Simplest form, and which fraction is bigger

Split­ting a restau­rant bill, read­ing a cricket run rate, com­par­ing your test score with a friend's: every one of these is a frac­tion in dis­guise, and the same frac­tion can wear many cos­tumes. To com­pare two of them fairly, you first want each one in its neat­est out­fit. We call that out­fit the sim­plest, or stan­dard, form. Once you have it, decid­ing which of two ratio­nal num­bers is big­ger becomes almost easy.

Think of half a cake. You can call it $\displaystyle \frac{1}{2}$ of the cake. Cut the same cake into four equal pieces and it becomes $\displaystyle \frac{2}{4}$; cut it into six and it is $\displaystyle \frac{3}{6}$. You still have exactly as much cake to eat. Only the slic­ing changed.

So one frac­tion can go by many names. Maths sim­ply agrees to use the tidi­est of them as its proper name.

Sim­plest form#

You met ratio­nal num­bers in Ratio­nal num­bers. A ratio­nal num­ber is any num­ber you can write as $\displaystyle \frac{p}{q}$. $\displaystyle \frac{2}{3}$ is one. So is $\displaystyle -\frac{7}{8}$. Here $p$ is the top num­ber and $q$ is the bot­tom one.

Both $p$ and $q$ are inte­gers, which just means whole num­bers with no bits bro­ken off. An inte­ger can be a count­ing num­ber, it can be zero, and it can carry a minus sign. The bot­tom num­ber $q$ is never $0$, though. After all, you can­not cut a cake into no pieces!

A ratio­nal num­ber is in sim­plest form when two things are true.

1. $p$ and $q$ have no com­mon divi­sor except $1$. That means no num­ber except $1$ divides both of them.
2. $q$, the bot­tom num­ber, is pos­i­tive. Pos­i­tive means above zero, with no minus sign on it.

Sim­plest form has a sec­ond name. It is also called stan­dard form. The two names mean the same thing.

Why the two parts must share noth­ing#

A com­mon divi­sor is a num­ber that divides both parts. Look at $\displaystyle \frac{10}{15}$. Both $10$ and $15$ divide by $5$. So this frac­tion can still be made smaller.

Pic­ture a choco­late bar snapped into $15$ equal pieces, and you take $10$ of them. Now push the pieces together in groups of five. The whole bar makes $3$ groups, and you are hold­ing $2$ of those groups. In other words, you hold $\displaystyle \frac{2}{3}$ of the bar.

Noth­ing was added and noth­ing was taken away. It is the same choco­late; we have only changed what we count as one piece.

Now try to cut $\displaystyle \frac{2}{3}$ down again. You can­not. The only num­ber that divides both $2$ and $3$ is $1$. So $\displaystyle \frac{2}{3}$ is in sim­plest form.

Tak­ing the same divi­sor out of the top and the bot­tom is called can­celling. In $\displaystyle \frac{2}{3}$ there is noth­ing left to can­cel.

Here are three more num­bers that are already in sim­plest form.

Check each pair. $5$ and $6$ share only $1$. So do $7$ and $8$. So do $3$ and $14$. None of them can be can­celled any fur­ther.

Why the bot­tom must be pos­i­tive#

A minus sign can sit in three places. You can write $\displaystyle \frac{-2}{3}$, or $\displaystyle \frac{2}{-3}$, or $\displaystyle -\frac{2}{3}$. All three mean the same amount. Stan­dard form keeps the sign off the bot­tom.

So $\displaystyle \frac{-2}{3}$ and $\displaystyle -\frac{2}{3}$ are both fine. The sign is off the bot­tom in each one. Only $\displaystyle \frac{2}{-3}$ needs fix­ing.

Why fuss about where the sign sits? Very soon you will com­pare frac­tions by look­ing at their top num­bers, and that trick only works when every bot­tom num­ber is pos­i­tive. Mul­ti­ply­ing by a neg­a­tive num­ber turns the order round, and a pos­i­tive bot­tom stops that from hap­pen­ing by acci­dent. You will see exactly how a lit­tle later.

