The counting numbers $1$, $2$, $3$, ... are called the natural numbers. They tell you how many elements (things) there are in a given finite set. Zero can be included as a natural number because it tells you how many things there are in an empty set - it is the answer to questions like "How many prime numbers are there between $200$ and $210$?" (try it yourself!).

If we count backwards from $10$ (say) we don't have to stop at zero, but can continue into the negative whole numbers $-1$, $-2$, $-3$, and so on. Any whole number, whether positive or negative, is called an integer. So all the natural numbers are integers, but not every integer is a natural number.

Now, when we set out the integers on a line, there are gaps between them, with many more numbers (not integers) in the gaps. For example, between 0 and 1 we find the numbers $\frac{1}{2}$,$\frac{3}{4}$,$\frac{2}{5}$ and so on.

Similarly $\frac{10}{7}$ lies between $1$ and $2$, $\frac{29}{13}$ between $2$ and $3$, and so on. You probably call a number like $\frac{2}{5}$ or $\frac{10}{7}$ a fraction, a word that shares the same Latin root with 'fragment' - something broken. This can be a problem, because although we can easily think of $\frac{2}{5}$ as a bit less than half of something, we don't normally think of $\frac{10}{7}$ as part (a fragment) of something else. Such `improper' fractions (greater than $1$) contradict our intuition. On the other hand, any fraction makes sense if we think of it simply as the ratio of two whole numbers. For this reason, another name for a fraction (proper or improper) is a rational number.

So is a rational number the same as a fraction? Well, it wouldn't do much harm (here, anyway) if we just said "yes", so you could ignore the rest of this paragraph. Two fractions like $\frac{2}{3}$ and $\frac{4}{6}$ are obviously different, but said to be equal or equivalent fractions. On the other hand, each of them represents the same rational number. They occupy the same point on the number line. In a sense, a rational number (the actual ratio) consists of a whole family of equivalent fractions.

Two games

Here are two games to play with a friend.

Guess my number. Your friend chooses an integer between, say, $0$ and $100$. You have to find out that number by asking questions, but your friend can only answer 'yes' or 'no'.

What strategy would you use? Move ahead to the next piece of text when you've thought about it.

A good strategy is to halve (or roughly halve) the number of possibilities at each stage. So ask "Is it less than $50$?" If the answer was "Yes", then ask "Is it less than $25$", and so on.

Guess my fraction. The rules are the same as those for 'guess my number', but this time your friend has chosen a fraction between, say, $0$ and $1$.

What strategy would you use? Move ahead to the next piece of text when you've thought about it.

The halving strategy won't work (unless your friend chose a fraction like $\frac{1}{2}$ or $\frac{5}{8}$) because there are an infinite number of possibilities! In 'guess my number', there were $100$ possibilities to start with, and every time you ask a question you halve (or nearly halve) the number of possibilities. In the end, it's bound to come down to just one. The nasty thing about an infinite number of possibilities is that even when you halve that number, it's still infinite!

In practice, you might want to agree that your friend chooses a fraction with denominator less than $50$ (say), and so the number of possibilities is finite (about $2500$, and a lot less if you agree that the fraction must be in its lowest terms (i.e. it can't be 'cancelled down'). Nevertheless, the halving strategy still won't work in general - unless you deal with the numerator and denominator separately, which rather undermines the point of choosing a fraction in the first place. A different strategy is suggested in the next section.

Mediants

Pick any two (positive) rational numbers (fractions, if you prefer that name) one or both can be greater than 1 if you like. Now make another rational by adding the numerators (top numbers) and adding the denominators (bottom numbers). The resulting rational is called the mediant of the two you began with. So the mediant of $\frac{2}{5}$ and $\frac{11}{7}$ is $\frac{13}{12}$.

Now arrange your two rational numbers and their mediant in order, from least to greatest. One way to do this is to find the decimal equivalent of each of your fractions, to $2$ or $3$ decimal places. That would be $0.4$, $1.57$ and $1.08$ for my example, so the correct order (least to greatest) $\frac{2}{5}$, $\frac{13}{12}$, $\frac{11}{7}$.

