There Is No Such Thing as the Next Rational Number

There isn't one — and that is the most important fact in this chapter.

After the integer $3$ comes $4$, with nothing in between. But name any rational number you like as the next one after $\frac{1}{3}$, and the average of it and $\frac{1}{3}$ lies between them, so your candidate was not next after all. The argument works every time, which means there are infinitely many rational numbers between any two.

So rationals have no gaps to step across. This page covers the first part of the ICSE Class 8 Mathematics chapter on rational numbers: what they are, how to write and place them, how to compare them, and how to produce as many as you want between any two.

A rational number is a number that can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.

So $\frac{3}{4}$, $-\frac{5}{7}$, $\frac{0}{9}$, $7$ (which is $\frac{7}{1}$) and $-2$ are all rational. Every integer is a rational number, since any integer $n$ can be written $\frac{n}{1}$.

**Why $q\ne 0$.** Division by zero has no meaning, so $\frac{5}{0}$ is not a number at all — rational or otherwise. This single restriction causes almost every special case in the chapter.

Standard form. A rational number is in standard form when:

- the denominator is positive, and
- $p$ and $q$ have no common factor other than $1$ — that is, the fraction is in lowest terms

Worked example 1. Express $\frac{-42}{56}$ in standard form.

The denominator is already positive. Now find the HCF of $42$ and $56$, which is $14$:

$$\frac{-42÷14}{56÷14}=\frac{-3}{4}$$

Worked example 2. Express $\frac{35}{-49}$ in standard form.

The denominator is negative, so first multiply numerator and denominator by $-1$:

$$\frac{35}{-49}=\frac{-35}{49}$$

Then divide by the HCF of $35$ and $49$, which is $7$:

$$\frac{-35÷7}{49÷7}=\frac{-5}{7}$$

Placing rationals on a number line. Positive rationals lie to the right of $0$ and negative ones to the left.

To mark $\frac{3}{5}$: divide the unit length from $0$ to $1$ into 5 equal parts and count 3 of them from $0$.

To mark $-\frac{7}{4}$: since $\frac{7}{4}=1\frac{3}{4}$, the point lies between $-1$ and $-2$. Divide the length from $-1$ to $-2$ into 4 equal parts and count 3 of them from $-1$.

The rule in general. The denominator tells you how many parts to divide each unit into, and the numerator tells you how many parts to count. Converting an improper fraction to a mixed number first tells you which pair of integers the point falls between, which is the step that makes negative rationals easy.

Either convert them to a common denominator, or convert them to decimals. Both work; the common denominator is exact and is what a written answer should show.

Worked example 1. Which is greater, $-\frac{3}{4}$ or $-\frac{5}{7}$?

The LCM of $4$ and $7$ is $28$:

$$-\frac{3}{4}=\frac{-21}{28},-\frac{5}{7}=\frac{-20}{28}$$

Since $-21<-20$:

$$-\frac{3}{4}<-\frac{5}{7}$$

Checking with decimals: $-0.75$ against $-\mathrm{0.714...}$, and $-0.75$ is indeed the smaller. Both methods agree.

The trap here is the negative sign. With positive fractions, the one with the bigger numerator over the same denominator is bigger. With negatives it is the reverse: $\frac{21}{28}>\frac{20}{28}$, but $\frac{-21}{28}<\frac{-20}{28}$. Forgetting this is the single commonest error in the chapter.

Worked example 2. Arrange in ascending order: $\frac{2}{3}$, $-\frac{1}{2}$, $\frac{5}{6}$, $-\frac{3}{4}$.

The LCM of $3,2,6$ and $4$ is $12$:

$$\frac{2}{3}=\frac{8}{12},-\frac{1}{2}=\frac{-6}{12},\frac{5}{6}=\frac{10}{12},-\frac{3}{4}=\frac{-9}{12}$$

Now order the numerators: $-9<-6<8<10$. So

$$-\frac{3}{4}<-\frac{1}{2}<\frac{2}{3}<\frac{5}{6}$$

Two shortcuts worth knowing. A negative rational is always less than a positive one, so the negatives can be separated out first and ordered among themselves. And zero is greater than every negative and less than every positive.

When decimals are the better tool. If the denominators are awkward and give a large LCM, converting to decimals is faster. But a recurring decimal must be taken to enough places to decide — comparing $0.714$ with $0.75$ needs three places, and rounding to one place would have made them look equal.

Two methods: the mean method for one or two numbers, and equivalent fractions when many are wanted.

The mean method. The average of two numbers always lies between them, so their mean is a rational number in the gap.

Worked example 1. Insert a rational number between $\frac{1}{3}$ and $\frac{1}{2}$.

$$\text{mean}=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{2}\right)=\frac{1}{2}\left(\frac{2}{6}+\frac{3}{6}\right)=\frac{1}{2}\times \frac{5}{6}=\frac{5}{12}$$

Checking: $\frac{1}{3}=\frac{4}{12}$ and $\frac{1}{2}=\frac{6}{12}$, and $\frac{4}{12}<\frac{5}{12}<\frac{6}{12}$. Correct.

Worked example 2 — with negatives. Insert a rational number between $-\frac{2}{3}$ and $-\frac{1}{2}$.

$$\text{mean}=\frac{1}{2}\left(-\frac{4}{6}-\frac{3}{6}\right)=\frac{1}{2}\times \left(-\frac{7}{6}\right)=-\frac{7}{12}$$

Checking: $-\frac{2}{3}=-\frac{8}{12}$ and $-\frac{1}{2}=-\frac{6}{12}$, and $-\frac{8}{12}<-\frac{7}{12}<-\frac{6}{12}$. Correct.

The equivalent-fractions method, for inserting many at once. To insert $n$ rational numbers:

- Make the two denominators equal.
- Multiply numerator and denominator of both by $(n+1)$.
- The numbers with numerators strictly between the two new numerators are your answers.

Worked example 3. Insert five rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$.

First make the denominators equal, using LCM $6$:

$$\frac{1}{3}=\frac{2}{6},\frac{1}{2}=\frac{3}{6}$$

The numerators $2$ and $3$ leave no integers between them, so multiply both by $n+1=6$:

$$\frac{2}{6}=\frac{12}{36},\frac{3}{6}=\frac{18}{36}$$

Now the numerators $12$ and $18$ have exactly five integers between them, giving

$$\frac{13}{36},\frac{14}{36},\frac{15}{36},\frac{16}{36},\frac{17}{36}$$

Checking the two ends: $\frac{12}{36}=\frac{1}{3}$ and $\frac{18}{36}=\frac{1}{2}$, so all five do lie in the required gap.

**Why multiply by $(n+1)$ rather than $n$.** Multiplying by $k$ makes the numerators differ by $k$ instead of $1$, and two numerators differing by $k$ have exactly $k-1$ integers between them. To get $5$ numbers you need a difference of $6$ — hence $n+1$.

And the point the whole section illustrates. Nothing stops you asking for a hundred numbers between $\frac{1}{3}$ and $\frac{1}{2}$; you would simply multiply by $101$. Since the method never runs out, there are infinitely many rational numbers between any two rational numbers — which is the property that makes rationals unlike integers.