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    (Original post by joostan)
    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
    Where did x come from, can I just use what you gave me, I feel awfully confused now
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    (Original post by Robbie242)
    Where did x come from, can I just use what you gave me, I feel awfully confused now
    Yeah- just where x is an irrational number e.g \pi or e etc.
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    (Original post by joostan)
    ...
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    Lemma
    if x is irrational, then x^{1/3} is irrational. This is trivial to show by contradiction - if x^{1/3} = p/q, (p,q) \in \mathbb{Z} \Rightarrow x = p^{3}/q^{3} but x is irrational, hence contradiction

    We will prove 5^{\frac{1}{3^{n}}} is irrational \forall n \in \mathbb{N}

    5^{1/3} = p/q \Rightarrow p^{3} = 5 q^{3} \Rightarrow p^{3} \equiv 0 \pmod{5} \Rightarrow p \equiv 0 \pmod{5}

    This implies q \equiv 0 \pmod{5} hence contradiction, so we've proved it for n = 1.

    Assume 5^{1/3^{k}} is irrational.

    Consider 5^{1/3^{k+1}} = \left( 5^{1/3^k} \right)^{1/3} by the inductive hypothesis, 5^{1/3^{k}} is irrational and hence by lemma \left( 5^{1/3^k} \right)^{1/3} is irrational, thereby completing the inductive step, thus the result is true for all n by the principle of mathematical induction.
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    (Original post by Hunarench95)
    Ok whenever you have a integration you must differentiate your 'u'

    u=x^2+5
    du/dx=2x. Therefore du=2x

    Take a factor of a half out of your integral. giving 1/2Int(2x(x^2+5)^0.5)dx

    You can then swap the 2x*dx with du

    Giving 1/2Int(u^0.5)du

    Integrating that gives 1/2{(2u^(1.5))/3 + c}

    Such in your 'u'

    giving ((x^2+5)^1.5)/3 + c
    thankyou so much!
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    (Original post by Felix Felicis)
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    Lemma
    if x is irrational, then x^{1/3} is irrational. This is trivial to show by contradiction - if x^{1/3} = p/q, (p,q) \in \mathbb{Z} \Rightarrow x = p^{3}/q^{3} but x is irrational, hence contradiction

    We will prove 5^{\frac{1}{3^{n}}} is irrational [tez]\forall n \in \mathbb{N}[/tex]

    5^{1/3} = p/q \Rightarrow p^{3} = 5 q^{3} \Rightarrow p^{3} \0mod5 \Rightarrow a \0mod5

    This implies q \0mod5 hence contradiction, so we've proved it for n = 1.

    Assume 5^{1/3^{k}} is irrational.

    Consider 5^{1/3^{k+1}} = \left( 5^{1/3^k} \right)^{1/3} by the inductive hypothesis, 5^{1/3^{k}} is irrational and hence by lemma \left( 5^{1/3^k} \right)^{1/3} is irrational, thereby completing the inductive step, thus the result is true for all n by the principle of mathematical induction.
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    Not convinved of the necessity of the modular arithmetic or the lemma, but an outstanding proof nonetheless
    You also seem to have abused the \Rightarrow
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    (Original post by joostan)
    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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    Were looking to prove the f(n)=\frac{5}{3^n} is irrational for all natural numbers n

    it was seen true for n=1, assume true for n=k such that:
    f(k)=\frac{5}{3^k}

    if true for n=k, assume true for n=k+1 such that
    f(k+1)=\frac{5}{3^{k+1}}

    so if we assume \frac{5}{3^{k+1}}=\frac{a}{b}
    by cross multiplying
    a(3^{k+1})=5b

    a(3(3^k))=5b

    let 3^k=x

    a(3x)=5b
    3ax=5b

    No idea how to proceed

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    (Original post by joostan)
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    Not convinved of the necessity of the modular arithmetic or the lemma, but an outstanding proof nonetheless
    You also seem to have abused the \Rightarrow
    I was on a wild goosechase with Mayman's question
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    (Original post by Robbie242)
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    Were looking to prove the f(n)=\frac{5}{3^n} is irrational for all natural numbers n

    it was seen true for n=1, assume true for n=k such that:
    f(k)=\frac{5}{3^k}

    if true for n=k, assume true for n=k+1 such that
    f(k+1)=\frac{5}{3^{k+1}}

    so if we assume \frac{5}{3^{k+1}}=\frac{a}{b}
    by cross multiplying
    a(3^{k+1})=5b

