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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.



    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:

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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.

    Well, you're actually right.
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    (Original post by Robbie242)
    What about the upsidedown A?
    For all
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    (Original post by Robbie242)
    What about the upsidedown A?
    For all
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    (Original post by joostan)
    Yep
    :excited: Can I have a similar one later on as a step up and to check if I get it properly lols
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    (Original post by joostan)
    For all
    Prove that
    5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}
    so its basically saying Prove that 5^{\frac{1}{3^n}} is in the set of rationals is not rational for all natural numbers n

    totally confused, do I have to prove it is not rational for all natural numbers n
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    (Original post by tigerz)
    :excited: Can I have a similar one later on as a step up and to check if I get it properly lols
    Sure:
    Why not have a go at Felix's problem:
    Prove that: \log_32 is irrational
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    (Original post by Robbie242)
    Prove that
    5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}
    so its basically saying Prove that 5^{\frac{1}{3^n}} is in the set of rationals is not rational for all natural numbers n

    totally confused, do I have to prove it is not rational for all natural numbers n
    Yes to the bold
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    Proof that π+e is an irrational.
    I really don't know if that is easy or hard, just thought about it.
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    (Original post by joostan)
    Yes to the bold
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    Setting up an expression, f(n)=5^{\frac{1}{3n}}
    If its seen true for n=1, then assume true for n=k such that
    f(k)=5^{\frac{1}{3k}}
    and if rational then 5^{\frac{1}{3k}}=\frac{a}{b}

    For positive integers n

    If also true for n=k, then also true for n=k+1 such that

    f(k+1)=5^{\frac{1}{3(k+1)}}
    f(k+1)=5^{\frac{1}{3k+3}}

    5^{\frac{1}{3k+3}}=\frac{a}{b}+1

    No idea honestly

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    (Original post by joostan)
    Sure:
    Why not have a go at Felix's problem:
    Prove that: \log_32 is irrational
    Okays Ima eat dinner and feed my niece then i'll attempt it, woohoo logs
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    (Original post by MAyman12)
    Proof that π+e is an irrational.
    I really don't know if that is easy or hard, just thought about it.
    I don't think there is even a proof for this - there is a set of about 5 combinations of pi and e, which have not been proven rational or irrational, only that only so many can be rational or irrational.
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    (Original post by MAyman12)
    Proof that π+e is an irrational.
    I really don't know if that is easy or hard, just thought about it.
    Are you trolling :curious: It's not known where \pi + e is irrational...
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    (Original post by DJMayes)
    I don't think there is even a proof for this - there is a set of about 5 combinations of pi and e, which have not been proven rational or irrational, only that only so many can be rational or irrational.
    (Original post by Felix Felicis)
    Are you trolling :curious: It's not known where \pi + e is irrational...
    I just thought about it, didn't even search it.
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    (Original post by MAyman12)
    Proof that π+e is an irrational.
    I really don't know if that is easy or hard, just thought about it.
    Assuming that we know both \pi and e are rational, and that \pi and e are transcendental you can easily prove that either \pi+e or e\pi or both are irrational
    But a proper solution can't be found :cool:
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    (Original post by Robbie242)
    Spoiler:
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    Setting up an expression, f(n)=5^{\frac{1}{3n}}
    If its seen true for n=1, then assume true for n=k such that
    f(k)=5^{\frac{1}{3k}}
    and if rational then 5^{\frac{1}{3k}}=\frac{a}{b}

    For positive integers n

    If also true for n=k, then also true for n=k+1 such that

    f(k+1)=5^{\frac{1}{3(k+1)}}
    f(k+1)=5^{\frac{1}{3k+3}}

    5^{\frac{1}{3k+3}}=\frac{a}{b}+1

    No idea honestly

    Sorry - the tex was probably hard to read.
    It's 1/3^n
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    (Original post by joostan)
    Sorry - the tex was probably hard to read.
    It's 1/3^n
    aha, any general say 3 points I should aim to do when tackling this question, and maybe I can figure the rest out?

    Am I aiming to find a contradiction for n=k+1?
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    (Original post by Robbie242)
    aha, any general say 3 points I should aim to do when tackling this question, and maybe I can figure the rest out?

    Am I aiming to find a contradiction for n=k+1?
    No - it's induction you want. Your basis case is n=1, which you've already proved
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    (Original post by joostan)
    No - it's induction you want. Your basis case is n=1, which you've already proved
    Right but was I on the right path last time? I set it equal to a/b when we are trying to prove it to be irrational? Or is the contradiction itself the inductive proof?
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    (Original post by Robbie242)
    Right but was I on the right path last time? I set it equal to a/b when we are trying to prove it to be irrational? Or is the contradiction itself the inductive proof?
    You may be able to do so, but a standard proof by induction is all that's required.
    Basis case. (already done)
    Assume true for n=k
    Show that it's consistent with n=k+1
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    (Original post by Robbie242)
    Right but was I on the right path last time? I set it equal to a/b when we are trying to prove it to be irrational? Or is the contradiction itself the inductive proof?
    Also I should've told you that for any irrational x:
    \sqrt[3]{x} \not\in \mathbb{Q}
    Though it's trivially simple to deduce
 
 
 
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