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Use proof by contradiction to prove that log2(3) is irrational. Watch

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    ??? Saw this in a C3 Solomon paper. How do I do it?
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    What is log2(3)?
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    \log_2 3 = \frac{p}{q} \Leftrightarrow 3^q = 2^p...
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    (Original post by Drederick Tatum)
    What is log2(3)?
    Sorry, the 2 is the base of the log
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    (Original post by SimonM)
    \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.
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    Expand it out in terms of p/q
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    (Original post by 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
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    Oh OK, I guess I have some more revision to do on logs lol
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    You may as well assume p and q are positive integers (in what simon M has said) since the log(3) > log(2) = 1.
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    An answer, in two stages: stop after first if you think you can see rest:
    Stage 1:
    Spoiler:
    Show

    So as SimonM did above, we do this:
    (Original post by SimonM)
    \log_2 3 = \frac{p}{q} \Leftrightarrow 3^q = 2^p...
    What we are doing here is assuming that \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, 3^q = 2^p (by rearrangement really).

    Stage 2:
    Spoiler:
    Show

    Note that 3 and 2 are mutually prime: i.e. they share no factors (and are in fact both actual prime numbers). But these numbers, 3^q and  2^p, are meant to be equal and must therefore have the same, unique, prime number expansion. However if they don't share any factors... So p and q do not exist: there are no integers p and q for which the equality is true.

    Hence our original assumption led to a contradiction: so \log_2 3 is not rational (so it is irrational).

    Bingo!
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    (Original post by Salavant)
    An answer, in two stages: stop after first if you think you can see rest:
    Stage 1:
    Spoiler:
    Show

    So as SimonM did above, we do this:

    What we are doing here is assuming that \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, 3^q = 2^p (by rearrangement really).

    Stage 2:
    Spoiler:
    Show

    Note that 3 and 2 are mutually prime: i.e. they share no factors (and are in fact both actual prime numbers). But these numbers, 3^q and  2^p, are meant to be equal and must therefore have the same, unique, prime number expansion. However if they don't share any factors... So p and q do not exist: there are no integers p and q for which the equality is true.

    Hence our original assumption led to a contradiction: so \log_2 3 is not rational (so it is irrational).

    Bingo!
    Thanks for that I'm still very baffled on why it came up on the C3 paper though, we've never been through this
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    (Original post by 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.
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    (Original post by usainlightning)
    The solomon papers where madeon the old syllabus, proof by contradiction isn't on it anymore.
    Oh right that explains it
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    i stil do not understand
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    (Original post by 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 \dfrac{p}{q}, where p,q are integers (q \ne 0).

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

    By the definition of logarithms, this gives 3 = 2^{p/q}, and raising both sides to the power of q gives 3^q=2^p. This can only happen if p=q=0, but we can't have q=0 so this can't be true, so the assumption can't be true, so it must be false; hence the proposition is true.
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    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?
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    (Original post by 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/Fundame..._of_arithmetic
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    (Original post by 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/Fundame..._of_arithmetic
    aha! Eureka! Thank you :yes:
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    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).
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    (Original post by 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.
 
 
 
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