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    proof that

    sqrt(2), sqrt(3) and sqrt(5) cannot all be part of the same geometric progression


    i will define sq(2) to be the first term

    then let the common difference of the arithmetic progression be d

    then

    sq(3)=sq(2)+md and sq(5)=sq(2)+nd where (m,n) are posotive integers

    so then

    (sq(3)-sq(2))/m=(sq(5)-sq(3))/n

    and so

    sq(3)(m+n)=msq(5)+nsq(2) now i square this then the LHS is an integer where as the RHS is irrational hence we arrive at a contradiction, so all three terms cannot be part of the same Arithmetic progression.

    Is this correct?
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    Doesn't the question ask to show they cannot be part of the same geometric progression?
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    I know this isn't particularly helpful, but could you please use LaTeX for this sort of stuff?
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    geometric progression therefore
    root(2)=a
    root(3)=ar
    root(5)=ar^2

    therefore r=\frac{\sqrt{3}}{\sqrt{2}}=\fra  c{\sqrt{5}}{\sqrt{3}} since this isn't true (shown by multiplying both sides by root 2root3) there exists no r satisfying the progression.
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    (Original post by Totally Tom)
    geometric progression therefore
    root(2)=a
    root(3)=ar
    root(5)=ar^2

    therefore r=\frac{\sqrt{3}}{\sqrt{2}}=\fra  c{\sqrt{5}}{\sqrt{3}} since this isn't true (shown by multiplying both sides by root 2root3) there exists no r satisfying the progression.
    It doesn't need to be consecutive does it? Then:

    r=r^n (\frac{\sqrt{3}}{\sqrt{2}})=r^m (\frac{\sqrt{5}}{\sqrt{3}}) where n and m are integers from 0-inf.

    That vaguely right?
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    (Original post by Totally Tom)
    geometric progression therefore
    root(2)=a
    root(3)=ar
    root(5)=ar^2

    therefore r=\frac{\sqrt{3}}{\sqrt{2}}=\fra  c{\sqrt{5}}{\sqrt{3}} since this isn't true (shown by multiplying both sides by root 2root3) there exists no r satisfying the progression.
    What if \sqrt{3} is the 50th or 75th or 3000th term instead of the 2nd term, as you assume here?
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    (Original post by The Bachelor)
    What if \sqrt{3} is the 50th or 75th or 3000th term instead of the 2nd term, as you assume here?
    jup, just realised.
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    no the question says arithmetic
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    (Original post by psanghaLFC)
    proof that

    sqrt(2), sqrt(3) and sqrt(5) cannot all be part of the same geometric progression
    wat.
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    whoops, it should be arithmetic progression, but i suppose i could just try and prove both
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    Let sq(3)=r^nsq(2) and sq(5)=r^msq(3) then

    (sq(3)/sq(2))^(1/n)=(sq(5)/sq(3))^1/m

    so

    sq(3)^m/sq(2)^m=sq(5)^n/sq(3)^n

    so that would mean

    sq(3)^(m+n)=sq(2)^n*sq(5)^m

    so after squaring both sides we would see two cases

    LHS rational and RHS rational
    LHS rational and RHS irrational

    in case two it leads to a contradiction, and in case 1 the only prime factors of the RHS are 2 and 5 and of the LHS they are 3 which cannot be the case.
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    (Original post by psanghaLFC)
    Let sq(3)=r^nsq(2) and sq(5)=r^msq(3) then

    (sq(3)/sq(2))^(1/n)=(sq(5)/sq(3))^1/m

    so

    sq(3)^m/sq(2)^m=sq(5)^n/sq(3)^n

    so that would mean

    sq(3)^(m+n)=sq(2)^n*sq(5)^m

    so after squaring both sides we would see two cases

    LHS rational and RHS rational
    LHS rational and RHS irrational

    in case two it leads to a contradiction, and in case 1 the only prime factors of the RHS are 2 and 5 and of the LHS they are 3 which cannot be the case.
    Seeing as no 'expert' has replied, I may as well have a pop at this proof...

    x and y are integer values.

     \sqrt{2} + xd = \sqrt{3} \Rightarrow xd = \sqrt{3} - \sqrt{2}

     \sqrt{3} + yd = \sqrt{5} \Rightarrow yd = \sqrt{5} - \sqrt{3}

     x(\sqrt{5} - \sqrt{3}) = y(\sqrt{3} - \sqrt{2})

     x\sqrt{5} + y\sqrt{2} = (x+y)\sqrt{3}

    Square both sides

     5x^2  + 2y^2 + 2\sqrt{10}xy = 3(x+y)^2

    Since x and y must be integers, x^2, y^2, xy are integers. Thus 2\sqrt{10}xy is irrational. Hence the left hand side is always irrational while the right hand side is always rational, hence the equality cannot hold. Proof by contradiction?

    Edit: sorry about that, I admit I was stupid enough to overlook his first solution. Ignore, or if anything, it makes it easier to read.
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    yes, thats what he did in his first post

    it's just he doesn't latex anything so its horrible to read through-most people don't bother.
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    (Original post by Totally Tom)
    yes, thats what he did in his first post

    it's just he doesn't latex anything so its horrible to read through-most people don't bother.
    lol

    wasted 10 minutes of my life!
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    (Original post by Totally Tom)
    yes, thats what he did in his first post

    it's just he doesn't latex anything so its horrible to read through-most people don't bother.
    Haha, my point exactly. OP, take note!
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    Is there some Latex code which you must follow? Or do you highlight it or something after to latex it?
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    Maybe read the forum rules?

    (maths forum rules.)
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    (Original post by psanghaLFC)
    Is there some Latex code which you must follow? Or do you highlight it or something after to latex it?
    *Mini* guide to latex:

    Enclose everything within ]latex[ ]/latex[ tags (flip brackets). Attempt to type normally. Things that need to be should be grouped with {}. ax^3+by^{3x+1} is ax^3+by^{3x+1}. \frac{a}{b} is a/b, \sqrt{x} is obvious. That's enough to get by with.
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    (Original post by pyrolol)
    *Mini* guide to latex:

    Enclose everything within ]latex[ ]/latex[ tags (flip brackets). Attempt to type normally. Things that need to be should be grouped with {}. ax^3+by^{3x+1} is ax^3+by^{3x+1}. \frac{a}{b} is a/b, \sqrt{x} is obvious. That's enough to get by with.
    @pyrolol: if you want to tell someone how to use a tag, use the [noparse] tag; e.g. [noparse][latex][/noparse] will appear as [latex].

    More LaTeX advice. Read the Guide to Posting, and download the short LaTeX guide at tobi.oetiker.ch/lshort/lshort.pdf

    If someone else posts something in LaTeX and you'd like to know how it works, you can use the "quote" button to see what the raw LaTeX was that they used.

    Most importantly:

    Preview before posting. It's really easy to end up with completely unintelligible garbage because you missed out a closing } or similar.
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    (Original post by DFranklin)
    @pyrolol: if you want to tell someone how to use a tag, use the [noparse] tag; e.g. [noparse][latex][/noparse] will appear as [latex].

    More LaTeX advice. Read the Guide to Posting, and download the short LaTeX guide at tobi.oetiker.ch/lshort/lshort.pdf

    If someone else posts something in LaTeX and you'd like to know how it works, you can use the "quote" button to see what the raw LaTeX was that they used.

    Most importantly:

    Preview before posting. It's really easy to end up with completely unintelligible garbage because you missed out a closing } or similar.
    Nested [noparse] tags! *brain explodes*.
 
 
 
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