Very simple math question

AFB

Lifer
Jan 10, 2004
10,718
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0
Why is

((-5)^.5) * ((-5)^.5) == -5

But when you multiply under the radical

(-5 * -5)^.5 == (25)^1/2 == 5 (Yes, can also be -5)



This is what I was told, no idea why.
 

AFB

Lifer
Jan 10, 2004
10,718
3
0
Originally posted by: notfred
you can't take the square root of a negative number.

Yes, I know. You actually bring up a better example.

Why is

(i)^2 == -1

Because i = (-1)^1/2

Well, normally to simplify a set of radicals of the same powere that are being multiplied, you multiply under the radical.

e.g., (5)^.5 * (125)^.5 == (5 * 125)^.5 == (625)^.5 = 25

So, wouldn't two negitives cancel out when you multiply them?

(-5)^.5 * (-125)^.5 == (-5 * -125)^.5 == (625)^.5 = 25

Or in our case

(-1)^.5 * (-1)^.5 == (-1 * -1)^.5 == (1)^.5 = 1
 

elbosco

Senior member
Jul 17, 2004
907
0
71
(-5)^½ * (-5)^½ = (-1)^½ * (5)^½ * (-1)^½ * (5)^½

(-1)^½ = ?

? * ? = -1 and (5)^½ * (5)^½ = 5

-1 * 5 = -5
 

her209

No Lifer
Oct 11, 2000
56,336
11
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Originally posted by: amdfanboy
(-1)^.5 * (-1)^.5 == (-1 * -1)^.5 == (1)^.5 = 1
I don't think that's correct.

(-1)^.5 * (-1)^.5 = (-1)^(.5+.5) = -1

 

Soccer55

Golden Member
Jul 9, 2000
1,660
4
81
Originally posted by: amdfanboy
Originally posted by: notfred
you can't take the square root of a negative number.

Yes, I know. You actually bring up a better example.

Why is

(i)^2 == -1

Because i = (-1)^1/2

Well, normally to simplify a set of radicals of the same powere that are being multiplied, you multiply under the radical.

e.g., (5)^.5 * (125)^.5 == (5 * 125)^.5 == (625)^.5 = 25

So, wouldn't two negitives cancel out when you multiply them?

(-5)^.5 * (-125)^.5 == (-5 * -125)^.5 == (625)^.5 = 25

Or in our case

(-1)^.5 * (-1)^.5 == (-1 * -1)^.5 == (1)^.5 = 1

But what are you actually doing when you take the square root of a number? You're looking for a number (or numbers) that, when squared, is (are) equal to the number under the radical.

For example, sqrt(25) = +5 and -5 because (-5)^2 = 25 and 5^2=25. I think what you're having problems with is that you're only looking at the principle (positive) root. Not looking at the negative solution leads you to the seeming contradiction that you've arrived at.

-Tom
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
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www.slatebrookfarm.com
Here's the reason:

Sqrt(a)*sqrt(b) = sqrt(a*b) If and only if a and b >= 0

in your case, since a and b are less than 0, you can't multiply them together.


edit: Since you typically learn operations with radicals a year (or more) before you learn about imaginary and complex numbers, teachers will often leave that if and only if part out of it when defining the operations with square roots.

edit edit:

Another weird example while we're at it..
In your earlier math text, it probably says that x^2+7 cannot be factored.


However, now that you know about imaginary numbers and irrational numbers,
x^2+7 can be factored as:
(x + i*sqrt(7) ) ( x - i*sqrt(7) )