Rational numbers

Frac­tions help you share a cake; neg­a­tive num­bers help you read a tem­per­a­ture below zero. Ratio­nal num­bers bring the two together into one fam­ily. Every time you write a mark as $\displaystyle \frac{18}{25}$, a loss as $\displaystyle -\frac{1}{4}$ of a kilo­gram, or a whole num­ber like $6$, you are writ­ing a ratio­nal num­ber. This les­son says exactly which num­bers belong, why the bot­tom can never be zero, and how two frac­tions that look dif­fer­ent can be the same num­ber.

What a ratio­nal num­ber is

You already know frac­tions like $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{3}{4}$. A ratio­nal num­ber is that same idea, made wider.

We use let­ters when any num­ber will do. Here $p$ is what­ever num­ber is on top. And $q$ is what­ever num­ber is on the bot­tom. So a ratio­nal num­ber is any num­ber you can write as $\displaystyle \frac{p}{q}$.

Here $p$ and $q$ must both be inte­gers. An inte­ger is a whole num­ber, with no bits of one. It can be a count­ing num­ber like $4$. It can be zero. It can be a num­ber with a minus sign like $-7$.

One rule comes with this. The bot­tom num­ber $q$ can never be zero. We write that as $q \neq 0$.

Remem­ber. A ratio­nal num­ber is any num­ber you can write as $\displaystyle \frac{p}{q}$. Here $p$ and $q$ are inte­gers, and $q \neq 0$.

All these num­bers together make a set. A set is just a col­lec­tion. This one is called $Q$. When­ever you see $Q$, it means the whole col­lec­tion of these num­bers.

Here are some of the num­bers in $Q$.

Each of these num­bers has its own place on the num­ber line. The frac­tions fall between the whole num­bers, and the neg­a­tive ones fall to the left of zero.

Whole num­bers are ratio­nal num­bers too

Look at that list again. It holds $1$, $0$, $2$ and $-3$. None of them looks like a frac­tion. They are ratio­nal num­bers all the same.

Think about shar­ing cakes. $\displaystyle \frac{6}{2}$ means 6 cakes shared between 2 chil­dren. Each child gets 3.

Now share those 6 cakes between 1 child. That child gets all 6. So $\displaystyle \frac{6}{1}$ is just $6$.

Every whole num­ber works this way. Put it over $1$ and you have a frac­tion.

One num­ber below has a minus sign. A minus can sit on the top, on the bot­tom, or out in front. All three mean the same num­ber. We usu­ally write it out in front, because that is eas­i­est to read.

Remem­ber. Every whole num­ber is a ratio­nal num­ber. Write it over $1$.

Zero is a ratio­nal num­ber

Zero fits the rule as well. You can write $\displaystyle 0 = \frac{0}{1} = \frac{0}{2}$.

Pic­ture a cake cut into 2 equal pieces. Take none of them. Your plate is empty. So $\displaystyle \frac{0}{2}$ is $0$.

Zero on top is fine. Zero on the bot­tom is the one thing we can­not have. The next part shows why.

Why the bot­tom can­not be zero

The bot­tom num­ber tells you how many equal pieces the whole is cut into. Cut a cake into 4 pieces and the bot­tom is 4. Now try to cut a cake into 0 pieces. You can­not. There would be no pieces to hand out.

Divi­sion also asks a ques­tion. $\displaystyle \frac{6}{2}$ asks what num­ber times 2 gives 6. The answer is 3.

Now look at $\displaystyle \frac{6}{0}$. It asks what num­ber times 0 gives 6. No num­ber does. Any­thing times 0 gives 0. So there is no answer.

Remem­ber. $q$ can never be $0$. Noth­ing times $0$ gives $6$, so $\displaystyle \frac{6}{0}$ has no answer at all.

Frac­tions that are worth the same

Two frac­tions can look dif­fer­ent and still be worth the same. Here is a way to see that.

Take three choco­late bars. They are all the same length. Lay them one under the other.

