 4 are related.
An element of {1, 3, 5} is odd and an element of {2, 4} is even, so their difference is odd, and they are not related.
Answer: shown
(i) Reflexive: |a − a| = 0 = 4 × 0. Symmetric: |a − b| = |b − a|. Transitive: if a − b = 4m and b − c = 4n, then a − c = 4(m + n). So R is an equivalence relation.
Elements related to 1: |1 − b| is a multiple of 4, so b = 1, 5, 9 (b = −3 and b = 13 are outside A). The set is {1, 5, 9}.
(ii) a = a (reflexive); a = b ⇒ b = a (symmetric); a = b and b = c ⇒ a = c (transitive). An equivalence relation.
Elements related to 1: only b = 1. The set is {1}.
Answer: (i) {1, 5, 9} (ii) {1}
(i) R = {(1, 2), (2, 1)} in {1, 2, 3}. It is symmetric; (1, 1) ∉ R; (1, 2), (2, 1) ∈ R but (1, 1) ∉ R.
(ii) R = {(a, b) : a < b} in R. If a < b and b < c then a < c (transitive). a < a is false (not reflexive), and a < b does not give b < a (not symmetric).
(iii) R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} in {1, 2, 3}. Reflexive and symmetric, but (1, 2), (2, 3) ∈ R while (1, 3) ∉ R.
(iv) R = {(a, b) : a ≤ b} in R. Reflexive and transitive, but 1 ≤ 2 while 2 ≤ 1 fails.
(v) R = {(1, 1), (2, 2), (1, 2), (2, 1)} in {1, 2, 3}. Symmetric and transitive, but (3, 3) ∉ R so it is not reflexive.
Answer: one example for each part, as above
Let OP denote the distance of P from the origin O.
Reflexive: OP = OP. Symmetric: OP = OQ ⇒ OQ = OP. Transitive: OP = OQ and OQ = OR ⇒ OP = OR. So R is an equivalence relation.
The points Q related to P are those with OQ = OP: they all lie at the fixed distance OP from O. That is the circle with centre O and radius OP, which passes through P.
Answer: equivalence relation; the circle through P with centre O
Reflexive: a triangle is similar to itself. Symmetric: if T₁ ~ T₂ then T₂ ~ T₁. Transitive: if T₁ ~ T₂ and T₂ ~ T₃ then T₁ ~ T₃. So R is an equivalence relation.
T₁ and T₃: 3/6 = 4/8 = 5/10 = 1/2, so the sides are proportional and T₁ ~ T₃.
T₂ and T₁: 5/3, 12/4, 13/5 are not equal, so they are not similar. Likewise T₂ and T₃: 5/6, 12/8, 13/10 are not equal.
Answer: T₁ and T₃ are related
Reflexive: P has the same number of sides as P. Symmetric: if P₁ and P₂ have the same number of sides, so do P₂ and P₁. Transitive: equal side counts are preserved along a chain. So R is an equivalence relation.
T has 3 sides, so the elements related to T are exactly the polygons with 3 sides: all triangles.
Answer: the set of all triangles
Reflexive: a line is parallel to itself (same slope). Symmetric: L₁ ∥ L₂ ⇒ L₂ ∥ L₁. Transitive: L₁ ∥ L₂ and L₂ ∥ L₃ ⇒ L₁ ∥ L₃, since all have the same slope. So R is an equivalence relation.
Lines parallel to y = 2x + 4 have slope 2, so they have the form y = 2x + c.
Answer: {y = 2x + c : c ∈ R}
Reflexive: (1, 1), (2, 2), (3, 3), (4, 4) are all in R.
Not symmetric: (1, 2) ∈ R but (2, 1) ∉ R.
Transitive: checking every pair (a, b), (b, c): (1, 3), (3, 2) give (1, 2) ∈ R; (1, 2), (2, 2) give (1, 2) ∈ R; (3, 3), (3, 2) give (3, 2) ∈ R; all others involve a repeated pair (a, a). So R is transitive.
Answer: (B)
A pair (a, b) is in R when b > 6 and a = b − 2.