Try it with a couple of your own examples. What do you notice? Move ahead to the next piece of text when you think you know what always happens.

The mediant always lies between the two rational you started with. Not normally mid-way between them, but more than one and less than the other. [You might like to prove that the mediant of two rationals is the same as their average only when they have the same denominator, or when they are equivalent].

This suggests a possible strategy for 'guess my fraction': close down the interval to which the fraction belongs by using the mediant (not the average, as in the unsuccessful halving strategy). Here's an example.

Suppose you choose $\frac{12}{19}$ (which is between $0$ and $1$ and in lowest terms).

I know that your fraction is between $\frac{0}{1}$ and $\frac{1}{1}$, so I ask if it's less than $\frac{1}{2}$ (which is their mediant). You reply 'no' (later you'll need a calculator to compare my suggestions with $\frac{12}{19} = 0.6315...$).

I now know that your fraction is between $\frac{1}{2}$ and $\frac{1}{1}$, so I ask if it's less than their mediant $\frac{2}{3}$. You reply 'yes'

I now know that your fraction is between $\frac{1}{2}$ and $\frac{2}{3}$ , so I ask if it's less than their mediant $\frac{3}{5}$. You reply 'no'

I now know that your fraction is between $\frac{3}{5}$ and $\frac{2}{3}$, so I ask if it's less than their mediant $\frac{5}{8}$. Again, you reply 'no'

I now know that your fraction is between $\frac{5}{8}$ and $\frac{2}{3}$. Is it less than $\frac{7}{11}$? "Yes"

I now know that your fraction is between $\frac{5}{8}$ and $\frac{7}{11}$. Is it less than $\frac{12}{19}$?

Well, no, because $\frac{12}{19}$ is the fraction you thought of in the first place, and I've 'guessed' it. So we should agree at the beginning of this game that if the 'guessing' player actually names the fraction he or she is trying to find, then s/he has successfully finished the game.

Notice how the sequence of mediants $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{5}$, $\frac{5}{8}$, $\frac{7}{11}$, $\frac{12}{19}$ successfully homes in on the fraction that was chosen.

Now try it yourself! Suppose I've chosen $\frac{13}{20}$. Starting with $\frac{0}{1}$ and $\frac{1}{1}$, use the method above to produce a sequence of mediants that homes in on $\frac{13}{20}$. When you've done it, look at the next piece of text.

You should have got a sequence like that shown in the table below, which was produced on a computer using a spreadsheet. Using the spreadsheet just speeds up the process. The numerator and denominator of each fraction are shown in adjacent columns, so the first row shows $\frac{0}{1}$, $\frac{1}{1}$ their mediant $\frac{1}{2}$.

 LOWER BOUND MEDIANT UPPER BOUND $0$ $1$ $1$ $2$ $1$ $1$ $1$ $2$ $2$ $3$ $1$ $1$ $1$ $2$ $3$ $5$ $2$ $3$ $3$ $5$ $5$ $8$ $2$ $3$ $5$ $8$ $7$ $11$ $2$ $3$ $7$ $11$ $9$ $14$ $2$ $3$ $9$ $14$ $11$ $17$ $2$ $3$ $11$ $17$ $13$ $20$ $2$ $3$

You probably noticed that the early mediants $\frac{1}{2}$ and $\frac{2}{3}$ each provided new lower and upper bounds, followed by a long 'run' in which each new mediant provided a new lower bound while the upper bound stayed fixed at $\frac{2}{3}$ until the mediant $\frac{13}{20}$ (the target number) appeared.

So the mediants that start the 'runs' can be shown as:

 $1$ $2$ $2$ $3$ $13$ $20$

Similarly, the table of mediants in a search for the rational $\frac{10}{23}$ would look like this (you can check it with your calculator!)

