Equivalence Relation R On Real Numbers: Explained

Hey guys! Today, we're diving deep into the fascinating world of equivalence relations, specifically focusing on a relation R defined on the set of real numbers. We'll break down the properties that make R an equivalence relation and then explore how to determine the equivalence class for each real number. Buckle up, it's gonna be a fun ride!

Proving R is an Equivalence Relation

Okay, so the first thing we need to do is demonstrate that the relation R, defined as ∀(x,y) ∈ ℝ², xRy ⇔ x² - y² = 3(x-y), is indeed an equivalence relation. For this to be true, R must satisfy three key properties: reflexivity, symmetry, and transitivity. Let's tackle each one step by step.

Reflexivity: Showing that xRx for all x in R

Reflexivity means that every element must be related to itself. In our case, we need to show that xRx holds true for all real numbers x. So, let's plug x into our relation:

xRx ⇔ x² - x² = 3(x - x)

Simplifying this, we get:

0 = 3(0)

0 = 0

This is clearly true! Therefore, R is reflexive because every real number x is related to itself under the relation R. This is a crucial first step in proving that R is an equivalence relation. Understanding reflexivity is fundamental because it sets the stage for the next two properties.

Symmetry: Showing that if xRy, then yRx

Symmetry means that if x is related to y, then y must also be related to x. In other words, the relation works both ways. Let's assume xRy and see if we can prove yRx.

Assume xRy, which means:

x² - y² = 3(x - y)

We can rewrite the left side as a difference of squares:

(x - y)(x + y) = 3(x - y)

Now, let's rearrange the equation to try and get it into a form that shows yRx. Subtract 3(x - y) from both sides:

(x - y)(x + y) - 3(x - y) = 0

Factor out (x - y):

(x - y)(x + y - 3) = 0

This equation holds if either (x - y) = 0 or (x + y - 3) = 0. Let's consider the second case, x + y - 3 = 0. This implies:

x + y = 3

Now, let's look at yRx. We want to show that y² - x² = 3(y - x).

y² - x² = (y - x)(y + x)

Since x + y = y + x, we can substitute 3 for (x + y):

(y - x)(3) = 3(y - x)

This is exactly what we wanted to show! Therefore, if xRy, then yRx. This confirms the symmetric property of the relation R. Symmetry ensures that the relationship between two numbers is bidirectional.

Transitivity: Showing that if xRy and yRz, then xRz

Transitivity is the final property we need to verify. It states that if x is related to y, and y is related to z, then x must also be related to z. This is the trickiest of the three properties to prove.

Assume xRy and yRz. This means:

x² - y² = 3(x - y) and y² - z² = 3(y - z)

We want to show that x² - z² = 3(x - z).

From x² - y² = 3(x - y), we have (x - y)(x + y) = 3(x - y). This gives us two possibilities:

1. x - y = 0, which means x = y
2. x + y = 3

Similarly, from y² - z² = 3(y - z), we have (y - z)(y + z) = 3(y - z). This gives us two possibilities:

1. y - z = 0, which means y = z
2. y + z = 3

Let's consider the different cases:

• Case 1: x = y and y = z. Then x = z, so x² - z² = x² - x² = 0 and 3(x - z) = 3(x - x) = 0. Thus, x² - z² = 3(x - z).
• Case 2: x = y and y + z = 3. Since x = y, we have x + z = 3, so x = 3 - z. Then x² - z² = (3 - z)² - z² = 9 - 6z + z² - z² = 9 - 6z. Also, 3(x - z) = 3(3 - z - z) = 3(3 - 2z) = 9 - 6z. Thus, x² - z² = 3(x - z).
• Case 3: x + y = 3 and y = z. Since y = z, we have x + z = 3, so x = 3 - z. This is the same as Case 2.
• Case 4: x + y = 3 and y + z = 3. Then x = 3 - y and z = 3 - y, so x = z. This reduces to Case 1.

In all cases, we've shown that x² - z² = 3(x - z). Therefore, R is transitive. Transitivity ensures that the relationship extends consistently across multiple elements.

Conclusion: R is an Equivalence Relation

Since R is reflexive, symmetric, and transitive, we can confidently conclude that R is an equivalence relation on the set of real numbers. Woohoo! Now, let's move on to figuring out the equivalence classes.

Determining the Equivalence Class for Each Real Number

Alright, now that we know R is an equivalence relation, let's figure out the equivalence class for each real number. The equivalence class of a real number x, denoted as [x], is the set of all real numbers y that are related to x by R. In other words:

[x] = {y ∈ ℝ | xRy}

Recall that xRy if and only if x² - y² = 3(x - y). We can rewrite this as (x - y)(x + y) = 3(x - y), which leads to two possibilities:

1. x - y = 0, which means y = x
2. x + y = 3, which means y = 3 - x

Therefore, the equivalence class of x consists of x itself and 3 - x. We can write this as:

[x] = {x, 3 - x}

Examples of Equivalence Classes

Let's look at a few examples to make this crystal clear:

• Example 1: x = 1

  [1] = {1, 3 - 1} = {1, 2}

  So, the equivalence class of 1 is the set containing 1 and 2.

• Example 2: x = 0

  [0] = {0, 3 - 0} = {0, 3}

  The equivalence class of 0 is the set containing 0 and 3.

• Example 3: x = 1.5

  [1.5] = {1.5, 3 - 1.5} = {1.5, 1.5} = {1.5}

  Interestingly, when x = 1.5, the equivalence class only contains 1.5 because 3 - 1.5 = 1.5. This happens when x = 3 - x, which implies x = 1.5. This is a special case where the equivalence class contains only one element.

General Interpretation

In general, for any real number x (except 1.5), its equivalence class will contain two distinct elements: x and 3 - x. These two numbers are related to each other under the relation R. When x is 1.5, it's related only to itself, forming a singleton equivalence class. Understanding the equivalence classes helps us partition the set of real numbers into subsets where elements within each subset are