Why is the decimal expansion of log2(3) infinite?

log2_of_3_never_ends.json

https://arbital.com/p/log2_of_3_never_ends

by Nate Soares Jun 20 2016 updated Jul 4 2016

Because 2 and 3 are relatively prime.

[summary: It takes more than one but less than two [binary_digit binary digits] to encode a 3-digit, so $~$\log_2(3)$~$ must be between 1 and 2. (Wait, what?). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so $~$10 \cdot \log_2(3)$~$ must be between 15 and 16, which means $~$1.5 < \log_2(3) < 1.6.$~$ It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so $~$1.58 < \log_2(3) < 1.59.$~$ And so on. Because no power of 3 is ever equal to any power of 2, $~$10^n \cdot \log_2(3)$~$ will never quite be a whole number, no matter how large $~$n$~$ is.]

$~$\log_2(3)$~$ starts with

1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360

and goes on indefinitely. Why is it 1.58… in particular? Well, it takes more than one but less than two [binary_digit binary digits] to encode a 3-digit, so $~$\log_2(3)$~$ must be between 1 and 2. (Wait, what?). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so $~$10 \cdot \log_2(3)$~$ must be between 15 and 16, which means $~$1.5 < \log_2(3) < 1.6.$~$ It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so $~$1.58 < \log_2(3) < 1.59.$~$ And so on. Because no power of 3 is ever equal to any power of 2, $~$10^n \cdot \log_2(3)$~$ will never quite be a whole number, no matter how large $~$n$~$ is.

Thus, $~$\log_2(3)$~$ has no finite decimal expansion, because $~$3$~$ is not a rational [-power] of $~$2$~$. Using this argument, we can see that $~$\log_b(x)$~$ is an integer if (and only if) $~$x$~$ is a power of $~$b$~$, and that $~$\log_b(x)$~$ only has a finite expansion if some power of $~$x$~$ is a power of $~$b.$~$

Comments

Joe Zeng

$~$8$~$ is not a power of $~$4$~$, but $~$\log_4 8$~$ is $~$1.5$~$. The only thing you prove with $~$3$~$ is not a power of $~$2$~$ is that $~$log_2 3$~$ is not an integer.

Nate Soares

Fixed, thanks.