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    (Original post by DJMayes)
    What problem is this? Based on that line it sounds very familiar to a STEP question I've seen before.
    Check joostan's post above

    He gave another hint about thinking of factorising x, y, z into primes or something so that's the approach I'm trying
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    (Original post by Robbie242)
    My attempt:
    Spoiler:
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    \log_{3} 2 if assumed to be irrational then \log_{3} 2=\frac{a}{b}

    it is given that Expoteniating to both sides

    3^{\frac{a}{b}}=2

    But since 3 to any power which is \geq 0 does not give 2, this then is assumed to be a negative power such that \frac{a}{b}<0



    and then I'm stuck any advice felix?
    I disagree with your last part.
    Spoiler:
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    negative powers end up with a fraction less than 1, 3^0 = 1, 3^1 = 3, therefore I'd say that the fraction would have to be between 0 and 1, so it's not going to be a negative power.... If that makes sense
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    (Original post by Zaphod77)
    I disagree with your last part.
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    negative powers end up with a fraction less than 1, 3^0 = 1, 3^1 = 3, therefore I'd say that the fraction would have to be between 0 and 1, so it's not going to be a negative power.... If that makes sense
    Ah yeah definitely my bad!
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    (Original post by joostan)
    ...
    Have I missed something in the question? It seems fairly straightforward...

    Spoiler:
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     xy = z^2+1

     x=a^2+b^2

     y = c^2+d^2

     z=ac+bd

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

     a^2c^2+b^2c^2+a^2d^2+b^2d^2 = a^2c^2+b^2d^2+2abcd+1

     b^2c^2+a^2d^2 = 2abcd+1

     (bc-ad)^2 = 1

     bc = ad +1

    From this we can easily find solution sets, such as b = 2, c = 11, a = 3, d = 7., with x=13, y= 170, z = 47, z^2+1 = 2210 = 13x170

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    (Original post by joostan)
    There's a solution above if you're tempted
    Hint:
    Spoiler:
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    Get a power on both sides
    I have got here but I'm kinda stuck on what to do next, I noticed 4 is not a prime number so I can't exploit the same method I used for for the cube root, 2 is a square number, perhaps this is important>?
    My attempt:

    Spoiler:
    Show


    \log_{3} 2 if assumed to be irrational then \log_{3} 2=\frac{a}{b}

    it is given that Expoteniating to both sides

    3^{\frac{a}{b}}=2

    a/b is this is assumed to be a positive power such that 0<\frac{a}{b}<1

    If we square both sides it is given that

    9^\frac{2a}{b}=2^2

    Equating the powers

    \frac{2a}{b}=4

    2a=4b
    a=2b
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    (Original post by Robbie242)
    I have got here but I'm kinda stuck on what to do next, I noticed 4 is not a prime number so I can't exploit the same method I used for for the cube root, 2 is a square number, perhaps this is important>?

    My attempt:

    Spoiler:
    Show


    \log_{3} 2 if assumed to be irrational then \log_{3} 2=\frac{a}{b}

    it is given that Expoteniating to both sides

    3^{\frac{a}{b}}=2

    a/b is this is assumed to be a positive power such that 0<\frac{a}{b}<1

    If we square both sides it is given that


    9^\frac{2a}{2b}=2^2

    2a=8b -> a=4b

    Equating the powers

    \frac{2a}{2b}=4

    a=4b
    Some of your working is not true. After the bold I recommend you take a look
    EDIT: Ignore that - missed the 2 undeneath That doesn't seem to help much though
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    (Original post by joostan)
    Some of your working is not true. After the bold I recommend you take a look
    is it 2a/b ?
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    (Original post by DJMayes)
    Have I missed something in the question? It seems fairly straightforward...

    Spoiler:
    Show


     xy = z^2+1

     x=a^2+b^2

     y = c^2+d^2

     z=ac+bd

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

     a^2c^2+b^2c^2+a^2d^2+b^2d^2 = a^2c^2+b^2d^2+2abcd+1

     b^2c^2+a^2d^2 = 2abcd+1

     (bc-ad)^2 = 1

     bc = ad +1

    From this we can easily find solution sets, such as b = 2, c = 11, a = 3, d = 7., with x=13, y= 170, z = 47, z^2+1 = 2210 = 13x170

    Almost identical to my solution, though apparently proof by example is not permitted
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    (Original post by joostan)
    Some of your working is not true. After the bold I recommend you take a look
    EDIT: Ignore that - missed the 2 undeneath That doesn't seem to help much though
    I noticed if I equated them withotu squaring I would have got a=2b anyway anymore guidance?
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    Why does my calculator straightaway rationalise the denominator i messed up a question due to not knowing root3/3 is just 1/root3 but i didnt know that upon closer inspection my calculator just rationalised the denominator do all calculators do this id prefer not to as im bad at realising when its been rationalised or not forgive me for being such a noob i just dnt kno anything bout calculators im self teaching so have no teachers or any help
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    (Original post by Robbie242)
    is it 2a/b ?
    Check my edit - Sorry
    \frac{2a}{2b} = \frac{a}{b} - this doesn't really help
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    (Original post by Robbie242)
    Depends what type of parents honestly :P My mum went insane when I got a few B's and C's at GCSE Lol, along with my A*'s in media but those don't really matter.

