Equivalent fractions, simplifying fractions: Visual walkthrough

We build every symbol before using it. Here are the only three we need:

Step 1 — Draw the whole and take $\frac{a}{b}$ of it

WHAT. Start with one whole bar. Cut it into $b$ equal parts. Shade $a$ of them. That shaded amount is the fraction $\frac{a}{b}$ .

WHY. Before we can prove two fractions are equal, we must be able to see what a fraction is: not a pair of numbers, but a shaded amount. The numbers $a$ and $b$ are just instructions for producing that shading.

PICTURE. Below, $b = 3$ so the bar has $3$ equal cells; $a = 2$ so $2$ are amber. The shaded strip is our fixed "amount" — remember its length, because the whole proof is this length never changing.

Figure — Equivalent fractions, simplifying fractions

$the shaded amount \frac{a}{b} = a = 2, b = 3 \frac{2}{3}$

Step 2 — Re-cut every piece into $k$ smaller pieces

WHAT. Take the same bar and draw $k - 1$ extra cuts inside each of the $b$ cells. No cell is added or removed — each is just chopped finer.

WHY. This is the only thing we are allowed to do. We are not eating any bar, not adding any bar — we are only relabelling how many cuts exist. Because we changed nothing physical, the shaded amount is forced to stay identical. That "forced to stay" is the heart of the whole idea.

PICTURE. Here $k = 2$ : each of the $3$ big cells gets one extra cut, becoming $2$ thin cells. Watch the amber strip — its length has not moved a millimetre.

Step 3 — Count the new total pieces: the denominator becomes $b \times k$

WHAT. Count how many small cells the whole bar now has. Each of the $b$ big cells split into $k$ , so the total is $b \times k$ .

WHY. We use multiplication here, not addition, and that choice is not arbitrary. " $b$ groups, each with $k$ items" is the exact definition of $b \times k$ . Multiplication is the tool for "so-many-copies-of-the-same-size", which is precisely equal cutting.

PICTURE. $3$ big cells $\times 2$ cuts each $= 6$ small cells. Count them in the figure.

$new denominator = big cells b \times cuts per cell k = 3 \times 2 = 6$

Step 4 — Count the new shaded pieces: the numerator becomes $a \times k$

WHAT. Now count only the shaded small cells. Each shaded big cell also split into $k$ , and there were $a$ shaded big cells, so we now hold $a \times k$ small shaded cells.

WHY. Same reasoning, applied only to the amber region: " $a$ shaded groups, each split into $k$ " $= a \times k$ . The same $k$ multiplies both counts — because we made the same cut everywhere. This is the secret reason numerator and denominator must be multiplied by the same number.

PICTURE. $2$ shaded big cells $\times 2 = 4$ shaded small cells (amber), out of $6$ total.

$new numerator = shaded big cells a \times cuts per cell k = 2 \times 2 = 4$

Step 5 — Read both bars side by side: the amounts coincide

WHAT. Put the Step 1 bar above the Step 4 bar and align them. The amber strips end at exactly the same place.

WHY. Two fractions are equal precisely when they shade the same amount of the same whole. The pictures shade the same amount — so the fractions naming them must be equal. That is the whole proof, stated as a picture.

PICTURE. Top: $\frac{2}{3}$ . Bottom: $\frac{4}{6}$ . One dashed vertical line touches the right edge of both amber strips.

$\frac{a}{b} = \frac{a \times k}{b \times k} ⟺ \frac{2}{3} = \frac{4}{6}$

Step 6 — The algebra behind the picture (the $\frac{k}{k} = 1$ view)

WHAT. We now say in symbols what the picture showed. The value of $\frac{a}{b}$ is the number $x$ obeying $b \cdot x = a$ . Multiply that whole equation by $k$ : $bk \cdot x = ak$ . So the same $x$ also fits denominator $bk$ and numerator $ak$ .

WHY. A picture convinces the eye; algebra convinces for every $a, b, k$ at once, including ones too big to draw. The tool we lean on is that ==multiplying by $\frac{k}{k} = 1$ never changes a value== — and $\frac{k}{k} = 1$ is just " $k$ small pieces make one big piece", i.e. Step 2 in symbols.

PICTURE. The bar labelled with $x$ = "size of one part" makes the equation $b \cdot x = a$ visible: $b$ copies of the part-length $x$ stack up to the shaded amount $a$ -parts... and re-cutting scales both counts by $k$ while $x$ stays fixed.

