Numbers between rationals

Rational numbers behave very differently from whole numbers and integers when it comes to "what lies between."

Integers are discrete. Between $3$ and $4$ there is nothing — no other integer fits. Each integer has an immediate next-door neighbour.

Rationals are dense. Between any two distinct rationals $a$ and $b$ (with $a < b$), at least one more rational number always exists — and once you find it, you can repeat the trick on the new, smaller gap, forever. The simplest way to produce that in-between number is to average the two endpoints:

where: $a$ and $b$ are the two given rational numbers with $a<b$, and $\text{mean}$ is a new rational number lying exactly halfway between them. This midpoint always satisfies $a < \text{mean} < b$, because halving a positive gap can never overshoot either end. Once you have the midpoint, you have two smaller gaps — $a$ to $\text{mean}$, and $\text{mean}$ to $b$ — and each of those can be halved again, so the process never runs out of new numbers to produce. When you need several rational numbers at once rather than one at a time, averaging repeatedly is slow. A faster route is the common-denominator method: rewrite $a$ and $b$ as equivalent fractions over a large enough shared denominator so that several whole-number numerators fall strictly between them, then read those numerators off as your answer. It helps to see why the two methods never contradict each other. Both are just different ways of naming rational numbers that satisfy the same inequality, $a < x < b$. The averaging method finds one such $x$ directly by arithmetic; the common-denominator method widens the "resolution" of the fractions involved so that several valid values of $x$ become visible as whole-number numerators simultaneously. Neither method is more correct than the other — they are simply suited to different questions, one number versus a specified count of numbers. It is also worth contrasting this with what happens for integers, to see exactly where the break occurs. Between the integers $2$ and $5$, the only integers are $3$ and $4$ — a short, closed list, because integers jump in fixed steps of $1$ with nothing in between consecutive values. Rational numbers have no such fixed step size: however small a gap you choose between two rationals, that gap can always be split again, because the average of any two rationals is itself rational. This is the essence of why "the numbers between $a$ and $b$," for rationals, is never a question with a finite answer, however close together $a$ and $b$ happen to be.

Find a rational number between $\dfrac{2}{5}$ and $\dfrac{3}{5}$, then find a further number between that result and $\dfrac{2}{5}$, and finally find three rational numbers between $\dfrac{1}{4}$ and $\dfrac{2}{4}$ using the common-denominator method.

Step 1. Average $\dfrac{2}{5}$ and $\dfrac{3}{5}$: $\dfrac{\frac{2}{5}+\frac{3}{5}}{2} = \dfrac{\frac{5}{5}}{2} = \dfrac{1}{2}$. Check: $0.4 < 0.5 < 0.6$, so $\dfrac{1}{2}$ correctly lies between them.

Step 2. Average $\dfrac{2}{5}$ and $\dfrac{1}{2}$: $\dfrac{\frac{2}{5}+\frac{1}{2}}{2} = \dfrac{\frac{4}{10}+\frac{5}{10}}{2} = \dfrac{\frac{9}{10}}{2} = \dfrac{9}{20}$. Check: $0.4 < 0.45 < 0.5$, so $\dfrac{9}{20}$ lies strictly between $\dfrac{2}{5}$ and $\dfrac{1}{2}$, confirming the halving can continue indefinitely.

Step 3. Now use the common-denominator method on $\dfrac{1}{4}$ and $\dfrac{2}{4}$. Over denominator $4$ the numerators are $1$ and $2$, with nothing strictly between, so scale up: to fit at least $3$ numbers, multiply by a number bigger than $3+1=4$; use $5$. This gives $\dfrac{1}{4}=\dfrac{5}{20}$ and $\dfrac{2}{4}=\dfrac{10}{20}$.

Step 4. List the numerators strictly between $5$ and $10$: these are $6, 7, 8, 9$, so any three of $\dfrac{6}{20}, \dfrac{7}{20}, \dfrac{8}{20}, \dfrac{9}{20}$ answer the question. Final answers: $\dfrac{1}{2}$ lies between $\dfrac{2}{5}$ and $\dfrac{3}{5}$; $\dfrac{9}{20}$ lies between $\dfrac{2}{5}$ and $\dfrac{1}{2}$; and $\dfrac{6}{20}, \dfrac{7}{20}, \dfrac{8}{20}$ are three rational numbers between $\dfrac{1}{4}$ and $\dfrac{2}{4}$.

Competitive stretch

Notice that $\dfrac{9}{20}$ from Step 2 also equals $\dfrac{9}{20}$ over the enlarged denominator used in Step 3 style working ($\dfrac{9}{20}$ itself, unscaled) — showing the averaging method and the common-denominator method must always agree wherever their ranges overlap, since both simply name rational numbers that satisfy the same inequality. A quick sanity check for such answers is to compare numerators directly once every fraction shares the same denominator: since $\dfrac{1}{4}=\dfrac{5}{20}$ and $\dfrac{2}{4}=\dfrac{10}{20}$, any numerator strictly between $5$ and $10$ — such as $6$, $7$ or $8$ — is guaranteed to give a fraction strictly between the two originals, without needing to convert anything to a decimal at all.

• Thinking only a fixed, finite number of rationals can be found between two given rationals, as with integers — density guarantees the supply is always infinite.
• Forgetting to convert to a common denominator before comparing or listing "numbers between," which produces numerators that do not actually fall in the right range.
• Slipping arithmetically when averaging fractions with unlike denominators — always convert to a common denominator first, add the numerators, then divide by $2$.
• Assuming the midpoint trick works only once — it can be reapplied to any newly found number to keep generating more values without limit.

Ask the tutor: Want to try finding ten rational numbers between two fractions of your own choice, or double-check a density argument step by step?