Lesson Overview
Subject
Mathematics
Chapter
1 — Relations
Reference
NCERT
Dates
23–26 Feb 2026
How to Use Each Question
📝 Tab 1 — Workspace
Click the Workspace tab under any question → ruled lines appear for students to write their working by hand or on screen.
💡 Tab 2 — Step Solution
Manual: Click "Next Step ▶" to reveal one step at a time like a PPT transition.
Auto Play: Steps reveal automatically at your chosen speed (1s / 2s / 3s).
Auto Play: Steps reveal automatically at your chosen speed (1s / 2s / 3s).
🎬 Tab 3 — Solution Clip
Click the upload zone → select your MP4/MOV video from your computer → video plays right inside the lesson plan. Add a caption too.
QR Code Slots
Each Day tab has 4 independent QR code slots in the right sidebar — label them Video 1 (Concept intro), Video 2 (Worked example), Video 3 (Practice), Video 4 (Board question). Paste any URL and press Enter or click Generate.
⚠ Common Misconceptions
✗Confusing symmetric with antisymmetric
✗Empty relation is vacuously S & T — but not Reflexive unless A = ∅
✗Transitivity — only pairs already in R need checking
✗Equivalence class [a] = set of elements, not ordered pairs
DAY 1 · MON 23 FEB 2026
Day 1 — Types of Relations
Types of Relations
BinaryAny subset R ⊆ A × A
IdentityIₐ = { (a,a) : a ∈ A }
EmptyR = ∅ — vacuously Symmetric & Transitive
UniversalR = A×A — RST all hold
CountTotal relations from A to B: 2^(n(A)·n(B))
Q1
Write the Cartesian product A×B and find n(A×B), given A={1,2,3} and B={4,5}.
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
💻 Code
Working Space
0 / 3
● Auto
✎ Type your own solution steps. Check KEY for conclusion/gold highlight. Click Apply when done.
1
Definition: A×B = {(a,b) : a∈A, b∈B} — all ordered pairs with first element from A, second from B.
2
A = {1,2,3}, B = {4,5}. Pair each a∈A with every b∈B:
1 → (1,4), (1,5) | 2 → (2,4), (2,5) | 3 → (3,4), (3,5)
1 → (1,4), (1,5) | 2 → (2,4), (2,5) | 3 → (3,4), (3,5)
3
A×B = {(1,4), (1,5), (2,4), (2,5), (3,4), (3,5)}
4
Formula: n(A×B) = n(A) × n(B) = 3 × 2 = 6
5
∴ A×B = {(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)}, n(A×B) = 6. Total possible relations from A to B = 26 = 64.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM supported
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Q2
Write the Identity relation on A={p,q,r}. Is this relation reflexive? Justify.
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
💻 Code
Working Space
0 / 3
● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Definition: Identity relation on A is IA = {(a,a) : a∈A} — every element related to itself only.
2
A = {p, q, r} → IA = {(p,p), (q,q), (r,r)}
3
Reflexive check: A relation is reflexive if (a,a)∈R for every a∈A.
4
Verify: (p,p)∈IA ✓ | (q,q)∈IA ✓ | (r,r)∈IA ✓ — every element of A satisfies the condition.
5
∴ Yes, Identity relation IA = {(p,p),(q,q),(r,r)} is Reflexive.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Q3
Let A={2,3,5,7}. Is R={(a,b): a²=b, a,b ∈ A} an empty relation? Justify.
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space
0 / 4● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: A = {2,3,5,7}, R = {(a,b): a²=b, a,b∈A}. We need to find if any pair (a,b) satisfies a²=b with both a and b in A.
2
Check a=2: a²=4, but 4∉A ✗
3
Check a=3: a²=9, 9∉A ✗ | a=5: a²=25, 25∉A ✗ | a=7: a²=49, 49∉A ✗
4
No element of A has its square also in A → no ordered pair (a,b) satisfies the given condition.
5
∴ R = ∅ (empty set). Yes, R is an Empty Relation on A.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Q4
Question not set — click ✎ Edit Q to add
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space
0 / 0● Auto
✎ Add Text steps or Image steps. Check KEY for gold. Click ✓ Apply.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
Q5
For the relation 'is similar to' on all triangles, state all applicable properties from {S,T,R,E,N}.
