1 = -1

[SLD]

1 = -1

Proof:

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

QED!

Prove me wrong!

 

[barbos]

1 = -1

Proof:

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

QED!

Prove me wrong!

SQR(1) != 1

SQR(1)={1,-1}

Also, i != SQR(-1), but SQR(-1)={i,-i}

 

[steve_bank]

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

 

[Angra Mainyu]

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.

I just proved you wrong.

Show

So, there is something wrong with your computation, and the only candidate is the SQR function, not defined for negative numbers (though if you want to try to define such a function ).

 

[steve_bank]

If your calculator has a complex number mode it will take sqrt(-1).

 

[Angra Mainyu]

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=sqrt(xy). This is actually proven by SLD's argument.

 

[steve_bank]

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=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.

 

[Wiploc]

Prove me wrong!

Division by zero is undefined.

 

[Angra Mainyu]

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=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=sqrt(xy). This is actually proven by SLD's argument.

 

[steve_bank]

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.

 

[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.
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=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.

 

[Angra Mainyu]

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=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.

 

[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=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.

 

[Angra Mainyu]

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=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.

 

[Angra Mainyu]

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=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=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.

 

[steve_bank]

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.

 

[Angra Mainyu]

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.

As I already proved.

 

[steve_bank]

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

 

[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=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?

 

[Angra Mainyu]

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.