Let R={(x,y)∈R×R│⟦3x⟧=⟦3y⟧}, where ⟦x⟧ ┤ denotes the greatest integer less than or equal to x.
Show that R is an equivalence relation on R.
Determine R⁄R .
To show that R is an equivalence relation on R, we need to prove three properties: reflexivity, symmetry, and transitivity.
1. Reflexivity:
For any x ∈ R, we need to show that (x, x) ∈ R. In this case, we have ⟦3x⟧ = ⟦3x⟧ because the greatest integer less than or equal to x is always equal to itself. Therefore, (x, x) ∈ R, and R is reflexive.
2. Symmetry:
For any (x, y) ∈ R, we need to show that if (x, y) ∈ R, then (y, x) ∈ R. If (x, y) ∈ R, it means ⟦3x⟧ = ⟦3y⟧. Since the greatest integer function is symmetric, ⟦3y⟧ = ⟦3x⟧. Therefore, (y, x) ∈ R, and R is symmetric.
3. Transitivity:
For any (x, y), (y, z) ∈ R, we need to show that if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R. If (x, y) ∈ R, it means ⟦3x⟧ = ⟦3y⟧. Similarly, if (y, z) ∈ R, it means ⟦3y⟧ = ⟦3z⟧. By transitivity of equality, we can conclude that ⟦3x⟧ = ⟦3z⟧, which implies (x, z) ∈ R. Therefore, R is transitive.
Since R satisfies all three properties (reflexivity, symmetry, and transitivity), we can conclude that R is an equivalence relation on R.
Now let's determine R⁄R, the set of equivalence classes of R.
The equivalence class [x] of an element x ∈ R is defined as [x] = {y ∈ R | (x, y) ∈ R}. In this case, for any x ∈ R, [x] contains all y such that ⟦3x⟧ = ⟦3y⟧.
Let's take an example to illustrate this:
If we take x = 1, then [1] = {y ∈ R | ⟦3 × 1⟧ = ⟦3y⟧}, which means [1] = {y ∈ R | 3 = ⟦3y⟧}. The elements in this equivalence class [1] are all the real numbers y such that 3 is the greatest integer less than or equal to 3y. We can see that [1] = {y ∈ R | 1 ≤ y < 4/3}.
Similarly, we can find the equivalence classes for other elements in R.
By determining all the equivalence classes, we can determine R⁄R, which is the set of all equivalence classes of R.