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    (Original post by Robbie242)
    One thing I'm algebraically confused on is how 5^{\frac{1}{3^{(k+1)}}} = \left(5^{\frac{1}{3^k}}\right)^{  \frac{1}{3}}

    Could you explain this if you don't mind, I know you multiply the inner power by the outer one but can't seem to reach that exact solution
    Sure:
    \dfrac{1}{3^{k+1}}=\dfrac{1}{3} \times \dfrac{1}{3^k}
    Rearrange a little bit more and voíla
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    Jheeze Robbie's on this irrationality maths shiz!

    time for me to attempt Felix's prove  \log_{3}2 is irrational :/
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    (Original post by Felix Felicis)
    I reckon I've a rubix complex Either that or I've been watching too much House :ninja:


    I saw DJ's solution but it's the exact same as the one joostan/ justinawe proposed, it only uses complex numbers
    The question relies on one specific result for generating sets, although there are probably other ways you can obtain the result to the ones we've been through. However, you cannot really go further with it due to the infinite number of potential solutions. By the way, I'm not entirely convinced your use of modular arithmetic is correct:

    Spoiler:
    Show


    You have:
     5q^3 = p^3
     \Rightarrow p^3 \equiv 0 (mod 5)
     \Rightarrow p \equiv 0 (mod 5)
     \Rightarrow q^3 \equiv 0 (mod 5)
     \Rightarrow q \equiv 0 (mod 5)

    And then you get the simplification contradiction.  p^3mod5 is absolutely meaningless as a statement, and I think you'd have been better writing up your solution without the modular arithmetic.



    (Also, you think you've been watching too much? Started watching it a month or two ago, now halfway through the fifth season. )
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    (Original post by joostan)
    Sure:
    \dfrac{1}{3^{k+1}}=\dfrac{1}{3} \times \dfrac{1}{3^k}
    Rearrange a little bit more and voíla
    Not entirely sure actually, if I multiplied them I'd get the same solution so should I perhaps try dividing by 1 third?
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    (Original post by Robbie242)
    Not entirely sure actually, if I multiplied them I'd get the same solution so should I perhaps try dividing by 1 third?
    Well in the power you get the above result, then you can split it up as I did.
    Then using the cubic root result, you get the desired proof.
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    \log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2 I somehow managed to simplify this to: 2b=3a is this correct?
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    (Original post by joostan)
    Well in the power you get the above result, then you can split it up as I did, using the cubic root result.
    eh I've only managed to prove \frac{1}{3^{k}}=\frac{1}{3^{k}} my algebra is weak
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    Gonna disappear from this thread for a bit I think :ninja:

    I'll be back before my S1 exam, I don't think there are any maths exams before that, but if there are good luck! (actually there won't be, it's half term :lol:)

    Anyway, adios!
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    (Original post by tigerz)
    \log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2 I somehow managed to simplify this to: 2b=3a is this correct?
    Try raising your 3^\frac{a}{b}=2 to the power of b, and then you may be able to draw some conclusions
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    (Original post by tigerz)
    \log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2 I somehow managed to simplify this to: 2b=3a is this correct?
    2^b = 3^a

    (Original post by L'Evil Fish)
    Gonna disappear from this thread for a bit I think :ninja:

    I'll be back before my S1 exam, I don't think there are any maths exams before that, but if there are good luck! (actually there won't be, it's half term :lol:)

    Anyway, adios!
    Aufwiedersehen :ninja:
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    (Original post by Robbie242)
    eh I've only managed to prove \frac{1}{3^{k}}=\frac{1}{3^{k}} my algebra is weak
    Spoiler:
    Show

    Given that:
    \sqrt[3]{5} \not\in \mathbb{Q}

\mathrm{\ Assume \ that \ the \ statement} u_n =  5^{\frac{1}{3^n}}\not\in \mathbb{Q} \mathrm{\ true \ for} \ n =  k
    \Rightarrow \ u_{k+1} = 5^{\frac{1}{3^{k+1}}} = \left(5^{\frac{1}{k}}\right)^{ \frac{1}{3}}
    As the cubic root of an irrational number is irrational too. Therefore, by induction: As \ u_1 , u_k \mathrm{\  and} \ u_{k+1} \not\in \mathbb{Q} So \ u_n \not\in  \mathbb{Q} \mathrm{ \ at \ least} \ \forall n \in \mathbb{N}, n \not=  0
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    (Original post by joostan)
    2^b = 3^a



    Aufwiedersehen :ninja:
    nvm managed to figure it out, the visual thing you gave me makes sense when multiplying the powers, your just expressing it in a different way.
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    (Original post by Robbie242)
    nvm managed to figure it out, the visual thing you gave me makes sense when multiplying the powers, your just expressing it in a different way.
    Coolio :congrats:
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    (Original post by Robbie242)
    Try raising your 3^\frac{a}{b}=2 to the power of b, and then you may be able to draw some conclusions
    (Original post by joostan)
    2^b = 3^a

    Aufwiedersehen :ninja:
    Tex error,my bad thats what I meant :P right so I can continue from there
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    (Original post by L'Evil Fish)
    Gonna disappear from this thread for a bit I think :ninja:

    I'll be back before my S1 exam, I don't think there are any maths exams before that, but if there are good luck! (actually there won't be, it's half term :lol:)

    Anyway, adios!
    I should do the same to get this revision done that I really should've done today! :-/ See you some time in the future


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    Can anyone run through the basics of induction and then give me some relatively easy questions so I can work at it, then slowly bump up the difficulty so I'm able to try a few STEP questions on the topic, I just feel like I've forgotten too much of it to do anything that would be any good :-/


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    Anyone who is doing/has done D2, in game theory, when you are working out the value of the game in a 3x2 matrix and you have 3 intersection points on a probability graph, how do you know which one is optimal?
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    And for anyone who did S2 last week, for the question on the number of games in stock, did you end up with 12 or 13? No one seems to know which one is the correct answer
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    \log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2 this means 2^b=3^a
    We know that 2^b will be even so can be represented by 2w
    We also know that 3^a will be odd so is  2x+1
    This means 2^b=3^a\equiv 2w=2x+1\therefore w=x+1

    URM i'm lost now, I need to link it back to the original question but confused :confused:
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    (Original post by MathsNerd1)
    Can anyone run through the basics of induction and then give me some relatively easy questions so I can work at it, then slowly bump up the difficulty so I'm able to try a few STEP questions on the topic, I just feel like I've forgotten too much of it to do anything that would be any good :-/
    Posted from TSR Mobile
    Basis case.
    Inductive step. (n=k)
    n=k+1

    e.g.
    Prove that:
    \displaystyle\sum_{r=1}^n r = \frac{n}{2}(n+1)
 
 
 
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