# One equals minus one?

1. Apr 26, 2010

### olek1991

I think that I have proof of 1 being -1 and I can't find any flaw in it.
Could you please take a look?

-1=i² =>
(-1)²=(i²)² =>
1 = i^4 => take the square root both sides
1 = i²

i² = -1 v i² = 1

Thus proving
1 = -1

Last edited by a moderator: Apr 26, 2010
2. Apr 26, 2010

### Cyosis

To confuse you a little more can you find the mistake: $2=\sqrt{4}=\sqrt{(-2)^2}=-2$.

3. Apr 26, 2010

### olek1991

Yea I know those too xD
Does that mean that it is correct? (but just not used since it's crazy)

Edit: the √(-2)² is not -2, but 2 btw :P
You probably meant (-2)² = √4 = 2²

Last edited: Apr 26, 2010
4. Apr 26, 2010

### Cyosis

No I meant exactly what I wrote, the root cancelling the square. We can't do this because we have defined taking the square root of a real number to be a positive value. This is called the principal square root. For the complex numbers this principal root is defined as $\sqrt{z}=\sqrt{|z|}e^{i \pi/2}$. In general for complex numbers it is not even true that $\sqrt{zw}=\sqrt{z}\sqrt{w}$.

5. Apr 26, 2010

### Gigasoft

$$a^{bc}=\left(a^b\right)^c$$ is not generally true. For example $$\left(\left(-1\right)^2\right)^{\frac 1 2}\neq-1$$. You should be careful with this rule when the base is not a positive real number and the exponent is not an integer.

6. Apr 26, 2010

### olek1991

I really don't get that
Could you dumb it down a little? (I'm a collage student)

7. Apr 26, 2010

### Cyosis

We can write every complex number z in the form $$z=|z|e^{i \theta}$$ with |z| the distance between z and the origin and $\theta$ the angle between the x-axis and |z| (polar coordinates). If you have had some complex numbers you should know this representation of a complex number. From this it follows that $i=e^{i \pi/2}$ and $i^4=e^{2 \pi i}$. Now taking the square root of i^4 we get $$\sqrt{i^4}=e^{i \pi}=-1$$.

8. Apr 26, 2010

### wisvuze

No that's not what it means, all of our mathematical foundations would be bogus if we ever said "it's true, but it's too crazy.. so it's pretty much false".
Sqrt(x) is a function (input/output relationships are unique), so given a number (perhaps 9), Sqrt(9) will map to 3.. never -3. If Sqrt(9) could be either -3 OR 3, it wouldn't be a function. Even though (-3)^2 = 9 = (3)^2, the root function is defined to take positive values and produce positive values.

Edit: the √(-2)² is not -2, but 2 btw :P
You probably meant (-2)² = √4 = 2²

This is exactly what you kind of said.. sqrt( (-2)^2 ) is indeed 2 since (-2)^2 gives us 4, and by the definition of the function, we will get the positive possible "root" only.

"You probably meant (-2)² = √4 = 2²" You probably made some typing mistake here.. (-2)^2 = sqrt(4) = 2^2?? 4 = 2 = 4? I don't know