Modular arithmetic — definition, addition, multiplication, congruence: Visual walkthrough

The parent note stated the rules:

$a \equiv b (mod n), c \equiv d (mod n) \Rightarrow a + c \equiv b + d, a c \equiv b d (mod n) .$

Here we build them from nothing, with a picture for every single step. By the end you will see why reducing before you add or multiply can never change the answer.

We assume only that you can divide whole numbers and get a remainder. Everything else — the wrap-around, the "buckets", the rules — we grow from that one seed.

Step 1 — What "remainder" really means: the number line, chopped

WHAT. Take any whole number and any fixed number $n$ (we'll use $n = 5$ ). The Division Algorithm says we can always write

$a = q \cdot n + r, 0 \leq r < n .$

$a$ ::: the number we start with
$n$ ::: the modulus — how far we count before wrapping
$q$ ::: the quotient — how many complete jumps of size $n$ we made
$r$ ::: the remainder — what's left over, always squeezed into $0, 1, \dots, n - 1$

WHY. This is the only fact we are allowed to assume. Every later claim reduces back to it. The remainder $r$ is the "address" of $a$ on a circle of $n$ slots — nothing more.

PICTURE. Look at the number line below. Every step of size $n = 5$ is a colored tick. The number $a = 13$ sits after $2$ full jumps (that's $q = 2$ ) with $3$ left over (that's $r = 3$ ).

Figure — Modular arithmetic — definition, addition, multiplication, congruence

Step 2 — Wrapping the line into a clock

WHAT. Bend that number line so slot $n$ lands back on slot $0$ . Numbers that differ by a multiple of $n$ now sit on the same spot.

WHY. This is the visual heart of the whole subject. The claim " $a \equiv b (mod n)$ means $a - b$ is a multiple of $n$ " becomes obvious: if $a$ and $b$ land on the same point of the circle, you must have taken a whole number of extra laps to get from one to the other.

$a - b = k n$

$a - b$ ::: the gap between the two numbers
$k$ ::: how many whole laps separate them (any integer, positive or negative)
$k n$ ::: that gap expressed as complete trips around the circle

PICTURE. On the $n = 5$ clock, watch $3$ , $8$ , and $13$ all pile onto the same slot. Their differences ( $8 - 3 = 5$ , $13 - 8 = 5$ ) are exactly one lap.

Step 3 — The core object: a number is its "bucket" + a leftover

WHAT. Rewrite any two congruent numbers transparently. If $a \equiv b (mod n)$ , then $a$ is just $b$ with some whole laps added:

$a = b + k_{1} n, c = d + k_{2} n .$

$b, d$ ::: the "small" representatives (the slot we care about)
$k_{1}, k_{2}$ ::: how many extra laps $a$ and $c$ carry beyond $b$ and $d$
$k_{1} n, k_{2} n$ ::: the invisible part — full laps that don't move the slot

WHY. This form is the whole trick. The laps ( $k_{1} n$ , $k_{2} n$ ) are invisible on the clock. We will show that when we add or multiply, all the lap-terms collect into another whole multiple of $n$ — which is still invisible.

PICTURE. $a = 13$ is drawn as the visible short arrow to slot $3$ ( $= b$ ) plus $2$ ghost laps ( $k_{1} = 2$ ). The ghost laps are dashed — they exist but don't change where you point.

Step 4 — Addition: the ghost laps add up to more ghost laps

WHAT. Add the two boxed expressions from Step 3:

$a + c = (b + k_{1} n) + (d + k_{2} n) = (b + d) + integer (k_{1} + k_{2}) n .$

$b + d$ ::: the visible part — where the two small arrows point together
$(k_{1} + k_{2})$ ::: total ghost laps, still a whole number
$(k_{1} + k_{2}) n$ ::: still a whole multiple of $n$ → still invisible

WHY. Because $(a + c) - (b + d) = (k_{1} + k_{2}) n$ is a multiple of $n$ , the definition from Step 2 fires:

$a + c \equiv b + d (mod n) .$

We reduced first (to $b, d$ ), then added — and it matched. The laps couldn't sabotage us because whole laps plus whole laps are still whole laps.

PICTURE. Two arrows on the clock placed head-to-tail. The dashed ghost laps of each simply extend the trip by complete circles — the final arrowhead lands on the same slot whether or not you keep the ghost laps.

