1 = -1
Proof:
-1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1
QED!
Prove me wrong!
1 = -1
Proof:
-1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1
QED!
Prove me wrong!
As in algebra texts is
sqrt(1) =+-1.
sqrt(1) = 1
SQRT(-1) = i, A complex number [0 + i]. i^2 = -1. i^3 = 1...
sqrt(4) = 2
sqrt(-4) = sqrt(-1*4) = [0 + i2] or just i2
Last edited by steve_bank; 12-08-2019 at 03:37 AM.
1=\=-1.
I just proved you wrong.
If your calculator has a complex number mode it will take sqrt(-1).
You can define a function on the real line (or in the complex field, for that matter) and call it "sqrt". However, it will not be an extension of the function defined on [0,+\infty) with the property that sqrt(x)sqrt(y)=sqrt(xy). This is actually proven by SLD's argument.
All numbers are complex. It is the top category. Reals are complex numbers with zero imaginary. My HP calculator does complex numbers as does most software packages. It is just trigonometry.
I
In software you can code the sqr function nay way you like. If you limit t to solely reals sqr(-1) is undefined.
The OP argument is counter to any high school algebra text. The quadratic equation. If the discriminant is negative the solution is complex, if not the solution is real.
Again, you can define a function on the real line (or in the complex field, for that matter) and call it "sqrt". However, it will not be an extension of the sqrt function defined on [0,+\infty) with the property that sqrt(x)sqrt(y)=sqrt(xy). This is actually proven by SLD's argument.
sqr(-4 * 9) = sqr(-1) * sqr(4) * sqr(9) = [0 + i6]
[0 + i6]^2= [0 + i6] * [0 + i6] = i^2 * 6^2 = -1*36 = -4 * 9
sqr(-1) = i, i^2 = sqr(-1) A definition.