2
$\begingroup$

I know the standard injection is to consider the binary expansion of a number in the interval, but I was wondering if it is possible to create an injection using a decimal expansion.

To that effect first denote the decimal expansion of a $r\in (0,1)$ as $r=0.d_1d_2d_3d_4\dots$. Choose the decimal expansion of all $x\in (0,1)$ as above except that $\forall n\in \mathbb{N} \exists m>n$ such that $d_m\in \{0,1,2,3,4,5,6,7,8\}$. (This does not exclude any element in the interval).

Now consider the function $f:(0,1)\to p(\mathbb{N})$ where $f(x)=\{2^{d_1},3^{d_2},5^{d_3},\dots,p_i^{d_i},\dots\}$. I believe that Euclid's theorem and the fundamental theorem of arithmetic ensure that $f$ is an injection, but I would like some confirmation.

$\endgroup$
10
  • 1
    $\begingroup$ (MInor typo: you want $p_i^{d_i}$ in the expression for $f(x)$.) This almost works, but you need to increase the exponents by $1$ so as not to get a bunch of $1$s in $f(x)$. $\endgroup$ – Brian M. Scott Apr 5 '16 at 21:12
  • $\begingroup$ Sorry: I slipped up with the first comment, so please check the revised version! $\endgroup$ – Brian M. Scott Apr 5 '16 at 21:15
  • $\begingroup$ Damnit I had actually considered that last night while falling asleep but completely forgot to put it in now (not just trying to save face I promise :) ) $\endgroup$ – K.Power Apr 5 '16 at 21:19
  • $\begingroup$ :-) I know the feeling! $\endgroup$ – Brian M. Scott Apr 5 '16 at 21:20
  • $\begingroup$ Oh well my lapse is your gain of at least $10/357000$ reputation points. Thanks for the help. I'll accept your answer when the time restriction ends. $\endgroup$ – K.Power Apr 5 '16 at 21:23
1
$\begingroup$

This almost works, but if you have more than one $0$ digit, you’ll get multiple copies of $1$ in the description of $f(x)$. To avoid this problem, define

$$f(x)=\left\{p_i^{d_i+1}:i\in\Bbb Z^+\right\}\;.$$

Added: As Henning Makholm notes below in the comments, the original idea of setting

$$f(x)=\left\{p_i^{d_i}:i\in\Bbb Z^+\right\}$$

actually does work; it just isn’t quite as elegant. If no multiple of some prime $p_i$ appears in $f(x)$, then $d_i=0$, so the $i$-th digit is still reconstructible. In that case the $1\in f(x)$ doesn’t actually tell us anything that we can’t already learn from $f(x)\setminus\{1\}$.

$\endgroup$
4
  • 1
    $\begingroup$ Is that actually a problem? The digit positions whose prime doesn't appear with positive power must be 0, so we can still reconstruct the original decimal fraction from the set, and the original mapping is in fact injective. $\endgroup$ – hmakholm left over Monica Apr 5 '16 at 22:00
  • $\begingroup$ @Henning: You’re right; it’s just slightly less slick. $\endgroup$ – Brian M. Scott Apr 5 '16 at 22:01
  • $\begingroup$ I was just about to comment asking for an example of why my original method wouldn't work. Your edit explains why I couldn't think of one. Thanks again. $\endgroup$ – K.Power Apr 6 '16 at 13:34
  • 1
    $\begingroup$ @K.Power: You're welcome. Sorry for the confusion; not one of my better efforts, but at least we got it sorted eventually! $\endgroup$ – Brian M. Scott Apr 6 '16 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.