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    (Original post by joostan)
    Not quite, because a and b are constants not variables, so saying a=0 is not permissible.
    If you consider what type of number 3^a is going to be, and what type 2^b is going to be what can you deduce?
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    3 is a prime number

    whilst 2 is also a prime number

    -> how can I utilise this

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    (Original post by Robbie242)
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    3 is a prime number

    whilst 2 is also a prime number

    -> how can I utilize this

    If I were to say 2 is even, what would you say about 2^a?
    And 3? And 3^b?
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    (Original post by joostan)
    Dunno what he's after, but yeah, Felix says he's getting somewhere.
    Well, alternatively:

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     (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1

     (a+ib)(a-ib)(c+id)(c-id) = (ac+bd+i)(ac+bd-i)

     (a+ib)(c-id) = (ac+bd)+(bc-ad)i

     (a-ib)(c+id) = (ac+bd)+(ad-bc)i

    This then gives us the same thing as the RHS, provided that  ad-bc = \pm 1 , which is exactly the result we derived using the other method.



    Also:

    (Original post by joostan)
    If I were to say 2 is even, what would you say about 2^a?
    And 3? And 3^b?
    I believe that Robbie's idea of prime numbers is actually a better way of tackling the question, because it can be extended to cases where both numbers are odd or both even. for example:

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    Prove that  log_5(7) is irrational.

    Assume the opposite i.e.  log_5(7) = \frac{a}{b}

    And we then get to:

     7^b = 5^a

    Now, your odd and even ideas won't work here. However, you can say that, no matter how many times you multiply 5 by itself, it will never gain a factor of 7 (and vice-versa) so as to derive the contradiction you're after.

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    (Original post by joostan)
    If I were to say 2 is even, what would you say about 2^a?
    And 3? And 3^b
    2^a is also even
    3^b is not even ?

    I'm trying to raise 3 to powers and it keeps coming out odd, is this the contradiction?
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    (Original post by Robbie242)
    2^a is also even
    3^b is not even ?

    I'm trying to raise 3 to powers and it keeps coming out odd, is this the contradiction?
    Precisely
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    (Original post by joostan)
    Precisely
    Got there in the end ahah! its good I'm getting into these problems though, the alevel ones often guide you a lot
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    (Original post by DJMayes)
    Well, alternatively:

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     (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1

     (a+ib)(a-ib)(c+id)(c-id) = (ac+bd+i)(ac+bd-i)

     (a+ib)(c-id) = (ac+bd)+(bc-ad)i

     (a-ib)(c+id) = (ac+bd)+(ad-bc)i

    This then gives us the same thing as the RHS, provided that  ad-bc = \pm 1 , which is exactly the result we derived using the other method.

    Basically it's the exact same method, only the complex numbers introduce a greater probability of an algebraic slip up
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    (Original post by Robbie242)
    Got there in the end ahah! its good I'm getting into these problems though, the alevel ones often guide you a lot
    There's a nice induction one with irrationals if you're interested
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    (Original post by joostan)
    There's a nice induction one with irrationals if you're interested
    Go on then , I'll at least get past the basis case and the assumption step xd
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    (Original post by joostan)
    In essence this is a solution.
    \dfrac{a}{b}
    was said to be in it's simplest form, which it isn't if both are a multiple of 5
    This is a contradiction, so therefore the cubed root of 5 is irrational.
    OOHH, so basically in there type of questions you prove that a/b isn't its simplest form, this is means its a contradiction and therefore is irrational? Ahh, so I kind of solved it? :/
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    (Original post by joostan)
    Basically it's the exact same method, only the complex numbers introduce a greater probability of an algebraic slip up
    I'm not saying it's any better (Although it did actually feel marginally quicker, possibly because I knew what I was going for) but it is an alternative approach using complex numbers which might've been the one Mayman was thinking of.
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    (Original post by Felix Felicis)
    Ahh, muy bien

    3 hours later on that xy = z^{2} + 1 problem, I think I'm onto something
    If it took you that much. That means I could never do it
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    (Original post by Robbie242)
    Go on then , I'll at least get past the basis case and the assumption step xd
    You've already got your basis case.
    Prove that:
    5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}
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    (Original post by DJMayes)
    Well, alternatively:

    Spoiler:
    Show


     (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1

     (a+ib)(a-ib)(c+id)(c-id) = (ac+bd+i)(ac+bd-i)

     (a+ib)(c-id) = (ac+bd)+(bc-ad)i

     (a-ib)(c+id) = (ac+bd)+(ad-bc)i

    This then gives us the same thing as the RHS, provided that  ad-bc = \pm 1 , which is exactly the result we derived using the other method.



    Also:



    I believe that Robbie's idea of prime numbers is actually a better way of tackling the question, because it can be extended to cases where both numbers are odd or both even. for example:

    Spoiler:
    Show


    Prove that  log_5(7) is irrational.

    Assume the opposite i.e.  log_5(7) = \frac{a}{b}

    And we then get to:

     7^b = 5^a


    Now, your odd and even ideas won't work here. However, you can say that, no matter how many times you multiply 5 by itself, it will never gain a factor of 7 (and vice-versa) so as to derive the contradiction you're after.

    Thanks DJ! I will probably use this one as it seems a bit more universal,
    i.e. I'd say for my one, no matter how many times you multiply 2 by itself, it will never gain a factor 3, therefore this is a contradiction and hence log_{3} 2 is irrational
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    (Original post by DJMayes)
    I'm not saying it's any better (Although it did actually feel marginally quicker, possibly because I knew what I was going for) but it is an alternative approach using complex numbers which might've been the one Mayman was thinking of.
    Yeah, I guess
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    (Original post by joostan)
    You've already got your basis case.
    Prove that:
    5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}
    Explain this notation Lol
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    (Original post by tigerz)
    OOHH, so basically in there type of questions you prove that a/b isn't its simplest form, this is means its a contradiction and therefore is irrational? Ahh, so I kind of solved it? :/
    Yep
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    (Original post by Robbie242)
    Explain this notation Lol
    N is the set of natural numbers.
    Q is the set of rationals.
    The \in means in

    Apologies for misleading you with the primes, you're right it works better
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    (Original post by Robbie242)
    Explain this notation Lol
    Not entirely sure what all the inequalities are there for, but the final part is saying "is not rational for all natural numbers n".

    Also, I would question how appropriate this is for this thread, given that the question you're being asked is in fact Q3 of STEP II, 2003...
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    (Original post by joostan)
    N is the set of natural numbers.
    Q is the set of rationals.
    The \in means in

    Apologies for misleading you with the primes, you're right it works better
    What about the upsidedown A?
 
 
 
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