Fractions & the Rationals

The defect, third verse, same as the first

We've done this dance twice. $\mathbb{N}$ couldn't answer $3-5$, so we forged negatives and got $\mathbb{Z}$. Now $\mathbb{Z}$ can't answer $3 \div 5$ — there's no integer you can multiply by $5$ to land on $3$. Same defect, same damn cure: invent the missing numbers.

Recall the master move from The Laws of Arithmetic: division is the inverse question for multiplication. $3 \div 5$ asks "what times $5$ gives $3$?" To answer it in general, we need, for every nonzero $b$, a number that undoes multiplication by $b$.

The multiplicative inverse, $1/b$

Multiplication has its own identity — the do-nothing element under $\times$ is $1$, since $a \cdot 1 = a$. (Compare $0$, the additive identity from last lesson. Every system has these.)

For each nonzero integer $b$, we define its multiplicative inverse, written $1/b$ or $b^{-1}$, by:

$b \cdot \frac{1}{b} = 1.$

That equation isn't a fact to discover — it IS the definition of $1/b$. The reciprocal of $5$ is the number that multiplies $5$ back to the identity $1$. And once we have all these reciprocals, a general fraction is just:

$\frac{a}{b} = a \cdot \frac{1}{b}.$

So $3/5$ means "$3$ copies of one-fifth." Division is solved: $3 \div 5 = 3 \cdot (1/5) = 3/5$. The whole collection of these — all $a/b$ with integers $a$ and nonzero $b$ — is $\mathbb{Q}$, the rationals (from quotient; $\mathbb{Z}$ was taken).

Drag the point and find the fraction-marks between the integers — the line is filling in:

Why $2/4 = 1/2$: equivalent fractions

Here's a thing that should genuinely bother you — it bothered me until I worked through the proof: $2/4$ and $1/2$ are different symbols, yet the same number. Why?

A fraction $a/b$ is the answer to "$a$ split into $b$ equal parts." If you scale both the count and the number of parts by the same nonzero factor $k$, you've described the same quantity — twice as many pieces, each half as big:

$\frac{a}{b} = \frac{a \cdot k}{b \cdot k}.$

Proof from the definition: $\frac{ak}{bk} = ak \cdot \frac{1}{bk}$, and since $\frac{1}{bk} = \frac{1}{b}\cdot\frac{1}{k}$ (both are inverses, and $bk \cdot \frac{1}{b}\frac{1}{k} = 1$), we get $ak \cdot \frac{1}{b}\frac{1}{k} = a \cdot \frac{1}{b} \cdot \big(k \cdot \frac{1}{k}\big) = \frac{a}{b}\cdot 1 = \frac{a}{b}$. So $2/4 = (1\cdot 2)/(2\cdot 2) = 1/2$. Scaling top and bottom together is invisible to the number itself. Reducing a fraction is just running this backwards: cancel the common factor.

Adding fractions: derive the common denominator, don't chant it

Why can't you just add tops and bottoms? Because $\frac{1}{2} + \frac{1}{3} \ne \frac{2}{5}$ — a half plus a third is clearly more than $2/5$ (which is less than a half). Checking that took two seconds, and the Federation still lets students do it for years. The honest reason: you can only add pieces of the same size. Fifths add to fifths the way apples add to apples.

So to add $\frac{a}{b} + \frac{c}{d}$, first make the piece-sizes match. Use equivalence to rewrite both over the common denominator $bd$:

$\frac{a}{b} = \frac{ad}{bd}, \qquad \frac{c}{d} = \frac{cb}{bd}.$

Now the pieces are the same size ($1/bd$ each), so we just count them — and here is where distributivity does the work:

$\frac{ad}{bd} + \frac{cb}{bd} = (ad)\cdot\frac{1}{bd} + (cb)\cdot\frac{1}{bd} = (ad + cb)\cdot\frac{1}{bd} = \frac{ad + cb}{bd}.$

That middle step — pulling out the common factor $\frac{1}{bd}$ — is distributivity, the same engine from two lessons ago. The common-denominator rule isn't a rule; it's distributivity in a costume. Concretely $\frac{1}{2}+\frac{1}{3} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}$.

Multiplying and dividing

Multiplication is the easy one — multiply across:

$\frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}.$

(Because $a\cdot\frac1b \cdot c\cdot\frac1d = ac \cdot \frac{1}{bd}$, regrouping by commutativity and associativity.)

Division is where everyone parrots "flip and multiply" with absolutely no idea why, and it makes me want to flip a specimen tray. Here's why. Dividing by $\frac{c}{d}$ means multiplying by its multiplicative inverse. And the inverse of $\frac{c}{d}$ is $\frac{d}{c}$, because

$\frac{c}{d}\cdot\frac{d}{c} = \frac{cd}{dc} = 1.$

It multiplies back to the identity, so it IS the inverse — by definition. Therefore:

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

"Flip and multiply" is "multiply by the reciprocal," which is what dividing has always meant. Derived, not decreed.

$\mathbb{Q}$ is dense — and that's a trap

One stunning property: between any two distinct rationals there's another rational — just average them. Between $\frac13$ and $\frac12$ sits $\frac{5}{12}$; between those, another; forever. We say $\mathbb{Q}$ is dense. There are no gaps you can see; the line looks completely full.

It is not full. Hold that thought — it detonates spectacularly in Irrationals & the Real Line, where I show you a hole the rationals can't fill. I've been waiting since the first lesson to detonate that one.

Decimals are fractions in disguise

Finally: $0.25$ is not a different kind of number. It's $\frac{25}{100} = \frac14$ — a fraction whose denominator is a power of ten. Every terminating decimal is a fraction over $10^n$. And the repeaters? $0.\overline{3} = \frac13$ exactly (a denominator that won't divide evenly into a power of ten just repeats forever). Terminating or repeating — both are rational. A decimal that does neither would not be a fraction at all... but that's a problem for a later lesson, and oh hell, is it a good one. Go run the gauntlet.