2

Prove that every rational number $r$ can be written as $r=\frac{p}{q}$, where $p$, $q$ $\in \mathbb{Z}$, $q \gt 0$ and $p$, $q$ are relatively prime. Moreover the integers $p$ and $q$ are uniquely determined by $r$.

My try: Let $R$ be a relation on $Z \times Z$ such that $(m,n) \sim (m',n')$ if $mn'=nm'$ where $n \ne 0$ and so is $n'$.

This is an equivalence relation. Now I denote such a class by $[m,n]=\frac{m}{n}$, where $m$ and $n$ are in its lowest form.

Git Gud
  • 31,356
tattwamasi amrutam
  • 12,802
  • 5
  • 38
  • 73
  • What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. – user1729 Aug 06 '14 at 10:30
  • @user1729 I am not sure how to proceed.. Do I have to use equivalence classes?? – tattwamasi amrutam Aug 06 '14 at 10:36
  • You should edit your question to talk about equivalence classes - this shows that you are thinking about the question! (You want to show that each equivalence class has a representative of this form, and that this representative is unique.) – user1729 Aug 06 '14 at 10:38
  • I was just wondering ..... what is a rational number anyways? – John Joy Aug 06 '14 at 12:34
  • @JohnJoy Per the second paragraph of the question, it's an equivalence class of the elements of $\mathbb Z\times \mathbb Z$. The goal is to show it contains a pair of relatively prime numbers. –  Aug 06 '14 at 13:24
  • @900 900 sit-ups a day It seems to me that if we define a rational number as an equivalence class of the elements of Z×Z and we denote such a class by $\frac{m}{n}$, haven't we just defined a rational number as a number that can be written as $\frac{m}{n}$? Is it really necessary to prove something that we have just defined? – John Joy Aug 06 '14 at 14:42
  • @JohnJoy The purpose of the problem is not the form $\frac m n$, but rather the properties that $m$ and $n$ must satisfy. These aren't immediate from the definition. – Git Gud Aug 06 '14 at 15:12

0 Answers0