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Sometimes the rational numbers $\mathbb{Q}$ are defined via equivalent classes $[(a.b)]\subset\mathbb{Z}\times\mathbb{Z}$ of integers. In general we have $(a_1,b_1)\sim (a_2,b_2):\Leftrightarrow a_1b_2=a_2b_1$.

How does such a class $[(a,b)]$ looks like if $\gcd(a,b)=1$?

  • Are you looking at the subset of $\mathbb{Z}\times\mathbb{Z}$ of pairs $(a,b)$ such that $\gcd(a,b)=1$ or just the equivalence class in $\mathbb{Z}\times\mathbb{Z}$ provided $\gcd(a,b)=1$? – Michael Burr Sep 29 '17 at 10:00
  • Also, rational numbers are equivalence classes of $\mathbb{Z}\times(\mathbb{Z}\setminus{0})$ to avoid $\frac{0}{0}$. – Michael Burr Sep 29 '17 at 10:01
  • Hint: how did you learn to simplify fractions back in Elementary school? – John Coleman Sep 29 '17 at 10:02

1 Answers1

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The class looks like this:

$$\{\dots(-2a, -2b), (-a, -b), (a, b), (2a, 2b), (3a, 3b)\dots\}$$

Also, every equivalence class $C$ can be written as $[(a,b)]$ such that $\gcd(a,b)=1$. This is pretty easy to show and good practice!

5xum
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  • But I thought that every class looks like ${...(-2a,-2b),(-a,-b),(a,b),...}$? –  Sep 29 '17 at 10:06
  • @R.A. How is that different from what I wrote? – 5xum Sep 29 '17 at 10:08
  • It is not at all but in my excercise I shall answer two questions: How does a general class looks like and how does a class with the gcd condition looks like. So the answer is that there is no difference? –  Sep 29 '17 at 10:10
  • @R.A. Well, there's a difference in that if $\gcd(a,b)=d\neq 1$, then the equivalnce class also contains $(\frac{a}d, \frac bd)$ which is not in ${\dots(-2a, -2b), (-a, -b), (a, b), (2a, 2b), (3a, 3b)\dots}$. – 5xum Sep 29 '17 at 10:12
  • @R.A. But the point of the exercise is that all equivalence classes contain a representative $(a,b)$ such that $\gcd(a,b)=1$. – 5xum Sep 29 '17 at 10:12