Question

Prove that the relationship \(X \sim Y\), defined by \(X \sim Y\) if \(Y\) can be expressed in the form $$ Y=\frac{a X+b}{c X+d} $$ with \(a, b, c\) and \(d\) as integers, is an equivalence relation on the set of real numbers R. Identify the class that contains the real number \(1 .\)

Short answer

Reflexive, symmetric, and transitive properties hold. The equivalence class containing 1 includes all numbers that can be written as a fraction \(\frac{a+b}{c+d}\) with integers \(a, b, c,\) and \(d\).

Step-by-step solution

Define Reflexivity

Check if \(X \sim X\) for any real number X. To do this, set \(a = d = 1\) and \(b = c = 0\) in the relation \(Y=\frac{a X+b}{c X+d}\). This simplifies to \(Y = X\). Hence, \(X \sim X\).

Define Symmetry

Check if \(X \sim Y\) implies that \(Y \sim X\). If \(Y=\frac{a X+b}{c X+d}\), we need to find if \(X\) can be written in a similar form in terms of \(Y\). By solving for \(X\) in terms of \(Y\), we get \(X = \frac{d Y - b}{-c Y + a}\). Therefore, \(Y \sim X\).

Define Transitivity

Check if \(X \sim Y\) and \(Y \sim Z\) implies that \(X \sim Z\). If \(Y = \frac{a X + b}{c X + d}\) and \(Z = \frac{e Y + f}{g Y + h}\), substitute \(Y\) into the second equation to get \(Z = \frac{e(\frac{a X + b}{c X + d}) + f}{g(\frac{a X + b}{c X + d}) + h}\). Simplifying, we get \(Z\) in terms of \(X\), thus proving \(X \sim Z\).

Identify the class that contains the number 1

To identify the class that contains 1, set \(X = 1\). The class will be all numbers \(Y\) such that \(Y = \frac{a \cdot 1 + b}{c \cdot 1 + d} = \frac{a + b}{c + d}\). This means any number that can be written as a ratio of integers \(a + b\) and \(c + d\) is in the equivalence class of 1.

Key concepts

Reflexivity

Reflexivity means that every element is related to itself. To verify this for our relation, we need to check if any real number, say X, is related to itself.
By definition, we have that X is related to Y if and only if we can express Y as \(\frac{aX + b}{cX + d}\).
For reflexivity, set a = 1, b = 0, c = 0, and d = 1, then the expression becomes Y = \(\frac{1 \times X + 0}{0 \times X + 1}\) = X. Hence, X is related to itself, which means our relation is reflexive.

Symmetry

A relation is symmetric if whenever X is related to Y, then Y is also related to X.
Given that Y is related to X if Y = \(\frac{aX + b}{cX + d}\), we need to express X in terms of Y in the same form.
Solve for X from the equation: \ Y = \frac{aX + b}{cX + d}.\ This can be rearranged to find X in terms of Y. Therefore, \(X = \frac{dY - b}{-cY + a}\). Hence, the relation is symmetric because you can switch the places of X and Y and still preserve the form of the relation.

Transitivity

Transitivity means that if X is related to Y and Y is related to Z, then X must be related to Z.
First, assume X is related to Y, which means Y = \(\frac{aX + b}{cX + d}\) and Y is related to Z such that Z = \(\frac{eY + f}{gY + h}\).
To prove transitivity, substitute Y into the equation for Z: Z = \(\frac{e(\frac{aX + b}{cX + d}) + f}{g(\frac{aX + b}{cX + d}) + h}\). Simplifying this expression, you'll get Z in terms of X. This shows that Z is related to X, completing the transitivity requirement.

Equivalence Class

An equivalence class groups all elements related to a particular element. To identify the equivalence class containing the number 1, set X = 1.
Then, by our established relation, Y will be written as Y = \(\frac{a \times 1 + b}{c \times 1 + d}\) = \(\frac{a + b}{c + d}\).
This means the equivalence class of 1 includes all numbers that can be expressed as a ratio of integers. Any pair of integers (a, b) and (c, d) will determine a number in the equivalence class of 1, creating a vast set of related numbers.