Exercise 1.1 answers: Relations and Functions

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(i) 3x − y = 0

1. Since y = 3x and y must also be in A = {1, 2, …, 14}, R = {(1, 3), (2, 6), (3, 9), (4, 12)}.
2. Reflexive: (a, a) would need 3a − a = 2a = 0, so a = 0, which is not in A. So (1, 1) ∉ R: not reflexive.
3. Symmetric: (1, 3) ∈ R, but for (3, 1) we need 3 × 3 − 1 = 0, and 3 × 3 − 1 = 8. So (3, 1) ∉ R: not symmetric.
4. Transitive: (1, 3) ∈ R and (3, 9) ∈ R, so (1, 9) is needed. But 3 × 1 − 9 = −6 ≠ 0, so (1, 9) ∉ R: not transitive.

AnswerR is not reflexive, not symmetric and not transitive.

(ii) y = x + 5 and x < 4

1. x < 4 and x is a natural number, so x can only be 1, 2 or 3. That gives R = {(1, 6), (2, 7), (3, 8)}.
2. Reflexive: (1, 1) ∉ R, because 1 ≠ 1 + 5: not reflexive.
3. Symmetric: (1, 6) ∈ R, but (6, 1) ∉ R, because 6 is not less than 4: not symmetric.
4. Transitive: a chain needs a pair (a, b) and a pair (b, c). The second entries 6, 7, 8 are never first entries (those are only 1, 2, 3), so there is no chain at all.
5. With no chain, there is no missing shortcut, so R is transitive.

AnswerR is not reflexive and not symmetric, but it is transitive.

(iii) y is divisible by x

1. Every number divides itself, so (a, a) ∈ R for every a in {1, 2, 3, 4, 5, 6}: reflexive.
2. 2 is divisible by 1, so (1, 2) ∈ R; but 1 is not divisible by 2, so (2, 1) ∉ R: not symmetric.
3. If y is divisible by x and z is divisible by y, then z is divisible by x (for example, 2 divides 4 and 4 divides 8, so 2 divides 8): transitive.

AnswerR is reflexive and transitive, but not symmetric.

(iv) x − y is an integer

1. For any two integers x and y, x − y is an integer, so every pair of integers is in R.
2. Reflexive: x − x = 0 is an integer.
3. Symmetric: if x − y is an integer, so is y − x = −(x − y).
4. Transitive: if x − y and y − z are integers, so is x − z = (x − y) + (y − z).

AnswerR is reflexive, symmetric and transitive.

(v)(a) x and y work at the same place

1. Here every person in A is taken to work at some place, which is how the question is meant.
2. Reflexive: every person works at the same place as themselves.
3. Symmetric: if x works at the same place as y, then y works at the same place as x.
4. Transitive: if x and y work at the same place, and y and z work at the same place, then x and z work at that same place.

AnswerR is reflexive, symmetric and transitive.

(v)(b) x and y live in the same locality

1. Every person in A lives in some locality.
2. Reflexive: every person lives in the same locality as themselves.
3. Symmetric: if x lives in the same locality as y, then y lives in the same locality as x.
4. Transitive: if x and y share a locality, and y and z share a locality, then x and z live in that same locality.

AnswerR is reflexive, symmetric and transitive.

(v)(c) x is exactly 7 cm taller than y

1. Reflexive: nobody is 7 cm taller than themselves (the difference is 0 cm, not 7 cm): not reflexive.
2. Symmetric: if x is 7 cm taller than y, then y is 7 cm shorter than x, not taller: not symmetric.
3. Transitive: if x is 7 cm taller than y and y is 7 cm taller than z, then x is 7 + 7 = 14 cm taller than z, not 7 cm: not transitive.

AnswerR is not reflexive, not symmetric and not transitive.

(v)(d) x is wife of y

1. Reflexive: nobody is their own wife: not reflexive.
2. Symmetric: if x is the wife of y, then y is the husband of x, not the wife of x: not symmetric.
3. Transitive: a chain would need x to be the wife of y AND y to be the wife of z. The first makes y a husband, the second makes y a wife, and in a marriage the same person cannot be both. So no such chain exists.
4. A demand about every chain is met when there is no chain at all, so R is transitive.

AnswerR is not reflexive and not symmetric, but it is transitive.

(v)(e) x is father of y

1. Reflexive: nobody is their own father: not reflexive.
2. Symmetric: if x is the father of y, then y is the child of x, not the father: not symmetric.
3. Transitive: a chain really can happen here: a grandfather is the father of a man, and that man is the father of a son.
4. But the grandfather is not the father of the son, so the shortcut pair is missing: not transitive.

