Addition, subtraction, multiplication, division of fractions: Visual walkthrough

Step 1 — What does a fraction even look like?

WHAT. A fraction $\frac{a}{b}$ is a recipe: cut one whole into $b$ equal pieces (the denominator = piece-size number), then take $a$ of those pieces (the numerator = how-many number).

WHY start here. We cannot divide fractions until we agree on what a single fraction is as a physical amount. Division will just be a question about these amounts.

PICTURE. The bar below is one whole. Cut into $b = 4$ equal slices; shade $a = 3$ of them. That shaded region is the number $\frac{3}{4}$ .

Step 2 — What is the question division asks?

WHAT. We are about to divide one fraction by another, so we need two fractions with their own names. Write the first (the pile) as $\frac{a}{b}$ and the second (the scoop) as $\frac{c}{d}$ , where — exactly as in Step 1 — $c$ is the numerator of the scoop (how many pieces it holds) and $d$ is its denominator (its piece-size), with $d \neq = 0$ .

The symbol $\div$ means "how many of these fit into that?" So

$pile a pieces of size 1/ b \frac{a}{b} \div scoop c pieces of size 1/ d \frac{c}{d} = how many \frac{c}{d} -scoops fit inside \frac{a}{b} .$

WHY this framing. Whole-number division already means this: $6 \div 2$ asks "how many 2's fit in 6?" (answer 3). We keep the same meaning and just let the numbers be fractions — no new rule invented, only extended.

PICTURE. We have $\frac{3}{4}$ of a bar (so $a = 3, b = 4$ ). Our scoop is $\frac{1}{8}$ of a bar (so $c = 1, d = 8$ ). Count how many $\frac{1}{8}$ -tiles line up under the $\frac{3}{4}$ region.

Step 3 — The easy warm-up: same-size pieces

WHAT. First do the case where both fractions already have the same denominator, e.g. $\frac{6}{8} \div \frac{1}{8}$ .

WHY. When pieces are identical in size, "how many fit?" is just "how many of its pieces sit inside my pieces?" — pure counting, no cleverness. This gives us a trustworthy anchor.

PICTURE. Both the pile and the scoop are measured in eighths. The pile has $6$ eighths; each scoop is $1$ eighth. So exactly $6$ scoops fit:

$\frac{6}{8} \div \frac{1}{8} = \frac{6}{1} = 6.$

Let me name the general shape before writing the formula. Give the pile's numerator the letter $p$ (how many pieces I hold), the scoop's numerator the letter $q$ (how many pieces the scoop holds), and the shared denominator the letter $n$ (the one piece-size both are cut into). In our example $p = 6$ , $q = 1$ , and $n = 8$ . With those names fixed, the counting rule reads:

Step 4 — Force the pieces to match (the golden-rule trick)

WHAT. Real problems have different denominators, like $\frac{3}{4} \div \frac{2}{5}$ . We use the golden rule to re-cut both bars into the same tiny piece, then fall back on Step 3.

WHY. Step 3 only works when denominators agree. The parent note's common denominator idea rescues us: rewrite each fraction over a shared bottom without changing its value.

PICTURE — why $b d$ aligns the pieces, step by step. Look at the figure. A number that is a multiple of both $4$ and $5$ is exactly a length that a whole row of quarter-marks and a whole row of fifth-marks both land on cleanly. The surest such number is their product $4 \times 5 = 20$ : cut the bar into $4$ big pieces then slice each into $5$ (that's $20$ ), or into $5$ then each into $4$ (also $20$ ) — both routes reach the same $20$ tiny tiles, so every quarter-line and every fifth-line falls on a tile edge. That is what "shared piece" means: a grid fine enough that both original cuts are visible in it.

Now re-cut both bars into these twentieths, using $\frac{a}{b} = \frac{a \cdot k}{b \cdot k}$ on each. (This is legal only because $b = 4 \neq = 0$ and $d = 5 \neq = 0$ , so the new bottoms stay valid.)

$\frac{3}{4} = \frac{3 \cdot 5}{4 \cdot 5} = \frac{15}{20}, \frac{2}{5} = \frac{2 \cdot 4}{5 \cdot 4} = \frac{8}{20} .$

Term by term: the top $3$ becomes $3 \cdot 5 = 15$ (more, smaller pieces), the bottom $4$ becomes $4 \cdot 5 = 20$ (finer cut); same for the scoop $\frac{2}{5} \to \frac{8}{20}$ .

