[myriam (OP)]

Some thing wrong right here ....  []

1= -(-1)
1= -[(-1)²]½
1= -(1)½
1=-√1
1=-1

[JP]

Some thing wrong right here ....  []

1= -(-1)
1= -[(-1)²]½

Since [(-1)²]½=(1)½=1, line 2 doesn't follow from line 1.

[myriam (OP)]

Some thing wrong right here ....  []

1= -(-1)
1= -[(-1)²]½

Since [(-1)²]½=(1)½=1, line 2 doesn't follow from line 1.

good try JP but no , this isn't the right answer

[imatfaal]

finding a root gives two possibilities +/-1; and in situations like this when the square root is extracted, it is the negative root −1, not the positive root, (which is absurd for this equation).  which leaves the next line of your equation as 1=-(-1).

[myriam (OP)]

sorry imatfaal
The answer is more simple than you think
Give it an other try



 

[imatfaal]

as JP and I have both given valid answers from different perspectives - your correct answer better be a good one :-)

[myriam (OP)]

well , lets give a chance to some other people to try also
till this Friday I'll see how much easy  or difficult maths are   

[graham.d]

I rather think imatfaal had it. Maybe it could be expressed more simply but you could show the faulty reasoning as follows:


let a=b=1
a=b
a²=b²
a²-b²=0
(a-b)(a+b)=0

so, as a≠0, b≠0
a+b=0 iff a-b≠0, which is false
a-b=0 iff a+b≠0, which is true

so a-b=0
a=b
so a=1 and b=1


Taking the square root give two answers (±1) and one is false because you are creating a quadratic equation then factorising to get the roots. The factorisation that gives you the false answer involves dividing by zero.

[myriam (OP)]

you showed that 1≠-1 which is true ,so you can do something easier,
you could show the faculty finding the error in my equations

try again graham.d

[graham.d]

The square root of 1 does not have one answer. It can be +1 or -1. You chose it (incorrectly) to be +1. I thought I showed, reasonable rigorously, why this choice was erronious because your solving of the quadratic equation involved a division by zero. Is there a fault in the proof? I give up.

This is why I'm a physicist and engineer and not a mathematician :-)

[Geezer]

1= -(-1)
1= -[(-1)²]½
1= -(1)½
1=-√1
1=-1[/b]

Line 2 is invalid. The power 1/2 has to be inside the square bracket which makes the square brackets redundant and the power 2/2 which is of course, 1.

[myriam (OP)]

for all real numbers
√x² =|x|
    = x if x≥0
    =-x if x≤0


I don't see the error in the line 4  ,  √1 = 1 since  1>0 already

But the really error is no there

Geezer it is great you found the faulty line but the reasoning is not this,
find the theorem that demonstrates the error in line 2

just to tell you there are other faulty proof for 1=-1 like for example

-1=i.i=√-1 .√-1 =√(-1.-1) = √1 = 1  but where is the fault
this is an other challenge

I'm not telling the answer rapidly []  ,  think about it   

[Geezer]

1= -(-1)
1= -[(-1)²]½

1 = -(-1)
  = -[-(1²)½]

[myriam (OP)]

ok, where is the theorem telling that i can't  write  1= -[(-1)²]½  ??

[Geezer]

ok, where is the theorem telling that i can't  write  1= -[(-1)²]½  ??

No theoerm required. You are substituting something that is not equal to -1. Show me how you get from my line 2 to your line 2.

[myriam (OP)]

ok, where is the theorem telling that i can't  write  1= -[(-1)²]½  ??

No theoerm required. You are substituting something that is not equal to -1. Show me how you get from my line 2 to your line 2.

oh yes there is a very important theoerme there

[Geezer]

Well, I suppose if you want to try to take the square root of a negative number, that's up to you.

[myriam (OP)]

 []

but do you know that  if x is any positive number, then the principal square root of −x is √-x =i.√x

The theorem is ..uhm ... not now, maybe someone is about to find it , lets wait   []

[Geezer]

Yes, but it was j in my day.

[JP]

... then you all realized how silly it was to talk about jmaginary numbers.   []