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Is it possible to have a number that extends to the left of the decimal point in mirror image of an irrational number? Such as <...95141.30000...>, to write pi as a mirror image.

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Isn't this a rather infinite "number"? It's larger than $1$, $10$, $100$, $1000$, etc. –  t.b. Jun 23 '11 at 0:45
I don't know how to think of this kind of expression. Does it make any sense to write a number with an infinite expansion of digits on the left of the decimal? I first thought of this in trying to get the reciprocal to an irrational (which can be approximated by getting the reciprocal of a nearby rational number). But specify the irrational expansion as a ratio of (say) 3.14159.../1.000000..., and then flip numerator-denominator. What ARE those values, as counting numbers, in the num/denom places? –  user12404 Jun 23 '11 at 0:51
10-adic numbers... –  yoyo Jun 23 '11 at 0:53
10-adic numbers, OK, but Emphasis not real numbers –  GEdgar Jun 23 '11 at 1:05
In case the OP (or any other reader) is interested and not aware of the construction, 10-adic numbers (as mentioned above by yoyo and GEdgar) are explained in the Introduction section of this Wikipedia entry. –  Willie Wong Jun 25 '11 at 15:28

1 Answer 1

No. Why? Let us assume that we don't claim that the result is a 'number,' as it's infinite (problematic). We could make this more precise, saying that we want some sort of way of placing a 'mirror' in the middle of $\pi$ s.t. we have something like $ ... \alpha _3 \alpha _2 \alpha _ 1 | \beta _1 \beta _2 \beta _3 ... $ and such that $\alpha _i = \beta _i$.

But even this is not very meaningful, and you included the key problem in your question. Eventually, you will have an infinite string of zeroes on one side - but not possible on the other side (as the expansion of pi does not terminate).

I wrote that knowing that it's hardly sensible to describe, not because I enjoy fancying things that are awkward (which may or may not be true), but because there is an interesting related fact. For any finite number $k$, there exists a place to put the 'mirror' in the expansion of $\pi$ such that $\alpha _i = \beta _i \quad \forall \; i \in [0, k]$. And that's pretty cool, and even related to the question.

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Your last paragraph may be correct, but I am pretty sure that nobody has managed to prove it yet. It's related to the normality of $\pi$, which is an open question. –  TonyK Jul 25 '11 at 19:04
@TonyK: you know, I was happy to write that at the time because I had read that statement in a paper just a few days beforehand. But I don't remember where - and you're certainly right that it's related to the normality of pi. But it's not as strong as saying pi is normal (at least, not at first sight). –  mixedmath Jul 25 '11 at 19:10
Yes, there are certainly non-normal numbers that contain palindromes of arbitrary length. But has $\pi$ really been proven to contain palindromes of arbitrary length? I would love to see a reference! Is it perhaps related to the Bailey-Borwein-Plouffe formula (en.wikipedia.org/wiki/…)? –  TonyK Jul 26 '11 at 16:44

protected by Qiaochu Yuan Sep 23 '11 at 16:36

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