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I was reading the article/wiki here When I came across this quote

ObviousFact?: examples:

2+2=4 for most people

Those with higher mathematical knowledge may disagree - not in the Z3 algebraic group. No, 2+2 is still 4 in Z3, it just also happens that 4=1. But this is really insignificant, since 4 is usually defined to be 2+2 or 3+1.

Could someone give me a rough idea for a layman what this person meant when they said that 2+2 != 4 in the Z3 algebraic group?

I'd like to understand the reference some so I can use it one day

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    $\begingroup$ Google "modular arithmetic" $\endgroup$
    – DonAntonio
    May 30 '14 at 19:03
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    $\begingroup$ That doesn't make any sense. It's self-contradictory. One moment it's telling you that $2+2$ isn't $4$ in $Z_3$, the next it concedes "well, $2+2$ is still $4$, but it's is also $1$" in $Z_3$ (which is correct). Also, while groups are part of modern aka abstract algebra, the term algebraic group is much more sophisticated - all we want to do here is call it a group. $\endgroup$
    – blue
    May 30 '14 at 19:03
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    $\begingroup$ @seaturtles Self-contradictory? Maybe you shouldn't overstate your case. It's just "popular math", written in a way to attract attention. You should read this cum grano salis, I think. $\endgroup$
    – user144248
    May 30 '14 at 20:54
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    $\begingroup$ In reference to 2+2=4 it says "not in the Z3 algebraic group" and also says "2+2 is still 4 in Z3." Regardless of writing style, that's a contradiction. $\endgroup$
    – blue
    May 30 '14 at 21:09
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    $\begingroup$ @seaturtles Yes, if read literally, then that's a contradiction. If interpreted literally, many everyday propositions do not satisfy mathematical rigor. (Do you happen to know the famous work Logic and conversation?) I think what these guys wanted to say is, "$2+2\neq 4$ - but no, hold, $2+2=4$ but also $4=1$". In any case, to say that this is a "self-contradictory text" is a bit far-fetched. $\endgroup$
    – user144248
    May 30 '14 at 21:39
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Think of $Z_3$ as a clock with only $3$ hours, (i.e. $0,1,2$). So if you are at $2$ o'clock and you go $2$ hours forward, you will be back at $1$ o'clock. This is the best way to think of it in my opinion.

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    $\begingroup$ That is $\mathbb{Z}_4$. $\mathbb{Z}_3$ is $\{0, 1, 2\}$. $\endgroup$
    – 6005
    May 30 '14 at 19:03
  • $\begingroup$ oopsy, I'll edit it. I said $Z_3$ and then provided an example in $Z_4$. Such is life. $\endgroup$
    – Rocket Man
    May 30 '14 at 19:04
  • $\begingroup$ So this Z group. This just means the only numbers available to use within that mathematical framework are all integers below n? $\endgroup$ May 30 '14 at 19:07
  • $\begingroup$ Yes, non-negative integers, including $0$, since any negative integer can be congruent to some element in $\mathbb Z_n = \{0, 1, \cdots, n-1\}$, modulo $n$. $\endgroup$
    – amWhy
    May 30 '14 at 19:09
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    $\begingroup$ @AnotherUser You are allowed to use integers above $n$, as long as you understand them to be alternative names for smaller integers. For example, in $Z_3$ you can use 4 as much as you like, as long as you understand that it is another name for 1. It is a remarkable fact that you can do this and never get mixed up. For example, what is $4×4$? It is 16, but this must be the same as $1×1=1$, so 16 must be another name for 1. But 16 is another name for 1 in this system, because if you start at 0 and go 16 hours forward on the clock, you end at 1. $\endgroup$
    – MJD
    May 30 '14 at 19:17
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Addition in $\mathbb Z_n$ is modulo $n$. In your case, $n = 3$. So $2 + 2 = 4 = 1$ modulo $3$. That means that $1$ is the remainder of $4$ when divided by $3$,

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    $\begingroup$ The elements of $\mathbb Z_n$ are by convention the smallest non-negative integers that exhaust all possible remainders when each integer in $\mathbb Z$ is divided by $n$. $$\mathbb Z_n = \{0, 1, 2, \cdots n-1\}.$$ Every integer is equivalent to one of these elements, modulo $n$. $\endgroup$
    – amWhy
    May 30 '14 at 19:30
  • $\begingroup$ Note that even in this case you are taking representatives of cosets and $4$ is in the same coset as $1$ $\endgroup$ May 30 '14 at 19:48
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$\mathbb{Z}_{3}$ has exactly $3$ elements.

You can denote them e.g. as $\overline{0},\overline{1},\overline{2}$ where $\bar{i}$ stands for $i+3\mathbb{Z}=\left\{ i+3n\mid n\in\mathbb{Z}\right\} $.

In this context $\overline{4}=\overline{1}$ and a nice way to describe the addition on $\mathbb{Z}_{3}$ is simply $\overline{i}+\overline{j}=\overline{i+j}$.

Then $\overline{2}+\overline{2}=\overline{4}=\overline{1}$ and for convenience the bars are quite often left out: $2+2=4=1$.

The symbols $1$ and $4$ should be interpreted here as labels that cover exactly the same mathematical object: set $\left\{ 1+3n\mid n\in\mathbb{Z}\right\} =\left\{ 4+3n\mid n\in\mathbb{Z}\right\} $.

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