# A number Theory question : Proof of a mod

Let $a,\ b$ be integers with $b < 0$. Show that $a\pmod b \in (b,0]$

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I didn't know what you meant with "E". The natural assumption is that it is $\in$. Please make sure the edit is correct. –  EuYu Oct 7 '12 at 1:43
This seems like a definition. What are you using to define $\text{mod}$? –  Alexander Gruber Oct 7 '12 at 2:44

If $a$ and $b \neq 0$ are integers, then "$a \equiv c \pmod {|b|}$" is equivalent to "$|b|$ divides $a-c$". We know there always exists an integer $c \in \{0,1,\ldots,|b|-1\}$ such that $a \equiv c \pmod {|b|}$ by the Division Algorithm (see also Euclidean Division).
Not many people talk about $a \equiv d \pmod b$ for negative $b$ (since $b$ and $-b$ differ only by a unit, and thus $b$ divides an integer $x$ if and only if $-b$ divides $x$), but if we defined it as meaning "$b$ divides $a-d$" it would be consistent, and you can verify your claim.
Suppose $a \equiv c \pmod {|b|}$ where $b$ is negative and $c \in \{0,1,\ldots,|b|-1\}$. Then:
• If $c=0$, then $a \equiv 0 \pmod b$.
• If $c>0$, then $a \equiv b-c \pmod b$, where $b-c \in \{b+1,b+2,\ldots,-1\}$.
In some sense, $\{b+1,b+2,\ldots,0\}$ forms a complete residue system modulo $b$. But, we would avoid this notation since $\pmod b$ arithmetic is the same thing and $\pmod {|b|}$, since $\mathbb{Z}/b\mathbb{Z} = \mathbb{Z}/|b|\mathbb{Z}$.