prove irrationality of logs (1 Viewer)

jimmysmith560

Le Phénix Trilingue
Moderator
Joined
Aug 22, 2019
Messages
4,550
Location
Krak des Chevaliers
Gender
Male
HSC
2019
Uni Grad
2022
Hey, here is an example of the same question, the difference being the need to prove instead of . Working out may vary, but I thought I'd include this so you might gain an idea of how to approach your question.

1632329329601.png
1632329353671.png

I hope this helps! :D
 

mr.habibbi

Member
Joined
Oct 27, 2019
Messages
41
Gender
Male
HSC
2021
Hey, here is an example of the same question, the difference being the need to prove instead of . 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! :D
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.
 

username_2

Active Member
Joined
Aug 1, 2020
Messages
116
Gender
Male
HSC
2020
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
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
Expanding slightly, you can go from where and are integers and where seeking integers and straight to there being a contradiction if:
  1. one of and is odd and the other is even, as does the proof above
  2. and are both prime
  3. and 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 is irrational, I could get the contradiction from as 9 and 17 are co-prime. Alternatively, I could re-write 9 as 32 to get . However, getting the result from co-primes is necessary in some cases, such as proving that is irrational.
 

mr.habibbi

Member
Joined
Oct 27, 2019
Messages
41
Gender
Male
HSC
2021
Expanding slightly, you can go from where and are integers and where seeking integers and straight to there being a contradiction if:
  1. one of and is odd and the other is even, as does the proof above
  2. and are both prime
  3. and 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 is irrational, I could get the contradiction from as 9 and 17 are co-prime. Alternatively, I could re-write 9 as 32 to get . However, getting the result from co-primes is necessary in some cases, such as proving that is irrational.
would proving ln2 irrational be the same?
 

username_2

Active Member
Joined
Aug 1, 2020
Messages
116
Gender
Male
HSC
2020
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
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
It might be possible to prove the irrationality of based on being transcendental, though. That is, assume is rational and demonstrate that it leads to the conclusion that 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. and are the two most famous transcendental numbers.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top