Now put $\displaystyle \frac{10}{-15}$ into sim­plest form. First find the biggest num­ber that divides both $10$ and $15$. That num­ber is $5$. This biggest one has a short name. It is called the H.C.F.

Put $\displaystyle \frac{10}{-15}$ into sim­plest form

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

So divide the top and the bot­tom by $5$. On the top, $10 \div 5 = 2$. On the bot­tom, $-15 \div 5 = -3$. That gives $\displaystyle \frac{2}{-3}$. Then lift the minus sign to the front. The answer is $\displaystyle -\frac{2}{3}$.

A num­ber with two minus signs

Put $\displaystyle \frac{-36}{-48}$ into sim­plest form. The H.C.F of $36$ and $48$ is $12$. Because both parts are neg­a­tive, divide both by $-12$. That can­cels the com­mon divi­sor and makes the bot­tom pos­i­tive in one step.

$$\displaystyle \begin{aligned}\frac{-36}{-48} &= \frac{(-36) \div (-12)}{(-48) \div (-12)} \\ &= \frac{3}{4}\end{aligned}$$

Check: $3$ and $4$ share only $1$, and the bot­tom is pos­i­tive. So $\displaystyle \frac{3}{4}$ is in sim­plest form. Two minus signs, one on top and one below, always make a pos­i­tive num­ber.

Don't worry if the H.C.F does­n't jump out at you. You can can­cel in stages instead. For $\displaystyle \frac{36}{48}$, divide by $2$ to get $\displaystyle \frac{18}{24}$, by $2$ again to get $\displaystyle \frac{9}{12}$, and then by $3$ to get $\displaystyle \frac{3}{4}$. You land on the same answer; the H.C.F sim­ply gets you there in one jump.

Remem­ber. A num­ber is in sim­plest form when two things are true. The top and bot­tom share no divi­sor but $1$. And the bot­tom num­ber is pos­i­tive. Can­celling by the H.C.F does the first job. Mov­ing the minus sign up to the front does the sec­ond.

Which frac­tion is big­ger#

Three sym­bols do a lot of work from here on. $<$ means 'is smaller than'. $>$ means 'is big­ger than'. The mouth always opens towards the big­ger num­ber. And $\Rightarrow$ is short for 'so'.

Some frac­tions are easy to com­pare. Take $\displaystyle \frac{3}{9}$ and $\displaystyle \frac{5}{9}$. Both are made of ninths, so the pieces are the same size, and five pieces obvi­ously beat three. So $\displaystyle \frac{5}{9} > \frac{3}{9}$.

Now try $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{2}{9}$. Thirds and ninths are dif­fer­ent sizes, so it is like weigh­ing $1$ big slice against $2$ small ones. Just count­ing slices won't set­tle it.

That gives us two jobs to do first: put every num­ber into stan­dard form, and then give them all the same bot­tom num­ber.

For that you need one bot­tom num­ber that every bot­tom fits into. Any such num­ber will do, and mul­ti­ply­ing the bot­toms together always works. The small­est one, though, keeps the arith­metic light. That small­est one is called the L.C.M.

Where neg­a­tive num­bers sit#

Zero is the mid­dle of the num­ber line. Num­bers to the right of zero are above zero. Num­bers to the left are below zero. We put a minus sign on those.

So $\displaystyle -\frac{2}{3}$ means two thirds of a step to the left of zero. And $\displaystyle -\frac{4}{3}$ is a whole step and one more third to the left. So it is fur­ther out still. Fur­ther left always means smaller.

Putting three num­bers in order#

Write $\displaystyle \frac{1}{3}$, $\displaystyle \frac{-2}{9}$ and $\displaystyle \frac{-4}{3}$ in ascend­ing order. Ascend­ing means small­est first. Descend­ing means biggest first.

All three are in stan­dard form already. The bot­tom num­bers are $3$, $9$ and $3$. The L.C.M of $3$ and $9$ is $9$. So write every num­ber in ninths.

The three top num­bers are $3$, $-2$ and $-12$. Put them on a num­ber line. $-12$ sits far to the left. $-2$ comes next. $3$ is over on the right.