1. Cut the first bar into 2 equal parts. Shade 1 part. You have shaded $\displaystyle \frac{1}{2}$.
2. Cut the sec­ond bar into 4 equal parts. Shade 2 parts. You have shaded $\displaystyle \frac{2}{4}$.
3. Cut the third bar into 6 equal parts. Shade 3 parts. You have shaded $\displaystyle \frac{3}{6}$.

Now look down the three bars. The shaded parts all end at the same place. Each one cov­ers half its bar. The pieces are dif­fer­ent sizes, but the amount is the same.

Remem­ber. $\displaystyle \frac{1}{2} = \frac{2}{4} = \frac{3}{6}$. Num­bers like these are called equiv­a­lent ratio­nal num­bers.

The same thing hap­pens with a full bar. Shade all 2 parts, or all 4 parts, or all 6 parts. Each time the whole bar is shaded. So $\displaystyle 1 = \frac{2}{2} = \frac{4}{4} = \frac{6}{6}$.

The first rule: mul­ti­ply the top and the bot­tom

How do you get from $\displaystyle \frac{1}{2}$ to $\displaystyle \frac{2}{4}$? You cut each piece in two. Now there are twice as many pieces. Each piece is half the size. The shaded amount has not moved.

So you mul­ti­ply the top by 2 and the bot­tom by 2.

From a half to quar­ters

$$\displaystyle \begin{aligned}&\frac{1}{2} \\ &= \frac{1 \times 2}{2 \times 2} \\ &= \frac{2}{4}\end{aligned}$$

Sixths work the same way. This time you cut each half into 3 pieces.

From a half to sixths

$$\displaystyle \begin{aligned}&\frac{1}{2} \\ &= \frac{1 \times 3}{2 \times 3} \\ &= \frac{3}{6}\end{aligned}$$

One rule cov­ers both of those. The let­ter $m$ just means the num­ber you pick. In the first work­ing, $m$ was 2. In the sec­ond, $m$ was 3.

Remem­ber. $\displaystyle \frac{p}{q} = \frac{p \times m}{q \times m}$, where $m \neq 0$.

Why must $m$ not be zero? Mul­ti­ply­ing the bot­tom by 0 would make it 0. That is not allowed. The top would turn into 0 as well.

The rule holds with minus signs too. Take $\displaystyle -\frac{5}{3}$ and dou­ble the top and the bot­tom.

Dou­bling the top and the bot­tom

$$\displaystyle \begin{aligned}&-\frac{5}{3} \\ &= \frac{-5 \times 2}{3 \times 2} \\ &= \frac{-10}{6}\end{aligned}$$

So $\displaystyle -\frac{5}{3}$ and $\displaystyle \frac{-10}{6}$ are the same num­ber. They are only cut into dif­fer­ent pieces.

The sec­ond rule: divide the top and the bot­tom

You can travel the other way as well. Glue the small pieces back into big­ger ones.

Look at the third bar again. It has 6 pieces, and 3 of them are shaded. Glue the pieces in threes. You now have 2 pieces, and 1 of them is shaded. The shaded amount stays where it was, so $\displaystyle \frac{3}{6}$ becomes $\displaystyle \frac{1}{2}$.

The let­ter $m$ means the num­ber you pick here too. Glu­ing in threes makes $m$ equal to 3.

Remem­ber. $\displaystyle \frac{p}{q} = \frac{p \div m}{q \div m}$, where $m \neq 0$.

Why must $m$ not be zero here? You can­not share the pieces into 0 groups. There would be noth­ing to count.

This time $m$ has to divide both num­bers with noth­ing left over. Take $\displaystyle \frac{10}{-15}$. Both 10 and 15 split into 5s. Pick the biggest num­ber that works, and you get the tidi­est answer in one go.

Divid­ing the top and the bot­tom by 5

$$\displaystyle \begin{aligned}&\frac{10}{-15} \\ &= \frac{10 \div 5}{-15 \div 5} \\ &= \frac{2}{-3} \\ &= -\frac{2}{3}\end{aligned}$$

The last step only tidies the minus sign. A minus on the bot­tom means the same as a minus in front. Here the top is pos­i­tive. So both forms say the num­ber sits below zero.

