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Math Help - An indredible discovery or a bad mistake ?

  1. #1
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    An indredible discovery or a bad mistake ?

    Let a_i>1 a real number, let b_i=\frac{a_i}{a_i-1}. We have (a_i-1)(b_i-1)=1and a_i\sqrt{b_i-1}=\sqrt{a_ib_i} and of course b_i\sqrt{a_i-1}=\sqrt{a_ib_i}, we will useequalities like x-y=\frac{x^2-y^2}{x+y}, with x+y\neq{0}. We have a_ib_i=a_i-1+b_i-1+2=a_i-1+b_i-1+2\sqrt{(a_i-1)(b_i-1)}=(\sqrt{a_i-1}+\sqrt{b_i-1})^2Thus \sqrt{a_ib_i}=\sqrt{a_i-1}+\sqrt{b_i-1}But \frac{\sqrt{a_i-1}}{\sqrt{b_i-1}}(\frac{b_i-2}{a_i-2})=\frac{a_i}{b_i}(\frac{-b_i}{a_i})=-1 =\frac{\sqrt{a_ib_i}-2\sqrt{a_i-1}}{\sqrt{a_ib_i}-2\sqrt{b_i-1}}We deduce \frac{2\sqrt{b_i-1}-2\sqrt{a_i-1}}{\sqrt{a_ib_i}-2\sqrt{b_i-1}}=-2 =\frac{2(b_i-a_i)}{2\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1})}Or 2(b_i-a_i)=-4\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1}) =-4\sqrt{a_ib_i}\frac{a_ib_i-4(b_i-1)}{\sqrt{a_ib_i}+2\sqrt{b_i-1}} =-4(\sqrt{a_i-1}+\sqrt{b_i-1})\frac{a_i-3b_i+4}{\sqrt{a_i-1}+3\sqrt{b_i-1}}Or 2(b_i-a_i)(\sqrt{a_i-1}+3\sqrt{b_i-1})=-4(\sqrt{a_i-1}+\sqrt{b_i-1})(a_i-3b_i+4) =2\sqrt{a_ib_i}+6b_i\sqrt{b_i-1}-2a_i\sqrt{a_i-1}-6\sqrt{a_ib_i} =-4a_i\sqrt{a_i-1}+12\sqrt{a_ib_i}-16\sqrt{a_i-1}-4\sqrt{a_ib_i}+12b_i\sqrt{b_i-1}-16\sqrt{b_i-1} =6b_i\sqrt{b_i-1}-2a_i\sqrt{a_i-1}-4\sqrt{a_ib_i} =-4a_i\sqrt{a_i-1}+12b_i\sqrt{b_i-1}-8\sqrt{a_ib_i}And -6b_i\sqrt{b_i-1}+2a_i\sqrt{a_i-1}+4\sqrt{a_ib_i}=0 =-6b_i\sqrt{b_i-1}+2a_i\sqrt{a_i-1}+4\sqrt{a_i-1}+4\sqrt{b_i-1}=0Or \sqrt{b_i-1}(-6b_i+4)=\sqrt{a_i-1}(-4-2a_i)And \sqrt{\frac{a_i-1}{b_i-1}}=\frac{a_i}{b_i}=\frac{-6b_i+4}{-4-2a_i}=\frac{2-3b_i}{-2-a_i}Or -2a_i-a_i^2=2b_i-3b_i^2And 3b_i^2-a_i^2=2a_i+2b_i=2a_ib_iAnd a_i^2+2a_ib_i-3b_i^2=(a_i-b_i)(a_i+3b_i)=0If we take a_i=3 and b_i=\frac{3}{2}, it is absurd ! We dit not multiply or divide by zero !
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: An indredible discovery or a bad mistake ?

    I think there is an error in this line:

    Quote Originally Posted by Henrygibbs View Post
    We deduce \frac{2\sqrt{b_i-1}-2\sqrt{a_i-1}}{\sqrt{a_ib_i}-2\sqrt{b_i-1}}=-2 =\frac{2(b_i-a_i)}{2\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1})}
    If you substitute a = 3 and b = 3/2 you get -2 = -2 = -1.
    Thanks from emakarov
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  3. #3
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    Re: An indredible discovery or a bad mistake ?

    We have \frac{2\sqrt{b-1}-2\sqrt{a-1}}{\sqrt{ab}-2\sqrt{b-1}} =\frac{2(\sqrt{b-1}-\sqrt{a-1})(\sqrt{b-1}+\sqrt{a-1})}{(\sqrt{b-1}+\sqrt{a-1})(\sqrt{ab}-2\sqrt{b-1})} =\frac{2(b-a)}{\sqrt{ab}(\sqrt{ab}-2\sqrt{b-1})}Where is the mistake ? Numerically, there is perhaps one but not algebraically ! Ah ! You have put a 2 below, we have -2=-2=-2 !
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: An indredible discovery or a bad mistake ?

    Quote Originally Posted by Henrygibbs View Post
    We have \frac{2\sqrt{b-1}-2\sqrt{a-1}}{\sqrt{ab}-2\sqrt{b-1}} =\frac{2(\sqrt{b-1}-\sqrt{a-1})(\sqrt{b-1}+\sqrt{a-1})}{(\sqrt{b-1}+\sqrt{a-1})(\sqrt{ab}-2\sqrt{b-1})} =\frac{2(b-a)}{\sqrt{ab}(\sqrt{ab}-2\sqrt{b-1})}Where is the mistake ? Numerically, there is perhaps one but not algebraically ! Ah ! You have put a 2 below, we have -2=-2=-2 !
    I agree - the "2 below" in your original post is an error. But you also continue to include that "2 below" in subsequent steps. I think that is the source of your error.
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  5. #5
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    Re: An indredible discovery or a bad mistake ?

    Ok, it was there ! Thank you very much !
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  6. #6
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    Re: An indredible discovery or a bad mistake ?

    if one proves something well-known to be true to be false, one should question the proof, first.
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