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space
0 / 4● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Reflexive (R): Every triangle is similar to itself (all three angles are equal to themselves, AA criterion) ∴ △A ~ △A ✓
2
Symmetric (S): If △A ~ △B, then corresponding angles of △B match △A as well ∴ △B ~ △A ✓
3
Transitive (T): If △A ~ △B and △B ~ △C, then corresponding angles of △A and △C are equal ∴ △A ~ △C ✓
4
Since R, S, T all hold → Equivalence Relation (E). It is NOT a universal relation (not every pair of triangles is similar). It is NOT empty. So N does not apply here.
5
∴ Applicable properties: R, S, T, E.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Day 1 — Homework
HW 1
Write Universal relation on A={1,3,5}; verify using a−b<5.HW 2
How many relations possible from A={1,2} to B={a,b,c}?📎 Assignments — Day 1 (Optional · Viewable by students)
A1
A2
A3
Day 1 — Video Resources
Exit Ticket
1Write one example each: empty relation and universal relation on A={1,2,3}
2Is Identity relation on A={a,b} an Equivalence Relation?
ISC Board Tags — Day 1
1–2 MDefinitionBEP Core
DAY 2 · TUE 24 FEB 2026
Day 2 — Reflexive, Symmetric & Transitive
RST Definitions
Reflexive(a,a) ∈ R ∀ a ∈ A | Count: 2^(n²−n)
Symmetric(a,b)∈R ⟹ (b,a)∈R ⟺ R=R⁻¹ | Count: 2^((n²+n)/2)
Transitive(a,b)∈R and (b,c)∈R ⟹ (a,c)∈R
Ex 1
R={(x,y): 3x−y=0} on A={1,2,…,14}. Check R for Reflexive, Symmetric, Transitive.
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space
0 / 5● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: R = {(x,y): 3x−y=0} on A={1,2,...,14}, i.e. y=3x. List R: {(1,3),(2,6),(3,9),(4,12)} since 5×3=15∉A.
2
Reflexive? For (a,a)∈R: need 3a−a=0 → 2a=0 → a=0. But 0∉A. So (1,1)∉R, (2,2)∉R, … → NOT Reflexive ✗
3
Symmetric? (1,3)∈R since 3(1)−3=0. Check (3,1): 3(3)−1 = 8 ≠ 0 → (3,1)∉R → NOT Symmetric ✗
4
Transitive? (1,3)∈R and (3,9)∈R. Check (1,9): 3(1)−9 = −6 ≠ 0 → (1,9)∉R → NOT Transitive ✗
5
∴ R is neither Reflexive, nor Symmetric, nor Transitive.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Ex 2
R={(x,y): y is divisible by x} on A={1,2,3,4,5,6}. Which of R/S/T hold?
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space
0 / 4● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: R={(x,y): x|y (x divides y)} on A={1,2,3,4,5,6}. Partial list: {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(5,5),(6,6)}
2
Reflexive: Every natural number divides itself → a|a for all a∈A → (a,a)∈R ∀a∈A ✓
3
Symmetric? (1,2)∈R since 1|2. Check (2,1): does 2 divide 1? No (2 > 1) → (2,1)∉R → NOT Symmetric ✗
4
Transitive: x|y ⟹ y=kx for some k∈ℕ. y|z ⟹ z=my. Then z=m(kx)=mkx ⟹ x|z → (x,z)∈R ✓
5
∴ R is Reflexive and Transitive but NOT Symmetric.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Ex1A · Q6
R={(a,b): a ≤ b} on ℝ. Show R is Reflexive and Transitive but NOT Symmetric.
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space
0 / 4● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: R = {(a,b): a≤b} on ℝ. (a,b)∈R means a is less than or equal to b.