In order, small­est first

$$\displaystyle \begin{aligned}-12 &< -2 < 3 \\ \frac{-12}{9} &< \frac{-2}{9} < \frac{3}{9} \\ \frac{-4}{3} &< \frac{-2}{9} < \frac{1}{3}\end{aligned}$$

In the last line each num­ber goes back to the name it started with. The order does­n't budge, because the num­bers only changed clothes, not size.

Descend­ing order is the same work read back­wards. Read the top num­bers the other way round. They go $3$, then $-2$, then $-12$.

In order, biggest first

$$\displaystyle \begin{aligned}3 &> -2 > -12 \\ \frac{3}{9} &> \frac{-2}{9} > \frac{-12}{9} \\ \frac{1}{3} &> \frac{-2}{9} > \frac{-4}{3}\end{aligned}$$

Find­ing the great­est of three#

Which is great­est: $\displaystyle \frac{2}{-3}$, $\displaystyle \frac{5}{6}$ or $\displaystyle \frac{3}{2}$?

Put them in stan­dard form first. The minus sign in $\displaystyle \frac{2}{-3}$ is on the bot­tom. Move it up. That num­ber is $\displaystyle \frac{-2}{3}$. The other two are already tidy.

The bot­tom num­bers are now $3$, $6$ and $2$. The L.C.M of $3$, $6$ and $2$ is $6$. So write all three in sixths.

Change each one into sixths

$$\displaystyle \begin{aligned}\frac{-2}{3} &= \frac{-2 \times 2}{3 \times 2} = \frac{-4}{6} \\ &\frac{5}{6} \text{ is in sixths already} \\ \frac{3}{2} &= \frac{3 \times 3}{2 \times 3} = \frac{9}{6}\end{aligned}$$

Now read the top num­bers: $-4$, $5$ and $9$. Read­ing them from small­est to biggest gives $-4$, then $5$, then $9$.

The order

$$\displaystyle \begin{aligned}\frac{-4}{6} &< \frac{5}{6} < \frac{9}{6} \\ \frac{-2}{3} &< \frac{5}{6} < \frac{3}{2}\end{aligned}$$

So $\displaystyle \frac{3}{2}$ is the great­est. A quick sense check agrees: it is the only one of the three that is big­ger than $1$.

A shorter way: cross prod­ucts#

When you only have two num­bers to com­pare, there is a quicker path. We'll try it on $\displaystyle \frac{4}{5}$ and $\displaystyle \frac{3}{7}$, doing it the long way first so you can see exactly what the short­cut is copy­ing.

Both $5$ and $7$ fit into $35$. So cut both cakes into $35$ pieces. One of those pieces is called a thirty-fifth.

Both in thirty-fifths

$$\displaystyle \begin{aligned}\frac{4}{5} &= \frac{4 \times 7}{5 \times 7} = \frac{28}{35} \\ \frac{3}{7} &= \frac{3 \times 5}{7 \times 5} = \frac{15}{35} \\ \frac{28}{35} &> \frac{15}{35} \Rightarrow \frac{4}{5} > \frac{3}{7}\end{aligned}$$

Look at the two top num­bers you got. The $28$ came from $4 \times 7$. The $15$ came from $3 \times 5$. Each one is a frac­tion's top times the other frac­tion's bot­tom. Those two answers are called the cross prod­ucts.

The two cross prod­ucts

$$\displaystyle \begin{aligned}4 \times 7 &= 28 \\ 3 \times 5 &= 15 \\ 28 &> 15 \Rightarrow \frac{4}{5} > \frac{3}{7}\end{aligned}$$

The bot­tom num­ber was $35$ for both, and a bot­tom that both share can't decide any­thing. Only the two cross prod­ucts mat­ter, so the short way sim­ply skips writ­ing the bot­toms down.

Now here is the rule, writ­ten once for every pair. The let­ters $p$, $q$, $r$ and $s$ stand for any num­bers you like. Take $\displaystyle \frac{p}{q}$ and $\displaystyle \frac{r}{s}$, both in stan­dard form. Mul­ti­ply $p$ by $s$. Then mul­ti­ply $r$ by $q$.

