# Addition of rational numbers exercises

This page con­sists of ex­er­cises de­signed to help you get to grips with the ad­di­tion of ra­tio­nal num­bers.

# Core exercises

Us­ing the in­stant rule from the text (which is ac­tu­ally a bit un­wieldy here): $$\frac{1}{10} + \frac{1}{5} = \frac{1 \times 5 + 10 \times 1}{10 \times 5} = \frac{5+10}{50} = \frac{15}{50}$$$No­tice that this can ac­tu­ally be made sim­pler: it is the same thing as $$\frac{3}{10}$$, be­cause when we take $$\frac{3}{10}$$ and split each $$\frac{1}{10}$$-block into five equal pieces, we ob­tain $$15$$ copies of the $$\frac{1}{50}$$-block. Alter­na­tively, one could spot that $$\frac{1}{10}$$-blocks ac­tu­ally can already be used to make $$\frac{1}{5}$$-blocks: $$\frac{1}{5} = \frac{2}{10}$$. There­fore we’re ac­tu­ally try­ing to find $$\frac{1}{10} + \frac{2}{10}$$, which is easy: it’s $$\frac{3}{10}$$. <div><div> ## Se­cond ex­er­cise: $$\frac{1}{15} + \frac{1}{10}$$ Us­ing the in­stant rule from the text: $$\frac{1}{10} + \frac{1}{15} = \frac{1 \times 15 + 10 \times 1}{10 \times 15} = \frac{25}{150} = \frac{1}{6}$$$

Were you ex­pect­ing a big de­nom­i­na­tor, or at least a mul­ti­ple of 5? From this ex­am­ple, we can see that the fi­nal an­swer of a ra­tio­nal ad­di­tion prob­lem can have a de­nom­i­na­tor which doesn’t even seem re­lated to the oth­ers.

Alter­na­tively, one could spot that $$\frac{1}{30}$$-blocks will make up—$$\frac{1}{10}$$and $$\frac{1}{15}$$-blocks. Then we are ac­tu­ally try­ing to find $$\frac{3}{30} + \frac{2}{30} = \frac{5}{30}$$; it’s a bit eas­ier to see that $$\frac{5}{30} = \frac{1}{6}$$ than it is to see that $$\frac{25}{150} = \frac{1}{6}$$. <div><div>

## Third ex­er­cise: $$\frac{1}{10} + \frac{1}{15}$$

You might find this ex­er­cise a lit­tle fa­mil­iar…

Us­ing the in­stant rule from the text: $$\frac{1}{15} + \frac{1}{10} = \frac{1 \times 10 + 15 \times 1}{15 \times 10} = \frac{25}{150} = \frac{1}{6}$$$You may no­tice that we’ve ba­si­cally done the same calcu­la­tions as in the sec­ond ex­er­cise. In fact, ad­di­tion doesn’t care which way round the num­bers go: $$\frac{1}{10} + \frac{1}{15} = \frac{1}{15} + \frac{1}{10}$$, even if we don’t already know that that num­ber is $$\frac{1}{6}$$. This is in­tu­itive from the fact that ad­di­tion is the idea of “place the ap­ples next to each other and count up the to­tal”: just putting the ap­ples down in a differ­ent or­der doesn’t change the to­tal amount of ap­ple. <div><div> ## Fourth ex­er­cise: $$\frac{0}{5} + \frac{2}{5}$$ No­tice that both the de­nom­i­na­tors are the same (namely $$5$$), so we can just com­bine the $$\frac{1}{5}$$-sized pieces straight away. We have $$0$$ pieces and $$2$$ pieces, so the to­tal is $$2$$ pieces. That is, the an­swer is $$\frac{2}{5}$$. <div><div> ## Fifth ex­er­cise: $$\frac{0}{7} + \frac{2}{5}$$ If you spot that this is “no $$\frac{1}{7}$$-pieces” next to “two $$\frac{1}{5}$$-pieces”, then you might just im­me­di­ately write down that the an­swer is $$\frac{2}{5}$$ be­cause there aren’t any $$\frac{1}{7}$$-pieces to change the an­swer; and you’d be cor­rect. To use the in­stant rule from the text: $$\frac{0}{7} + \frac{2}{5} = \frac{0 \times 5 + 2 \times 7}{5 \times 7} = \frac{0 + 14}{35} = \frac{14}{35}$$$

