prove irrationality of logs (1 Viewer)

mr.habibbi

Member
hey all,

need a bit of help with this question

jimmysmith560

Le Phénix Trilingue
Moderator
Hey, here is an example of the same question, the difference being the need to prove $\bg_white \log _{10}\left(5\right)$ instead of $\bg_white \log _5\left(13\right)$. Working out may vary, but I thought I'd include this so you might gain an idea of how to approach your question.

I hope this helps!

mr.habibbi

Member
Hey, here is an example of the same question, the difference being the need to prove $\bg_white \log _{10}\left(5\right)$ instead of $\bg_white \log _5\left(13\right)$. Working out may vary, but I thought I'd include this so you might gain an idea of how to approach your question.

View attachment 32250
View attachment 32251

I hope this helps!
thanks for the suggestion but your question shows an odd and even value 10 and 5, the question i uploaded shows 5 and 13 where they are both odd.

Active Member
thanks for the suggestion but your question shows an odd and even value 10 and 5, the question i uploaded shows 5 and 13 where they are both odd.
Hmmm a slight modification to the blue underlined statement to account for all log irrationalities would be the following:
As 5 and 13 have no common factors (as they are both prime number), there exists no real number where a = b.

This is applicable to the 5 10 case if we do the following (although the above solution is much more reliable):
5^b = 10^a
As 10 = 2*5

5^b = 2^a * 5^a

5^(b-a) = 2^a

As b and a are integers, b-a must also be an arbitrary integer q

Hence, 5^q = 2^a

And then then you can use the statement that I have mentioned. Hope this helps and point any mistakes too.

CM_Tutor

Moderator
Moderator
Expanding slightly, you can go from $\bg_white a^x = b^y$ where $\bg_white a$ and $\bg_white b$ are integers and where seeking integers $\bg_white x$ and $\bg_white y$ straight to there being a contradiction if:
1. one of $\bg_white a$ and $\bg_white b$ is odd and the other is even, as does the proof above
2. $\bg_white a$ and $\bg_white b$ are both prime
3. $\bg_white a$ and $\bg_white b$ are co-prime, which means that they have no common factor except 1 - which covers many of the cases under 1 and 2 as well
So, if I was asked for $\bg_white \log_9{17}$ is irrational, I could get the contradiction from $\bg_white 9^y = 17^x$ as 9 and 17 are co-prime. Alternatively, I could re-write 9 as 32 to get $\bg_white 3^{2y} = 17^x$. However, getting the result from co-primes is necessary in some cases, such as proving that $\bg_white \log_{33}{35}$ is irrational.

mr.habibbi

Member
Expanding slightly, you can go from $\bg_white a^x = b^y$ where $\bg_white a$ and $\bg_white b$ are integers and where seeking integers $\bg_white x$ and $\bg_white y$ straight to there being a contradiction if:
1. one of $\bg_white a$ and $\bg_white b$ is odd and the other is even, as does the proof above
2. $\bg_white a$ and $\bg_white b$ are both prime
3. $\bg_white a$ and $\bg_white b$ are co-prime, which means that they have no common factor except 1 - which covers many of the cases under 1 and 2 as well
So, if I was asked for $\bg_white \log_9{17}$ is irrational, I could get the contradiction from $\bg_white 9^y = 17^x$ as 9 and 17 are co-prime. Alternatively, I could re-write 9 as 32 to get $\bg_white 3^{2y} = 17^x$. However, getting the result from co-primes is necessary in some cases, such as proving that $\bg_white \log_{33}{35}$ is irrational.
would proving ln2 irrational be the same?

Active Member
would proving ln2 irrational be the same?
I actually don't think you can prove ln(2) being irrational with the exact layout used in ext2 because ln is basically log(e) where e is a transcendental number (i.e. not an integer) so dont worry too much about.

CM_Tutor

Moderator
Moderator
It might be possible to prove the irrationality of $\bg_white \ln{2}$ based on $\bg_white e$ being transcendental, though. That is, assume $\bg_white \ln{2}$ is rational and demonstrate that it leads to the conclusion that $\bg_white e$ can't be transcendental, thereby generating a contradiction.

A transcendental number, for any who don't know, is a number such that there is no polynomial of any finite degree with rational coefficients where the number is a root. $\bg_white e$ and $\bg_white \pi$ are the two most famous transcendental numbers.