Proofs for -1 = 1: Exploring the Problem

[T@P]

heres a little problem that at a first glance is real:

[tex] \frac{1}{-1} = \frac{-1}{1}[/tex]

so
[tex] \sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}} [/tex]

by splitting it the square root into two parts...

[tex] \frac{i}{1} = \frac{1}{i} [/tex]
and [tex] i^2 = 1 [/tex]

-1 = 1

wonder if there are any more similar "proofs"?

 

[AKG]

You can't split the square root into two parts. There are plenty of similar "proofs". You can search the web for them, and there are a number of them on this site alone.

 

[mathlete]

You can not take the square root of a negative number.

 

[Tom McCurdy]

You can not take the square root of a negative number.

[tex] \sqrt{-1}=i [/tex]

imaginay numbers allow for negitive sqroots

he just violated a law in the way he split up his negitive signs.

 

[Tom McCurdy]

http://mathworld.wolfram.com/ImaginaryNumber.html
imagianary numbers

 

[Tom McCurdy]

http://en.wikipedia.org/wiki/Imaginary_numbers

more explanation on how you can take sqrt of negitive numbers

 

[shmoe]

A thinly veiled version of the same, though the fallacy is perhaps more transparent:

Euler's formula tells us:

[tex]e^{i\theta}=\cos(\theta)+i\sin(\theta)[/tex]

So we see that:

[tex]e^{-i\pi}=e^{i\pi}[/tex]

taking roots gives:

[tex](e^{-i\pi})^{1/2}=(e^{i\pi})^{1/2}[/tex]
[tex]e^{-i\frac{\pi}{2}}=e^{i\frac{\pi}{2}}[/tex]

Using Euler's formula again and we get:

[tex]-i=i[/tex]

or -1=1

 

[Gokul43201]

Here's another (though this one cheats in a different way) :

[tex]1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1} = i^2 = -1[/tex]