Now both are in twentieths, so apply Step 3 ( $p = 15, q = 8, n = 20$ ):

$\frac{15}{20} \div \frac{8}{20} = \frac{15}{8} .$

Step 5 — Watch the numbers turn into "flip and multiply"

WHAT. Redo Step 4 with letters instead of numbers to expose the pattern. Take $\frac{a}{b} \div \frac{c}{d}$ (with $b \neq = 0, d \neq = 0$ ), common denominator $b d$ :

$\frac{a}{b} = \frac{a d}{b d}, \frac{c}{d} = \frac{c b}{b d} .$

WHY. If the same thing happens for every $a, b, c, d$ , we get a shortcut law and never need to draw twentieths again.

PICTURE (the reveal). Both are now over $b d$ , so by Step 3 the denominators cancel and we divide the numerators:

$\frac{a}{b} \div \frac{c}{d} = \frac{a d}{b d} \div \frac{c b}{b d} = \frac{a d}{c b} .$

Now look at the bottom $c b$ . Multiplying two numbers gives the same answer in either order. The figure makes this concrete: an array of dots with $c$ rows and $b$ columns holds the same number of dots as one with $b$ rows and $c$ columns — you just turn the grid a quarter-turn; not one dot is added or removed. That is commutativity, $c \times b = b \times c$ , so we may rewrite $c b$ as $b c$ :

$\frac{a d}{c b} = \frac{a d}{b c} .$

Read the final fraction slowly:

$\frac{a d}{b c} = first, unchanged \frac{a}{b} \times second, upside-down \frac{d}{c} .$

$a$ stays on top → the first fraction is untouched ("Keep").
$d$ (was the bottom of the second) is now on top.
$c$ (was the top of the second) is now on the bottom.
So the second fraction got flipped: $\frac{c}{d} \to \frac{d}{c}$ ("Flip"), and $\div \to \times$ ("Change").

Step 6 — Case check: dividing by something bigger than one whole

WHAT. Two scoop-size cases decide whether the answer beats the pile. Compare the scoop $\frac{c}{d}$ to one whole:

Scoop smaller than $1$ (like Step 4's $\frac{2}{5}$ ): more than one fits, and the count can even exceed the pile — Step 4 gave $\frac{15}{8} > \frac{3}{4}$ .
Scoop bigger than $1$ : fewer than one whole scoop fits, so the count is below $1$ .

Take a scoop bigger than one whole: $\frac{1}{2} \div \frac{3}{2}$ , where $\frac{3}{2} = 1 \frac{1}{2} > 1$ .

WHY. A good rule must survive every case. The honest measuring stick is the scoop versus $1$ , not the scoop versus the pile.

PICTURE. The scoop $\frac{3}{2}$ is longer than a whole bar; the pile $\frac{1}{2}$ is only part of one bar, so far less than one scoop is used:

$\frac{1}{2} \div \frac{3}{2} = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3} < 1. ✓$

The $\frac{1}{3}$ means "one-third of a scoop fits" — exactly what the picture shows.

WHAT. What if the pile itself is zero, i.e. $a = 0$ , so $\frac{0}{b} \div \frac{c}{d}$ ?

WHY. Zero is the quiet edge case people forget. "How many scoops fit into nothing?" — none, because there is nothing to scoop. So the answer must be $0$ , and our formula must agree.

PICTURE. An empty pile: no matter how small the scoop, zero of them fit. Applying the rule:

$\frac{0}{b} \div \frac{c}{d} = \frac{0}{b} \times \frac{d}{c} = \frac{0 \cdot d}{b \cdot c} = \frac{0}{b c} = 0. (c \neq = 0)$

The numerator $0 \cdot d = 0$ forces the whole thing to $0$ — the formula and the picture agree.

Step 8 — Edge cases: divide by $1$ , and the forbidden zero scoop

WHAT. Two more boundary situations for the scoop.

(a) Divide by $\frac{c}{c} = 1$ . The scoop is one whole. Exactly $\frac{a}{b}$ of them fit: $\frac{a}{b} \div 1 = \frac{a}{b} \times \frac{1}{1} = \frac{a}{b} .$ The flip does nothing because $1$ flipped is still $1$ .

(b) Divide by $\frac{0}{d} = 0$ . The scoop has no size. "How many zero-sized scoops fill a real pile?" You can never finish — no answer exists.

WHY. Flipping $\frac{0}{d}$ demands $\frac{d}{0}$ , and a fraction with zero on the bottom is undefined — you cannot cut a whole into $0$ pieces. The formula itself blocks you: the reciprocal simply doesn't exist.

