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.
Asked
Active
Viewed 529 times
4
-
3Isn't this a rather infinite "number"? It's larger than $1$, $10$, $100$, $1000$, etc. – t.b. Jun 23 '11 at 00:45
-
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 00:51
-
310-adic numbers... – yoyo Jun 23 '11 at 00:53
-
910-adic numbers, OK, but Emphasis not real numbers – GEdgar Jun 23 '11 at 01:05
-
1In 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
-
@WillieWong The Wikipedia article for numbers in general has an easier introduction to p-adic numbers. – Socowi Dec 28 '22 at 11:25
-
This question boils down to "What is a number"? See Is infinity a number? for a similar (but different!) discussion. – Socowi Dec 28 '22 at 11:25
1 Answers
2
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.
davidlowryduda
- 91,687
-
3Your 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). – davidlowryduda 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 (http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula)? – TonyK Jul 26 '11 at 16:44