    a(3(3^k))=5b

    No idea how to proceed

    It's
    5^{1/3^n}
    I'm not much good at this problem setting
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    (Original post by joostan)
    It's
    5^{1/3^n}
    I'm not much good at this problem setting
    fml ok
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    (Original post by Felix Felicis)
    I was on a wild goosechase with Mayman's question
    Lol. How much time did you say you wasted? :lol:
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    (Original post by joostan)
    Lol. How much time did you say you wasted? :lol:
    Like 4 hours :rofl:
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    (Original post by Felix Felicis)
    Like 4 hours :rofl:
    o.O
    I'd never do that :nah:
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    (Original post by Felix Felicis)
    I was on a wild goosechase with Mayman's question
    DJ answered it btw
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    (Original post by Felix Felicis)
    Like 4 hours :rofl:
    That's commitment!


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    (Original post by joostan)
    It's
    5^{1/3^n}
    I'm not much good at this problem setting
    Spoiler:
    Show

    Were looking to prove the f(n)=5^{\frac{1}{3^n}} is irrational for all natural numbers n

    it was seen true for n=1, assume true for n=k such that:
    f(k)=5^{\frac{1}{3^k}}

    if true for n=k, assume true for n=k+1 such that
    f(k+1)=5^{\frac{1}{3^{k+1}}}

    so if we assume
    5^\frac{1}{3^{k+1}}=\frac{a}{b}

    by multiplying both sides by b
    5b^{\frac{1}{3^{k+1}}}=a


    And then I'm stuck, not sure exactly what contradiction I'm supposed to meet
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    (Original post by joostan)
    o.O
    I'd never do that :nah:
    (Original post by MathsNerd1)
    That's commitment!


    Posted from TSR Mobile
    I reckon I've a rubix complex Either that or I've been watching too much House :ninja:

    (Original post by MAyman12)
    DJ answered it btw
    I saw DJ's solution but it's the exact same as the one joostan/ justinawe proposed, it only uses complex numbers
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    (Original post by Robbie242)
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    Were looking to prove the f(n)=5^{\frac{1}{3^n}} is irrational for all natural numbers n

    it was seen true for n=1, assume true for n=k such that:
    f(k)=5^{\frac{1}{3^k}}

    if true for n=k, assume true for n=k+1 such that
    f(k+1)=5^{\frac{1}{3^{k+1}}}

    so if we assume
    5^\frac{1}{3^{k+1}}=\frac{a}{b}

    by multiplying both sides by b
    5b^{\frac{1}{3^{k+1}}}=a


    And then I'm stuck, not sure exactly what contradiction I'm supposed to meet
    Spoiler:
    Show

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

    As the cubed root of any irrational number is irrational . . .
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    (Original post by joostan)
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    5^{\frac{1}{3^{(k+1)}}} = \left(5^{\frac{1}{3^k}}\right)^{  \frac{1}{3}}

    As the cubed root of any irrational number is irrational . . .
    Spoiler:
    Show


    so could I say if we assume rationality, \frac{a}{b}=5^{\frac{1}{3^{(k+1)  }}}

    But 5^{\frac{1}{3^{(k+1)}}} = \left(5^{\frac{1}{3^k}}\right)^{  \frac{1}{3}}

    And as the cube root of any irrational number is irrational, true for n=k+1

    Seen true for n=1, if true for n=k, then also seen true for n=k+1, therefore seen true that 5^{\frac{1}{3^n}} is irrational for all natural numbers n by mathematical induction.

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    (Original post by Robbie242)
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    so could I say if we assume rationality, \frac{a}{b}=5^{\frac{1}{3^{(k+1)  }}}

    But 5^{\frac{1}{3^{(k+1)}}} = \left(5^{\frac{1}{3^k}}\right)^{  \frac{1}{3}}

    And as the cube root of any irrational number is irrational, true for n=k+1

    Seen true for n=1, if true for n=k, then also seen true for n=k+1, therefore seen true that 5^{\frac{1}{3^n}} is irrational for all natural numbers n by mathematical induction.

    Precisely
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    (Original post by joostan)
    Precisely
    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
 
 
 
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