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What is the minimum value of $\frac{2^a}{3^b}$ if it's at least 1 (that is if $2^a>3^b$) and $a$ and $b$ are positive integers?

Also, the opposite of the problem too: minimum of $\frac{3^a}{2^b}$ if $3^a>2^b$.

toorelevant
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    It can be as close as 1 as you want. The fraction part of $n\log_3 2$ can be arbitrarily close to 0. – Hw Chu Feb 15 '18 at 21:15
  • The concept of infimum and supremum may be helpful if I understand what your underlying confusing is correctly. https://en.wikipedia.org/wiki/Infimum_and_supremum – siegehalver Feb 15 '18 at 21:22
  • Oops, forgot to mention $a$ and $b$ are positive integers. – toorelevant Feb 15 '18 at 21:34
  • Something linked to this (see comments): https://math.stackexchange.com/questions/2539458/n-0-to-n-i-and-n-0k-cdot2j-ton-ik-cdot3i-same-glide .Didn't had time yet to look at it :( – Collag3n Feb 16 '18 at 18:15

1 Answers1

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It can never be equal to $1$ but it can be as arbitrarily close to $1$ as you like.

$\log_3 2$ is an irrational number that is larger than $0$.

And can find a natural number $m$ and $n$ so that $\frac mn < \log_3 2 < \frac {m+1}n$

We can fine-tune this as close as we want by taking larger and larger $n$ and the larger then $n$ get the larger the $m$ gets and the closer and finer tuned we get.

So $m < n\log_3 2 < m+ 1$

$3^m < 3^{n\log_3 2} = (3^{\log_3 2})^n = 2^n < 3^m + 1$

$1 < \frac {2^n}{3^m} < 1 + \frac 1{3^m}$

Now as we said before $n$ and $m$ can be as large as we want, so $\frac 1{3^m}$ can be as close to $0$ (but never equal) as we want.

So $\frac {2^n}{3^m}$ can be as close to $1$ (but never equal) as we want.

fleablood
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