Miscellaneous Exercise answers: Relations and Functions

Open NCERT p. 15One way to think about it

1. f really lands in the target: |x| < 1 + |x|, so the size of x/(1 + |x|) is less than 1. So −1 < f(x) < 1.
2. Sign: the bottom 1 + |x| is always positive, so f(x) has the same sign as x: positive when x > 0, 0 when x = 0, negative when x < 0.
3. One-one: suppose f(x₁) = f(x₂). By the sign fact, x₁ and x₂ are either both ≥ 0 or both < 0.
4. If both are ≥ 0, then |x| = x: x₁/(1 + x₁) = x₂/(1 + x₂). Cross-multiply: x₁ + x₁x₂ = x₂ + x₁x₂, so x₁ = x₂.
5. If both are < 0, then |x| = −x: x₁/(1 − x₁) = x₂/(1 − x₂). Cross-multiply: x₁ − x₁x₂ = x₂ − x₁x₂, so x₁ = x₂. So f is one-one.
6. Onto: take any y with −1 < y < 1.
7. If 0 ≤ y < 1, let x = y/(1 − y). This x is ≥ 0, and 1 + x = 1/(1 − y), so f(x) = x/(1 + x) = y.
8. If −1 < y < 0, let x = y/(1 + y). This x is < 0, and 1 − x = 1/(1 + y), so f(x) = x/(1 − x) = y.
9. Every y between −1 and 1 is reached, so f is onto. Hence f is one-one and onto.

In shortf is one-one and onto.

Watch this explained “Trim until it is onto”, 18:28 into Onto as the demand that nothing in the codomain be left unhit

Open NCERT p. 15One way to think about it

1. Suppose f(x₁) = f(x₂), so x₁³ = x₂³, which means x₁³ − x₂³ = 0.
2. Factor: x₁³ − x₂³ = (x₁ − x₂)(x₁² + x₁x₂ + x₂²).
3. The second factor x₁² + x₁x₂ + x₂² can be rewritten as (x₁ + x₂/2)² + 3x₂²/4, which is always ≥ 0, and is 0 only when x₁ = x₂ = 0.
4. So except in that one case, the second factor is positive, which forces x₁ − x₂ = 0, giving x₁ = x₂.
5. When x₁ = x₂ = 0, they are equal anyway. So in every case x₁ = x₂, and f is injective.

In shortf is injective (one-one).

Watch this explained “Same rule, different domain”, 7:29 into What one-one asks of distinct inputs, and what many-one allows

Open NCERT p. 15Matches NCERT’s answer

1. Reflexive: every set is a subset of itself, so A ⊂ A and A R A for every A in P(X). R is reflexive.
2. Transitive: if A ⊂ B and B ⊂ C, every element of A is in B, and so in C. So A ⊂ C, and R is transitive.
3. Symmetric: X is non-empty, so the empty set φ and X itself are two different members of P(X).
4. φ ⊂ X, so φ R X. But X is not a subset of φ, because X has at least one element and φ has none. So X R φ is false.
5. So R is not symmetric, and therefore R is not an equivalence relation on P(X).

AnswerNo, R is not an equivalence relation on P(X). It is reflexive and transitive but not symmetric.

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. 15Matches NCERT’s answer

1. The set {1, 2, ..., n} is finite, and the function goes from this set back to itself.
2. For a function from a finite set to itself, being onto is exactly the same as being one-one — each forces the other.
3. So counting onto functions is the same as counting one-one functions from the set to itself.
4. A one-one function from a finite set to itself must send the n elements to n different elements — that is exactly a rearrangement (permutation) of the set.
5. The number of rearrangements of n things is n! (n factorial).

Answern! (n factorial) onto functions.

Watch this explained “Use the theorem instead of counting”, 17:51 into Bijections, and why on a finite set either half implies the other

Open NCERT p. 15Matches NCERT’s answer

1. Work out f at every point of A: f(−1) = 1 − (−1) = 2, f(0) = 0, f(1) = 1 − 1 = 0, f(2) = 4 − 2 = 2.
2. Work out g at every point of A: g(−1) = 2|−1.5| − 1 = 3 − 1 = 2, g(0) = 2|−0.5| − 1 = 1 − 1 = 0.
3. g(1) = 2|0.5| − 1 = 1 − 1 = 0, g(2) = 2|1.5| − 1 = 3 − 1 = 2.
4. f and g give the same output at every point of A: 2, 0, 0, 2.
5. Two functions with the same domain are equal when they agree at every input, so f = g.

AnswerYes, f and g are equal.

Watch this explained “The one place you can check everything”, 2:38 into What one-one asks of distinct inputs, and what many-one allows

Open NCERT p. 16Matches NCERT’s answer

1. Reflexive forces (1,1), (2,2), (3,3) to be in the relation.
2. The relation must already contain (1,2) and (1,3). Symmetric then forces (2,1) and (3,1) in too.
3. That is 7 pairs fixed. Only (2,3) and (3,2) are still undecided, and symmetric means they must come as a pair — both in, or both out.
4. If both are left out: (2,1) and (1,3) are both in the relation, so transitivity would need (2,3) — but it is not there. This relation fails transitivity.
5. If both are put in: the relation becomes the universal relation (all 9 pairs), which is always transitive — so this one does not satisfy 'not transitive'.
6. So exactly one relation is reflexive and symmetric but not transitive.

Answer(A) 1

Watch this explained “Given a pattern, how many relations make it?”, 15:27 into Reflexive, symmetric and transitive as three demands that can fail independently

Open NCERT p. 16Matches NCERT’s answer

1. An equivalence relation on {1, 2, 3} matches exactly one way of splitting the set into non-overlapping boxes (a partition).
2. The possible partitions of {1, 2, 3} are: all separate {1}{2}{3}; {1,2}{3}; {1,3}{2}; {2,3}{1}; and all together {1,2,3} — 5 in total.
3. The relation must contain (1, 2), which means 1 and 2 must be in the same box.
4. Only two of the five partitions put 1 and 2 in the same box: {1,2}{3} and {1,2,3}.
5. So exactly 2 equivalence relations contain (1, 2).

Answer(B) 2

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