 LOWER BOUND MEDIANT UPPER BOUND $0$ $1$ $1$ $2$ $1$ $1$ $0$ $1$ $1$ $3$ $1$ $2$ $1$ $3$ $2$ $5$ $1$ $2$ $2$ $5$ $3$ $7$ $1$ $2$ $3$ $7$ $4$ $9$ $1$ $2$ $3$ $7$ $7$ $16$ $4$ $9$ $3$ $7$ $10$ $23$ $7$ $16$

The mediants that start the 'runs' are $\frac{1}{2}$, $\frac{3}{7}$ and $\frac{10}{23}$ (the target). Each of these is closer to the target than the previous one - they are called 'convergents' because they converge to the target. Fortunately, they can be found without having to evaluate all the other mediants, by using a process involving 'continued fractions'. But that's another story.

RECURRING DECIMALS

You probably know that rational numbers can be 'converted' to decimal numbers by dividing the numerator by the denominator. For example, $\frac{5}{8} = 0.625$ and $\frac{2}{3} = 0.666666...$ So the decimal equivalent of $\frac{5}{8}$ terminates with 3 digits after the decimal point, whereas the decimal equivalent of $\frac{2}{3}$ does not terminate - it goes on forever. It recurs in a very obvious way. Similarly $\frac{10}{27} = 0.370 370 370...$ recurs and so does $\frac{9}{14} = 0.6 428571 428571 428571...$. Notice that the recurring pattern for $\frac{10}{27}$ begins immediately after the decimal point, whereas the pattern for $\frac{9}{14}$ does not. In either case, the decimal is called a recurring decimal, so e.g. $0.4 21951 21951 21951...$ is a recurring decimal.

Now, it is fairly obvious that every terminating decimal is equal to some rational number: for example, $0.3756$ is equal to $\frac{3756}{10000}$, which we could cancel down if we wanted. It is also true, but not quite so obvious, that every rational number is equal to either a terminating or a recurring decimal, as in the examples above. That is because, when you do the 'long division' for $\frac{9}{14}$ (for example), there are only $13$ possible remainders ($1$ to $13$) if it does not divide exactly. After a while, one of the remainders you've had already must come around again, so the pattern recurs. What is even less obvious is that every recurring decimal is also equal to a rational number. We can use mediants to find out which one.

Take the example above $0.4 21951 21951 21951...$ It must lie between $\frac{0}{1}$ and $\frac{1}{1}$, and we just proceed as before. The table of mediants will look like this:

 $0$ $1$ $1$ $2$ $1$ $1$ $0$ $1$ $1$ $3$ $1$ $2$ $1$ $3$ $2$ $5$ $1$ $2$ $2$ $5$ $3$ $7$ $1$ $2$ $2$ $5$ $5$ $12$ $3$ $7$ $5$ $12$ $8$ $19$ $3$ $7$ $8$ $19$ $11$ $26$ $3$ $7$ $8$ $19$ $19$ $45$ $11$ $26$ $8$ $19$ $27$ $64$ $19$ $45$ $27$ $64$ $46$ $109$ $19$ $45$ $27$ $64$ $73$ $173$ $46$ $109$ $27$ $64$ $100$ $237$ $73$ $173$ $100$ $237$ $173$ $410$ $73$ $173$

It turns out that $0.4 21951 21951 21951... = \frac{173}{410}$. There are no long 'runs' of particular upper or lower bounds, and the mediants approach (shown) is pretty efficient. The convergents (see above) are shaded. Each convergent is closer to the target than the previous one. (When there are long runs, the continued fractions method, which only picks out the convergents, is more efficient because it cuts out a lot of steps.)

THE SQUARE ROOT OF TWO

The square root of $2$, $\sqrt2$, is a number between $1$ and $2$. By Pythagoras' Theorem, it is the length of the diagonal of a square with side $1$. Its decimal form begins:

$1.41421 35623 73095 04880 16887 24209 69807 85697...$

which shows no sign (yet) of recurring. That suggests that it may not be equal to any rational number because, as we noted earlier, every rational number is equal to either a terminating or a recurring decimal. Of course, we can't yet be sure that the decimal written above will not eventually terminate or recur.

What happens if we try to find a 'target' rational for $\sqrt2$ using the mediants method? First, try it yourself with a calculator. Start with $\frac{1}{1}$ and $\frac{2}{1}$, because $\sqrt2$ lies between $1$ and $2$. Move ahead to the next piece of text when you've thought about it.

