   Chapter 2.5, Problem 36E

Chapter
Section
Textbook Problem

Let x and y be integers. Prove that if there is an equivalence class [ a ] modulo n such that x   ∈   [ a ] and y   ∈   [ a ] , then ( x ,   n ) = ( y ,   n ) .

To determine

To prove: If there is an equivalence class [a] modulo n, such that x[a] and y[a], then (x,n)=(y,n), where x and y are integers.

Explanation

Given information:

x and y are integers.

Formula used:

i) Definition: Equivalence class:

Let R be an equivalence relation on the non-empty set A. For each aA, the set

[a]={xA|xRa} is called the equivalence class containing a.

ii) Definition: Congruence Modulo n

Let n be a positive integer, n>1. For integers x and y, x is congruent to y modulo n, if and only if xy is a multiple of n. We write xy(modn) to indicate that x is congruent to y modulo n.

iii) Definition: An integer d is a greatest common divisor of a and b if all these conditions are satisfied:

1. d is a positive integer.

2. d|a and d|b.

3. c|a and c|b imply c|d.

Proof:

Let x,y,a, such that x[a] and y[a].

x[a] implies that xa(modn) and y[a] implies that ya(modn)

By transitivity and symmetry,

xy(modn)

By using definition, xy is multiple of n.

Let xy=nq for q.

This implies y=xnq and x=y+nq.

Let d1=(x,n) and d2=(y,n)

By using definition,

d1|x and d1|n, then d1 divides any linear combination of x and n

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