2
Reflexive: For any a∈ℝ, a≤a is always true → (a,a)∈R for all a∈ℝ ✓
3
Symmetric? Counter-example: a=1, b=2 → 1≤2 so (1,2)∈R. But 2≤1 is false → (2,1)∉R → NOT Symmetric ✗
4
Transitive: (a,b)∈R and (b,c)∈R → a≤b and b≤c. By transitivity of ≤ on ℝ: a≤c → (a,c)∈R ✓
5
∴ R is Reflexive and Transitive but NOT Symmetric. Hence R is NOT an Equivalence Relation.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Day 2 Homework
HW 1
R={(a,b): a≤b²} on ℝ. Show R is neither R, S, nor T.HW 2
Write a relation which is: (a) only T (b) only S (c) only R and T (d) only S and R.📎 Assignments — Day 2 (Optional · Viewable by students)
A1
A2
A3
Day 2 — Video Resources
Exit Ticket
1Symmetric but neither R nor T — give one example.
2R={(1,2),(2,1)} on {1,2,3}: Is R symmetric? Reflexive? Transitive?
Mark Scheme Note
State property → assume (a,b)∈R → show consequence → conclude. Every step earns marks.
DAY 3 · WED 25 FEB 2026
Day 3 — Equivalence Relations & Classes
Equivalence Relation
DefinitionR is ER ⟺ Reflexive AND Symmetric AND Transitive
Class [a]{b ∈ A : (b,a) ∈ R} — set of all elements related to a
PartitionClasses are mutually disjoint; their union = A
ISC NoteAlways write conclusion sentence — 1 mark awarded by ISC for it
Ex 7
R={(a,b): (a−b) divisible by 5} on ℤ. Prove R is ER. Write equivalence classes [0],[1],[2].
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space (12 lines)
0 / 6● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: R = {(a,b): 5|(a−b)} on ℤ, i.e. (a−b) is divisible by 5.
2
Reflexive: a−a=0=5×0, so 5|(a−a) for every a∈ℤ → (a,a)∈R ✓
3
Symmetric: (a,b)∈R → 5|(a−b) → a−b=5k for some k∈ℤ → b−a=5(−k) → 5|(b−a) → (b,a)∈R ✓
4
Transitive: (a,b)∈R and (b,c)∈R → 5|(a−b) and 5|(b−c) → 5|[(a−b)+(b−c)] → 5|(a−c) → (a,c)∈R ✓
5
Conclusion: Since R is Reflexive, Symmetric and Transitive → R is an Equivalence Relation on ℤ.
6
Equivalence Classes: [0]={…,−10,−5,0,5,10,…} | [1]={…,−9,−4,1,6,11,…} | [2]={…,−8,−3,2,7,12,…}
7
There are exactly 5 distinct equivalence classes: [0],[1],[2],[3],[4] — they partition ℤ.
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
CBSE 2024
R on A={−4,…,4}, R={(x,y): x+y divisible by 2}. Show R is ER; write equivalence class [2].
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space (12 lines)
0 / 5● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: A={−4,−3,−2,−1,0,1,2,3,4}, R={(x,y): 2|(x+y)} i.e. (x+y) is divisible by 2.
2
Reflexive: x+x = 2x, which is always divisible by 2 → (x,x)∈R ∀x∈A ✓
3
Symmetric: 2|(x+y) → x+y = 2k → y+x = 2k (same sum) → 2|(y+x) → (y,x)∈R ✓
4
Transitive: 2|(x+y) and 2|(y+z). Adding: (x+y)+(y+z) = x+2y+z. Since 2|2y, we get 2|(x+z) → (x,z)∈R ✓
5
∴ R is an Equivalence Relation on A.
6
[2] = {x∈A : (x,2)∈R} = {x : x+2 divisible by 2} = {x : x is even} = {−4,−2,0,2,4}
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Day 3 Homework
HW 1
Prove R={(L₁,L₂): L₁∥L₂} on all lines in XY-plane is an ER. Find all lines related to y=2x+4.HW 2
aRb iff (a+b) is even, a,b∈ℤ. Prove R is an ER.📎 Assignments — Day 3 (Optional · Viewable by students)
A1
A2
A3
Day 3 — Video Resources
Exit Ticket
1What is [3] under R={(a,b): 3|(a−b)} on ℤ?
2Can two equivalence classes overlap? Prove or disprove.
