1 = -1

[Angra Mainyu]

You appear to be arguing theory without knowing algebra and exponents.-1^3 = -1`*-1*-1 = -1.

The first sentence is false, and it should be obvious that it is false. It may appear to you that that is so, because you believe that I'm saying something very different from what I'm saying. But that is not on me.

The second sentence has the problem of lacking parentheses, but it is obviously true that (-1)^3=(-1)*(-1)*(-1)=-1 (well, it is obvious to me anyway).

cube root (-27_^1/3 is -3. square root (-3)^1/2 is not defined.

That is correct.

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.

Of course I reject nothing of the sort.

Clarify, exactly what is it you are trying to prove or assert?

I already have, and this seems pointless because you are failing to follow, but here goes again: There is no function f from a domain D into the complex numbers with the following properties:

1. D contains the real line.
2. For all nonnegative real x, f(x)=sqrt(x).
3. For all x,y in D, f(xy)=f(x)f.
4. For all x in D, (f(x))^2=x.

This is the sort of function that would extend the usual sqrt function, defined for nonnegative reals, and keeping some nice properties used in SLD's proof.

By the way, SLD's proof is actually a proof by contradiction that there is no such function. It's just that SLD (just for fun, because he understands it) reached the opposite conclusion, i.e., that 1=-1. I took this thread also to be for fun (hence, my first reply), but you made it contentious for some reason.

 

[steve_bank]

The proof violates the rules of complex numbers.

You can not get 1=-1 by properly applying the rules of imaginary numbers any motr tjan you can get -1=1 by properly applying algebra. The proof is based on operations not defined for imaginary numbers.

You argued -1^1/N is not valid for n = 1,3,5.... do you still?

1. D contains the real line.
2. For all nonnegative real x, f(x)=sqrt(x).
3. For all x,y in D, f(xy)=f(x)f.
4. For all x in D, (f(x))^2=x.

The problem is 2. What is sqrt(x)? How is that defined? Based ion the definition of sqrt(x) why is it not defined for negative numbers?

In 3 are you invoking composite functions for some reason?

2 and 3 is just convoluted hand waving. That is derived and proven by the laws of exponents. [ab]^1/2 = [a]^1/2] * ^1/2.

All's you have done is say in a roundabout way the square root of a positive real is a positive real, but we already know that. We know the square root of negative number has an imaginary solution. In algebra square root of a negative number is not an issue.

So what is your point?

 

[Angra Mainyu]

The proof violates the rules of complex numbers.

You can not get 1=-1 by properly applying the rules of imaginary numbers any motr tjan you can get -1=1 by properly applying algebra. The proof is based on operations not defined for imaginary numbers.

It is of course a proof that the operation in question is not defined.

You argued -1^1/N is not valid for n = 1,3,5.... do you still?

I never did any of the sort.

The problem is 2. What is sqrt(x)? How is that defined? Based ion the definition of sqrt(x) why is it not defined for negative numbers?

It is the standard square root function, defined on the set of nonnegative real numbers.

All's you have done is say in a roundabout way the square root of a positive real is a positive real, but we already know that.

No, I did not say that (in a roundabout way or not). It is of course true, but I didn't think it was worth mentioning, so I did not say it.

In 3 are you invoking composite functions for some reason?

No.

So what is your point?

Sorry, I have dedicated a considerable amount of time to explain that, but you continue to attribute to me claims and arguments not related to what I said, and continue to raise objections to things that are not related to what I said, so I'm afraid I'm giving up.

 

[steve_bank]

A proof that the square root of a real negative number is not possible.

p1. The square root-of a number y equal to the square root of x is y such that y multiplied times it self equals x.
p22. For y a real square root of a real negative number, y multiplied by y must be negative.
p3. For a real number +y , y*y is positive.
p4. For a real number -y , y*y is positive.
c Therefore no real number solution exists for the square root of a negative real number.

The square root of a real number is limited to positive reals because a solution to the square root of negative number using reals does not exist. It is not because it is theoretically defined as such.

Anyone who uses algebra would not see the square root of a real number limited to positive reals, they would understand imaginary numbers. If it were limited to positive reals it would not have much value in engineering.

Take a look at the Quadratic Equation which provides the roots, real and imaginary, of a 2nd order polynomial. Widely used in engineering. General solutions to roots of a polynomial of any order involve both real and imaginary roots.

So when you say sqr(x) is limited to positive reals because it is defined that way it has no merit. You can define it that way in the context of same application, but not globally.I still do not see the point you are trying to make.

 

[Artemus]

1 = -1

Proof:

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

QED!

Prove me wrong!

The controversy over the existence of the  Zenneck wave was due to this sign error. As others have noted, the square root operator must be considered a  multi-valued function as soon as negative numbers are added into the mix. (The fact that the term "multl-valued function" gives some mathematicians conniptions gives this electrical engineer a certain sort of perverse pleasure.)

 

[Jimmy Higgins]

1 = -1

Proof:

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

QED!

Prove me wrong!

I'm late to the party, but shouldn't: -1j = i² (using j for imaginary notation)

 

[steve_bank]

And there I was thinking that for once I had the last word on a thread.

 

[Don2 (Don1 Revised)]

1 = -1

Proof:

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

QED!

Prove me wrong!

The controversy over the existence of the  Zenneck wave was due to this sign error. As others have noted, the square root operator must be considered a  multi-valued function as soon as negative numbers are added into the mix. (The fact that the term "multl-valued function" gives some mathematicians conniptions gives this electrical engineer a certain sort of perverse pleasure.)

*heart palpatations* It's a relation. *breathing in and out*

 

[SLD]

1 = -1

Proof:

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

QED!

Prove me wrong!

The controversy over the existence of the  Zenneck wave was due to this sign error. As others have noted, the square root operator must be considered a  multi-valued function as soon as negative numbers are added into the mix. (The fact that the term "multl-valued function" gives some mathematicians conniptions gives this electrical engineer a certain sort of perverse pleasure.)

*heart palpatations* It's a relation. *breathing in and out*

It seems to me to be simply the last equals sign in the proof. Artemus is right. The square root of 1 is not 1, but either 1 or -1. That’s all we need to know to show the logical flaw.

But there is something more to this issue. 1 can still be equal to -1 depending on the topological space one finds oneself. Rotate a vector (1,0) through 2 pi. Should come back to (1,0). But on a Möbius strip it’s (-1,0). (Basically) Topology rules!

SLD

 

[steve_bank]

Topology is beyond me. Are you saying in topology 1 can algebraically equal -1? Or are you talking about vectors and direction, analogues to unit vectors?

 

[SLD]

A vector of 1,0 becomes -1,0 through. A rotation of 2*pi

 

[steve_bank]

View attachment 26070

A vector of 1,0 becomes -1,0 through. A rotation of 2*pi

Then -1 != 1. Direction changes with sign.