An indredible discovery or a bad mistake ?
Let 
a real number, let 
. We have (b_i-1)=1)
and 
and of course 
, we will useequalities like 
, with 
. We have (b_i-1)}=(\sqrt{a_i-1}+\sqrt{b_i-1})^2)
Thus 
But =\frac{a_i}{b_i}(\frac{-b_i}{a_i})=-1)
We deduce 
}{2\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1})})
Or =-4\sqrt{a_ib_i}(\sqrt{a_ib_i}-2\sqrt{b_i-1}))
}{\sqrt{a_ib_i}+2\sqrt{b_i-1}})
\frac{a_i-3b_i+4}{\sqrt{a_i-1}+3\sqrt{b_i-1}})
Or (\sqrt{a_i-1}+3\sqrt{b_i-1})=-4(\sqrt{a_i-1}+\sqrt{b_i-1})(a_i-3b_i+4))
And 
Or =\sqrt{a_i-1}(-4-2a_i))
And 
Or 
And 
And (a_i+3b_i)=0)
If we take 
and 
, it is absurd ! We dit not multiply or divide by zero !
Re: An indredible discovery or a bad mistake ?
I think there is an error in this line:
Quote:
Originally Posted by
Henrygibbs 
We deduce
If you substitute a = 3 and b = 3/2 you get -2 = -2 = -1.
Re: An indredible discovery or a bad mistake ?
We have 
(\sqrt{b-1}+\sqrt{a-1})}{(\sqrt{b-1}+\sqrt{a-1})(\sqrt{ab}-2\sqrt{b-1})})
}{\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 !
Re: An indredible discovery or a bad mistake ?
Quote:
Originally Posted by
Henrygibbs We have
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.
Re: An indredible discovery or a bad mistake ?
Ok, it was there ! Thank you very much !
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.
