Proof that by multipliying by $10^k$ decimal point moves to the right Trying to solve the primer below but don't really understand what is needed to be done:
$$Proof\ that\ if\ {p\over{q}}\ is\ a\ rational\ number\ such\ that\ p \in N\ and\ q \in N\ that\ corresponds\ to\ an\ infinite\ periodic\ decimal\ \alpha, then\ 10^k\cdot{p\over{q}}\ corresponds\ to\ decimal\ number\ such\ is\ \alpha\ with\ decimal\ point\ moved\ k\ positions\ to\ the\ right.$$
I don't quite understand how to do that without applying multiplication rules(is it even what's needed to be done here?).
 A: You can define a decimal number by $z + \sum_{i=1}^\infty a_i\cdot 10^{-i}$ where $z$ is an integer and the $a_i$s are integers between 0 and 9 (inclusive). In this definition, $z$ is the part before the decimal point, and the sum is the part after it. We see that the sum is bounded above by $\sum_{i=1}^\infty 9\cdot 10^{-i} = 9\sum_{i=1}^\infty 10^{-i} = \frac{9}{10}\sum_{i=0}^\infty 10^{-i} = \frac{9}{10}\cdot \frac{1}{1-10^{-1}} = \frac{9}{10}\cdot \frac{1}{1-\frac{1}{10}} = \frac{9}{10}\cdot \frac{1}{\frac{9}{10}} = 1$ so the sum will always converge (increasing and bounded above implies convergence; Monotone Convergence Theorem). Since it converges, we are allowed to say that $c\sum(\text{something}) = \sum c(\text{something})$; i.e. we can bring constants in and out. Then $10^k\cdot (z + \sum_{i=1}^\infty a_i 10^{-i}) = 10^k\cdot z + 10^k\cdot \sum_{i=1}^\infty a_i 10^{-i} \\ = 10^k\cdot z + \sum_{i=1}^\infty a_i 10^{-i+k}  \\ = 10^kz + 10^{k-1}a_1 + ... + 10^{k-k}a_k +\sum_{i=1}^\infty a_{k+i}10^{-i}$
The first part is a sum of integers (which is an integer) and the sum is the rest of the decimal digits. We see that the digits themselves did not change, but the location of the decimal (right before the sum) has changed.