AnswerR is not reflexive, not symmetric and not transitive.

Watch this explained “Rules read straight off, and classified”, 10:45 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 5One way to think about it

1. To break reflexive, find one real number a with a > a². Try a = 1/2: is 1/2 ≤ (1/2)² = 1/4? No, so (1/2, 1/2) is not in R.
2. To break symmetric, find a, b with a ≤ b² true but b ≤ a² false. Try a = 1, b = 2: 1 ≤ 4 is true, but 2 ≤ 1 is false.
3. To break transitive, find a, b, c with a ≤ b² and b ≤ c² true, but a ≤ c² false. Try a = 3, b = −2, c = 1.
4. Check: 3 ≤ (−2)² = 4 is true. −2 ≤ 1² = 1 is true. But 3 ≤ 1² = 1 is false.
5. One failing case for each demand is enough to rule it out.

In shortR is neither reflexive (a = 1/2 fails), nor symmetric (a=1, b=2 fails), nor transitive (a=3, b=−2, c=1 fails).

Watch this explained “One symbol changes, twice”, 12:31 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 5Matches NCERT’s answer

1. Write out R by using b = a + 1 for each a in {1,...,6}: R = {(1,2),(2,3),(3,4),(4,5),(5,6)}.
2. Reflexive needs (a,a) in R for every a, but a + 1 is never equal to a, so no (a,a) is ever in R.
3. Symmetric would need (2,1) since (1,2) is in R, but 1 ≠ 2 + 1, so (2,1) is not in R.
4. Transitive would need (1,3) since (1,2) and (2,3) are both in R, but 3 ≠ 1 + 1 = 2, so (1,3) is not in R.

AnswerR is neither reflexive, nor symmetric, nor transitive.

Watch this explained “Passing only the first”, 6:57 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 5One way to think about it

1. Reflexive: every real number a satisfies a ≤ a, so (a,a) is in R for every a.
2. Transitive: if a ≤ b and b ≤ c, then a ≤ c always follows for real numbers, so the shortcut pair is always there.
3. Symmetric: pick a = 1, b = 2. Then 1 ≤ 2 is true, so (1,2) is in R, but 2 ≤ 1 is false, so (2,1) is not in R.
4. One failing pair is enough to rule out symmetric.

In shortR is reflexive and transitive, but not symmetric (1 ≤ 2 does not give 2 ≤ 1).

Watch this explained “One symbol changes, twice”, 12:31 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 5Matches NCERT’s answer

1. Reflexive would need a ≤ a³ for every real a. Try a = 1/2: is 1/2 ≤ 1/8? No, so (1/2, 1/2) is not in R.
2. Symmetric would need: whenever a ≤ b³, also b ≤ a³. Try a = 1, b = 2: 1 ≤ 8 is true, but 2 ≤ 1 is false.
3. Transitive would need: a ≤ b³ and b ≤ c³ together forcing a ≤ c³. Try a = 100, b = 5, c = 2.
4. Check: 100 ≤ 5³ = 125 is true. 5 ≤ 2³ = 8 is true. But 100 ≤ 2³ = 8 is false.

AnswerR is not reflexive, not symmetric, and not transitive.

Watch this explained “One symbol changes, twice”, 12:31 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 6One way to think about it

1. Symmetric: R has only two pairs, (1,2) and (2,1), and each one's swap is also in R. So symmetric holds.
2. Reflexive would need (1,1), (2,2) and (3,3) all in R, but none of them is in R: not reflexive.
3. Transitive would need (1,1) in R, because (1,2) and (2,1) are both in R and chain to give (1,1).
4. Since (1,1) is not in R, transitive fails.

In shortR is symmetric, but neither reflexive nor transitive.

Watch this explained “The argument that trips almost everybody”, 7:57 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 6One way to think about it

1. Reflexive: a book has the same number of pages as itself, so (x,x) is in R for every book x.
2. Symmetric: if book x has the same page count as book y, then book y has the same page count as book x.
3. Transitive: if x and y match in page count, and y and z match too, then x and z have that same page count.
4. All three demands hold, so R is an equivalence relation.

In shortR is reflexive, symmetric and transitive, so it is an equivalence relation.

