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Is it possible to write any rational number (say 1 or 2 or .15) using a number system that was base pi instead of a number system that used a rational number as its base?

Tyler Hilbert
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3 Answers3

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EDIT : The issue : The general setup for a given base $b$ is to write a given Real number $a$ as $a_0 b^0 + a_1b^1+....+a_k b^k$ with $a_k$ an integer less than the floor of $b$.

EDIT 2: STILL needs some work, which I am doing right now. Will be back soon to rewrite, or, if necessary, delete. Comments are welcome.

In our case, for $\pi$ we want to represent ( I assume) a Real number $a$ as $ \pi^0 a_0+ \pi^1 a_1+....+ \pi^n a_n ; a_j < \pi$

at each stage, you choose the largest number $a_k$ , so that $$a_0 \pi^0+a_1\pi^1+...+a_k \pi^k \leq a $$.

This is a monotone, non-decreasing sequence of Real numbers bounded above by $a$, so it will converge to its $lub=a $.

I think we need as a basis a Real number $x; |x|>1$.

gary
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  • You need to impose some more constraints on the coefficients $a_i$. Otherwise your process will terminate at stage $0$ with $a_0 = a$. – Rob Arthan Aug 06 '17 at 23:57
  • @RobArthan: Please note I used the constraint $a_j < \pi$ in the top line. – gary Aug 07 '17 at 00:01
  • But there is no largest real number meeting the constraint $a_j < \pi$. – Rob Arthan Aug 07 '17 at 00:07
  • No, that is a general condition on the coefficients. I mean we find the sum as described above with $a_j < \pi$ and $ a_0 \pi^0+...+a_k \pi^k \leq a $. I am not requiring the largest Real number less than $\pi$. – gary Aug 07 '17 at 00:10
  • So what constraints are you imposing on the coefficients? Integers? Integers less than $\pi$? Or what? If you allow $a_k$ to be any rational or real number there won't be a largest $a_k$ satisfying your conditions. – Rob Arthan Aug 07 '17 at 00:13
  • Part of the problem is I am not clear on the requirements of the OP for a number to be a base. But I see your point. This clearly would not work with integer coefficients, but why not for Rational $a_j$? – gary Aug 07 '17 at 00:21
  • With rational coefficients you still don't get a largest $a_j$ satisfying the conditions. – Rob Arthan Aug 07 '17 at 00:31
  • @RobArthan: Ok, let me think it through more carefully. I will edit to this effect. – gary Aug 07 '17 at 00:34
  • See my first comment for the OP. You will need to allow negative powers of $\pi$ also. – GEdgar Aug 07 '17 at 12:03
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We can write rational numbers in irrational bases; easily shown by $100$ base $\sqrt 2$ is 2. There are in fact problems and papers that I have seen that use base phi, $\frac{\sqrt 5 + 1}{2}$.

However, because pi is transcendental, it is not the solution of any polynomial with rational coefficients. This means that no rational number can be written as a sum of powers of pi, which is equivalent to the fact that you can not write any rational number base pi.

Edit: Of course, you can always write the numbers non negative integers less than or equal to the base in that base, like 2 base pi is still $2\cdot \pi^0 = 2$. I was talking about everything else.

D.R.
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    $1_{\pi}$ is rational. – Harry Aug 06 '17 at 23:31
  • @Harry: what's $1_{\pi}$? – Rob Arthan Aug 06 '17 at 23:39
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    $1_{\pi}$ is $1$ in base $\pi$. It can be expressed as the ratio of two integers in base $\pi$ (namely $1$ and $1$), so it's rational. – Harry Aug 06 '17 at 23:41
  • @Harry: Your terminology is a bit confused: the number $1$ is independent of the base you represent it in. Something like $1_{10}$ means the representation of $1$ in base $10$. So $1_{10}$ or $1_{\pi}$ aren't numbers but rather strings of symbols. However you are quite right that the string of symbols "1" will represent $1 = 1 \times \pi^0$ base $\pi$ in any formulation of "base $\pi$" numbers that allows $1$ as a digit. – Rob Arthan Aug 06 '17 at 23:49
  • @RobArthan I suppose I was confused. Are you saying that, for example, $45_{\pi}$ is ($1$) the representation of $45$ in base $\pi$ ($1110.30022 \dots$) or ($2$) $4\pi+5$? Or am I missing the point completely? – Harry Aug 07 '17 at 00:10
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    @Harry: I was a little confused too. On reflection, there is a choice of terminology: $S_b$ can reasonably be either an operation on a string of base $b$ digits $S$ that returns an integer or an operation on an integer $S$ that returns a string of base $b$ digits. You were using the former interpretation. – Rob Arthan Aug 07 '17 at 00:29
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This isn't an answer. But it's too big to be a comment.

There is a problem with non-integer bases. Let $B$ be an integer and let $b=B-1$. Then $$(0.bbbbbb\dots)_B = 1_B$$ just as we would expect.

In base $\pi$, however, $\pi - 1 < 3$. So it shouldn't come as a surprise that $$(0.333333\dots)_B = \dfrac{3}{\pi-1} \approx 1.40083_{10} > 1_\pi$$

So ordering numbers base $\pi$ is more complicated than ordering numbers in an integer base. It also seems that some numbers can be expressed in more than one, non-trivial, way.

Steven Alexis Gregory
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