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One equals minus one?

by olek1991
Tags: equals, minus
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olek1991
#1
Apr26-10, 02:56 PM
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=i =>
(-1)=(i) =>
1 = i^4 => take the square root both sides
1 = i

i = -1 v i = 1

Thus proving
1 = -1
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Cyosis
#2
Apr26-10, 03:06 PM
HW Helper
P: 1,495
To confuse you a little more can you find the mistake: [itex]2=\sqrt{4}=\sqrt{(-2)^2}=-2[/itex].
olek1991
#3
Apr26-10, 03:07 PM
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) is not -2, but 2 btw :P
You probably meant (-2) = √4 = 2

Cyosis
#4
Apr26-10, 03:41 PM
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 [itex]\sqrt{z}=\sqrt{|z|}e^{i \pi/2}[/itex]. In general for complex numbers it is not even true that [itex]\sqrt{zw}=\sqrt{z}\sqrt{w}[/itex].
Gigasoft
#5
Apr26-10, 03:49 PM
P: 59
[tex]a^{bc}=\left(a^b\right)^c[/tex] is not generally true. For example [tex]\left(\left(-1\right)^2\right)^{\frac 1 2}\neq-1[/tex]. You should be careful with this rule when the base is not a positive real number and the exponent is not an integer.
olek1991
#6
Apr26-10, 03:51 PM
P: 17
Quote Quote by Cyosis View Post
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 [itex]\sqrt{z}=\sqrt{|z|}e^{i \pi/2}[/itex]. In general for complex numbers it is not even true that [itex]\sqrt{zw}=\sqrt{z}\sqrt{w}[/itex].
I really don't get that O.o
Could you dumb it down a little? (I'm a collage student)
Cyosis
#7
Apr26-10, 03:59 PM
HW Helper
P: 1,495
We can write every complex number z in the form [tex]z=|z|e^{i \theta}[/tex] with |z| the distance between z and the origin and [itex]\theta[/itex] 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 [itex]i=e^{i \pi/2}[/itex] and [itex]i^4=e^{2 \pi i}[/itex]. Now taking the square root of i^4 we get [tex]\sqrt{i^4}=e^{i \pi}=-1[/tex].
wisvuze
#8
Apr26-10, 04:03 PM
P: 367
Quote Quote by olek1991 View Post
Yea I know those too xD
Does that mean that it is correct? (but just not used since it's crazy)

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

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


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