# One equals minus one?

by olek1991
Tags: equals, minus
 P: 17 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
 HW Helper P: 1,495 To confuse you a little more can you find the mistake: $2=\sqrt{4}=\sqrt{(-2)^2}=-2$.
 P: 17 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
HW Helper
P: 1,495

## One equals minus one?

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 $\sqrt{zw}=\sqrt{z}\sqrt{w}$.
 P: 59 $$a^{bc}=\left(a^b\right)^c$$ 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.
P: 17
 Quote by 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 $\sqrt{zw}=\sqrt{z}\sqrt{w}$.
I really don't get that O.o
Could you dumb it down a little? (I'm a collage student)
 HW Helper P: 1,495 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 $$\sqrt{i^4}=e^{i \pi}=-1$$.
P: 367
 Quote by 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
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

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