Where's the fallacy in this proof? (that 1 - 1 equals 2)

[noobishnoob]

I found this "proof" but I can't seem to find the fallacious step.

 

[pka]

I found this "proof" but I can't seem to find the fallacious step.

HINT: The rules for exponents are among the most miss-understood in all of mathematics.
Can you spot one is that so-called proof?

 

[JeffM]

PKA has given you the correct answer, but perhaps his hint is too obscure.

\(\displaystyle \{-\ 1)^{1/2}\}^2 = (\pm i)^2 = -\ 1.\)

\(\displaystyle \{(-\ 1)^2\}^{1/2} = (1)^{1/2} = 1.\)

Quite clearly \(\displaystyle (a^2)^{1/2} \not \equiv (a^{1/2})^2 \text { for every } a \in \mathbb R.\)

Was your book wrong, or did you forget something in your book that may have appeared inconsequential?

 

[lookagain]

\(\displaystyle \{-\ 1)^{1/2}\}^2 = (\pm i)^2 = -\ 1.\)

Other than missing an open parenthesis (by typo?), the plus or
minus symbol should not be there:

\(\displaystyle \{(-1)^{1/2}\}^2 \ = \ (i)^2 \ = \ -1.\)

\(\displaystyle (-1)^{1/2} \ = \ \sqrt{-1} \ = \ i \)

 

[Steven G]

Other than missing an open parenthesis (by typo?), the plus or
minus symbol should not be there:

\(\displaystyle \{(-1)^{1/2}\}^ \ = \ (i)^2 \ = \ -1.\)

\(\displaystyle (-1)^{1/2} \ = \ \sqrt{-1} \ = \ i \)

You missed a square

 

[Steven G]

HINT: The rules for exponents are among the most miss-understood in all of mathematics.
Can you spot one is that so-called proof?

I had always thought that you could replace one expression for an equivalent one UNTIL I looked at exponents carefully!

 

[lookagain]

You missed a square

Thank you. My expression had "^" at the end, followed by nothing instead of "^2" at the end in the Latex format.

I went back and edited it.

 

[Denis]

Thank you. My expression had "^" at the end, followed by nothing instead of "^2" at the end in the Latex format.
I went back and edited it.

To the corner for 4^2 minutes...with no smartphone!!