Step 5 — Multiplication: why the extra terms are still multiples of $n$

WHAT. Multiply the two boxed forms. This is scarier — there are cross-terms:

$a c = (b + k_{1} n) (d + k_{2} n) = b d + b k_{2} n + d k_{1} n + k_{1} k_{2} n^{2} .$

Now factor $n$ out of everything except $b d$ :

$a c = b d + n (one big integer b k_{2} + d k_{1} + k_{1} k_{2} n) .$

$b d$ ::: the visible product — the slot we actually want
$b k_{2} n, d k_{1} n$ ::: cross-terms, each carrying a factor $n$
$k_{1} k_{2} n^{2}$ ::: double-lap term, carrying two factors of $n$
the bracket ::: a single whole number → the whole correction is a multiple of $n$

WHY. Every term that isn't $b d$ has at least one factor of $n$ , so $a c - b d = n \cdot (integer)$ . By Step 2's definition:

$a c \equiv b d (mod n) .$

That's the multiplication rule — no leap of faith, just " $n$ divides the leftover."

PICTURE. A rectangle of area $a c$ split into four tiles: the small $b d$ corner (colored) plus three strips/tiles that each contain a full width or height of $n$ — visibly "wrappable" chunks that disappear on the clock.

Step 6 — Powers: multiplication rule, used again and again

WHAT. Since $a \equiv b$ lets us replace $a$ by $b$ inside any product, do it $k$ times:

$a^{k} = k a \cdot a \dots a \equiv k b \cdot b \dots b = b^{k} (mod n) .$

WHY. No new idea — the power rule is Step 5 applied repeatedly. This is what makes huge exponents tame (and is the doorway to Fermats Little Theorem and Eulers Theorem).

PICTURE. A chain: each link is one multiplication where $a \to b$ is swapped. The congruence survives every link.

Step 7 — The degenerate cases (never leave the reader stranded)

WHAT & WHY. We must handle the "boring but sneaky" inputs.

Case $n = 1$ . Every number leaves remainder $0$ . The clock has one slot. So $a \equiv b (mod 1)$ is always true — the rules hold trivially (everything is congruent to everything).

Case a remainder of $0$ . If $a$ is an exact multiple of $n$ (like $10 mod 5$ ), it lands on slot $0$ — the "top of the clock." It behaves like any other slot; $0$ is a legitimate representative.

Case negative numbers. $- 3 mod 7$ is not $- 3$ . Walk backwards around the clock: from $0$ , three steps back lands on slot $4$ . Formally add laps until you're in range:

$- 3 \equiv - 3 + 7 = 4 (mod 7) .$

Here $k_{1}$ from Step 3 is simply negative — the ghost laps run counter-clockwise, but the algebra of Steps 4–5 never assumed $k_{1}, k_{2} \geq 0$ , so everything still works.

Case reducing then dividing — a warning. Adding and multiplying always survive reduction. Division does not, unless $g cd (a, n) = 1$ (see Modular Inverses, Linear Congruences). That's a different operation and lives on a different page.

PICTURE. Three mini-clocks side by side: the one-slot $n = 1$ clock; the $mod 5$ clock with slot $0$ highlighted at the top; and the $mod 7$ clock showing $- 3$ walked counter-clockwise onto slot $4$ .

The one-picture summary

Everything compresses to one idea: whole laps are invisible. Add or multiply the visible small arrows; every leftover collects into multiples of $n$ that vanish on the clock.

Recall Feynman retelling — the whole walkthrough in plain words

Chop the number line every $n$ steps and roll it into a circle. A number's only real identity is which slot it lands on — its remainder. Every extra trip around the circle is a "ghost lap": it moves you, but not your slot.

Write each number as small part + ghost laps. When you add two numbers, the small parts combine and the ghost laps combine into more ghost laps — still whole circles, still invisible — so the slot you reach is the same as if you'd added just the small parts. When you multiply, you get a few cross-terms, but every one of them secretly contains a full width or height of $n$ , so they're wrappable tiles that disappear too. Powers are just multiplication done over and over, so they inherit the same magic.

Negative numbers just walk backwards; slot $0$ is the top of the clock; and if $n = 1$ the clock has a single slot so everything is congruent. The one caution: this only covers $+$ , $\times$ , and powers — division is a different beast needing $g cd (a, n) = 1$ .

One sentence: reduce first, then compute — the ghost laps can never bite you.

Prerequisites and neighbours: parent topic · Division Algorithm · GCD and LCM · Modular Inverses · Linear Congruences · Fermats Little Theorem · Eulers Theorem · Chinese Remainder Theorem · Quadratic Residues · RSA Cryptography.

Recall Quick self-test

Why can we reduce $a$ and $c$ before adding? ::: Because the leftover $(a + c) - (b + d) = (k_{1} + k_{2}) n$ is a multiple of $n$ , so both land on the same slot. Why does multiplication survive reduction? ::: Every term of $a c$ except $b d$ contains a factor of $n$ , so $a c - b d$ is a multiple of $n$ . What is $- 3 mod 7$ ? ::: $4$ , by walking three steps backward (adding one lap of $7$ ). When does this argument fail? ::: For division — you also need $g cd (a, n) = 1$ .