But that is the same as $$\frac{2}{5}$$ (sim­ply ex­pressed with each $$\frac{1}{5}$$-piece sub­di­vided fur­ther into sev­enths). <div><div>

# Ex­ten­sion exercises

Th­ese ex­er­cises are meant to be harder and to stretch your con­cep­tual un­der­stand­ing. Give them a proper go, but don’t worry too much if you don’t get the same an­swers as me. Mine are, in a tech­ni­cal sense, “right”, but no mat­ter what you end up with, you will de­rive a lot of benefit from try­ing to work out what the an­swers are your­self with­out hav­ing been told ex­actly how. The learn­ing of math­e­mat­ics is much more about think­ing and un­der­stand­ing (usu­ally guided by ex­am­ples) than it is about just re­peat­edly car­ry­ing out calcu­la­tions.

## First ex­ten­sion ex­er­cise: $$\frac{1}{5} + \frac{-1}{10}$$

Yes, it is pos­si­ble to add a nega­tive num­ber of chunks. Try us­ing the in­stant rule and see what hap­pens.

Us­ing the in­stant rule from the text: $$\frac{1}{15} + \frac{-1}{10} = \frac{1 \times 10 + 15 \times (-1)}{15 \times 10} = \frac{10 - 15}{150} = \frac{-5}{150} = \frac{-1}{30}$$\$

What has hap­pened here? What have we “re­ally done” with our chunks of ap­ple? Have a think; we’ll see a lot more of this when we get to sub­trac­tion. <div><div>

## Se­cond ex­ten­sion ex­er­cise: what ra­tio­nal num­ber must we add to $$\frac{7}{8}$$ to ob­tain $$\frac{13}{8}$$?

You’re look­ing for an an­swer that looks like $$\frac{a}{b}$$ where there are in­te­gers in place of $$a$$ and $$b$$.

Some­thing you can do to make this ques­tion eas­ier is to no­tice that both the num­bers have the same chunk-size (namely $$\frac{1}{8}$$), so we might try adding some num­ber of $$\frac{1}{8}$$-chunks. Then we’re try­ing to get from $$7$$ chunks to $$13$$ chunks, so we need to add $$6$$ chunks.

That is, the fi­nal num­ber is $$\frac{6}{8}$$ (which is also $$\frac{3}{4}$$). <div><div>

## Third ex­ten­sion ex­er­cise: what ra­tio­nal num­ber must we add to $$\frac{7}{8}$$ to ob­tain $$\frac{13}{7}$$?

You’re look­ing for an an­swer that looks like $$\frac{a}{b}$$ where there are in­te­gers in place of $$a$$ and $$b$$.

Now, the num­bers are no longer of the same chunk-size, so we should make it so that they are of the same size.

The chunk-size to use is $$\frac{1}{8 \times 7} = \frac{1}{56}$$. The rea­son for this is the same as the rea­son­ing we saw when work­ing out how to add $$\frac{1}{8}$$ and $$\frac{1}{7}$$.

Then the two num­bers are $$\frac{7 \times 7}{7 \times 8} = \frac{49}{56}$$ and $$\frac{8 \times 13}{8 \times 7} = \frac{104}{56}$$, so the an­swer is that we need to get from $$49$$ to $$104$$; to do that, we need to add $$55$$ chunks of size $$\frac{1}{56}$$, so the an­swer is $$\frac{55}{56}$$. <div><div>

Parents: