An indredible discovery or a bad mistake ?

**Henrygibbs** · January 7th 2013, 08:39 AM

Let $a_i>1$ a real number, let $b_i=\frac{a_i}{a_i-1}$ . We have $(a_i-1)(b_i-1)=1$ and $a_i\sqrt{b_i-1}=\sqrt{a_ib_i}$ 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 $a_ib_i=a_i-1+b_i-1+2=a_i-1+b_i-1+2\sqrt{(a_i-1)(b_i-1)}=(\sqrt{a_i-1}+\sqrt{b_i-1})^2$ Thus $\sqrt{a_ib_i}=\sqrt{a_i-1}+\sqrt{b_i-1}$ 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 $-6b_i\sqrt{b_i-1}+2a_i\sqrt{a_i-1}+4\sqrt{a_ib_i}=0$ $=-6b_i\sqrt{b_i-1}+2a_i\sqrt{a_i-1}+4\sqrt{a_i-1}+4\sqrt{b_i-1}=0$ Or $\sqrt{b_i-1}(-6b_i+4)=\sqrt{a_i-1}(-4-2a_i)$ 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 $-2a_i-a_i^2=2b_i-3b_i^2$ And $3b_i^2-a_i^2=2a_i+2b_i=2a_ib_i$ And $a_i^2+2a_ib_i-3b_i^2=(a_i-b_i)(a_i+3b_i)=0$ If we take $a_i=3$ and $b_i=\frac{3}{2}$ , it is absurd ! We dit not multiply or divide by zero !

**ebaines** · January 7th 2013, 10:02 AM

I think there is an error in this line:

Originally Posted by Henrygibbs

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.

**Henrygibbs** · January 7th 2013, 10:27 AM

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 !

**ebaines** · January 7th 2013, 10:56 AM

Originally Posted by Henrygibbs

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.

**Henrygibbs** · January 7th 2013, 11:07 AM

Ok, it was there ! Thank you very much !

**Deveno** · January 7th 2013, 11:35 AM

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

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