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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.

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  • $\begingroup$ Hint: a number is irrational if and only if the digits never form a repeating pattern. Try finding the first say 20 digits. $\endgroup$ – Nate Aug 23 '13 at 15:38
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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 $$.\overbrace{0114456699}^x\ \overbrace{5449911660}^y\ \overbrace{0114456699}^x\ \overbrace{0114456699}^x\ \overbrace{5449911660}^y\ \overbrace{0114456699}^x\ldots .$$

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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.

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  • $\begingroup$ Hah I can't believe it was so easy $\endgroup$ – Ovi Aug 23 '13 at 15:39

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