Watch this explained “One machine, in different clothes”, 15:58 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 6One way to think about it

1. Reflexive: |a − a| = 0, and 0 is even, so (a,a) is in R for every a in {1,...,5}.
2. Symmetric: |a − b| and |b − a| are the same number (just a sign flip inside the modulus), so if one is even so is the other.
3. Transitive: if |a − b| is even and |b − c| is even, then a − b and b − c are both even numbers, so their sum (a − c) is even too.
4. So R is reflexive, symmetric and transitive — an equivalence relation.
5. In {1, 3, 5}, every pair of numbers differs by an even number (2 or 4), so all of 1, 3, 5 are related to each other.
6. In {2, 4}, the numbers differ by 2, which is even, so 2 and 4 are related to each other.
7. Between the two groups, e.g. |1 − 2| = 1, |3 − 4| = 1, |5 − 2| = 3 — every such difference is odd, so no element of {1,3,5} is related to any element of {2,4}.

In shortR is an equivalence relation; it splits {1,...,5} into the odd numbers {1,3,5} and the even numbers {2,4}, with no relation crossing between the two groups.

Watch this explained “Fifteen integers, and an even difference”, 1:08 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 6Matches NCERT’s answer

(i) |a − b| is a multiple of 4

1. A = {0, 1, 2, …, 12}.
2. Reflexive: |a − a| = 0 = 4 × 0, a multiple of 4, so (a, a) ∈ R for every a in A.
3. Symmetric: |b − a| = |a − b|, so if one is a multiple of 4, so is the other.
4. Transitive: if a − b = 4m and b − c = 4n, then a − c = (a − b) + (b − c) = 4(m + n), a multiple of 4.
5. So R is an equivalence relation.
6. Elements related to 1: we need |x − 1| to be a multiple of 4, so x − 1 = 0, 4, 8, 12, … or −4, −8, …, which gives x = 1, 5, 9, 13, … or x = −3, −7, ….
7. Of these, only 1, 5 and 9 lie in A (0 to 12).

AnswerThe set of elements related to 1 is {1, 5, 9}.

(ii) a = b

1. Reflexive: a = a for every a in A.
2. Symmetric: if a = b, then b = a.
3. Transitive: if a = b and b = c, then a = c.
4. So R is an equivalence relation.
5. The only element of A equal to 1 is 1 itself.

AnswerThe set of elements related to 1 is {1}.

Watch this explained “The boxes are not the same size”, 9:14 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 6Checked by computerAnswers can differ: one example

(i) Symmetric but neither reflexive nor transitive

1. Take A = {1, 2, 3} and R = {(1,2), (2,1)}.
2. Symmetric: the swap of each pair is also in R.
3. Reflexive fails: (1,1), (2,2), (3,3) are all missing.
4. Transitive fails: (1,2) and (2,1) are in R, so (1,1) is needed, but it is missing.

AnswerOne example (others are possible): R = {(1, 2), (2, 1)} on {1, 2, 3} is symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric

1. Take A = {1, 2, 3} and R = {(1,2)} — just one pair.
2. Transitive holds automatically, because there is no chain of two pairs to test at all.
3. Reflexive fails: none of (1,1), (2,2), (3,3) is in R.
4. Symmetric fails: (2,1) is not in R.

AnswerOne example (others are possible): R = {(1, 2)} on {1, 2, 3} is transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive

1. Take A = {1, 2, 3} and R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}.
2. Reflexive holds: all three diagonal pairs are in R.
3. Symmetric holds: (1,2)&(2,1) and (2,3)&(3,2) are both swapped pairs.
4. Transitive fails: (1,2) and (2,3) are in R, so (1,3) is needed, but it is missing.

AnswerOne example (others are possible): R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} on {1, 2, 3} is reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric

1. Take A = {1, 2, 3} and R = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)} — this is just a ≤ b on {1,2,3}.
2. Reflexive holds: all three diagonal pairs are in R.
3. Transitive holds: this is the usual ≤ order, which always chains correctly.
4. Symmetric fails: (1,2) is in R but (2,1) is not.

AnswerOne example (others are possible): R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)} on {1, 2, 3} is reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive

1. Take A = {1, 2, 3} and R = {(1,1)} — just the single pair (1,1).
2. Symmetric holds: the only pair's swap is itself.
3. Transitive holds: the only chain is (1,1) and (1,1), which gives (1,1), already in R.
4. Reflexive fails: (2,2) and (3,3) are missing.

AnswerOne example (others are possible): R = {(1, 1)} on {1, 2, 3} is symmetric and transitive but not reflexive.

