Is the Pi in reversed order Real number?

Question

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?

Answer 1

Any real number has, by the Archimedian Property only finitely many digits before the dot. So no, this cannot be a real number.

Answer 2

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).

Answer 3

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 $r_{n+2} > L$.