Why is -8 $\equiv$ 6 mod 7? I read in a book that $-8 \equiv 6 \bmod 7$ which means that $-8$ and $6$ leave the same remainder when divided by $7.$
The remainder when $-8$ is divided by $7$ is $-1.$ But when $6$ is divided by $7,$ isn't the remainder $6$?
I recognise that we can write $7\cdot1 - 1=6$, so from here it seems that the remainder is $-1.$ Why is the above reasoning (remainder${}= 6$) incorrect?
 A: The definition is $a\equiv b\mod n$ iff $n$ divides $a-b$.
Here $-8\equiv 6\mod 7$ since $7$ divides $-8-6 = -14$.

Another definition is that $a\equiv b\mod n$ iff both leave the same remainder modulo $n$. But note that the remainder is between $0$ and $n-1$.
Here $-8$ and $6$ have the same remainder $6$ modulo $7$, since $-8 = (-2)\cdot 7 + 6$.
A: Imagine a clock numbered 0,1,2,3,4,5,6 in the usual clockwise orientation. Then $7 \equiv 0 \pmod 7$: if you go 7 numbers clockwise, you're back where you started. Likewise, $-7 \equiv 0 \pmod 7$ (going a full rotation anticlockwise).
Then consider -8. That's a full rotation anticlockwise, plus one further anticlockwise rotation through angle $2\pi/7$. Where do you end up? At 6.
A: The definition that two numbers are congruent if they leave the same remainder on division by the modulus is only valid if we have some agreement as to what "remainder" means.  If we say "the same smallest positive remainder", then it works.
It's better to define congruence by saying that two numbers are congruent if their difference is divisible by the modulus.  Then there's no ambiguity.
A: Two integers are said to be equivalent modulo $n$ if their difference is divisible by $n$.
You have an unclear definition of remainder - but for any integer $m$ you can express $m=qn+r$ with $0\le r\lt n$, and with this consistent expression of the remainder you get $6$ in both cases.
Really when we work modulo $n$ we are not working with individual numbers, but with equivalence classes, with two numbers being equivalent if their difference is divisible by $n$. With this understanding $-1\equiv 6$ in your case, and the answers are the same up to multiples of $7$.
A: $...=-8=-1=6=13=...=7k+6(\text{mod} {7})$, $k\in Z$
The numbers of the form $7k+6$ are equal $(\text {mod}{7})$
It is the equivalence class $\text{C}[6]=\left\{...,-15,-8,-1,6,13,20,...\right\}$