$\frac{a}{b} = \frac{a}{b} \times 1 = \frac{a}{b} \times \frac{k}{k} = \frac{a \times k}{b \times k}$

Step 7 — Run it backwards: this is simplifying

WHAT. Read the arrow the other way. If a fraction already has cut-marks that group evenly — i.e. top and bottom share a common factor $k$ — you may erase those inner cuts, dividing both counts by $k$ .

WHY. Going forward (multiply) makes numbers bigger; going backward (divide) makes them smaller. Simplifying is literally the same picture played in reverse. You stop when no shared cut remains — that is when $g cd (a, b) = 1$ , the definition of lowest terms. Finding that largest shared $k$ is exactly the job of the GCD/HCF, often found via Prime factorisation.

PICTURE. Start at $\frac{4}{6}$ ; the $6$ small cells regroup into $3$ big cells (erase every 2nd cut, $k = 2$ ), landing back at $\frac{2}{3}$ , which has no shared cut left.

$\frac{a \div k}{b \div k} : \frac{4 \div 2}{6 \div 2} = \frac{2}{3} (g cd (2, 3) = 1 \Rightarrow done)$

Step 8 — The degenerate cases you must never trip on

WHAT & WHY. Every rule has edges. Here are all of them, each shown in the figure.

$k = 1$ (no new cuts). $\frac{a \times 1}{b \times 1} = \frac{a}{b}$ — you drew zero extra lines, so nothing changed. Harmless but useless.
$k = 0$ (forbidden). Slicing each piece into $0$ pieces would erase the bar; $\frac{a \times 0}{b \times 0} = \frac{0}{0}$ , which is undefined. This is exactly why the rule demands $k \neq = 0$ .
$a = 0$ (empty shading). $\frac{0}{b}$ shades nothing; cutting nothing finer still shades nothing: $\frac{0 \times k}{b \times k} = \frac{0}{bk}$ , still $0$ . Fine for any $b \neq = 0$ .
The trap — adding instead of cutting. $\frac{a + m}{b + m}$ does not re-cut; it lengthens the bar and shifts the strip. See the last panel: $\frac{2}{4} = 0.5$ but $\frac{2 + 1}{4 + 1} = \frac{3}{5} = 0.6$ — a different amount. Only $\times$ and $\div$ preserve the shading.

The one-picture summary

One bar, one fixed shaded amount, three labels for it — $\frac{1}{3} \to \frac{2}{6} \to \frac{4}{12}$ going right (multiply, cut finer) and back left (divide, simplify). The amber never moves; only the number of tick-marks does.

Recall Feynman: the whole walkthrough in plain words

I've got a chocolate bar. I break it into $b$ equal chunks and grab $a$ of them — that grab is my fraction, and it's really just an amount of chocolate. Now I take the very same bar and score every chunk into $k$ thinner slivers. I didn't add chocolate or eat any — I only drew more lines. So now the whole has $b \times k$ slivers and I'm holding $a \times k$ of them, but it's the exact same handful of chocolate. That's why $\frac{a}{b} = \frac{ak}{bk}$ : same amount, more tick-marks. Play the film backwards — erase the extra lines — and you're simplifying: keep erasing shared cuts until none are left, and that's lowest terms. The one rule I must never break: I'm allowed to cut (multiply) or un-cut (divide) both counts by the same $k$ , but I'm never allowed to just add chocolate to the top and bottom — that moves where the shading ends, so it's a different amount entirely.

Recall Quick self-test

Why does $k$ multiply the top and the bottom by the same number? ::: Because the same cut is made in every cell — shaded cells and unshaded cells split into $k$ alike. What picture-fact makes $\frac{a}{b} = \frac{ak}{bk}$ obvious? ::: The shaded strip length never changes when you only add cut-lines. Why is $k = 0$ banned? ::: Cutting each piece into $0$ pieces destroys the bar and gives the undefined $\frac{0}{0}$ . Simplifying is which direction of the film? ::: Backwards — dividing both counts by a shared factor until $g cd = 1$ . Why does $\frac{a + m}{b + m}$ fail? ::: Adding lengthens the bar and shifts the shading; only $\times / \div$ keep the amount fixed.

Connections

Equivalent fractions, simplifying fractions — the parent this page derives.
Factors, multiples and GCD/HCF — the shared cut $k$ you divide out when simplifying.
Prime factorisation — reliable route to the largest shared $k$ .
Adding and subtracting fractions — uses the forward move to match denominators.
Ratios and proportion — same "keep the ratio" principle as equal-cutting.
Decimals and percentages — $\frac{1}{3}$ , $\frac{2}{6}$ , $\frac{4}{12}$ all equal $0.333 \dots$
Number Systems – rational numbers — each rational is one such bar in lowest terms.