This spreadsheet shows what you should have found (and probably more!):

 $1$ $1$ $3$ $2$ $2$ $1$ $1$ $1$ $4$ $3$ $3$ $2$ $4$ $3$ $7$ $5$ $3$ $2$ $7$ $5$ $10$ $7$ $3$ $2$ $7$ $5$ $17$ $12$ $10$ $7$ $7$ $5$ $24$ $17$ $17$ $12$ $24$ $17$ $41$ $29$ $17$ $12$ $41$ $29$ $58$ $41$ $17$ $12$ $41$ $29$ $99$ $70$ $58$ $41$ $41$ $29$ $140$ $99$ $99$ $70$ $140$ $99$ $239$ $169$ $99$ $70$ $239$ $169$ $338$ $239$ $99$ $70$ $239$ $169$ $577$ $408$ $338$ $239$ $239$ $169$ $816$ $577$ $577$ $408$ $816$ $577$ $1393$ $985$ $577$ $408$ $1393$ $985$ $1970$ $1393$ $577$ $408$ $1393$ $985$ $3363$ $2378$ $1970$ $1393$ $1393$ $985$ $4756$ $3363$ $3363$ $2378$ $4756$ $3363$ $8119$ $5741$ $3363$ $2378$ $8119$ $5741$ $11482$ $8119$ $3363$ $2378$ $8119$ $5741$ $19601$ $13860$ $11482$ $8119$ $8119$ $5741$ $27720$ $19601$ $19601$ $13860$ $27720$ $19601$ $47321$ $33461$ $19601$ $13860$ $47321$ $33461$ $66922$ $47321$ $19601$ $13860$ $47321$ $33461$ $114243$ $80782$ $66922$ $47321$ $47321$ $33461$ $161564$ $114243$ $114243$ $80782$ $161564$1142432758071950251142438078227580719502539005027580711424380782275807195025665857470832390050275807$It is interesting to note that alternate mediants are 'convergents' which start a 'run' of upper and lower bounds (alternately) to$\sqrt2$. The mediants never stabilise on a target rational number. The table represents a process of persistently looking for a rational number equal to$\sqrt2$, but one which is doomed to search forever, but without success. We can begin to see why the process is doomed to fail by inspecting the sequence of convergents (shaded). How is each one related to the previous one? Look how they begin: $327517124129$If the convergent$\frac{a}{b}$is followed by$\frac{c}{d}$, you can see that$d=a+b$and then$c=b+d$, or$a+2b$. The pattern applies to the whole table above (I really ought to prove that it would apply no matter how far we extend the table, but I won't!), so that$\frac{a}{b}$is always followed by$\frac{a+2b}{a+b}$. If the sequence did eventually hit the target (a rational number exactly equal to$\sqrt2$, then at some stage$\sqrt2 = \frac{a+2b}{a+b}$. [Now we need to do some algebra.] It follows that$\sqrt2(a+b) = (a+2b)$. Squaring both sides of this equation gives$2(a^2+2ab+b^2) = a^2+4ab+4b^2$, or$2a^2+4ab+2b^2 = a^2+4ab+4b^2$, so$a^2 = 2b^2$, in which case$\frac{a}{b} = \sqrt2$. To recap: this shows that a particular convergent$(\frac{a+2b}{a+b})$can only be equal to$\sqrt2$if its previous one$(\frac{a}{b})$was also equal to$\sqrt2$. Since none of the initial mediants$\frac{3}{2}$,$\frac{7}{5}$, etc is exactly equal to$\sqrt2$, none of the subsequent ones can ever be. Nevertheless, it is worth noting that the mediants process will certainly get you closer and closer to$\sqrt2$, with some very good approximations. For example, if you square$\frac{8119}{5741}$you get$\frac{65918161}{32959081}$which equals$2 - \frac{1}{32959081}$. Or, the decimal equivalent of the last convergent listed above$(\frac{665857}{470832})$is$1.41421 35623 7469...$which differs from$\sqrt2$by less than$\frac{1}{10000000000}$. The discovery that the number line contains numbers like$\sqrt2$which are not rational was a big shock to the ancient Greeks, who believed that any two numbers (unlike human beings?) enjoy a rational relationship. Clearly this is not true of$\sqrt2$and$1$, the lengths of the diagonal and side of a square. Non-rational numbers like$\sqrt2$are called irrational numbers. Tradition says that Pythagoras first proved that$\sqrt2$is irrational, and that he sacrificed 100 oxen to celebrate his success. Pythagoras' proof is the one still usually taught today. The celebrated Cambridge mathematician G. H. Hardy regarded Pythagoras' proof as a model of mathematical beauty. I shall give a variation on it: If you're interested, you can read Pythagoras' original in Hardy's book A Mathematician's Apology. Suppose that$\sqrt2$is rational, so it can be written as$\frac{a}{b}$. We can arrange that$a$and$b$have no common factor (by 'cancelling down' if necessary). Since$\sqrt2 = \frac{a}{b}$, it follows that$2 = (\frac{a}{b})^2$. In other words,$a^2$is exactly double$b^2$.