In particular, this tells us that the mysterious decimal representation of $\frac 1{9801}$, too, is bound to repeat eventually. (In fact, it repeats after having cycled through all possible pairs of digits, except for the second-to-last:

$$\frac 1{9801}=0.\overline{000102030405\dots9799}.)$$

## Which representation is better?

Now comes the moment to address "decimal fixation" head-on. **Why should we continue dealing with fractions** when we now have decimals? The answer lies in comparing a few problems in either representation. Give the following "a" problems to one half of the class, and the "b" problems to the other, and have them work by hand:

**Problem 1a. **$\frac {26}{75}+\frac {77}{100}=?$

**Problem 1b. **$0.34\overline{6} + 0.77 = ?$

**Problem 2a. **$\frac{3}{8}\times\frac{4}{15} = ?$

**Problem 2b. **$0.375 \times 0.2\overline{6} = ?$

**Problem 3a. **Which is greater: $\frac{7}{11}$ or $\frac{13}{20}$?

**Problem 3b. **Which is greater: $0.\overline{63}$ or $0.65$?

By comparing the effort needed for each problem, it becomes apparent that the decimal representation is much better suited for **adding and comparing**, while **multiplication** is the strong suit of the fractional representation. This is why we math teachers insist on keeping the latter around (especially since multiplication is the most frequent operation in higher and applied math). So it turns out that the fractional representation **can often give us more useful information** about a number than its decimal representation, and vice versa.

## Moonwalking back to fractions

At this point, one can already see what **irrational numbers** are about: they are **decimals that fall in none of the three categories** listed in the theorem. Or at least, such decimals certainly have no fractional representation. But is this the only restriction that applies? What about very complicated repeating decimals such as $0.987\overline{14916253649}$ or $0.\overline{1234567891011112\dots9989991000}$: do they also have a fractional representation? Or are they irrational?

While the existence of irrational numbers can already be deduced here, and the question of how to transition from a decimal to a fractional representation casts no doubt on this result, it is worth investigating to gain a thorough understanding of two deep concepts:

- Fractions and decimals are
**but two different representations****: ****none is a truer form than the other.** They both have their advantages and drawbacks. (This is why from now on we speak of **rational numbers**, rather than fractions.) - Indeed, irrationals are
**precisely** the decimals that fit none of the aforementioned three categories.

So how do we represent a decimal as a fraction? A **terminating decimal** is easy: it is always an integer multiple of one of the "base-ten" fractions

$$0.1 = \frac 1{10},\quad 0.01 = \frac 1{100},\quad 0.001 = \frac 1{1000}, \quad\ldots$$

so that e. g.

$$3.285 = 3285\times 0.001 = 3285\times\frac 1{1000} = \frac{3285}{1000} = \frac{657}{200}.$$

The "base-ten" fractions (inverse powers of ten) are the **building blocks** of the terminating decimals.

**Repeating decimals** require a bit more work. But we can find the general method by looking at select examples alone:

$$0.\overline{2} = 2\times 0.\overline{1}=\frac 29$$

$$0.\overline{35} = 35\times 0.\overline{01} = \frac{35}{99}$$

$$0.\overline{123} = 123\times 0.\overline{001} = \frac{123}{999} = \frac{41}{333}$$

The primitive fractions $\frac 19$, $\frac1 {99}$, $\frac1 {999}$, … turn out to be the **building blocks of the purely repeating decimals**, just as $\frac 1{10}$, $\frac1 {100}$, $\frac1 {1000}$, … were the building blocks of the terminating decimals.

What about periods that don't start right after the decimal point? Let's do the easier case of just zeroes before the period first:

$$0.0\overline{4} = 4\times 0.0\overline{1} = 4\times\frac{0.\overline{1}}{10} = 4\times \frac 1{10}\times\frac 1{9} = \frac 4{90} = \frac{2}{45}$$

$$0.00\overline{234} = 234\times\frac {0.\overline{001}}{100} = 234\times \frac 1{100}\times\frac 1{999} = \frac{26}{11100}$$

The inverse powers of ten help their friends find the right place after the decimal point!

Lastly, decimal numbers with non-zero digits before the period can be **split into their non-repeating and purely repeating parts:**

$$1.1\overline{45} = 1.1 + 45\times 0.0\overline{01} = \frac{11}{10} + \frac{45}{990} = \frac{63}{55}.$$

We have thus established, by exhaustion, the **converse of the classification theorem** from above: