elementary number theory - What does $a$ and $b$ leaving equal remainders upon division by $m$ means

Theorem $3.1.3$

When $a$ and $b$ are nonnegative integers, the relationship $a\equiv b\text{(mod }m)$ is equivalent to $\underline{\text{$a$ and $b$ leaving equal remainders upon division by $m$}}$

(from UTM "A Readable Introduction to Real Mathmatics" Chapter 3)

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"is equivalent to $a$ and $b$ leaving equal remainders upon division by $m$"

I'm trying to understand what this means, is it saying:

Let $a,b\ge0$

$a\equiv b\mod m\leftrightarrow (a\equiv s\mod m\wedge b\equiv s\mod m)$, where $s$ is the same remainder

Thanks for your help.

Answer

It means what I think you think it means.

For any $a, b \ge 0$ there are unique integers $j,k,s,t$ so that $a = jm + s; 0\le s < m$ and $b=km + t; 0 \le t < m$. $t$ and $s$ are called the remainders of $a$ and $b$ respectively.

So that statement is saying:

$a \equiv b \pmod m \iff s =t$.

It's easy to verify.

Pf: If $a \equiv b \pmod m$ then $m|a-b = (jm + s)-(km+t) = (j-k)m + (s-t)$ so $m|s-t$ but $0\le t< m; 0\le s< m$ so $-m < t-s < m$ so if $m|s-t$ that means $s-t = 0$ and $t=s$.

If $t=s$ then $a-b = (jm+s)-(km+t)= (j-k)m + (s-t) = (j-k)m$. So $m|a-b$ and $a\equiv b\pmod m$

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This means you can define $a\equiv b \pmod m$ in two ways.

You can define it as $a\equiv b\pmod m$ if $m|a-b$.

Or

You can define it as $a\equiv b\pmod m$ if $a$ and $b$ have the same remainder when divided by $m$

You can also define it as $a \equiv b\pmod m$ if there is an integer $k$ so that $a = km + b$.