Can a be written k + b for all integers a and b?

Question

Instinctively I think yes, since $a \equiv b$ (modulo 1) for all $a$ and $b$.

Answer 1

One way to think about congruence modulo $c$ is in the following way: $$a \equiv b \text{ (mod c)}$$ if $c$ divides $b - a$.

Let's check that for any $a$ and $b$, $a \equiv b$ (mod $1$). We need to show $1$ divides $b - a$. But since $b$ and $a$ are integers, so is $b - a$. Clearly, since $1 \cdot (b - a) = (b - a)$, $1$ divides $b - a$, so $$b \equiv a \text{ (mod 1)}.$$

Answer 2

Yes. You are right. Here is a maybe incomplete proof that demonstrates the idea:

WLOG, $a=1*a \implies a=1*a+0 \implies a\equiv_10$.

Answer 3

As all integers $a$ are such $a = 1*a$ all integers are such $a = 0 \mod 1$ and any integers $a, b$ are such $a = b \mod 1$. The residue class(es) are/is simply {0} = $\mathbb Z$.

So as commented. Very boring.