Let $\displaystyle a_i>1$ a real number, let $\displaystyle b_i=\frac{a_i}{a_i-1}$. We have $\displaystyle (a_i-1)(b_i-1)=1$and $\displaystyle a_i\sqrt{b_i-1}=\sqrt{a_ib_i}$ and of course $\displaystyle b_i\sqrt{a_i-1}=\sqrt{a_ib_i}$, we will useequalities like $\displaystyle x-y=\frac{x^2-y^2}{x+y}$, with $\displaystyle x+y\neq{0}$. We have$\displaystyle a_ib_i=a_i-1+b_i-1+2=a_i-1+b_i-1+2\sqrt{(a_i-1)(b_i-1)}=(\sqrt{a_i-1}+\sqrt{b_i-1})^2$Thus$\displaystyle \sqrt{a_ib_i}=\sqrt{a_i-1}+\sqrt{b_i-1}$But$\displaystyle \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$$\displaystyle =\frac{\sqrt{a_ib_i}-2\sqrt{a_i-1}}{\sqrt{a_ib_i}-2\sqrt{b_i-1}}We deduce\displaystyle \frac{2\sqrt{b_i-1}-2\sqrt{a_i-1}}{\sqrt{a_ib_i}-2\sqrt{b_i-1}}=-2$$\displaystyle =\frac{2(b_i-a_i)}{2\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1})}$Or$\displaystyle 2(b_i-a_i)=-4\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1})$$\displaystyle =-4\sqrt{a_ib_i}\frac{a_ib_i-4(b_i-1)}{\sqrt{a_ib_i}+2\sqrt{b_i-1}}$$\displaystyle =-4(\sqrt{a_i-1}+\sqrt{b_i-1})\frac{a_i-3b_i+4}{\sqrt{a_i-1}+3\sqrt{b_i-1}}$Or$\displaystyle 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)$$\displaystyle =2\sqrt{a_ib_i}+6b_i\sqrt{b_i-1}-2a_i\sqrt{a_i-1}-6\sqrt{a_ib_i}$$\displaystyle =-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}$$\displaystyle =6b_i\sqrt{b_i-1}-2a_i\sqrt{a_i-1}-4\sqrt{a_ib_i}$$\displaystyle =-4a_i\sqrt{a_i-1}+12b_i\sqrt{b_i-1}-8\sqrt{a_ib_i}$And$\displaystyle -6b_i\sqrt{b_i-1}+2a_i\sqrt{a_i-1}+4\sqrt{a_ib_i}=0$$\displaystyle =-6b_i\sqrt{b_i-1}+2a_i\sqrt{a_i-1}+4\sqrt{a_i-1}+4\sqrt{b_i-1}=0Or\displaystyle \sqrt{b_i-1}(-6b_i+4)=\sqrt{a_i-1}(-4-2a_i)And\displaystyle \sqrt{\frac{a_i-1}{b_i-1}}=\frac{a_i}{b_i}=\frac{-6b_i+4}{-4-2a_i}=\frac{2-3b_i}{-2-a_i}Or\displaystyle -2a_i-a_i^2=2b_i-3b_i^2And\displaystyle 3b_i^2-a_i^2=2a_i+2b_i=2a_ib_iAnd\displaystyle a_i^2+2a_ib_i-3b_i^2=(a_i-b_i)(a_i+3b_i)=0If we take \displaystyle a_i=3 and \displaystyle b_i=\frac{3}{2}, it is absurd ! We dit not multiply or divide by zero ! 2. ## Re: An indredible discovery or a bad mistake ? I think there is an error in this line: Originally Posted by Henrygibbs We deduce\displaystyle \frac{2\sqrt{b_i-1}-2\sqrt{a_i-1}}{\sqrt{a_ib_i}-2\sqrt{b_i-1}}=-2$$\displaystyle =\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.

3. ## Re: An indredible discovery or a bad mistake ?

We have$\displaystyle \frac{2\sqrt{b-1}-2\sqrt{a-1}}{\sqrt{ab}-2\sqrt{b-1}}$$\displaystyle =\frac{2(\sqrt{b-1}-\sqrt{a-1})(\sqrt{b-1}+\sqrt{a-1})}{(\sqrt{b-1}+\sqrt{a-1})(\sqrt{ab}-2\sqrt{b-1})}$$\displaystyle =\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 !

4. ## Re: An indredible discovery or a bad mistake ?

Originally Posted by Henrygibbs
We have$\displaystyle \frac{2\sqrt{b-1}-2\sqrt{a-1}}{\sqrt{ab}-2\sqrt{b-1}}$$\displaystyle =\frac{2(\sqrt{b-1}-\sqrt{a-1})(\sqrt{b-1}+\sqrt{a-1})}{(\sqrt{b-1}+\sqrt{a-1})(\sqrt{ab}-2\sqrt{b-1})}$$\displaystyle =\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.

5. ## Re: An indredible discovery or a bad mistake ?

Ok, it was there ! Thank you very much !

6. ## 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.