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Given $\{\log_ab \mid a,b\in \mathbb N, \mathrm{gcd}(a,b)=1,a,b≥3\}$ does the sum of any two $\log_ab$ form an irrational or rational number?

I know that $\log_ab$ is irrational, but does the sum of any two $\log_ab$ form an irrational or rational number?

e.g. Is $\{\log_3(5)+log_7(11) \in \mathbb Q\}$ or is $\{\log_3(5)+\log_7(11) \in \mathbb R \setminus \mathbb Q\}$.

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    $\begingroup$ Thank you for adding the question in words. But do you rather want to ask if the set $\{\,\log_a b\mid a,b\in\mathbb N, \gcd(a,b)=1, a,b\ge 3\,\}$ is a subgroup of $\mathbb R$ under addition? $\endgroup$ Commented Jul 23, 2013 at 14:42
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    $\begingroup$ Try verifying the conditions for the group. Is it closed? Is it associative? Does it have an identity? What about inverses? $\endgroup$ Commented Jul 23, 2013 at 14:43
  • $\begingroup$ @Hagen - Yes I did not state the problem correctly. You are correct, what I was looking for was not is log_ab is group, but is log_ab closed under addition. $\endgroup$
    – KeithSmith
    Commented Jul 23, 2013 at 15:33
  • $\begingroup$ It's still not clear to me what's being asked. Are you wondering whether sums such as $\log_3(5) + \log_7(11)$ always come out irrational? Or whether they can always be put back into the form $\log_a(b)$ with $a,b$ integers satisfying $\mathrm{gcd}(a,b) = 1$ and $a,b \geq 3$? $\endgroup$
    – Mike F
    Commented Jul 23, 2013 at 16:35
  • $\begingroup$ I am asking if $log_3(5)+log_7(1)) always comes out irrational. $\endgroup$
    – KeithSmith
    Commented Jul 23, 2013 at 16:58

2 Answers 2

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Hint What is the identity of this operation? Is it an element in your set?

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  • $\begingroup$ Unfortunately, OP's edit to the question has made this a non-answer. $\endgroup$ Commented Jul 24, 2013 at 9:22
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No, for instance,

$log_5(7) - 2log_{25}(7) = \frac{log(7)}{log(5)} - 2\frac{log(7)}{log(25)} = \frac{log(7)}{log(5)} - \frac{2log(7)}{2log(5)} = \frac{log(7)}{log(5)}-\frac{log(7)}{log(5)} = 0$

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  • $\begingroup$ I see that a,b=5,7 for the first term. What are a and b for the second term? $\endgroup$
    – KeithSmith
    Commented Jul 23, 2013 at 20:25
  • $\begingroup$ sorry a=25, b=7 fixed for clarity. $\endgroup$
    – Rob Smith
    Commented Jul 24, 2013 at 9:18

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