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

Core exercises

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

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: \(\frac{1}{15} + \frac{1}{10}\)

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…

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: \(\frac{0}{5} + \frac{2}{5}\)

That is, the answer is \(\frac{2}{5}\). <div><div>

Fifth exercise: \(\frac{0}{7} + \frac{2}{5}\)

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.

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>