(A) (2, 4): b = 4 is not > 6. (B) (3, 8): b − 2 = 6 ≠ 3. (C) (6, 8): b = 8 > 6 and 8 − 2 = 6 = a ✓. (D) (8, 7): 7 − 2 = 5 ≠ 8.
Answer: (C)
One-one: f(x₁) = f(x₂) ⇒ 1/x₁ = 1/x₂ ⇒ x₁ = x₂.
Onto: for y ∈ R•, x = 1/y ∈ R• and f(1/y) = y.
If the domain is N, f : N → R• is still one-one (same argument), but it is not onto: for y = 2/3, 1/x = 2/3 gives x = 3/2 ∉ N.
Answer: f is one-one and onto; with domain N it is one-one but not onto, so the result is not true
(i) x₁² = x₂² with x₁, x₂ ∈ N gives x₁ = x₂, so injective. Not surjective: 2 is not a perfect square.
(ii) f(−1) = f(1) = 1, so not injective. Not surjective: −1 (or 2) is not a square of an integer.
(iii) f(−1) = f(1), so not injective. Not surjective: −1 is not a square of a real number.
(iv) x₁³ = x₂³ gives x₁ = x₂, so injective. Not surjective: 2 is not a cube of a natural number.
(v) x₁³ = x₂³ gives x₁ = x₂, so injective. Not surjective: 2 is not a cube of an integer.
Answer: (i) injective, not surjective (ii) neither (iii) neither (iv) injective, not surjective (v) injective, not surjective
Not one-one: f(1.2) = [1.2] = 1 and f(1.5) = [1.5] = 1, but 1.2 ≠ 1.5.
Not onto: [x] is always an integer, so a non-integer such as 1/2 in the co-domain R is not the image of any x.
Answer: neither one-one nor onto
Not one-one: f(−1) = 1 = f(1), but −1 ≠ 1.
Not onto: |x| ≥ 0 always, so −1 ∈ R is not the image of any x.
Answer: neither one-one nor onto
Not one-one: f(1) = 1 = f(2), but 1 ≠ 2.
Not onto: f takes only the values 1, 0 and −1, so 2 ∈ R is not the image of any x.
Note: the textbook PDF prints the third value as 1 (the minus sign is missing). The signum function takes −1 for x < 0, and the conclusion is the same either way: f(1) = f(2), and 2 is never a value.
Answer: neither one-one nor onto
The images are f(1) = 4, f(2) = 5, f(3) = 6. They are all different, so distinct elements of A have distinct images.
Hence f is one-one. (It is not onto, since 7 is not an image.)
Answer: f is one-one
(i) One-one: 3 − 4x₁ = 3 − 4x₂ ⇒ x₁ = x₂. Onto: for any real y, x = (3 − y)/4 satisfies f(x) = y. So f is bijective.
(ii) f(1) = f(−1) = 2, so not one-one. 1 + x² ≥ 1, so 0 is not in the range and f is not onto. Neither one-one nor onto.
Answer: (i) one-one and onto, so bijective (ii) neither one-one nor onto
One-one: f(a₁, b₁) = f(a₂, b₂) ⇒ (b₁, a₁) = (b₂, a₂) ⇒ b₁ = b₂ and a₁ = a₂ ⇒ (a₁, b₁) = (a₂, b₂).
Onto: any element (b, a) of B × A is the image of (a, b) ∈ A × B.
Answer: f is bijective
f(1) = (1 + 1)/2 = 1 and f(2) = 2/2 = 1, so f(1) = f(2) and f is not one-one.
Onto: for any m ∈ N, the odd number 2m − 1 gives f(2m − 1) = (2m − 1 + 1)/2 = m.
So f is onto but not one-one, hence not bijective.
Answer: not bijective (onto, but not one-one)
One-one: f(x₁) = f(x₂) ⇒ (x₁ − 2)(x₂ − 3) = (x₂ − 2)(x₁ − 3) ⇒ x₁x₂ − 3x₁ − 2x₂ + 6 = x₁x₂ − 3x₂ − 2x₁ + 6 ⇒ −x₁ = −x₂ ⇒ x₁ = x₂.