6M Mark Scheme
1M Reflexive + 1M Symmetric + 1M Transitive + 1M Conclusion + 2M for Equivalence Class
DAY 4 · THU 26 FEB 2026
Day 4 — ℕ×ℕ Equivalence & Review
Equivalence on ℕ×ℕ
Standard(a,b)R(c,d) ⟺ ad=bc (cross-multiplication form of a/b=c/d)
ISC 2024(a,b)R(c,d) iff a−c=b−d → same as a−b=c−d
Class[(2,6)]={(1,3),(2,6),(3,9),…} all pairs in ratio 1:3
Ex 13
R on ℕ×ℕ: (a,b)R(c,d) ⟺ ad=bc. Prove R is ER. Find [(2,6)].
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space (14 lines)
0 / 6● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: (a,b)R(c,d) ⟺ ad=bc on ℕ×ℕ. This is equivalent to a/b = c/d (rational number equality). e.g. (1,2)R(2,4) since 1×4=2×2=4.
2
Reflexive: (a,b)R(a,b)? Need a·b = b·a. True by commutativity of multiplication → (a,b)R(a,b) ✓
3
Symmetric: (a,b)R(c,d) → ad=bc → cb=da → (c,d)R(a,b) ✓
4
Transitive: (a,b)R(c,d): ad=bc …(i). (c,d)R(e,f): cf=de …(ii). Multiply: ad·cf = bc·de → acdf = bcde. Divide both sides by cd (c,d∈ℕ, so cd≠0): af = be → (a,b)R(e,f) ✓
5
∴ R is an Equivalence Relation on ℕ×ℕ.
6
[(2,6)]: (a,b)R(2,6) iff 6a=2b iff b=3a iff a/b=1/3 → [(2,6)] = {(1,3),(2,6),(3,9),(4,12),(5,15),…}
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Ex 19
R on A×A (A={1..9}): (a,b)R(c,d) iff a+d=b+c. Show R is ER. Find [(2,5)].
Replace question with typed text or upload an image (photo of textbook / handwritten question).
📝 Workspace
💡 Full Solution
🎬 Solution Clip
Working Space (14 lines)
0 / 6● Auto
✎ Add Text steps (type your solution) or Image steps (upload photo of handwritten work / textbook scan). Check KEY to highlight conclusion in gold. Click ✓ Apply when done.
1
Given: (a,b)R(c,d) ⟺ a+d=b+c on A×A, A={1,2,...,9}. Key insight: a+d=b+c ⟺ a−b=c−d (same difference between components).
2
Reflexive: (a,b)R(a,b)? a+b = b+a ✓ (commutative addition) → (a,b)R(a,b) ∀(a,b)∈A×A ✓
3
Symmetric: (a,b)R(c,d) → a+d=b+c → c+b=d+a → (c,d)R(a,b) ✓
4
Transitive: (a,b)R(c,d): a+d=b+c …(i). (c,d)R(e,f): c+f=d+e …(ii). Adding (i)+(ii): a+d+c+f = b+c+d+e → a+f=b+e → (a,b)R(e,f) ✓
5
∴ R is an Equivalence Relation on A×A.
6
[(2,5)]: (a,b)R(2,5) iff a+5=b+2 iff a−b=−3. All (a,b) in A×A with a−b=−3: [(2,5)] = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}
✓ Complete
Click to upload solution video
MP4 · MOV · AVI · WebM
💻 Code / Program
💻
No code added for this question.
Teacher will add code if applicable.
Day 4 Homework
HW 1
Complete all pending Ex1(A). NCERT Exemplar: aRb iff (a−b) divisible by n on ℤ — show R is ER.HW 2
Attempt Assertion–Reason Q21(i)–(vi) from textbook.📎 Assignments — Day 4 (Optional · Viewable by students)
A1
A2
A3
HW 3
Read Case Study Q24 (General Elections context) and answer all parts.Day 4 — Video Resources
Final Exit Ticket
1In one sentence: what makes a relation an equivalence relation?
2Write [(1,2)] under (a,b)R(c,d) iff a+d=b+c on ℕ×ℕ.
3How many equivalence relations on {1,2,3}? (Bell number)
Chapter Summary
R(a,a)∈R ∀a
S(a,b)∈R ⟹ (b,a)∈R
T(a,b),(b,c)∈R ⟹ (a,c)∈R
ER+S+T simultaneously
[a]Disjoint partition of A