    If they are likely to get disappointed, I'd just tell them now tbh, to either add shock value to if the results come out good, or to ensure they don't lose their **** on the day.
    Hey bro, Do you think there is a chance of me getting an A grade prediction for A2 maths considering those marks?
    Thanks
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    (Original post by Robbie242)
    I noticed if I equated them withotu squaring I would have got a=2b anyway anymore guidance?
    Spoiler:
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    Raise both sides to the power of b, rather than squaring
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    (Original post by Revisionbug)
    Why does my calculator straightaway rationalise the denominator i messed up a question due to not knowing root3/3 is just 1/root3 but i didnt know that upon closer inspection my calculator just rationalised the denominator do all calculators do this id prefer not to as im bad at realising when its been rationalised or not forgive me for being such a noob i just dnt kno anything bout calculators im self teaching so have no teachers or any help
    There's nothing wrong with a rationalised denominator.
    Personally I prefer it, but really it makes no difference unless it's a show that it's
    \dfrac{1}{\sqrt{3}}
    In which case remembering that the two forms are equivalent is necessary.
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    (Original post by joostan)
    I can't make any such promises
    How about . . .
    Prove that: \sqrt[3]{5} is irrational?

     \sqrt[3]{5}=\frac{a}{b}

     5=\frac{a^3}{b^3} \rightarrow  5b^3=a^3

     b=5w as it has to be a multiple of 5 if rational

     5=\frac{(5w)^3}{b^3} \therefore 5=\frac{125w^3}{b^3}

\rightarrow 5b^3=125w^3\rightarrow b^3=25w^3

    this means b^3 is a multiple of 5 so it should be able to simplify
    but if a and b was a multiple of 5 then it should be able to simplify further earlier on...

    uhm I dunno Clue please? i've confused myself :cry2:
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    (Original post by joostan)
    Almost identical to my solution, though apparently proof by example is not permitted
    I don't see why not - it's a valid (and the most sensible!) way of attaining a solution and he can't expect a comprehensive list of solutions as there appear to be infinitely many.
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    (Original post by tigerz)
     \sqrt[3]{5}=\frac{a}{b}

     5=\frac{a^3}{b^3} \rightarrow  5b^3=a^3

     b=5w as it has to be a multiple of 5 if rational

     5=\frac{(5w)^3}{b^3} \therefore 5=\frac{125w^3}{b^3}

\rightarrow 5b^3=125w^3\rightarrow b^3=25w^3

    this means b^3 is a multiple of 5 so it should be able to simplify
    but if a and b was a multiple of 5 then it should be able to simplify further earlier on...

    uhm I dunno Clue please? i've confused myself :cry2:
    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.
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    (Original post by DJMayes)
    I don't see why not - it's a valid (and the most sensible!) way of attaining a solution and he can't expect a comprehensive list of solutions as there appear to be infinitely many.
    Dunno what he's after, but yeah, Felix says he's getting somewhere.
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    (Original post by joostan)
    Spoiler:
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    Raise both sides to the power of b, rather than squaring
    Spoiler:
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    \log_{3} 2 if assumed to be irrational then \log_{3} 2=\frac{a}{b}

    it is given that Expoteniating to both sides

    3^{\frac{a}{b}}=2

    a/b is this is assumed to be a positive power such that 0<\frac{a}{b}<1 -> so a and b are positive integers

    By raising to the power of b, it is given that

    3^{a}=2^b

    If b is non-zero, and if we try a=0

    2^b=1

    b=0, but since b is a non-zero integer, this a contradiction, therefore log_{3} 2 is irrational ?

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    (Original post by Robbie242)
    Spoiler:
    Show

    \log_{3} 2 if assumed to be irrational then \log_{3} 2=\frac{a}{b}

    it is given that Expoteniating to both sides

    3^{\frac{a}{b}}=2

    a/b is this is assumed to be a positive power such that 0<\frac{a}{b}<1

    By raising to the power of b, it is given that

    3^{a}=2^b

    If b is non-zero, and if we try a=0

    2^b=1

    b=0, but since b is a non-zero integer, this a contradiction, therefore log_{3} 2 is irrational ?

    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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