# Is this number Transcendental? [closed]

I've been messing around with microtonal scales and I came up with this number and I was wondering if it is Transcendental.

U = 17.312340490667609 to 15 decimal places

U = 17.31234049066756088831909617202270564911975144983583120962539288317331063017422095862631947472207613396397021971583053106711722024602031571264794589868175203529444843861662192356841665196

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## closed as unclear what you're asking by 6005, Daniel W. Farlow, Shailesh, Leucippus, G. SassatelliJul 22 at 1:24

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No number for which you can give a finite decimal expansion is even irrational, let alone transcendental. Is $U$ some sort of approximation of a number that you can express in some other way? – Aaron Jun 3 '11 at 14:22
Since it is impossible to list infinitely many digits, you need to specify a mathematical expression for $U$. Was there a formula that you were using that you got $U$ out of? What was it? – Zev Chonoles Jun 3 '11 at 14:29
@Ginger Bill: I don't understand what "trial and error" means here. We need to know a mathematical expression, i.e. formula, for $U$, in order to answer this question. Presumably you used a computer when producing this number $U$; can you describe the process that you used to produce it? Perhaps started with an integer, then added smaller and smaller fractions? Also, as I said before, "only specifying a finite number of decimal places doesn't suffice to determine whether or not $U$ is transcendental", so while I appreciate your providing more digits of $U$, I'm afraid it doesn't help. – Zev Chonoles Jun 3 '11 at 14:43
@Ginger Bill: At this moment the "short" and "long" versions differ at the $13$th place after the decimal point, where presumably the $5$ is the long version is a typo. You must have at least an excellent algorithm for computing $U$. Is it possible to have a clear description of this? – André Nicolas Jun 3 '11 at 14:45
@Ginger Bill: The statement that a real number is transcendental requires a usually-complicated proof. Mathematicians took a long time to prove that $\pi$ is transcendental, and the proof is not so obvious! (See here). Can you describe why you think $U$ is transcendental? – Zev Chonoles Jun 3 '11 at 14:56

It looks as if the number is $12/(\ln 2)$. So it is transcendental.

It was the connection to music, and hence to $2^{1/12}$, that did it.

The transcendence of $\ln 2$ follows from the Lindemann-Weierstrass Theorem.

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$12/\ln(2) = 17.312340490667562...$ which disagrees with the number above. It is possible that the OP miscalculated the constant, but I'm not sure it is what you have written (although I would like it to be!). – JavaMan Jun 3 '11 at 15:06
@DJC: The much longer version added in the post actually agrees with what you wrote, as I pointed out the short and long versions disagree in the $13$th place after the decimal. – André Nicolas Jun 3 '11 at 15:13
@AndréNicolas How did you discover that 12/ln2 came close to the OP's number? – zerosofthezeta Aug 8 '13 at 1:17
Connection to music, $12$-tone scale made me suspect. The agreement to many places clinches it, unless something very weird is happening. – André Nicolas Aug 8 '13 at 1:23

No. Any transcendental number must have an infinite decimal expansion; your number has a finite decimal expansion, so it is in fact a rational number, specifically $$\small U=\frac{1731234049066756088831909617202270564911975144983583120962539288317331063017422095862631947472207 \atop 613396397021971583053106711722024602031571264794589868175203529444843861662192356841665196}{10^{185}}.$$

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This is only to 15 decimal places. I'll add a note. – Ginger Bill Jun 3 '11 at 14:24
+1: This answer is really funny. – Eric Naslund Jun 3 '11 at 16:09
Very nice. And I've never come across \atop before. Cool. – mixedmath Jun 3 '11 at 21:17
@Zev: I was just thinking about this answer and wishing I would could vote it up again! – JavaMan Jul 20 '11 at 3:29