Integer outputs of $y=x^2$ , do their last digits form an irrational? Let the domain of $y=x^2$ be the positive integers. I input consecutive positive integers from $[1, \infty)$ their last digits are $a, b, c, ...$ respectively. If I then make the number $z=\frac {a}{10}+\frac{b}{100} +\frac {c}{1000}+\cdot \cdot \cdot$ , is this number irrational? If not, is there any other combination (such as $\frac {c}{10}+\frac{f}{100} +\frac {q}{1000}+\cdot \cdot \cdot$, but still using all of $a, b, c, \cdot \cdot \cdot$) that could result in an irrational number? I have no idea how to solve this problem, it is just something I have wondered about for a long time.
 A: The result is rational, because the last digit of $x^2$ depends only on the last digit of $x$, which repeats with period $10$. Therefore your $a, b, c \ldots$ also repeat with period 10 (or with a period that's a divisor of 10).
If you consider rearrangments of the digits, it's easy enough to produce an irrational. Your sequence $a,b,c,\ldots$ contains infinitely many of each of the digits 1, 4, 5, 6, 9 and 0 (and no other digits), so any decimal fraction that's also made of only these digits, and infinitely many of each, can be viewed as a rearrangement of your sequence. For example,
$$ 0.14569014569001456900014569000014\ldots $$
where each sequence of zeroes is one digit longer than the previous one.
A: I will elaborate on Henning Makholm's answer.  The last digit of $x^2$ depends only on the last digit of $x$, and so must repeat with period 10.  In fact:
$$\begin{array}{r|cccccccccc}
ld(x)  & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
ld(x^2)& 0 & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1
\end{array}$$
so your number $z$ is exactly $$z = 0.\overline{1496569410}$$
which is equal to $$\frac{1496569410}{9999999999} = \frac{166285490}{1111111111}.$$
To answer your second question is easy.  To generate an irrational number, we need only avoid repeating the digits. Just just choose two sequences of $1496569410$ in different orders, say $ x= 0114456699$ and $y = 5449911660$, and append them in a non-periodic way, say $xyxxyxxxyxxxxy\ldots$ giving $$.\overbrace{0114456699}^x\ \overbrace{5449911660}^y\ \overbrace{0114456699}^x\ \overbrace{0114456699}^x\ \overbrace{5449911660}^y\ \overbrace{0114456699}^x\ldots .$$