PICTURE. Left: a full-length scoop lands exactly $\frac{a}{b}$ times (nothing to flip). Right: a zero-width scoop that can never cover anything, whose flip $\frac{d}{0}$ is off-limits.

Step 9 — Negatives: what a minus sign does to the flip

WHAT. Fractions can be negative — a minus sign may sit on the top, the bottom, or in front. First, all three mean the same number:

$\frac{- c}{d} = \frac{c}{- d} = - \frac{c}{d} .$

WHY. Division doesn't care where you park the sign, only the overall sign of the amount. When you flip a fraction, the sign rides along unchanged: the reciprocal of a negative is negative, because a number and its reciprocal must multiply to $+ 1$ , and $(-) \times (-) = +$ .

PICTURE. Put a negative pile on the left of $0$ on a number line; a negative scoop points left too. Same-sign pile and scoop → positive count of scoops; opposite signs → negative count.

Worked signs (magnitudes handled by the rule, sign by "like signs → $+$ , unlike → $-$ "): $\frac{- 3}{4} \div \frac{2}{5} = - \frac{3}{4} \times \frac{5}{2} = - \frac{15}{8}, \frac{- 3}{4} \div \frac{- 2}{5} = + \frac{3}{4} \times \frac{5}{2} = \frac{15}{8} .$

The one-picture summary

The storyboard below adds one new insight the step figures never showed side-by-side: the same division, solved two different ways, lands on the identical answer — proving the flip-shortcut is not a trick but a consequence. On the left, $\frac{3}{4} \div \frac{2}{5}$ is ground out the slow way (re-cut to twentieths, then count: $\frac{15}{20} \div \frac{8}{20} = \frac{15}{8}$ ); on the right, the fast way (Keep-Change-Flip: $\frac{3}{4} \times \frac{5}{2} = \frac{15}{8}$ ). Two roads, one destination — that agreement is the whole point of the derivation.

Recall Feynman retelling — say it like you'd tell a friend

Dividing fractions is just asking "how many of the little scoops fit into my pile?" If both are cut into the same size of piece, you just count — that's easy. When they're cut differently, you re-cut both into one shared tiny size (that's the golden rule, cutting every piece into more, smaller bits without changing the amount — allowed as long as no bottom is zero); the surest shared size is $b \times d$ , because slicing $b$ ways then $d$ ways lands every original cut-line on a tile edge. Once the piece-sizes match, the bottoms cancel and you divide the top counts; the leftover bottom $c b$ is the same as $b c$ because a grid of dots holds the same number turned sideways. Written with letters, the second fraction's top and bottom swap places — the scoop turns upside-down. That flipped scoop is the reciprocal, the number that undoes the original, so "divide by a fraction" becomes "multiply by its flip": Keep, Change, Flip. Then the edge cases: tiny scoops fit many times (answer big); scoops bigger than a whole fit less than once (answer under 1); an empty pile gives $0$ ; a zero-size scoop never fits, so dividing by zero has no answer; and a minus sign just rides along — count the minuses, odd means negative.

Recall Rebuild the derivation yourself

Start from $\frac{a}{b} \div \frac{c}{d}$ . ::: Rewrite over $b d$ : $\frac{a d}{b d} \div \frac{c b}{b d}$ . Now the denominators match — what happens? ::: They cancel; divide numerators: $\frac{a d}{c b}$ . Why may $\frac{a d}{c b}$ be written $\frac{a d}{b c}$ ? ::: Because multiplication commutes: $c b = b c$ . Rewrite $\frac{a d}{b c}$ as a product of two fractions. ::: $\frac{a}{b} \times \frac{d}{c}$ — the second one flipped. What is $\frac{0}{b} \div \frac{c}{d}$ (with $c \neq = 0$ )? ::: $0$ — an empty pile holds no scoops, so nothing fits. What breaks if $c = 0$ ? ::: The flip $\frac{d}{c} = \frac{d}{0}$ is undefined — you can't divide by zero. Sign of $\frac{- 3}{4} \div \frac{2}{5}$ ? ::: Negative: one minus sign, so $- \frac{15}{8}$ .

Connections

Addition, subtraction, multiplication, division of fractions — parent topic
Fractions - meaning and equivalent fractions — the golden rule lives here
LCM and HCF — smarter common denominators than $b d$
Reciprocals and multiplicative inverse — the "flip" is a reciprocal
Rational numbers — negatives and signs live here