Solution. In each case multiply by the two powers of \(10\) that place the repeating block in the same position, then subtract.

(a). Let \(x=4.\overline {3}\). Then \(10x=43.\overline {3}\) and \[10x-x=43.\overline {3}-4.\overline {3}=39 \implies 9x=39 \implies x=\frac {39}{9}=\frac {13}{3}.\]

(b). Let \(x=-0.2\overline {55}\). One digit precedes the block and the block has two digits, so use \(100x\) and \(10x\)... but the block here is \(55\) repeating from the second decimal place, so \[10x=-2.\overline {55},\quad 1000x=-255.\overline {55},\] and \(1000x-10x=-253\), giving \(990x=-253\) and \[x=-\frac {253}{990}=-\frac {23}{90},\] dividing above and below by \(11\).

(c). Let \(x=12.34\overline {11}\). Two digits precede the block and the block has two digits, so \[100x=1234.\overline {11},\quad 10000x=123411.\overline {11},\] and subtracting, \(9900x=122177\), so \[x=\frac {122177}{9900},\] which is already in lowest terms (\(9900=2^{2}\cdot 3^{2}\cdot 5^{2}\cdot 11\) and \(122177\) is divisible by none of \(2\), \(3\), \(5\) or \(11\)).

Solution. (a). By contradiction. Suppose \(\sqrt {3}=\frac {p}{q}\) with \(p,q\) integers having no common factor and \(q\neq 0\). Squaring, \(3q^{2}=p^{2}\), so \(p^{2}\) is divisible by \(3\); since \(3\) is prime this forces \(p\) to be divisible by \(3\), say \(p=3k\). Then \(3q^{2}=9k^{2}\), so \(q^{2}=3k^{2}\) and \(q\) is divisible by \(3\) as well. But then \(p\) and \(q\) share the factor \(3\), contradicting the assumption that they have none. Hence \(\sqrt {3}\) is irrational.

(b). This one needs no fresh work. Suppose \(\sqrt {2}-1\) were rational, say equal to \(r\). Then \[\sqrt {2}=r+1,\] and the sum of two rationals is rational, so \(\sqrt {2}\) would be rational — which was disproved earlier in this chapter. Hence \(\sqrt {2}-1\) is irrational.

Solution. Two digits precede the repeating block and the block has one digit, so multiply by \(10^{2}\) and \(10^{3}\) and subtract. Let \(x=-0.12\overline {3}\): \[100x=-12.\overline {3},\qquad 1000x=-123.\overline {3}.\] Subtracting, \(900x=-111\), so \[x=-\frac {111}{900}=-\frac {37}{300},\] dividing above and below by \(3\).