$a$and$b$are whole numbers, so each ends (in our usual whole number notation) in one of the digits$0$to$9$. Therefore, considering the squares of$0$to$9$, both$a^2$and$b^2$must end in$0$,$1$,$4$,$9$,$6$or$5$. But since$a^2$=$2b^2$,$a^2$must end in$0$,$8$or$2$. As we've seen$a^2$can't end in$8$or$2$, so it must end in$0$. Again, since$a^2=2b^2$, this must mean that$b^2$ends on$0$or$5$. To summarise:$b$ends in$0$or$5$, and$a$ends in$0$. Clearly then, if$\sqrt2= \frac{a}{b}$, both$a$and$b$would have to be multiples of$5$. But this must be ruled out, because we said that$a$and$b$would have no common factor. We must conclude that it can't be done - the attempt to express$\sqrt2$as a rational number has failed. Finale -$\pi\sqrt2$is a good number to focus on to prove that irrational numbers exist. In fact, there are plenty of irrational numbers. Georg Cantor (1845-1918) showed that, in a sense, most of the number line consists of irrational numbers! It is easy to show that$\sqrt n$is always irrational when$n$is not a perfect square$1$,$4$,$9$,$16$,$25$, etc. One of the most interesting irrational numbers is$\pi$, the ratio of the circumference of any circle to its diameter. Unfortunately, one cannot use a simple Pythagoras-type proof to show that$\pi$is irrational, but a quick look at the spreadsheet from the mediants/convergents approach is very illuminating. Since we know that$\pi$lies between$3$and$4$, we take$\frac{3}{1}$and$\frac{4}{1}$as our initial upper/lower bounds. $317241311037231134103311651343119616531227196312582272584715227471569222276922912922791291133622711336135432271354315750227157501795722717957201642272016422371227223712457822724578267852272678528992227289923119922731199333106227333106355113227333106688219355113$In fact,$\frac{3}{1}$starts a 'run' of lower bounds, and is the first convergent. It is the value for$\pi$given (or implied) in the Bible around 900BC (1 Kings 7, verse 23). The next convergent is$\frac{22}{7}$, without doubt the best-known approximation to$\pi$. So well-known, in fact, that many people believe (from memories of school) that$\frac{22}{7}$is$\pi$, rather than an approximation. The mediant/convergent$\frac{22}{7}$starts a long 'run' (length$15$) of upper bounds. The length of the run indicates what a good approximation it is. To improve on it, the denominator has to increase from$7$to$57$(the rational number is$\frac{179}{57}$, in fact, one of the mediants shown in the table above). That is to say, no rational number with denominator less than$57$is closer to$\pi$than$\frac{22}{7}$is. The next convergent is$\frac{333}{106}$, followed immediately by$\frac{355}{113}$. The Chinese Mathematician Tsu Chung Chih knew this approximation around 480 AD. An indication of how good an approximation this is may be gathered from the fact that it starts a 'run' of length$292$! Which is one reason why I have not shown the run above. In fact,$\frac{355}{113}= 3.14159 29203...$Whereas$\pi= 3.14159 26535...$The next convergent after that long 'run' would be$\frac{103993}{33102}\$, although the improvement as an approximation is not as great as you might expect - try it!