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For the following relation $R$ on the set $X$ determine whether it is $(i)$ reflexive, $(ii)$ symmetric and $(iii)$ transitive. Give proofs or counter examples. In the case where $R$ is an equivalence relation, describe the equivalence classes of $R$.

$X=\mathbb C,\; a\;R\;b\iff |a|=|b|$

$(i)$ The relation is reflexive since $|a|=|a|$ is true

$(ii)$ The relation is symmetric as if $|a|=|b|$ then $|b|=|a|$

$(iii)$ The relation is transitive if $|a|=|b|$ and $|b|=|c|$ then it follows that $|a|=|c|$

And this is as far as I get. I am struggling when it comes to identifying an equivalence class for this relation. I honestly don't even think I'd know where to begin for this. A nudge in the right direction would be greatly appreciated.

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  • $\begingroup$ Also, at this level, you should probably justify each property by citing the corresponding property of equality. $\endgroup$ – The Chaz 2.0 Nov 2 '15 at 19:10
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If I've understood correctly, you need to characterize the equivalence classes for the above equivalence relation on the set $X=\mathbb{C}$.

The definition of complex absolute value is very geometric. Given some complex number $a+bi$, $\lvert a+bi\rvert=\sqrt{a^2+b^2}$. It follows from here that an equivalence class on $\mathbb{C}$ with the relation of absolute value is actually a circle of radius corresponding to the absolute value of elements in the class, centered at the origin.

So, multiple different classes form a collection of concentric circles about the origin. Can you see why?

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  • $\begingroup$ You've understood perfectly. The explanation you have given really helps and if I understand you correctly then the equivalence class would be the set containing the element $z : z=\sqrt{x^2+y^2}$ and $x,y$ $\epsilon$ $\mathbb R$. I don't really know how to write this (assuming I'm correct) in the form of an answer, though. $\endgroup$ – Overclock Nov 2 '15 at 19:24
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    $\begingroup$ It should be sufficient to say that for some absolute value $\lvert x\rvert$, the equivalence class is precisely $\{z=a_z+b_zi\in \mathbb{C}: \sqrt{a_z^2+b_z^2}=x\}$. $\endgroup$ – Alekos Robotis Nov 2 '15 at 19:36
  • $\begingroup$ Absolutely fantastic, thank you very much for clarifying that. $\endgroup$ – Overclock Nov 2 '15 at 20:04
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What number(s) is $-3 + 4i$ equivalent to? What number(s) is $4.68$ equivalent to? What number(s) is $0$ equivalent to?

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  • $\begingroup$ -3 is equivalent to -3 and 3, 4.68 is equivalent to 4.68 and -4.68 and 0 is equivalent to 0. $-3+4i$ would be equivalent to $-3+4i$, 5 and -5 I think. That comes from $|-3+4i|$? $\endgroup$ – Overclock Nov 2 '15 at 19:12
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    $\begingroup$ -3 is also equivalent to $3(1/2 + i *\sqrt 3 /2)$. $\endgroup$ – fleablood Nov 2 '15 at 19:14
  • $\begingroup$ @Overclock, you're making progress. Yes, you should be using $| -3 + 4i|$, but write it as $\sqrt{(-3)^2 + 4^2)}$ and compare this to the distance formula in the xy plane. $\endgroup$ – The Chaz 2.0 Nov 2 '15 at 19:18
  • $\begingroup$ @TheChaz2.0 So having got to the point where I now understand that the equivalence class is the set containing $\sqrt{x^2+y^2} = z$ where $x,y$ $\epsilon$ $\mathbb R$ but I'm not entirely sure I'm on the right tracks in terms of summarising an answer. $\endgroup$ – Overclock Nov 2 '15 at 19:40
  • $\begingroup$ For each non-negative real number, you have an equivalence class. I would call these classes something like $S_i$, where $i \in \mathbb R$. $\endgroup$ – The Chaz 2.0 Nov 2 '15 at 19:44
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Hint:

You can find the answer if you think to complex number $a$ in polar form $a=|a| e^{i\theta}$. than you have an equivalence class for any value of $|a| \in \mathbb{R}$.

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