If I exchange infinitely many digits of $\pi$ and $e$, are the two resulting numbers transcendental?

Question

If I swap the digits of $\pi$ and $e$ in infinitely many places, I get two new numbers. Are these two numbers transcendental?

Answer 1

Nice question, Erin. Here is one quick easy thing to say.

If $\pi$ and $e$ disagree in infinitely many digits, then there are continuum many choices of the particular subset of those digits to swap, and so we get continuum many different numbers this way. Since there are only countably many algebraic numbers, it would follow that most of the time, yes, you do get transcendental numbers by doing this.

I'm unsure, however, whether one can say that all the resulting reals are transcendental. Perhaps we'll have to wait for some number theory experts to answer.

Lastly, if it happens (as seems unlikely) that all but finitely many digits of $\pi$ and $e$ are the same, then $\pi-e$ would be rational, and furthermore swapping the digits doesn't actually do anything except on those finitely many digits of difference, and so this won't affect transcendentality. In this case, there are only finitely many possible reals resulting, but they are all differing from the original reals by only finitely many digits, and so yes, they are all transcendental.

Answer 2

If, as is commonly believed, $\pi$ and $e$ are normal numbers, then one can use a counting argument (or entropy argument) to show that no possible transposition of $\pi$ and $e$ can produce a rational number. Indeed, if there was a rational number that could be made this way, then its digit expansion would eventually be periodic with some period $q$; by repeating this period, one can make $q$ large and even. If one looks at a given $q$-digit block of this periodic expansion, then either $\pi$ would have to share at least $q/2$ of its digits with this fixed block, or $e$ would. But if $\pi$ is normal, the former happens with density at most $\binom{q}{q/2} 10^{-q/2}$ among all the $q$-blocks, and if $e$ is normal, the latter happens with density at most $\binom{q}{q/2} 10^{-q/2}$. For $q$ large enough, the two densities sum to less than 1 (here we use the fact that the base is at least $4$ - not sure which way things will go in base $2$ or base $3$), and so one cannot actually match the given rational number.

[There ought to be some slick information theoretic way to reformulate the above argument, perhaps using the Shannon entropy inequalities, but I was not able to locate one.]

Settling the problem unconditionally looks to be at least as hard as making some major breakthrough on the normality of $\pi$ and $e$. Even ruling out a terminating decimal (i.e. that for all sufficiently large $k$, either the $k^{th}$ digit of $\pi$ or the $k^{th}$ digit of $e$ vanishes) is probably out of reach of current technology.

Answer 3

A variant of my previous answer. It is commonly believed that all irrational algebraic numbers are normal. If this is the case, then there can be at most two algebraic numbers (up to shifts by rationals) that can be obtained by transposing digits of $e$ and $\pi$.

To see this, suppose for sake of contradiction that there are three algebraic numbers $\alpha,\beta,\gamma$, no two of which differ by a rational, that can all be attained by transposing digits of $e$ and $\pi$. By the pigeonhole principle, we see that for each natural number $k$, at least one of the pairs $(\alpha,\beta)$, $(\alpha,\gamma)$, $(\beta,\gamma)$ agree at the $k^{th}$ digit. By the pigeonhole principle again, this means that one of these pairs agrees on a set of digits of (upper) density at least $1/3$. Without loss of generality we can assume that the pair $(\alpha,\beta)$ has this property, and that $\beta > \alpha$. But then, by long subtraction, the difference $\beta - \alpha$ will have digits $0$ or $9$ on a set of digits of upper density at least $1/3$, which contradicts the normality of $\beta-\alpha$. (Now I need the base to be at least seven!)

It might be possible to upgrade "up to shifts by rationals" in the above claim by "up to shifts by terminating decimals", but I have not strenuously attempted to do this. It's also worth noting that this is an example of an ineffective argument, in that no bound whatsoever is provided on the height of the algebraic numbers that might still be obtainable by transposing digits of $e$ and $\pi$, even if one had some quantitative normality bound on algebraic numbers depending on the height.

p.s. We can combine the two answers: if we assume that the sum of $\pi$ and an algebraic number, or the sum of $e$ and an algebraic number, is always normal, then the answer to the original question is positive: every transposition of $\pi$ and $e$ is transcendental. For if there was an algebraic number $\alpha$ that was achievable as a transposition, then it would have to share at least half its digits with $\pi$ or $e$, and hence one of $|\pi-\alpha|$ or $|e-\alpha|$ would have digits $0$ or $9$ on a set of upper density at least $1/2$, contradicting the normality of these numbers. (Now I need base at least five.)

Answer 4

Another easy thing to say is that there are at least 9 transcendental numbers in (0,1) such that any rearrangement of their digits place wise gives 9 other transcendental numbers.

Proof Start: list the countable many algebraic numbers in decimal form, including those with two representations (having terminating nines). Take the nth decimal digit after the point from the nth number, and don't use it in that place in any of the nine numbers you will form. Rest is left to the enthused reader.

Gerhard "It's Like Telling Old Jokes..." Paseman, 2017.03.23.

Answer 5

Note that the sum of the swapped numbers is the same as $\pi + e$ itself, meaning for them to be simultaneously algebraic would imply that $\pi + e$ is algebraic (an open problem as far as I know). I don't know if you say anything about the case where just one is algebraic.

Answer 6

Here's an argument that one should expect that the two numbers gotten this way must be transcendental. Really, what I am showing is that the locus of $(a,b)$ in $\mathbb{R}\times\mathbb{R}$ where one can get an algebraic number by swapping digits of $a$ out for digits of $b$ is of measure $0$.

I'm going to ignore difficulties coming from infinite ending sequences of $9$s (they form a measure $0$ set, so they don't change anything).

To do this, it suffices to show that this locus is a countable union of measure $0$ sets. As there are a countable number of algebraic numbers, we can restrict attention to a single one $\alpha$. Also, we can restrict to looking at $[0,1]\times[0,1]$ because an algebraic number stays algebraic when we add an integer.

Now, we have $2^n$ choices of how to split the first $n$ digits of $\alpha$, and each splitting gives us a set of measure $\frac{1}{10^n}.$ As $\displaystyle\lim_{n\rightarrow\infty}\frac{2^n}{10^n}=0,$ we see that our set has measure $0$, as desired.

Answer 7

I'd like to reiterate, though with a renewing idea that might be THE answer, on Terry's answer.

Pi and e are normal numbers, let's agree upon that. Now, pi = 3.141592653...7....7...7.....7..... e = 2.71828182845904...

Let's introduce a mapping function alpha(e) that maps every 7 in pi to each digit, after the decimal point, in e. Clearly, pi is normal, and hence, there would be enough 7s in the "reserve" to map to every number (after decimal point) in e. Now, let me swap all such digits alpha(e) returns to me.

This gives two new numbers, pi* and e*, of which e* is rational (2.77777....7...).

So, YES. But you can rather introduce a swap function like alpha (x), rather than choosing to swap infinitely.

Thank you.