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Pi is real number. We can enumerate any digit of Pi number.

Pi has infinite number of digits and it's enumeration in base 10 starts as follow:

Pi=3.14159265…

What if we create the new number n from digits of Pi in reversed order? What if we will put every digit of Pi from float into natural part and from natural part into float. Reverse transformation works like mirror, the closest digits from float point will be mapped at the same distance into natural part and vice versa.

The new number n will looks like this:

...56295141.3 = n

The n will have infinite number of digits. We can enumerate any digit of this new number, but it has no first digit, similar as Pi which has no last digit.

Why it is allowed creation of numbers of infinite depth when we are going to smaller regions as Pi, but not allowed to create numbers of infinite size as n, where we are going to larger regions?

Number n can be written as infinite sum. How can we go beyond the real numbers, if we know that basic property of Real numbers is that if a,b are Real numbers then a+b is also Real number? We can write n as 0.3+1+4*10+1*100+5*1000... At each point we are adding real number to real number and there is no point where we can go outside of real numbers. We can even divide this sum into smaller sub sums and each of that smaller part also should be within real numbers.

If n is real number then n without floating point part can be natural number. If we could create natural number of infinite size, then we can show that Cantor's diagonal proof will fail, because in such case we will be able to create natural number of infinite size. If natural numbers can have infinite size then, we can create with the same diagonal manipulation on infinite size natural numbers as Cantor did with infinite depth Real numbers.

Is the n within Real number set?

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closed as off-topic by Adam Hughes, Ferra, Claude Leibovici, user26857, Watson Oct 28 '16 at 16:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Adam Hughes, Ferra, Claude Leibovici, user26857, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ The limit of the sequence $0.3, 1.3, 41.3, 141.3, 5141.3, 95141.3,\dots$ is infinite and is thus not a real number. It is larger than any real number. $\endgroup$ – JMoravitz Oct 27 '16 at 21:47
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    $\begingroup$ @Kaynax: $n$ is a perfectly good $10$-adic number. $\endgroup$ – Rob Arthan Oct 27 '16 at 21:49
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    $\begingroup$ @MrYouMath What does that string even mean, though? Both $\pi$ and "reversed $\pi$" make sense in the appropriate metrics on $\mathbb{Q}$ - I don't know what the thing you've described is supposed to be. $\endgroup$ – Noah Schweber Oct 27 '16 at 22:24
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    $\begingroup$ @NoahSchweber: I don't know it either, I never studied math :). I just thought it would be interessting to start by $0.1$ then $0.41$ then $0.141$ and so forth ... and see what happens. $\endgroup$ – MrYouMath Oct 27 '16 at 22:26
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    $\begingroup$ There is no real number which when added to another real number results in a non-real number. You are correct in saying when you add two real numbers you get another real number. You are correct in saying that every number in the sequence I describe is real and that you can get the next number in the sequence by adding another real number. Again, this does not imply that the limit of the sequence is real. You cannot point to a specific place and say "here is where it breaks" because any finite point in the sequence you point to it is still real. $\endgroup$ – JMoravitz Oct 27 '16 at 23:20
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What you have doesn't make sense as a real number. However, within the $10$-adic numbers this does make sense. There aren't really any fancy properties one can divine about this number, but it certainly exists.

For any natural $n\geq2$, the $n$-adic numbers is like usual real numbers base $n$, except we allow infinitely many digits to the left and only finitely many digits to the right instead. Numbers are added and multiplied as "usual".

I should also add that the $10$-adic numbers aren't too common in use, for the simple reason that $10$ is not a prime. That fact gives the $10$-adic numbers some properties we like to avoid (specifically, division doesn't work).

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  • $\begingroup$ If we will use instead of base 10, some base with prime number, will we be able to divide such prime-adic number? $\endgroup$ – Ivan Sas Oct 27 '16 at 22:20
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    $\begingroup$ @IvanSas Yes, the $p$-adics for $p$ prime are much better. So working in base $p$ will be better for this. (I worry that $p=2$ is suboptimal here, as it so often is, but not greatly so - certainly all the basic algebraic properties still hold.) $\endgroup$ – Noah Schweber Oct 27 '16 at 22:21
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    $\begingroup$ The p-adics are not real numbers. This is made clear in at least 3 separate instances in the Wikipedia article. $\endgroup$ – Xodarap777 Oct 27 '16 at 23:42
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    $\begingroup$ @Xodarap It is also strongly hinted at in the first sentence of my answer. $\endgroup$ – Arthur Oct 28 '16 at 6:32
  • $\begingroup$ @Arthur I would word it differently than "isn't a usual real number"; $\pi$, $e$, and $1729$ are also very unusual real numbers! I'm not sure if it is obvious for everyone that the first sentence means that it is not a real number at all. $\endgroup$ – JiK Oct 28 '16 at 9:51
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Any real number has, by the Archimedian Property only finitely many digits before the dot. So no, this cannot be a real number.

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  • $\begingroup$ Is this really a result of the Archimedian property? What algebraic structure would allow such a "number"? $\endgroup$ – GFauxPas Oct 27 '16 at 21:50
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    $\begingroup$ @GFauxPas Are you familiar with the $10$-adics? $\endgroup$ – Dustan Levenstein Oct 27 '16 at 22:01
  • $\begingroup$ @GFauxPas A consequence of the Archimedian property is that for each positive real number $x$ you can find an integer $n$ such that $10^n \leq x < 10^{n+1}$. This means that $x$ has exactly $n$ digits..... $\endgroup$ – N. S. Oct 28 '16 at 0:11
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    $\begingroup$ @MichaelS That property does NOT imply finitely many digits after the dot.... $\endgroup$ – N. S. Oct 28 '16 at 0:36
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    $\begingroup$ @MichaelS The Archimedian property is a property of $\mathbb R$, you cannot start applying it to random sets.Moreover, the Archimedian property is about existence for ONE real number of a LARGER natural number, has nothing to do with infinitely many smaller numbers .... Finally, there is no such thing as infinitely many digits smaller than one digit, the only possible digits are $\{ 0,1,2,3,4,5,6,7,8,9\}$, and no matter what the first digit is (in this case 1) there can only be 9 digits smaller than it. $\endgroup$ – N. S. Oct 28 '16 at 0:44
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Let $(a_n)_{n \ge 0}$ be the sequence of digits of $\pi$: $a_0 = 3$, $a_1 = 1$, $a_2 = 4$, $a_3 = 1$, etc such that $$\pi = \sum_{k = 0}^\infty a_k 10^{-k}.$$

Now define the sequence $$r_n = \frac{1}{10} \sum_{k=0}^n a_k 10^k,$$ of which the first elements are $r_0 = 0.3$, $r_1 = 1.3$, $r_2 = 41.3$, etc.

Your question is whether $$\lim_{n \to \infty} r_n$$ exists.

The answer is no, as this is clearly a divergent sequence. For the proof, suppose you claim that $L$ is the limit, and it is a real number with $n$ digits before the decimal point (that is, let $n = \lceil \log_{10} L \rceil$). Then I claim that $r_{n+2} > L$.

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  • $\begingroup$ Pi is irrational and has no finite n number of digits, so the reversed one will also have not the finite number of digits. I don't know why it's allowed to have infinite sequence in floating point part, but finite one in natural part. We can do one to one mapping from Pi digits to the natural numbers, so the Pi itself is not bigger than natural numbers. It is like that, we can do infinitely many times zoom in closer and closer, but we can't do the opposite infinite process which for zoom out to the larger and larger. $\endgroup$ – Ivan Sas Oct 28 '16 at 16:27

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