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

First ex­er­cise: \(\frac{1}{10} + \frac{1}{5}\)

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>

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