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What is log2(3)?
Reply 2
log23=pq3q=2p\log_2 3 = \frac{p}{q} \Leftrightarrow 3^q = 2^p...
Reply 3
Drederick Tatum
What is log2(3)?


Sorry, the 2 is the base of the log
Reply 4
SimonM
log23=pq3q=2p\log_2 3 = \frac{p}{q} \Leftrightarrow 3^q = 2^p...


Whaaat? Are we supposed to have covered this in C3? I've never come across this type of question before.
Expand it out in terms of p/q
Reply 6
pimpman777
Whaaat? Are we supposed to have covered this in C3? I've never come across this type of question before.


I'm surprised it's on C3, but I don't think anything I've written is off syllabus
Reply 7
Oh OK, I guess I have some more revision to do on logs lol
You may as well assume p and q are positive integers (in what simon M has said) since the log(3) > log(2) = 1.
Reply 9
An answer, in two stages: stop after first if you think you can see rest:
Stage 1:

So as SimonM did above, we do this:
SimonM
log23=pq3q=2p\log_2 3 = \frac{p}{q} \Leftrightarrow 3^q = 2^p...

What we are doing here is assuming that log23\log_2 3 is a rational number: i.e. is the rato of two integers p and q. This implies that for some integer p, q, 3q=2p3^q = 2^p (by rearrangement really).

Stage 2:

Spoiler


Bingo!
Reply 10
Salavant
An answer, in two stages: stop after first if you think you can see rest:
Stage 1:

So as SimonM did above, we do this:

What we are doing here is assuming that log23\log_2 3 is a rational number: i.e. is the rato of two integers p and q. This implies that for some integer p, q, 3q=2p3^q = 2^p (by rearrangement really).

Stage 2:

Spoiler


Bingo!

Thanks for that :smile: I'm still very baffled on why it came up on the C3 paper though, we've never been through this
pimpman777
Oh OK, I guess I have some more revision to do on logs lol

The solomon papers where madeon the old syllabus, proof by contradiction isn't on it anymore.
Reply 12
usainlightning
The solomon papers where madeon the old syllabus, proof by contradiction isn't on it anymore.


Oh right that explains it
Reply 13
i stil do not understand
Reply 14
misstee91
i stil do not understand

Proof by contradiction is a method of proof whereby you assume the conclusion is false, and then show this assumption leads to something which can't be true (e.g. 1=0 or "2 is odd").

A number is rational if it is in the form pq\dfrac{p}{q}, where p,qp,q are integers (q0q \ne 0).

Piecing this together, we want to show that log23\log_2 3 is irrational; i.e. that it can't be written in the form pq\dfrac{p}{q} for any integers p,qp,q. So, we start our proof by assuming that there exist integers p,q (q nonzero) such that log23=pq\log_2 3 = \dfrac{p}{q}.

By the definition of logarithms, this gives 3=2p/q3 = 2^{p/q}, and raising both sides to the power of q gives 3q=2p3^q=2^p. This can only happen if p=q=0p=q=0, but we can't have q=0q=0 so this can't be true, so the assumption can't be true, so it must be false; hence the proposition is true.
(edited 13 years ago)
nuodai, would you be able to just explain the last stage of this? I don’t quite see why there are no integers p and q for which 2^p=3^q
10^4=100^2 but there the bases have common factors. Is that the point?
Plato's Trousers
but surely that only proves there is only one solution of 3^x=2^x (ie x=0). But it doesn't imply that there cannot be two different integers (p and q) for which 2^p=3^q does it?

Sorry, if I'm being thick


2^p=3^q=n

If p and q are not zero then n has two different prime factorisations. This is not possible. You might want to check the proof of that statement.

It's probably on this page:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
(edited 13 years ago)
Get me off the £\?%!^@ computer
2^p=3^q=n

If p and q are not zero then n has two different prime factorisations. This is not possible. You might want to check the proof of that statement.

It's probably on this page:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic


aha! Eureka! Thank you :yes:
(edited 13 years ago)
Rather than resort to FTA, in this case it is surely simpler to note that 2^p is even, while 3^q is odd.

(Which some will say is just a special case of the FTA, but I think what I've said is much more accessible at C1-C4 level).
(edited 13 years ago)
DFranklin
Rather than resort to FTA, in this case it is surely simpler to note that 2^p is even, while 3^q is odd.


Oh yes! Nice one.
(edited 13 years ago)

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