0

Consider the equivalence relation $\sim$ on $\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})$ defined by $(a,b) \sim (c,d)$ if $a \cdot d = b \cdot c$. Describe the equivalence classes in terms of familiar mathematical objects?

The above is a homework problem. I can see some patterns in the equivalence classes, however I'm not sure how to answer or approach the question given. Thanks!

clay
  • 2,713
  • 23
  • 34
  • Have you written down some examples of equivalent pairs? What do they remind you of? Can you describe the set of all pairs equivalent to (1,2) for example? – G Tony Jacobs Jul 07 '14 at 12:35
  • [((1,1),(2,2)),((1,2),(1,2)),((2,1),(2,1)),((2,2),(1,1))]. They look like a square grid of points. Is the answer they are looking for a square? – clay Jul 07 '14 at 12:37
  • No, you just want pairs that are equivalent to $(1,2)$. For example, you have $(1,2)\sim (1,2)$, $(1,2)\sim (2,4)$, $(1,2)\sim (3,6)$, etc. Do you see why those are true? – G Tony Jacobs Jul 07 '14 at 12:39
  • $(1,1)\not\sim (2,1)$, as you seem to be suggesting, since $2\neq 1$. – HSN Jul 07 '14 at 12:39
  • @HSN, I think the OP was listing several examples of equivalent pairs, but not saying that they're all equivalent to each other. Look at the use of parentheses. He's writing down part of the equivalence relation as a set of ordered pairs. – G Tony Jacobs Jul 07 '14 at 12:42
  • both of you are right. The pairs equivalent to (1,2) are (k,2k) like [(1,2),(2,4),(3,6),...]. So the equivalence class is like a straight line? – clay Jul 07 '14 at 12:55
  • Kind of, yes. The points lie along a straight line, but they're all integer pairs, and we exclude the origin. I think the solution you're looking for is not geometric. In the example you worked out, every pair of the form $(k,2k)$ is equivalent to $(1,2)$, just like every fraction of the form $k/(2k)$ is equivalent to $1/2$. Reinterpret your ordered pairs as fractions, and you'll find this equivalence relation very familiar. It will also be clear why $0$ is excluded in the second place. – G Tony Jacobs Jul 07 '14 at 13:20
  • O, I see. I was confused, since [] is often used to denote an equivalence class and as a result I didn't notice those extra brackets. My mistake. – HSN Jul 07 '14 at 14:06
  • Thanks G Tony Jacobs! Equiv classes are merely rational numbers. My brain should have figured that out faster. – clay Jul 07 '14 at 15:00

1 Answers1

1

The equivalence classes corespond to rational numbers with $\frac{a}{b}$ representing the class $(a,b)$.

Rene Schipperus
  • 39,526