Practice Misc. Exercise

Let x₁, x₂ ∈ [0, ∞) such that f(x₁) = f(x₂).
x₁/(1+x₁) = x₂/(1+x₂)
x₁(1+x₂) = x₂(1+x₁) ⇒ x₁ + x₁x₂ = x₂ + x₂x₁
Subtract x₁x₂ from both sides: x₁ = x₂.
Conclusion: f is one-one.

Let y ∈ [0, 1). We need to find x.
y = x / (1+x) ⇒ y(1+x) = x ⇒ y + xy = x
y = x - xy ⇒ y = x(1 - y) ⇒ x = y / (1 - y)
Since 0 ≤ y < 1, the denominator (1-y) is positive and ≤ 1.
Also y ≥ 0, so x ≥ 0. Thus x ∈ [0, ∞) exists for every y.
Conclusion: f is onto.

Let x₁, x₂ ∈ R such that f(x₁) = f(x₂).
x₁³ + 5 = x₂³ + 5

Subtract 5 from both sides: x₁³ = x₂³
Since the cube function preserves signs and magnitude uniquely for real numbers, taking the cube root gives:
x₁ = x₂

A ∪ B = B is mathematically equivalent to A ⊆ B (A is a subset of B).

Is A ⊆ A? Yes, every set is a subset of itself.
Reflexive: Yes.

If A ⊆ B, does it imply B ⊆ A? No.
Example: A={1}, B={1,2}. A ⊆ B but B ⊈ A.
Symmetric: No.

If A ⊆ B and B ⊆ C, then A ⊆ C. This is a standard property of sets.
Transitive: Yes.

A bijective function from a finite set to itself is essentially a permutation of its elements.
For a set with n elements, the number of bijections is n! (n factorial).

Here, n = 4 (elements are 1, 2, 3, 4).
Number of functions = 4! = 4 × 3 × 2 × 1.

Both functions have the same domain A = {-1, 1} and same codomain R.

For x = 1: f(1) = 1² = 1. g(1) = |1| = 1. (Match)
For x = -1: f(-1) = (-1)² = 1. g(-1) = |-1| = 1. (Match)

Since f(x) = g(x) for all x ∈ A, the functions are equal.

For Reflexive: Must contain {(1,1), (2,2), (3,3)}.
For Symmetric & containing (1,2): Must contain {(1,2), (2,1)}.

Current Base: R₁ = {(1,1), (2,2), (3,3), (1,2), (2,1)}. (This is transitive).
We need to add pairs to break transitivity but keep symmetry.
Option A: Add (2,3) and (3,2).
R₂ = Base ∪ {(2,3), (3,2)}.
Check Transitivity: We have (1,2) and (2,3). Is (1,3) in R₂? No.
So R₂ is Reflexive, Symmetric, Not Transitive.

Option B: Add (1,3) and (3,1).
R₃ = Base ∪ {(1,3), (3,1)}.
Check Transitivity: We have (2,1) and (1,3). Is (2,3) in R₃? No.
So R₃ is also Reflexive, Symmetric, Not Transitive.
Note: Adding all pairs creates the Universal relation, which IS transitive.

Reflexive: {(1,1), (2,2), (3,3)}.
Given: (1,2), (2,3).
Symmetric: Must add (2,1), (3,2).
Transitive: (1,2) & (2,3) ⇒ Must add (1,3). Symmetry ⇒ Must add (3,1).

The relation now contains: {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2), (1,3), (3,1)}.
This includes ALL possible pairs in A × A.
This is the Universal Relation.

There are no smaller equivalence relations containing both (1,2) and (2,3) because transitivity forces all other pairs to be included.