Onto: let y ∈ B, so y ≠ 1. From y = (x − 2)/(x − 3): y(x − 3) = x − 2, so x(y − 1) = 3y − 2 and x = (3y − 2)/(y − 1).
This is defined since y ≠ 1. Is x = 3? (3y − 2)/(y − 1) = 3 gives 3y − 2 = 3y − 3, which is impossible. So x ∈ A and f(x) = y.
Answer: f is one-one and onto
f(−1) = f(1) = 1, so f is many-one.
x⁴ ≥ 0, so −1 is not an image and f is not onto.
Answer: (D)
One-one: 3x₁ = 3x₂ ⇒ x₁ = x₂.
Onto: any real y equals f(y/3).
Answer: (A)
f(x) has the same sign as x, because 1 + |x| > 0.
One-one: suppose f(x₁) = f(x₂). Then x₁ and x₂ have the same sign (or both are 0).
If both are ≥ 0: x₁/(1 + x₁) = x₂/(1 + x₂) ⇒ x₁ + x₁x₂ = x₂ + x₁x₂ ⇒ x₁ = x₂.
If both are < 0: x₁/(1 − x₁) = x₂/(1 − x₂) ⇒ x₁ − x₁x₂ = x₂ − x₁x₂ ⇒ x₁ = x₂.
Onto: let y ∈ (−1, 1). If y ≥ 0, take x = y/(1 − y) ≥ 0: f(x) = x/(1 + x) = [y/(1 − y)] / [1/(1 − y)] = y.
If y < 0, take x = y/(1 + y) < 0 (since 1 + y > 0): 1 − x = 1/(1 + y), so f(x) = x/(1 − x) = [y/(1 + y)] × (1 + y) = y.
Answer: f is one-one and onto
Suppose f(x₁) = f(x₂), i.e. x₁³ = x₂³. Then x₁³ − x₂³ = (x₁ − x₂)(x₁² + x₁x₂ + x₂²) = 0.
The second factor is (x₁ + x₂/2)² + 3x₂²/4, which is zero only if x₁ = x₂ = 0. So in every case x₁ = x₂.
Answer: f is injective
Here A ⊂ B means A is a subset of B (equality allowed).
Reflexive: A ⊂ A. Transitive: A ⊂ B and B ⊂ C give A ⊂ C.
Not symmetric: ∅ ⊂ X, so (∅, X) ∈ R, but X ⊂ ∅ is false for non-empty X, so (X, ∅) ∉ R.
R is reflexive and transitive but not symmetric, so it is not an equivalence relation.
Answer: no, R is not symmetric
For a finite set, a function from the set to itself is onto if and only if it is one-one.
A one-one function from {1, 2, …, n} to itself is a permutation of the n symbols, and there are n! permutations (n choices for the image of 1, n − 1 for 2, and so on).
Answer: n!
f(−1) = 1 + 1 = 2, f(0) = 0, f(1) = 0, f(2) = 4 − 2 = 2.
g(−1) = 2 × 3/2 − 1 = 2, g(0) = 2 × 1/2 − 1 = 0, g(1) = 2 × 1/2 − 1 = 0, g(2) = 2 × 3/2 − 1 = 2.
f(a) = g(a) for every a ∈ A, so f and g are equal.
Answer: yes, f = g
Reflexive: (1, 1), (2, 2), (3, 3) must be in R. Symmetric: (2, 1) and (3, 1) must be in R as well. This gives R₀ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}.
R₀ is not transitive: (2, 1), (1, 3) ∈ R₀ but (2, 3) ∉ R₀.
The only pairs left are (2, 3) and (3, 2). Adding one without the other breaks symmetry, and adding both gives the universal relation, which is transitive.
So R₀ is the only such relation.
Answer: (A) 1
An equivalence relation containing (1, 2) also contains (2, 1), (1, 1), (2, 2), (3, 3). The smallest is R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.
Adding (2, 3) forces (3, 2) (symmetry) and then (1, 3), (3, 1) (transitivity), which gives the universal relation. Adding (1, 3) leads to the same.
So the equivalence relations are R₁ and the universal relation A × A.
Answer: (B) 2