1. Originally Posted by Angra Mainyu
Originally Posted by steve_bank

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.
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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.

What definition? I covered rings and fields as part of linear algebra but had no practical use for it.

2. Originally Posted by steve_bank
It is pointless to try to argue theory when you do not understand actual math that is use.
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,

-1=sqrt(-1))^2=sqrt((-1)^2)=sqrt(1)=1.

Thus, no such function exists.

3. Originally Posted by Angra Mainyu
Originally Posted by steve_bank
It is pointless to try to argue theory when you do not understand actual math that is use.
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,

-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.

4. Originally Posted by Angra Mainyu
Originally Posted by steve_bank
It is pointless to try to argue theory when you do not understand actual math that is use.
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,

-1=sqrt(-1))^2=sqrt((-1)^2)=sqrt(1)=1.

Thus, no such function exists.
I should have specified that D contains the negative real numbers.

5. Originally Posted by steve_bank
Originally Posted by Angra Mainyu
Originally Posted by steve_bank
It is pointless to try to argue theory when you do not understand actual math that is use.
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,

-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.

6. Originally Posted by SLD
1 = -1

Proof:

-1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1

QED!

Prove me wrong!
https://en.wikipedia.org/wiki/Imaginary_number

QR(-1)*SQR(-1) = SQR[(-1)(-1)] is not a valid operation in complex numbers.

7. Originally Posted by steve_bank
Originally Posted by SLD
1 = -1

Proof:

-1 = i^2 = SQR(-1)*SQR(-1) = SQR[(-1)(-1)] = SQR[(-1)^2)] = SQR(1) = 1

QED!

Prove me wrong!
https://en.wikipedia.org/wiki/Imaginary_number

QR(-1)*SQR(-1) = SQR[(-1)(-1)] is not a valid operation in complex numbers.

8. Take a calculator and calculate (-1)^1/3 and see what you get.

9. Originally Posted by Angra Mainyu
Originally Posted by steve_bank

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.
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?

10. Originally Posted by steve bank
Take a calculator and calculate (-1)^1/3 and see what you get.
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.

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