Whats wrong with this imaginary number problem?

[G01]

I saw this thing where someone proved that the imaginary number, i, the sqrt(-1) was equal to 1.

here it is:

i= sqrt(-1)

i^2 = [sqrt(-1)]^2

i^2 = sqrt(-1) * sqrt(-1)

i^2 = sqrt(-1*-1)

i^2 = sqrt(1)

i^2 = 1

so

i = 1

I know there's something wrong here but i can't figure it out. any help?

 

[HallsofIvy]

Isn't this in a "faq" somewhere? The "rule" \sqrt{a}\sqrt{b}= \sqrt{ab} does not hold for complex numbers.
On a more fundamental level, "defining" i to be \sqrt{-1}[/tex] causes a problem: in the complex numbers every number, including -1, has <b>two</b> square roots. Since the complex numbers are not an &quot;ordered field&quot; as the real numbers are, we can&#039;t just declare i to be &quot;the positive root&quot;. There are other ways of defining the complex numbers that avoid that problem.

 

[G01]

ok thanks a lot ivy

 

[fourier jr]

yeah you need to use i^2 = -1 as the definition of i

 

[fourier jr]

here's a proof that -1=1.
define i = \sqrt{-1}

then i=i

=> \sqrt{-1} = \sqrt{-1}

=> \sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}

=> \frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}

=> \sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}

=> 1=-1

 

[G01]

here's a proof that -1=1.
define i = \sqrt{-1}
then i=i
=> \sqrt{-1} = \sqrt{-1}
=> \sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}
=> \frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}
=> \sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}
=> 1=-1

I guess you can prove anything as long as you can define an y number as we want it to be. lol I get it the rule sqrt(ab)=sqrt(a)sqrt(b) doesn't work for complex numbers because i is defined as the sqrt of -1 and that rule would make it something else. Thanks a lot for all your help.

 

[masudr]

...
\sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}
1=-1

I'm afraid that's wrong since (as others have pointed out already)
\sqrt{-1} = \pm i, \sqrt{1} = \pm 1

and you have not accounted for that. Since that leads to various problems, it is not a good idea to define i = \sqrt{-1}.

 

[fourier jr]

yeah that's what i was trying to say. you've got to define i as the number with the property that i^2 = -1. that "proof" shows what happens when you try to define i as the square root of -1.

 

[HallsofIvy]

yeah you need to use i^2 = -1 as the definition of i

Even that's not sufficient. There are two complex numbers whose square is equal to -1. Which one do you mean?

A more standard way of defining the complex numbers is as pairs of real numbers, (a,b) with addition defined by (a,b)+ (c,d)= (a+c, b+d) and multiplication by (a,b)*(c,d)= (ac-bd, bc+ ad). Of course, the pairs (a,0) correspond to the real numbers. That way (0,1)*(0,1)= (0(0)-1(1),1(0)+0(1))= (-1, 0) and (0,-1)*(0,-1)= (0(0)-(-1)(-1),(-1)(0)+0(-1))= (-1,0) but now we can define i to be (0,1) rather than (0,-1). Of course, we can write
(a,b)= a(1,0)+ b(0,1) and since we are identifying (1, 0) with the real number 1 and (0,1) with i, a(1,0)+ b(0,1)= a+ bi.

 

[masudr]

I personally like this definition. We have the set of all matrices

\left(<br /> \begin{array}{clrr}<br /> x&amp;-y\\<br /> y&amp;x<br /> \end{array}<br /> \right)

such that x, y \in \mathbb{R}. Under addition and multiplication of matrices we have a field. Furthermore, associate every matrix with an entity I will call a "complex number," which I will also write in a more compact notation x+iy. The reason I choose this notation is because if I define i^2=-1, then the normal rules of manipulating algebraic expressions will still hold.

I prefer it because the multiplication of two complex numbers doesn't seem so arbitrary (although of course it is just as arbitrary, but I feel it's more aesthetic).