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I understand what the Modulo/Modulus is and how it operates and everything, but a textbook is using some notation that I just can't quite grasp. Basically, we have the solution

$(x,y) = (\pi($mod$(2\pi)),0)$

From the problem, clearly I can deduce the answers for x are $\pi, 3\pi, 5\pi....$ But I'm just not quite sure what this is notation is saying. I mean, it looks as if it's $\pi$ multiplied by mod$(2\pi)$! But I'm pretty sure that's not right! Additionally, when it says mod$(2\pi)$ I'm not clearly understanding that either. What am I taking the modulus of exactly, with respect to what and what? I mean usually if you say $3\pi$ mod $2\pi$ then the answer is $\pi$. How do I link that thought process to what is going on here? Thanks!

MJD
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Spaderdabomb
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    Extremely poor, bad, infelicitous choice of notation on the author’s part. What is being said is that $x\equiv\pi\pmod{2\pi}$, and $y=0$. In other words, $x$ is an odd multiple of $\pi$, $y$ is zero. – Lubin Jul 31 '13 at 12:10
  • infelicitous...looool. Ok yeah I figured, it is weird notation though, not sure I like it. So basically it's just letting me know that $\pi$ is the modulus answer, but x could be $3\pi, 5\pi...$ etc. – Spaderdabomb Jul 31 '13 at 12:31
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    Not quite: you’re treating “mod” as a function that gives a result, an “answer”. The other way to look at an expression of form $a\equiv b\pmod r$ is simply that $b-a$ is a multiple of $r$. Nothing more nor less than that. Similar concept, but very different in outlook. – Lubin Jul 31 '13 at 12:36
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    To add to the previous comment by @Lubin: when working with trigonometric functions (particularly when doing analytic geometry), it is not uncommon to give answers in $\mathbb{R}/2\pi\mathbb{Z}$, which is probably what is meant by mod $2\pi$. It still is bad notation to write it like this, though. Lubin, are you going to put your comments in an answer? – HSN Jul 31 '13 at 12:46
  • Ok yeah that makes sense, thanks! – Spaderdabomb Jul 31 '13 at 12:48

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