Fine, I left the condition that (sqrt(x))^2=x implicit before. My bad. So, let D be the domain of sqrt, and let's say that for all x in D, (sqrt(x)^2)=x and sqrt(x)sqrt(y)=sqrt(xy). Let us also stipulate that sqrt(x) takes the usual values for all non-negative real x. Then,Originally Posted by steve_bank
-1=sqrt(-1))^2=sqrt((-1)^2)=sqrt(1)=1.
Thus, no such function exists.
It does not exist using real numbers. Limited to reals there is no solution to the square root of a negative number.
sqr(-4 * 9) = sqr(-1) * sqr(4) * sqr(9) = [0 + i6]...i6 js on the imaginary not real axis.
[0 + i6]^2= [0 + i6] * [0 + i6] = i^2 * 6^2 = -1*36 = -4 * 9
In electrical theory and electronics in general square roots of negative numbers are routine. It requires complex numbers. Reals are a subset of complex numbers.
The solution is not on the real number line.
There are an infinite number of integer roots.
(x)^1/n n = 0,1,2,3....
(-1)^n has a real solution for n = 1,3,5.... and it is -1.
No, it doesn't exist as long as D contains the negative real numbers. For example, if D is the complex field, no such function exists (please take a careful look at the condition (sqrt(x)^2)=x and sqrt(x)sqrt(y)=sqrt(xy) and the proof that follows).
In proofs by contradiction (I prefer to call them by reductio ad absurdum), keeping the absurdity is not the right conclusion.
https://en.wikipedia.org/wiki/Imaginary_number
QR(-1)*SQR(-1) = SQR[(-1)(-1)] is not a valid operation in complex numbers.
Take a calculator and calculate (-1)^1/3 and see what you get.
You appear to be arguing theory without knowing algebra and exponents.-1^3 = -1`*-1*-1 = -1.
cube root (-27_^1/3 is -3. square root (-3)^1/2 is not defined.
This is fundamental to imaginary numbers. You can invoke theory as you please as long as you do not insist on rejecting the established well used rules of algebra, exponents, and complex numbers.
Clarify, exactly what is it you are trying to prove or assert?
Try busing exponents for roots, square root.
sqr(xy) = (xy)^1/2 = x^1/2 * y^1/2. Applying riles of exponents. Not a proof.
(sqrt(x)^2)=x-> (x^1/2)^2 -> x^2/2 = x^1 -> x...so what? Still high school math. Proof of what?
You would get -1, of course. Note that this is not relevant to the question of the thread, as it is not the square root.Originally Posted by steve bank