Watch this explained “All eight patterns, counted”, 14:02 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 6One way to think about it

1. Reflexive: a point P is always the same distance from the origin as itself.
2. Symmetric: if P is the same distance from the origin as Q, then Q is the same distance from the origin as P.
3. Transitive: if P and Q match in distance from the origin, and Q and S match too, then P and S match as well.
4. So R is reflexive, symmetric and transitive — an equivalence relation.
5. Now take a point P that is not the origin, at distance r from the origin (r > 0).
6. Every point related to P is at that same distance r from the origin.
7. The set of all points at a fixed distance r from the origin is, by definition, the circle of radius r centred at the origin, and it passes through P since P itself is at distance r.

In shortR is an equivalence relation, and the class of any point P ≠ (0,0) is the circle centred at the origin passing through P.

Watch this explained “A cut you can look at”, 11:38 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 6Matches NCERT’s answer

1. Reflexive: every triangle is similar to itself.
2. Symmetric: if a triangle X is similar to a triangle Y, then Y is similar to X.
3. Transitive: if X is similar to Y and Y is similar to Z, then X is similar to Z.
4. So R is an equivalence relation.
5. T1 (3, 4, 5) and T3 (6, 8, 10): 6/3 = 8/4 = 10/5 = 2, so all sides are in the same ratio and T1 is similar to T3.
6. T1 (3, 4, 5) and T2 (5, 12, 13): 5/3, 12/4 = 3 and 13/5 are not equal, so T1 and T2 are not similar.
7. T2 (5, 12, 13) and T3 (6, 8, 10): 6/5, 8/12 = 2/3 and 10/13 are not equal, so T2 and T3 are not similar.

AnswerT1 and T3 are related; T2 is not related to either of them.

Watch this explained “Two cuts of the same objects”, 12:52 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 6Checked by computer

1. Reflexive: a polygon has the same number of sides as itself.
2. Symmetric: if P1 and P2 have the same number of sides, so do P2 and P1.
3. Transitive: if P1 and P2 match in side count, and P2 and P3 match too, then P1 and P3 match as well.
4. So R is an equivalence relation.
5. T has sides 3, 4, 5, so T is a polygon with exactly 3 sides — a triangle.
6. Every polygon related to T must have exactly 3 sides too, and every 3-sided polygon is, by definition, a triangle.

AnswerThe set of all elements related to T is the set of every triangle in A (every polygon with 3 sides).

Watch this explained “Two cuts of the same objects”, 12:52 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 6Checked by computer

1. Reflexive: every line is parallel to itself.
2. Symmetric: if L1 is parallel to L2, then L2 is parallel to L1.
3. Transitive: if L1 is parallel to L2, and L2 is parallel to L3, then L1 is parallel to L3.
4. So R is an equivalence relation.
5. The line y = 2x + 4 has slope 2.
6. Two lines are parallel exactly when they have the same slope, so every line related to it has slope 2 too.
7. Any line of slope 2 can be written y = 2x + c for some real number c, and this includes the given line itself (c = 4).

AnswerThe set of all lines related to y = 2x + 4 is the family y = 2x + c, for every real number c.

Watch this explained “Lines, and a union that breaks”, 14:26 into Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

Open NCERT p. 7Matches NCERT’s answer

1. Reflexive: the set is {1, 2, 3, 4}, and (1, 1), (2, 2), (3, 3), (4, 4) are all in R, so R is reflexive.
2. Symmetric: (1, 2) ∈ R but (2, 1) ∉ R, so R is not symmetric.
3. Transitive: a chain that uses a pair like (a, a) just gives back the other pair of the chain, which is already in R.
4. The pairs with two different entries are (1, 2), (1, 3) and (3, 2). The only chain made of two of them is (1, 3) then (3, 2), which needs (1, 2), and (1, 2) ∈ R.
5. Every chain has its shortcut, so R is transitive.
6. So R is reflexive and transitive but not symmetric: option (B).

Answer(B) R is reflexive and transitive but not symmetric.

Watch this explained “Passing only the first”, 6:57 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 7Matches NCERT’s answer

1. For (2,4): does 2 = 4 − 2? Yes. But is 4 > 6? No. So (2,4) is not in R.
2. For (3,8): does 3 = 8 − 2 = 6? No. So (3,8) is not in R.
3. For (6,8): does 6 = 8 − 2 = 6? Yes. Is 8 > 6? Yes. Both conditions hold, so (6,8) is in R.
4. For (8,7): does 8 = 7 − 2 = 5? No. So (8,7) is not in R.

Answer(C) (6, 8) ∈ R.

Watch this explained “Two demands, checked one at a time”, 12:38 into A relation as a subset of a product set, and the two extreme cases