You have not changed the num­ber. You have writ­ten it with smaller num­bers.

More worked exam­ples

Write $\displaystyle \frac{-18}{24}$ in its sim­plest form.

The biggest num­ber that divides both $18$ and $24$ is $6$.

$$\displaystyle \begin{aligned}\frac{-18}{24} &= \frac{-18 \div 6}{24 \div 6} \\ &= \frac{-3}{4} = -\frac{3}{4}\end{aligned}$$

Are $\displaystyle \frac{4}{6}$ and $\displaystyle \frac{10}{15}$ equiv­a­lent?

Sim­plify both. Divid­ing by $2$ gives $\displaystyle \frac{4}{6} = \frac{2}{3}$. Divid­ing by $5$ gives $\displaystyle \frac{10}{15} = \frac{2}{3}$. Both reach the same sim­plest form, so yes, they are equiv­a­lent.

What is $\displaystyle \frac{-7}{-21}$?

Divide the top and the bot­tom by $-7$. The rule allows any $m$ except zero, and a neg­a­tive $m$ is fine.

$$\displaystyle \frac{-7}{-21} = \frac{-7 \div (-7)}{-21 \div (-7)} = \frac{1}{3}$$

A minus on the top and a minus on the bot­tom can­cel, so the num­ber is pos­i­tive.

Fill the gap: $\displaystyle \frac{3}{4} = \frac{9}{?} = \frac{?}{20}$

From $3$ to $9$ the top was mul­ti­plied by $3$, so the bot­tom is $4 \times 3 = 12$. From $4$ to $20$ the bot­tom was mul­ti­plied by $5$, so the top is $3 \times 5 = 15$. So $\displaystyle \frac{3}{4} = \frac{9}{12} = \frac{15}{20}$.

Your turn

Now try these

Each one is miss­ing a num­ber. Work it out, then check below.

Com­mon mis­takes

• Think­ing whole num­bers and zero are not ratio­nal. Each can be writ­ten over $1$.
• Allow­ing a zero on the bot­tom. $\displaystyle \frac{6}{0}$ has no answer at all.
• Chang­ing only the top or only the bot­tom. What­ever you do to one, you must do to the other.
• Adding the same num­ber to top and bot­tom. $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{1+1}{2+1} = \frac{2}{3}$ are not equal; only mul­ti­ply­ing or divid­ing keeps the value.
• Los­ing the minus sign when sim­pli­fy­ing, or writ­ing two minus signs where one pos­i­tive num­ber is meant.

Key terms

Inte­ger

A count­ing num­ber, zero, or a count­ing num­ber with a minus sign.

Ratio­nal num­ber

Any num­ber that can be writ­ten as $\displaystyle \frac{p}{q}$, where $p$ and $q$ are inte­gers and $q \neq 0$.

$Q$

The set of all ratio­nal num­bers.

Numer­a­tor and denom­i­na­tor

The top num­ber and the bot­tom num­ber of a frac­tion.

Equiv­a­lent ratio­nal num­bers

Ratio­nal num­bers that look dif­fer­ent but have the same value, such as $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{3}{6}$.

Sim­plest form

The form in which the top and bot­tom share no fac­tor other than $1$.

Answers

1. $\displaystyle \frac{1}{3} = \frac{3}{9}$, mul­ti­ply­ing top and bot­tom by $3$.
2. $\displaystyle \frac{2}{5} = \frac{8}{20}$, mul­ti­ply­ing by $4$.
3. $\displaystyle \frac{3}{4} = \frac{6}{8}$, mul­ti­ply­ing by $2$.
4. $\displaystyle \frac{6}{8} = \frac{3}{4}$, divid­ing by $2$.
5. $\displaystyle \frac{12}{18} = \frac{2}{3}$, divid­ing by $6$.
6. $\displaystyle \frac{-4}{7} = \frac{-12}{21}$, mul­ti­ply­ing by $3$.