1. ## 1 = -1

1 = -1

Proof:

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

QED!

Prove me wrong!

2. 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!
SQR(1) != 1

SQR(1)={1,-1}

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

3. 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

4. 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!

1=\=-1.

I just proved you wrong.

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 ).

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

6. Originally Posted by 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(y)=sqrt(xy). This is actually proven by SLD's argument.

7. Originally Posted by Angra Mainyu
Originally Posted by 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(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.

8. Originally Posted by SLD

Prove me wrong!
Division by zero is undefined.

9. Originally Posted by steve_bank
Originally Posted by Angra Mainyu
Originally Posted by 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(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.

10. 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.

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