# I've never understood why this is the case

• Feb 12th 2009, 07:50 PM
mitchelljk
I've never understood why this is the case
$\displaystyle \sqrt{4} \bullet \sqrt{9} = 2 \bullet 3 = 6$

When I do it this way --- $\displaystyle \sqrt{4 \bullet 9} = \sqrt{36} = 6$

I get the same answer either way I write it. However if i do this:

$\displaystyle \sqrt{-1} \bullet \sqrt{-1} = 1$

or if I write:

$\displaystyle \sqrt{-1 \bullet -1}$ Either way I get answer of 1 because

$\displaystyle -1 \bullet -1 = 1$ and $\displaystyle \sqrt{1}$ is just one.

However complex numbers say that $\displaystyle i^2$ is equal to $\displaystyle -1$

How can $\displaystyle i^2$ be equal to $\displaystyle -1$ if
$\displaystyle \sqrt{-1} \bullet \sqrt{-1}$ equals $\displaystyle \sqrt{1}$ or just $\displaystyle 1$?

By my logic $\displaystyle i^2$ should be just 1. So there is obviously a rule I am missing here that would lead me to believe this. If someone could explain to me why $\displaystyle \sqrt{4 \bullet 9} = \sqrt{36} = 6$ but $\displaystyle \sqrt{-1 \bullet -1}$ doesnt not equal $\displaystyle \sqrt{1}$ or just $\displaystyle 1$ . It would be much appreciated!

Sorry in advance if my code got messed up, I havent used it before.
• Feb 12th 2009, 08:09 PM
Reckoner
Quote:

Originally Posted by mitchelljk
How can $\displaystyle \iota^2$ be equal to $\displaystyle -1$ if
$\displaystyle \sqrt{-1} \bullet \sqrt{-1}$ equals $\displaystyle \sqrt{1}$ or just $\displaystyle 1$?

Simple. $\displaystyle \sqrt{-1}\sqrt{-1}$ does not equal 1.

Here's where you're going wrong: the property that you are using does not necessarily work for negative values. $\displaystyle \sqrt a\sqrt b = \sqrt{ab}\text{ for }a, b\geq0$, but not necessarily if $\displaystyle a$ or $\displaystyle b$ is negative.
• Feb 12th 2009, 08:17 PM
mitchelljk
Awesome thanks. I knew it would be something simple like that.
• Feb 12th 2009, 08:20 PM
Reckoner
No problem. Check out this page for an interesting "proof" that 2=1, using this same property incorrectly. They give a nice explanation of the error.
• Feb 14th 2009, 05:53 AM
stapel
Quote:

Originally Posted by mitchelljk
...there is obviously a rule I am missing here....

The rule is that, by adding the ability to take the square root of negative values, you lose the ability to swap the values around. If you allow for negatives inside the radical, then you must deal with that negative sign (that is, you must do the "i" part) first. :D