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Title says it all. Google doesn't have anything. Trying to get it to let me submit by typing more words. There is nothing really to be added but it wont let me submit without this.

sabertenn
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    Because $-1+3=2$. – Shaun Feb 05 '21 at 01:30
  • more generally, -1 - 2 is divisible by the modulo 3 – Maths explorer Feb 05 '21 at 01:32
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    It's worth noting that in some contexts, a = b mod c is not a computation of a function (i.e., giving the value a to the output of "b mod c") but just a statement that a - b is divisible by c. This is more common in math than in e.g. computer science but it is worth keeping in mind because it simplifies a lot. This is not to say that the choice of 0, 1, 2, ..., b-1 as representatives of a value of a mod b (when a is an integer and "a mod b" is a function output) is not a natural one, which may be what is going on here. You move a number by a multiple of b until you land in this interval. – leslie townes Feb 05 '21 at 01:39
  • By definition, $a \equiv b\pmod{c} \iff c|(a-b)$. Thus, $3|[(-1) - (2)].$ Further, per leslie townes comment above, $2$ is clearly the unique value $x$ in ${0,1,2}$ such that $3|[-1] - (x)].$ Further, (again) per leslie townes comment, it is very common practice for congruency classes $\pmod{n}$ to be represented by ${0,1,\cdots, (n-1)}.$ – user2661923 Feb 05 '21 at 01:57
  • Please, be honest and show some effort. Reading the second paragraph in the wikipedia article could be a better start. – PinkyWay Feb 05 '21 at 02:15
  • someone is mad they don't get laid – sabertenn Feb 05 '21 at 19:13

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If you ignore the meaning of $ -1 $ and consider it an arbitrary symbol, what properties does it have? By definition, it is the value such that $(-1) + 1 = 0$.

But, working mod 3, $2+1= 3 \equiv 0.$ But from earlier, we have $0 = (-1) + 1$. If you combine the two into $2+1 \equiv (-1)+1 \mod 3$ and drop the $+1$ from both sides, you get $2 \equiv -1 \mod 3$.

redroid
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