• If $p \times s > r \times q$ then $\displaystyle \frac{p}{q} > \frac{r}{s}$.
• If $p \times s < r \times q$ then $\displaystyle \frac{p}{q} < \frac{r}{s}$.
• If $p \times s = r \times q$ then $\displaystyle \frac{p}{q} = \frac{r}{s}$.

This is where the pos­i­tive bot­tom earns its keep. Mul­ti­ply­ing by a neg­a­tive bot­tom would turn the answer round. Stan­dard form stops that hap­pen­ing. The next sec­tion shows you why.

Cross prod­ucts with two neg­a­tive num­bers

Which is greater, $\displaystyle \frac{-3}{4}$ or $\displaystyle \frac{-5}{7}$? Both are in stan­dard form, with pos­i­tive bot­toms, so the rule can be used.

$$\displaystyle \begin{aligned}(-3) \times 7 &= -21 \\ (-5) \times 4 &= -20 \\ -21 \lt -20 &\Rightarrow \frac{-3}{4} \lt \frac{-5}{7}\end{aligned}$$

So $\displaystyle \frac{-5}{7}$ is the greater. Check with a com­mon bot­tom of $28$: the num­bers are $\displaystyle \frac{-21}{28}$ and $\displaystyle \frac{-20}{28}$, and $-20$ is fur­ther right on the num­ber line.

Watch what goes wrong if a bot­tom is left neg­a­tive. Com­pare $\displaystyle \frac{1}{-2}$ with $\displaystyle \frac{1}{3}$ with­out tidy­ing first. The cross prod­ucts are $1 \times 3 = 3$ and $1 \times (-2) = -2$. Since $3 \gt -2$ the rule would say $\displaystyle \frac{1}{-2}$ is the big­ger, which is false, because a neg­a­tive num­ber is smaller than a pos­i­tive one. Write it as $\displaystyle \frac{-1}{2}$ first and the cross prod­ucts become $-3$ and $2$, which give the right order.

Remem­ber. First put every num­ber into stan­dard form. Then give them the same bot­tom num­ber. Now every piece is the same size. So count­ing the pieces tells you the order, and the top num­bers are that count.

Two facts about signs#

A big­ger bot­tom gives a smaller piece#

Cut a cake into $2$ equal pieces, then cut an iden­ti­cal cake into $3$. Which piece would you rather have? The half, of course. The more pieces you cut, the smaller each one gets.

So $2 < 3$, but $\displaystyle \frac{1}{2} > \frac{1}{3}$. The order turns round. Call the two num­bers $a$ and $b$. They stand for any two num­bers you like.

Here is the first rule. If $a < b$ then $\displaystyle \frac{1}{a} > \frac{1}{b}$. It holds when $a$ and $b$ have the same sign.

It works for two neg­a­tives as well. Take $-3$ and $-2$. Both are below zero, so the signs match. Here $-3 < -2$ gives $\displaystyle -\frac{1}{3} > -\frac{1}{2}$.

You can read that line the other way round. $-2 > -3$, and $\displaystyle -\frac{1}{2} < -\frac{1}{3}$. It says the same thing.

Here is the sec­ond rule. If $a$ and $b$ have oppo­site signs, then $a < b$ gives $\displaystyle \frac{1}{a} < \frac{1}{b}$. This time the order does not turn round.

Try it with $-2$ and $3$. Here $-2 < 3$, and $\displaystyle -\frac{1}{2} < \frac{1}{3}$.

Here is the rea­son. $-2$ is below zero, so $\displaystyle -\frac{1}{2}$ is below zero too. And $3$ is above zero, so $\displaystyle \frac{1}{3}$ is above zero. One num­ber is on the left of zero and one is on the right. The left one is still the smaller, what­ever the pieces look like.

Mul­ti­ply­ing by a neg­a­tive turns the order round#

Start with $2 < 3$. Mul­ti­ply both by the same num­ber $m$. What hap­pens to the order?

If $m$ is pos­i­tive, noth­ing sur­pris­ing hap­pens. Dou­ble both and you get $4$ and $6$, and still $4 < 6$. The pair moved fur­ther apart, but the order held.

