2.1 Rational Numbers
- The set of rational numbers, denoted by \(\mathbb {Q}\), is \[\mathbb {Q}=\left \{\frac {a}{b}\ :\ a,b\in \mathbb {Z},\ b\neq 0\right \}.\]
- The decimal representation of a rational number either terminates, as \(\frac {4}{25}=0.16\), or repeats forever, as \(\frac {5}{7}=0.\overline {714285}\).
- A repeating block is marked with a bar over it. Note that \(\frac {5}{7}=0.\overline {714285}\) exactly, whereas \(0.714285714\) is only an approximation — a calculator display is rounded, and writing the rounded value with an equals sign is wrong.
- Conversely, when a number is given in decimal form and terminates or repeats, integers \(a\) and \(b\) can always be found with the number equal to \(\frac {a}{b}\). The method is shown below.
Theorem 2.2. A real number is rational if and only if its decimal expansion either terminates or is eventually periodic.
Note 2.3. The forward direction is visible in long division. Dividing \(a\) by \(b\), the only remainders possible are \(0,1,2,\dots ,b-1\) — just \(b\) of them. If a remainder of \(0\) ever appears the division stops and the decimal terminates; otherwise, within at most \(b\) steps some remainder must repeat, and from that point the whole calculation repeats, so the digits do too. That is also why the repeating block of \(\frac {a}{b}\) can never be longer than \(b-1\) digits: \(\frac {5}{7}\) repeats with period \(6\), and no seventh is worse.
The reverse direction is the trick used in the examples below.
Note 2.4. Why the method works. To convert a repeating decimal, multiply by the power of \(10\) that shifts the decimal point past one whole repeating block, then subtract the original. The repeating tails are then identical and cancel exactly, leaving a terminating number on the right and a whole-number multiple of \(x\) on the left.
Choosing the power is the only thing to get right: if the block has \(k\) digits and there are \(j\) non-repeating digits after the point, subtract \(10^{j}x\) from \(10^{j+k}x\). For \(0.354\overline {54}\) there is one digit before the block starts (\(j=1\)) and the block has two digits (\(k=2\)), so it is \(1000x-10x\) — which is exactly what the first example does.
Solution. Let \(x=0.354\overline {54}\). One digit stands between the point and the start of the repeating block and the block has two digits, so by the rule above the two multiples to take are \(10x\) and \(1000x\): \[10x=3.\overline {54},\qquad 1000x=354.\overline {54}.\] Both have the same repeating tail, so subtracting removes it entirely: \[\begin {array}{r@{\;}c@{\;}l} 1000x & = & 354.\overline {54}\\[2pt] -\ \ 10x & = & \phantom {00}3.\overline {54}\\[2pt] \hline 990x & = & 351 \end {array}\] Therefore \[x=\frac {351}{990}=\frac {39}{110},\] which is a ratio of two integers, so \(0.354\overline {54}\) is rational.
Example 2.6. Express \(3.8\overline {3}\) in the form \(\frac {a}{b}\) where \(a\) and \(b\) are integers.
Solution. Let \(x=3.8\overline {3}=3.8333\ldots \) Here one digit stands before the repeating block and the block is a single digit, so the multiples to take are \(10x\) and \(100x\): \[10x=38.\overline {3},\qquad 100x=383.\overline {3}.\] Subtracting, \[\begin {array}{r@{\;}c@{\;}l} 100x & = & 383.\overline {3}\\[2pt] -\ \ 10x & = & \phantom {0}38.\overline {3}\\[2pt] \hline 90x & = & 345 \end {array}\] Therefore \[x=\frac {345}{90}=\frac {23}{6}.\]
Solution. Let \(x=0.217\overline {217}=0.\overline {217}\). The repeating block begins immediately after the point and has three digits, so a single multiplication by \(10^{3}\) is enough: \[1000x=217.\overline {217}.\] Subtracting the original, \[\begin {array}{r@{\;}c@{\;}l} 1000x & = & 217.\overline {217}\\[2pt] -\ \ \ \ x & = & \phantom {000}0.\overline {217}\\[2pt] \hline 999x & = & 217 \end {array}\] Therefore \[x=\frac {217}{999},\] and since \(999=27\times 37\) and \(217=7\times 31\) share no factor, this is already in lowest terms.
Solution. Let \(x=4.75\), multiply by 100 both sides, we get \(100x=475\)
\(\implies \quad x=\frac {475}{100}\quad \implies \quad x=\frac {19}{4}\).
Solution. \( \begin {matrix} \text {Let}\quad x &= &-0.72\overline {72}\\ 100x & = & -72.72\\ \end {matrix} \)
\( \begin {matrix} 100x & = & -72.72\\ \underline {-x} & = & \underline {+ 0.72}\\ 99x & = & -72\\ \end {matrix} \)
\(\implies \qquad x=\frac {-72}{99}=-\frac {8}{11}\)
Always reduce the fraction at the end. Both \(-\frac {72}{99}\) and \(-\frac {8}{11}\) are correct, but only the second is in lowest terms.
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