$(x+y)\bmod n = (x \bmod n +y\bmod n)\bmod n\,$ [Congruence Sum, Product Rules]

Question

For all $n \in \mathbb{N}$, $n > 1$ and $x,y \in \mathbb{Z}$ prove that

\begin{equation} (x+y) \text{ mod } n = ((x \text{ mod } n)+(y \text{ mod } n))\text{ mod } n. \end{equation}

Would this be a proof if I say:

Let $x = x' \text{ mod } n$ and let $y=y' \text{ mod } n$. Then

\begin{equation} x+y = x'+y' \text{ mod } n. \end{equation}

Now, if we look at the current thing we want prove, this will follow:

\begin{equation} x+y \text { mod } n = (x \text{ mod } n + y \text{ mod } n) \text{ mod } n. \end{equation}

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I'm not sure if this is allowed or will count as a "proof" at all. Maybe you can show me some other ways of doing this, more precise ones?

Answer 1

Recall: same remainder $\!\iff\!$ congruent, i.e. with $\,\bar x = x\bmod n =$ remainder
we have $\ {\bar{\color{#0a0}a}\ =\ \bar{\color{#c00} b}^{\vphantom{i}}}\color{c00}\,\ \ \iff \ \ \,\color{#0a0}a\:\ \ \equiv\ \ \ \color{#c00}b \ \ \pmod{\!n},\:$
so $\ \overline{\color{#0a0}{x\!+\!y}}^{\vphantom{|^|}}\, =\, \overline{\color{#c00}{\bar x \!+\! \bar y}} \color{#c00}\iff \color{#0a0}{x\!+\!y}\:\!\equiv\:\!\color{#c00}{\bar x \!+\! \bar y}_{\vphantom{|_|}},\,$ which is true by Congruence Sum Rule.
The same proof also works for products by using the $ $ Congruence Product Rule, $ $ so

$$\bbox[1px,border:1px solid #c00]{\bbox[6px,border:1px solid #c00]{\begin{align} (x+y)\bmod n &\,=\, (x\bmod n + y\bmod n)\bmod n\\[.2em] (x\,*\,y)\bmod n &\,=\:\! (x\bmod n \,*\, y\bmod n)\bmod n\\[.4em] {\rm i.e.}\ \ \ \ \ \ \ \ \overline{x+y}\ &=\ \overline{\bar x+\bar y} \\ \overline{x\ *\ y}\ &=\ \overline{\bar x\ *\ \bar y}\\[.4em] {\rm i.e.}\ \ \ \ h(x+y) &\,=\, h(x)+_{\small n}h(y)\\[.2em] \ \ h(x\,*\,y) &\,=\, h(x)\,*_{\small n}h(y)\\ \end{align}}}\qquad\qquad$$

i.e. $\,h(x) := x\bmod n\,$ is a ring homomorphism from $\,\Bbb Z\,$ to $\,\Bbb Z_n\,$ which, being surjective (onto), enables transporting from $\,\Bbb Z\,$ to $\,\Bbb Z_n\,$ all the ring laws, e.g. associative, commutative, distributive.

Key Idea $ $ Generally, by proceeding as above, using said equivalence and congruence laws, erasing all $\!\bmod\!$ operations yields an equivalent congruence in any polynomial expression (i.e. where $\!\bmod\!$ appears in arguments of sums and products, but not in exponents). For example let's do that for a power analog of the above: $$\begin{align}(\color{#c00}{g^b\color{#bbb}{\bmod n})^a\color{#bbb}{\bmod n}} \,&=\, \color{#0a0}{(g^a \color{#bbb}{\bmod n})^b\color{#bbb}{\bmod n}}\\[.2em] \iff \color{#c00}{(g^b)^a} &\equiv\, \color{#0a0}{(g^a)^b}\!\!\pmod{\!n} \end{align}\qquad\qquad$$

Since congruence arithmetic inherits all common (ring) arithmetic laws (commutative, associative, distributive), as above, it is much easier to use our well-honed arithmetical intuition to first prove these properties in congruence equation form, then at the last moment - if need be - compute their normal form reps by applying a $\!\bmod\!$ operation.

See here for much further discussion of $\!\bmod\!$ as an operator vs. (congruence) relation.

See here for how to interpret the above as transporting the ring operations on cosets in the quotient ring $\,\Bbb Z/n\,$ to corresponding operations on their normal-form (remainder) reps in $\,\Bbb Z_n$.

Answer 2

I will assume that $x\pmod n$ stands for element of $\{0,1\ldots,n-1\}$ which I will denote with $x'$ such that $x = kn + x'$. But, it doesn't actually matter at all.

First, let me clarify definitions: $x\equiv x'\pmod n$ if and only if $n\mid x - x'$. Note that since we have Euclidean division (division with remainder) in $\Bbb Z$, for every $x\in\Bbb Z$, there is a unique number $x'\in\{0,\ldots,n-1\}$ such that $x = kn + x'$ for some $k\in\Bbb Z$ (which you denote by $x\pmod n$). But, $x-x' = kn$ which implies that $x\equiv x'\pmod n$. Conversely, if $x\equiv x'\pmod n$ and $x'\in\{0,1,\ldots,n-1\}$, then $x' = x\pmod n$. So, to prove your statement, I will use this more general notion of congruence.

I claim that whenever $x\equiv x'\pmod n$ and $y\equiv y'\pmod n$, then $x+y\equiv x' + y'\pmod n$ (i.e., addition is well defined in modular arithmetic). This will give your result as a special case.

So, $x-x' = kn$, $y - y' = ln$, with $k,l\in\Bbb Z$. Add this two to get $(x+y)-(x'+y') = (k+l)n$ which implies that $x + y\equiv x'+y'\pmod n$.