If $m$ is neg­a­tive, the order flips. Mul­ti­ply both by $-1$. You get $-2$ and $-3$. Now $-2 > -3$.

The num­ber line shows why. Both $2$ and $3$ sit right of zero. And $3$ is fur­ther from zero than $2$ is.

Mul­ti­ply­ing by $-1$ sends each num­ber to the other side of zero. Each one keeps its dis­tance from zero. So $-3$ lands fur­ther to the left than $-2$.

On a num­ber line, fur­ther left means smaller. So $-3 < -2$, which is the same as $-2 > -3$. The order turned round.

Here is the rule in short. Take the num­bers $a$ and $b$ again. If $m > 0$ then $a < b$ gives $m \times a < m \times b$. If $m < 0$ then $a < b$ gives $m \times a > m \times b$.

Remem­ber. When you mul­ti­ply both sides by a neg­a­tive num­ber, the order turns round. $2 < 3$, but $-2 > -3$.

Prac­tice#

Write each one in sim­plest form#

Com­pare these#

1. $$\displaystyle \text{Which is greater: } \frac{2}{5} \text{ or } \frac{3}{8}$$
2. $$\displaystyle \text{Which is greater: } \frac{-1}{2} \text{ or } \frac{-2}{5}$$
3. $$\displaystyle \text{Ascending order (smallest first): } \frac{1}{2}, \frac{-3}{4}, \frac{5}{8}$$
4. $$\displaystyle \text{Descending order (biggest first): } \frac{2}{3}, \frac{-1}{6}, \frac{1}{2}$$

Com­mon mis­takes#

• Can­celling by a com­mon divi­sor that is not the biggest, and then stop­ping. Always check that the new top and bot­tom share noth­ing but $1$.
• Leav­ing the minus sign on the bot­tom, as in $\displaystyle \frac{2}{-3}$. Stan­dard form needs a pos­i­tive bot­tom.
• Divid­ing only the top, or only the bot­tom, by the H.C.F. Both must be divided by the same num­ber.
• Com­par­ing top num­bers when the bot­tom num­bers are dif­fer­ent. Make the bot­toms the same first.
• Using cross prod­ucts before the bot­toms are pos­i­tive; the order can come out back­wards.
• Think­ing $-5$ is big­ger than $-4$ because $5$ is big­ger than $4$. On the num­ber line $-5$ is fur­ther left, so it is smaller.
• Writ­ing ascend­ing order when the ques­tion asks for descend­ing. Ascend­ing is small­est first.

Key terms#

Ratio­nal num­ber

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

Sim­plest (stan­dard) form

A ratio­nal num­ber whose top and bot­tom share no divi­sor but $1$, and whose bot­tom is pos­i­tive.

Com­mon divi­sor

A num­ber that divides both the top and the bot­tom.

H.C.F

The high­est com­mon fac­tor: the biggest num­ber that divides both.

L.C.M

The low­est com­mon mul­ti­ple: the small­est num­ber that each bot­tom num­ber divides into.

Cross prod­ucts

For $\displaystyle \frac{p}{q}$ and $\displaystyle \frac{r}{s}$, the two prod­ucts $p \times s$ and $r \times q$.

Ascend­ing and descend­ing order

Small­est first, and biggest first.

Answers#

Write each one in sim­plest form#

1. $\displaystyle \frac{12}{-18} = -\frac{2}{3}$
2. $\displaystyle \frac{-14}{21} = -\frac{2}{3}$
3. $\displaystyle \frac{9}{-24} = -\frac{3}{8}$
4. $\displaystyle \frac{-20}{25} = -\frac{4}{5}$

Com­pare these#

1. $\displaystyle \frac{2}{5}$ is greater, because $16 \gt 15$.
2. $\displaystyle \frac{-2}{5}$ is greater, because $-4 \gt -5$ in tenths.
3. $\displaystyle \frac{-3}{4} \lt \frac{1}{2} \lt \frac{5}{8}$
4. $\displaystyle \frac{2}{3} \gt \frac{1}{2} \gt \frac{-1}{6}$

The full work­ing for each ques­tion:

Answers#